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PHYSICS XII

CHAPTERS 1-7

BOOKLET-1

Contents: Page No.

Chapter 1 Electric Charges and Fields (Electrostatics Part 1) 1-17







c Electric Charge

Definition : Charge is the property associated with matter due to which it produces and experiences electrical and magnetic effects.

Origin of electric charge : It is known that every atom is electrically neutral, containing as many electrons as the number of protons in the nucleus.

Charged particles can be created by disturbing neutrality of an atom. Loss of electrons gives positive charge (as then np > ne) and gain of electrons gives negative charge (as then ne > np) to a particle. When an object is negatively charged it gains electrons and therefore its mass increases negligibly. Similarly, on charging a body with positive electricity its mass decreases. Change in mass of object is equal to n × me. Where, n is the

number of electrons transferred and em is the mass of electron Kg31101.9 −×= .

Type : There exists two types of charges in nature (i) Positive charge (ii) Negative charge

Charges with the same electrical sign repel each other, and charges with opposite electrical sign attract each other.

Unit and dimensional formula : Rate of flow of electric charge is called electric current i.e., dtdQ

i =

⇒ idtdQ = , hence S.I. unit of charge is – Ampere × sec = coulomb (C), smaller S.I. units are mC, µC, nC

)101,101,101( 963 CnCCCCmC −−− === µ . C.G.S. unit of charge is – Stat coulomb or e.s.u. Electromagnetic

unit of charge is – ab coulomb coulombabcoulombstatC101

1031 9 =×= . Dimensional formula [ ]ATQ =][

Note : ≅ Benjamin Franklin was the first to assign positive and negative sign of charge.

≅ The existence of two type of charges was discovered by Dufog.

≅ Franklin (i.e., e.s.u. of charge) is the smallest unit of charge while faraday is largest (1 Faraday = 96500 C).

Mas

s M′

Electron < Proton

Positively charged

+

M′ < M

Electron = Proton

Neutral

Mas

s M

Negatively charged

M′′ > M

–

Electron > Proton

+ – – – + +

Electric Charges and Fields (Electrostatics Part 1)

Chapter 1 1

c≅ The e.s.u. of charge is also called stat coulomb or Franklin (Fr) and is related to e.m.u.

of charge through the relation 10103chargeof esuchargeof emu

×=

Point charge : A finite size body may behave like a point charge if it produces an inverse square electric field. For example an isolated charged sphere behave like a point charge at very large distance as well as very small distance close to it’s surface.

Properties of charge

(i) Charge is transferable : If a charged body is put in contact with an uncharged body, uncharged body becomes charged due to transfer of electrons from one body to the other.

(ii) Charge is always associated with mass, i.e., charge can not exist without mass though mass can exist without charge.

(iii) Charge is conserved : Charge can neither be created nor be destroyed. e.g. In radioactive decay the uranium nucleus (charge e92+= ) is converted into a thorium nucleus (charge e90+= ) and emits an α -

particle (charge e2+= )

42

23490

23892 HeThU +→ . Thus the total charge is e92+ both before and after the decay.

(iv) Invariance of charge : The numerical value of an elementary charge is independent of velocity. It is proved by the fact that an atom is neutral. The difference in masses on an electron and a proton suggests that electrons move much faster in an atom than protons. If the charges were dependent on velocity, the neutrality of atoms would be violated.

(v) Charge produces electric field and magnetic field : A charged particle at rest produces only electric field in the space surrounding it. However, if the charged particle is in unaccelerated motion it produces both electric and magnetic fields. And if the motion of charged particle is accelerated it not only produces electric and magnetic fields but also radiates energy in the space surrounding the charge in the form of electromagnetic waves.

(vi) Charge resides on the surface of conductor : Charge resides on the outer surface of a conductor because like charges repel and try to get as far away as possible from one another and stay at the farthest distance from each other which is outer surface of the conductor. This is why a solid and hollow conducting sphere of same outer radius will hold maximum equal charge and a soap bubble expands on charging.

(vii) Charge leaks from sharp points : In case of conducting body no doubt charge resides on its outer surface, if surface is uniform the charge distributes uniformly on the surface and for irregular surface the distribution of charge, i.e., charge density is not uniform. It is maximum where the radius of curvature is minimum and vice versa. i.e., ∝σ ( )/R1 . This is why charge leaks from sharp points.

v

= constant

E

and B

but no Radiation

+ 0=v

E

+ v

≠ constant

E

, B

and Radiates energy

+


Chapter 1 2

c

(viii) Quantization of charge : When a physical quantity can have only discrete values rather than any value, the quantity is said to be quantised. The smallest charge that can exist in nature is the charge of an

electron. If the charge of an electron ( C19106.1 −×= ) is taken as elementary unit i.e. quanta of charge the charge on any body will be some integral multiple of e i.e.,

neQ ±= with ....3,2,1=n

Charge on a body can never be e32

± , e2.17± or e510−± etc.

Note : ≅ Recently it has been discovered that elementary particles such as proton or neutron are composed

of quarks having charge ( )3/1± e and ( )3/2± e. However, as quarks do not exist in free state, the

quanta of charge is still e.

≅ Quantization of charge implies that there is a maximum permissible magnitude of charge.

Comparison of Charge and Mass.

We are familiar with role of mass in gravitation, and we have just studied some features of electric charge. We can compare the two as shown below

Charge Mass

(1) Electric charge can be positive, negative or zero. (1) Mass of a body is a positive quantity.

(2) Charge carried by a body does not depend upon velocity of the body.

(2) Mass of a body increases with its velocity as

22

0

/1 cv

mm

−= where c is velocity of light in

vacuum, m is the mass of the body moving with velocity v and 0m is rest mass of the body.

(3) Charge is quantized. (3) The quantization of mass is yet to be established.

(4) Electric charge is always conserved. (4) Mass is not conserved as it can be changed into energy and vice-versa.

(5) Force between charges can be attractive or repulsive, according as charges are unlike or like charges.

(5) The gravitational force between two masses is always attractive.

Methods of Charging.

A body can be charged by following methods :

(1) By friction : In friction when two bodies are rubbed together, electrons are transferred from one body to the other. As a result of this one body becomes positively charged while the other negatively charged,

+

+

+

+ +

+

+ +

+ + +

+ +

+

+

+ +

+

+ + +

+

+ +

+

+ +

+

+ +

+


Chapter 1 3

ce.g., when a glass rod is rubbed with silk, the rod becomes positively charged while the silk negatively. However, ebonite on rubbing with wool becomes negatively charged making the wool positively charged. Clouds also become charged by friction. In charging by friction in accordance with conservation of charge, both positive and negative charges in equal amounts appear simultaneously due to transfer of electrons from one body to the other.

(2) By electrostatic induction : If a charged body is brought near an uncharged body, the charged body will attract opposite charge and repel similar charge present in the uncharged body. As a result of this one side of neutral body (closer to charged body) becomes oppositely charged while the other is similarly charged. This process is called electrostatic induction.

Note : ≅ Inducting body neither gains nor loses charge.

≅Induced charge can be lesser or equal to inducing charge (but never greater) and its

maximum value is given by

−−=

KQQ'

11 where Q is the inducing charge and K is the dielectric

constant of the material of the uncharged body. Dielectric constant of different media are shown below

Medium K

Vacuum / air

Water

Mica

Glass

Metal

1

80

6

5–10

∞

≅Dielectric constant of an insulator can not be ∞

≅For metals in electrostatics ∞=K and so ;QQ' −= i.e. in metals induced charge is equal and

opposite to inducing charge.

(3) Charging by conduction : Take two conductors, one charged and other uncharged. Bring the conductors in contact with each other. The charge (whether ve− or ve+ ) under its own repulsion will spread over both the conductors. Thus the conductors will be charged with the same sign. This is called as charging by conduction (through contact).

– –

– – –

– –

+ + + + + +

+ +

+ + +

+Q

– –

– – –

– – +

+ + +

+

+ + + + + +

+ +

+ + +

+Q Q′

⇒

– –

– – –

– –

+ + + + + +

+ +

+ + +

+Q

⇒ ⇒ – – – –

–

– –

– – –

– –

–

– – –


Chapter 1 4

c

Note : ≅ A truck carrying explosives has a metal chain touching the ground, to conduct away the charge

produced by friction.

Electroscope.

It is a simple apparatus with which the presence of electric charge on a body is detected (see figure). When metal knob is touched with a charged body, some charge is transferred to the gold leaves, which then diverges due to repulsion. The separation gives a rough idea of the amount of charge on the body. If a charged body brought near a charged electroscope the leaves will further diverge. If the charge on body is similar to that on electroscope and will usually converge if opposite. If the induction effect is strong enough leaves after converging may again diverge.

(1) Uncharged electroscope

(2) Charged electroscope

D

+

+ +

+ +

+ +

+

+ + + +

(A)

+

+ +

+ +

+ +

+

+ +

+ +

+

C (C)

– – – – –

– – – –

–

– –

– –

– –

–

– – –

–

–

– – – – – –

D (B)

– – – – –

– – – –

+

– –

+ +

+ +

+

– –

– –

–

C (B)

–

+ +

– –

– –

–

+ +

+ +

+

– – – – – –

D (C)

– – – – –

– – – –

Charging by conduction Charging by conduction

D

+

+ +

+

+ + +

+ + + +

+ +

+ +

+

+ + + +

(A)

⇒ ⇒

Uncharged

+ +

+ + + +

+ +

+ + +

+ + +

+ +

+ +

+

+ +

Bodies in contact

+ +

+ +

+ +

+ + + +

+ +

+ + +

+ + +

+ +

+ + + +

+ +

+ +

Both are positively charged

+ + +

+ +

+ + + +

+ +

+ +

Charged

+


Chapter 1 5

cConcepts

After earthing a positively charged conductor electrons flow from earth to conductor and if a negatively charged conductor is earthed then electrons flows from conductor to earth.

When a charged spherical conductor placed inside a hollow insulated conductor and connected if through a fine conducting wire the charge will be completely transferred from the inner conductor to the outer conductor.

Lightening-rods arrestors are made up of conductors with one of their ends earthed while the other sharp, and protects a building from lightening either by neutralising or conducting the charge of the cloud to the ground.

With rise in temperature dielectric constant of liquid decreases.

Induction takes place only in bodies (either conducting or non-conducting) and not in particles.

If X-rays are incident on a charged electroscope, due to ionisation of air by X-rays the electroscope will get discharged and hence its leaves will collapse. However, if the electroscope is evacuated. X-rays will cause photoelectric effect with gold and so the leaves will further diverge if it is positively charged (or uncharged) and will converge if it is negatively charged.

If only one charge is available than by repeating the induction process, it can be used to obtain a charge many times greater than it’s equilibrium. (High voltage generator)

Example: 1 A soap bubble is given negative charge. Its radius will [DCE 2000; RPMT 1997; CPMT 1997; MNR

1988]

(a) Increase (b) Decrease (c) Remain unchanged (d) Fluctuate Solution: (a) Due to repulsive force.

Example: 2 Which of the following charge is not possible

(a) C18106.1 −× (b) C19106.1 −× (c) C20106.1 −× (d) None of these

Solution: (c) ,106.1 20C−× because this is 101

of electronic charge and hence not an integral multiple.

Example: 3 Five balls numbered 1 to 5 balls suspended using separate threads. Pair (1,2), (2,4) and (4,1) show electrostatic attraction, while pair (2,3) and (4,5) show repulsion. Therefore ball 1 must be [NCERT 1980]

(a) Positively charged (b) Negatively charged (c) Neutral (d) Made of metal

Solution: (c) Since 1 does not enter the list of repulsion, it is just possible that it may not be having any charge. Moreover, since ball no. 1 is being attracted by 2 and 4 both. So 2 and 4 must be similarly charged, but it is also given that 2 and 4 also attract each other. So 2 and 4 are certainly oppositely charged.

+Q

+ –

–

– – –

– – –

e– + +

+

+ + +

+ + +

e–

Examples based on properties of charge


Chapter 1 6

c Since 1 is attracting 2, either 1 or 2 must be neutral but since 2 is already in the list of balls repelling each

other, it necessarily has some charge, similarly 4 must have some charge. It means that though 1 is attracting 2 and 4 it does not have any charge.

Example: 4 If the radius of a solid and hollow copper spheres are same which one can hold greater charge

[BHU 1999; KCET 1994; IIT-JEE 1974]

(a) Solid sphere (b)Hollow sphere (c) Both will hold equal charge (d) None of these

Solution: (c) Charge resides on the surface of conductor, since both the sphere having similar surface area so they will hold equal charge.

Example: 5 Number of electrons in one coulomb of charge will be

(a) 291046.5 × (b) 181025.6 × (c) 19106.1 × (d) 11109 ×

Solution: (b) By using eQ

nneQ =⇒= 1819 1025.6

106.11

×=×

=⇒ −n

Example: 6 The current produced in wire when 107 electron/sec are flowing in it [CPMT 1994]

(a) 1.6 × 10–26 amp (b) 1.6 × 1012 amp (c) 1.6 × 1026 amp (d) 1.6 × 10–12 amp

Solution: (d) ampt

netQ

i 12197 106.1106.110 −− ×=××===

Example: 7 A table-tennis ball which has been covered with a conducting paint is suspended by a silk thread so that it hangs between two metal plates. One plate is earthed. When the other plate is connected to a high voltage generator, the ball

(a) Is attracted to the high voltage plate and stays there

(b) Hangs without moving

(c) Swings backward and forward hitting each plate in turn

(d) None of these

Solution: (c) The table tennis ball when slightly displaced say towards the positive plate gets attracted towards the positive plate due to induced negative charge on its near surface.

The ball touches the positive plate and itself gets positively charged by the process of conduction from the plate connected to high voltage generator. On getting positively charged it is repelled by the positive plate and therefore the ball touches the other plate (earthed), which has negative charge due to induction. On touching this plate, the positive charge of the ball gets neutralized and in turn the ball shares negative charge of the earthed plate and is again repelled from this plate also, and this process is repeated again and again.

Here it should be understood that since the positive plate is connected to high voltage generator, its potential and hence its charge will always remain same, as soon as this plate gives some of its charge to ball, excess charge flows from generator to the plate, and an equal negative charge is always induced on the other plate.

+ + + +

+

+ + + +

–

–

– – – +

+ +

+ +


Chapter 1 7

c In 1 gm of a solid, there are 5 × 1021 atoms. If one electron is removed from everyone of 0.01%

atoms of the solid, the charge gained by the solid is (given that electronic charge is 1.6 × 10–19 C)

(a) + 0.08 C (b) + 0.8 C (c) – 0.08 C (d) – 0.8 C

Solution: (a) To calculate charge, we will apply formula Q = ne for this, we must have number of electrons. Here, number of electrons %01.=n of 5 × 1021

i.e. 100

01.105 21 ××=n 421 10105 −××= = 5 × 1017

So Q = 5 × 1017 × 1.6 × 10–19 = 8 × 10–2 = 0.08 C

Since electrons have been removed, charge will be positive i.e. Q = + 0.08 C

Coulomb’s Law.

If two stationary and point charges 1Q and 2Q are kept at a distance r, then it is found that force of

attraction

or repulsion between them is 2

21

r

QQF ∝ i.e.,

221

r

QkQF = ; (k = Proportionality constant)

(1) Dependence of k : Constant k depends upon system of units and medium between the two charges.

(i) Effect of units

(a) In C.G.S. for air ,1=k 2

21

r

QQF = Dyne

(b) In S.I. for air 2

29

0

1094

1C

mNk

−×==

πε,

221

0

.4

1r

QQF

πε= Newton (1 Newton = 105 Dyne)

Note : ≅ =0ε Absolute permittivity of air or free space = 2

2121085.8

mNC−

× −

=

mFarad

. It’s

Dimension is ][ 243 ATML−

≅ 0ε Relates with absolute magnetic permeability ( 0µ ) and velocity of light (c) according

to the following relation 00

1

εµ=c

(ii) Effect of medium

(a) When a dielectric medium is completely filled in between charges rearrangement of the charges inside the dielectric medium takes place and the force between the same two charges decreases by a factor of K

Tricky example: 1

Q2 Q1

r


Chapter 1 8

cknown as dielectric constant or specific inductive capacity (SIC) of the medium, K is also called relative permittivity εr of the medium (relative means with respect to free space).

Hence in the presence of medium 2

21

0

.4

1

r

QQKK

FF air

m πε==

Here εεεε == rK 00 (permittivity of medium)

(b) If a dielectric medium (dielectric constant K, thickness t) is partially filled between the charges then

effective air separation between the charges becomes )( Kttr +−

Hence force 2

21

0 )(41

Kttr

QQF

+−=

πε

(2) Vector form of coulomb’s law : Vector form of Coulomb’s law is

,ˆ.. 12221

12321

12 rr

qqKr

r

qqKF == where 12r is the unit vector from first charge to second charge along the line

joining the two charges.

(3) A comparative study of fundamental forces of nature

S.No. Force Nature and formula Range Relative

strength

(i) Force of gravitation between two masses

Attractive F = Gm1m2/r2, obey’s Newton’s third law of motion, it’s a conservative force

Long range (between planets and between electron and proton)

1

(ii) Electromagnetic force (for stationary and moving charges)

Attractive as well as repulsive, obey’s Newton’s third law of motion, it’s a conservative force

Long (upto few kelometers) 3710

(iii) Nuclear force (between nucleons)

Exact expression is not known till date. However in some cases empirical formula 0/

0rreU can

be utilized for nuclear potential energy 0U and 0r are constant.

Short (of the order of nuclear size 10–15 m)

1039

(strongest)

(iv) Weak force (for processes like β decay)

Formula not known Short (upto 10–15m) 1024

Note : ≅ Coulombs law is not valid for moving charges because moving charges produces magnetic field also.

≅ Coulombs law is valid at a distance greater than .10 15 m−

≅ A charge 1Q exert some force on a second charge 2Q . If third charge 3Q is brought near, the

force of 1Q exerted on 2Q remains unchanged.

≅ Ratio of gravitational force and electrostatic force between (i) Two electrons is 10–43/1. (ii) Two protons is 10–36/1 (iii) One proton and one electron 10–39/1.

≅ Decreasing order to fundamental forces nalGravitatioWeakneticElectromagNuclear FFFF >>>

Q2 Q1

r

K

Q2 Q1 K

r


Chapter 1 9

c(4) Principle of superposition : According to the principle of super position, total force acting on a

given charge due to number of charges is the vector sum of the individual forces acting on that charge due to all the charges.

Consider number of charge 1Q , 2Q , 3Q …are applying force on a

charge Q

Net force on Q will be

nnnet FFFFF

++++= −121 ..........

Concepts

Two point charges separated by a distance r in vacuum and a force F acting between them. After filling a dielectric medium having dielectric constant K completely between the charges, force between them decreases. To maintain the force as before

separation between them changes to Kr . This distance known as effective air separation.

Example: 8 Two point charges Cµ3+ and Cµ8+ repel each other with a force of 40 N. If a charge

of Cµ5− is added to each of them, then the force between them will become

(a) N10− (b) N10+ (c) N20+ (d) N20−

Solution: (a) Initially 2

121083r

kF−××

×= and Finally 2

121032

rk'F

−××−= so

41

−=F

'F ⇒ N'F 10−=

Example: 9 Two small balls having equal positive charge Q (coulomb) on each are suspended by two insulated string of equal length L meter, from a hook fixed to a stand. The whole set up is taken in satellite into space where there is no gravity (state of weight less ness). Then the angle between the string and tension in the string is

[IIT-JEE 1986]

(a) 2

2

0 )2(.

41

,180L

Qo

πε

(b) 2

2

0.

41

,90LQ

πε

(c) 2

2

0 2.

41

,180L

Qπε

(d) 20 4

.4

1,180

LQLo

πε

Solution: (a) In case to weight less ness following situation arises

So angle 180=θ and force ( )2

2

0 2.

41

L

QF

πε=

Example: 10 Two point charges 1 Cµ & Cµ5 are separated by a certain distance. What will be ratio of forces acting on these two

(a) 5:1 (b) 1:5 (c) 1:1 (d) 0

Solution: (c) Both the charges will experience same force so ratio is 1:1

r1

r2 r3

Q

Q1

Q2

Q3

Qn – 1

Qn

Examples based on Coulomb’s

+Q – Q

L L

+Q +Q 180o

L

+Q

L

+Q


Chapter 1 10

cExample: 11 Two charges of Cµ40 and Cµ20− are placed at a certain distance apart. They are touched

and kept at the same distance. The ratio of the initial to the final force between them is

(a) 1:8 (b) 1:4 (c) 1 : 8 (d) 1 : 1

Solution: (a) Since only magnitude of charges are changes that’s why 21qqF ∝ ⇒ 18

10102040

21

21

2

1 =××

==q'q'qq

FF

Example: 12 A total charge Q is broken in two parts 1Q and 2Q and they are placed at a distance R from each other. The maximum force of repulsion between them will occur, when

(a) RQ

QQRQ

Q −== 12 , (b) 3

2,

4 12Q

QQQ

Q −== (c) 4

3,

4 12Q

QQ

Q == (d)

2

,2 21

QQ

QQ ==

Solution: (d) Force between charges 1Q and 2Q ( )

211

221

R

QQQk

R

QQkF

−==

For F to be maximum, 01=

dQdF

i.e., ( )

02

211

1=

−

RQQQ

kdQd

or 2

,02 11Q

QQQ ==−

Hence 221Q

QQ ==

Example: 13 The force between two charges 0.06m apart is 5 N. If each charge is moved towards the other by 0.01m, then the force between them will become

(a) 7.20 N (b) 11.25 N (c) 22.50 N (d) 45.00 N Solution: (b) Initial separation between the charges = 0.06m Final separation between the charges = 0.04m

Since 21r

F ∝ ⇒ 2

1

2

2

1

=

rr

FF

⇒ 94

06.004.05 2

2=

=

F⇒ NF 25.112 =

Example: 14 Two charges equal in magnitude and opposite in polarity are placed at a certain distance apart and force acting between them is F. If 75% charge of one is transferred to another, then the force between the charges becomes

(a) 16F

(b) 169F

(c) F (d) F1615

Solution: (a)

Initially 2

2

rQ

kF = Finally 16

4.

2

2

Fr

Qk

'F =

=

Example: 15 Three equal charges each +Q, placed at the corners of on equilateral triangle of side a what

will be the force on any charge

=

041πε

k

(a) 2

2

akQ

(b) 2

22akQ

(c) 2

22akQ

(d) 2

23akQ

Solution: (d) Suppose net force is to be calculated on the charge which is kept at A. Two charges kept at B and C are applying force on that particular charge, with direction as shown in the figure.

r

– Q +Q

A B

r

– Q/4 +Q/4

A B


Chapter 1 11

c

Since 2

2

a

QkFFF cb ===

So, 60cos222CBCBnet FFFFF ++=

2

233

akQ

FFnet ==

Example: 16 Equal charges Q are placed at the four corners A, B, C, D of a square of length a. The magnitude of the force on the charge at B will be

(a) 20

2

4

3

a

Q

πε (b) 2

0

2

4

4

a

Q

πε (c)

20

2

4

2211

a

Q

πε

+ (d)

2

0

2

4

2

12

a

Q

πε

+

Solution: (c) After following the guidelines mentioned above

DCADACnet FFFFFF ++=+= 22

Since 2

2

a

kQFF CA == and

2

2

)2(a

kQFD =

+=+=

21

22

22

2

2

2

2

2

a

kQ

a

kQ

a

kQFnet

+=

2221

4 20

2

aQπε

Example: 17 Two equal charges are separated by a distance d. A third charge placed on a perpendicular bisector at x distance, will experience maximum coulomb force when

(a) 2

dx = (b)

2d

x = (c) 22

dx = (d)

32

dx =

Solution: (c) Suppose third charge is similar to Q and it is q

So net force on it Fnet = 2F cosθ

Where

+

=

4

.4

12

20 dx

QqF

πε and

4

cos2

2 dx

x

+

=θ

∴ 2/32

20

2/122

220

44

2

44

.4

12

+

=

+

×

+

×=d

x

Qqx

dx

x

dx

QqFnet

πεπε

for Fnet to be maximum 0=dx

dFnet i.e. 0

44

22/32

20

=

+

dx

Qqxdxd

πε

or 04

34

2/5222

2/322 =

+−

+

−−d

xxd

x i.e. 22

dx ±=

60o

60o

A

B C +Q +Q

+Q

FB FC

A

D C

B

FC

FAC FA

FD

+Q +Q

+Q

θ

B C Q Q

F F θ

q

x θ θ

2

d

2

d

4/22 dx + 4/22 dx +


Chapter 1 12

cExample: 18 ABC is a right angle triangle in which AB = 3 cm, BC = 4 cm and

2π

=∠ABC . The three

charges +15, +12 and – 20 e.s.u. are placed respectively on A, B and C. The force acting on B is

(a) 125 dynes (b) 35 dynes (c) 25 dynes (d) Zero

Solution: (c) Net force on B 22CAnet FFF +=

( )

dyneFA 203

12152 =

×=

( )

dyneFC 154

20122 =

×=

dyneFnet 25=

Example: 19 Five point charges each of value +Q are placed on five vertices of a regular hexagon of side L. What is the magnitude of the force on a point charge of value – q placed at the centre of the hexagon [IIT-JEE 1992]

(a) 2

2

LQ

k (b) 2

2

4LQ

k

(c) Zero (d) Information is insufficient

Solution: (a) Four charges cancels the effect of each other, so the net force on the charge placed at centre due to remaining fifth charge is

2

2

LQ

kF =

Example: 20 Two small, identical spheres having +Q and – Q charge are kept at a certain distance. F force acts between the two. If in the middle of two spheres, another similar sphere having +Q charge is kept, then it experience a force in magnitude and direction as

(a) Zero having no direction (b) 8F towards +Q charge

(c) 8F towards – Q charge (d) 4F towards +Q charge

Solution: (c) Initially, force between A and C 2

2

r

QkF =

When a similar sphere B having charge +Q is kept at the mid point of line joining A and C, then Net force on B is

CAnet FFF +=( ) ( )

Fr

kQ

r

kQ

r

Qk 88

22 2

2

2

2

2

2

==+= . (Direction is shown

in figure)

A

FC

FA

3 cm

4 cm

+15 esu

– 20 esu +12 esu B C

22CAnet FFF +=

r

A C

– Q +Q

B

r/2 r/2

+Q

FA FC

L +Q +Q

+Q

+Q +Q

– Q


Chapter 1 13

c Two equal spheres are identically charged with q units of electricity separately. When they are placed at

a distance 3R from centre-to-centre where R is the radius of either sphere the force of repulsion between them is

(a) 2

2

0.

41

R

qπε

(b) 2

2

0 9.

41

R

qπε

(c) 2

2

0 4.

41

R

qπε

(d) None of these

Solution: (a) Generally students give the answer 2

2

0 )3(41

R

qπε

but it is not true. Since the charges are not uniformly

distributed, they cannot be treated as point charges and so we cannot apply coulombs law which is a law for point charges. The actual distribution is shown in the figure above.

Electrical Field.

A positive charge or a negative charge is said to create its field around itself. If a charge 1Q exerts a force

on charge 2Q placed near it, it may be stated that since 2Q is in the field of 1Q , it experiences some force, or it

may also be said that since charge 1Q is inside the field of 2Q , it experience some force. Thus space around a

charge in which another charged particle experiences a force is said to have electrical field in it.

(1) Electric field intensity )(E

: The electric field intensity at any point is defined as the force

experienced by a unit positive charge placed at that point. 0q

F E

=

Where 00 →q so that presence of this charge may not affect the source charge Q and its electric field is

not changed, therefore expression for electric field intensity can be better written as 0

0q qF

LimE0

→=

(2) Unit and Dimensional formula : It’s S.I. unit –metercoulomb

Joulemetervolt

coulombNewton

×== and

C.G.S. unit – Dyne/stat coulomb.

Dimension : [ E ] =[ 13 −− AMLT ]

(3) Direction of electric field : Electric field (intensity) E

is a vector quantity. Electric field due to a positive charge is always away from the charge and that due to a negative charge is always towards the charge

(q0) F

P

+Q

+Q E

– Q E

+ +

+ +

+

+

+ +

+

+

+

+ +

+ +

+

+

+

+

+

+

+

+

Tricky example: 2


Chapter 1 14

c(4) Relation between electric force and electric field : In an electric field E

a charge (Q) experiences a force QEF = . If charge is positive then force is directed in the direction of field while if charge is

negative force acts on it in the opposite direction of field

(5) Super position of electric field (electric field at a point due to various charges) : The resultant electric field at any point is equal to the vector sum of electric fields at that point due to various charges.

...321 +++= EEEE

The magnitude of the resultant of two electric fields is given by

θcos2 2122

21 EEEEE ++= and the direction is given by

θθ

αcos

sintan

21

2

EEE+

=

(6) Electric field due to continuous distribution of charge : A system of closely spaced electric charges forms a continuous charge distribution

Continuous charge distribution

Linear charge distribution Surface charge distribution Volume charge distribution

In this distribution charge distributed on a line.

For example : charge on a wire, charge on a ring etc. Relevant parameter is λ which is called linear

charge density i.e.,

lengthcharge

=λ

RQπ

λ2

=

In this distribution charge distributed on the surface.

For example : Charge on a conducting sphere, charge on a sheet etc. Relevant parameter is σ which is

called surface charge density i.e.,

areacharge

=σ

24 R

Q

πσ =

In this distribution charge distributed in the whole volume of the body.

For example : Non conducting charged sphere. Relevant parameter is ρ which

is called volume charge density i.e.,

volumecharge

=ρ

3

34

R

Q

πρ =

To find the field of a continuous charge distribution, we divide the charge into infinitesimal charge

elements. Each infinitesimal charge element is then considered, as a point charge and electric field dE is determined due to this charge at given point. The Net field at the given point is the summation of fields of all

the elements. i.e., ∫= dEE

Electric Potential.

(1) Definition : Potential at a point in a field is defined as the amount of work done in bringing a unit positive test charge, from infinity to that point along any arbitrary path (infinity is point of zero potential).

Electric potential is a scalar quantity, it is denoted by V; 0q

WV =

+

+

+ + +

+

+

+ +

R

Q

Spherical shall

+ +

+ + +

+

+

+ +

R + + +

+ +

+

Q

Non conducting sphere

+ +

+

+ + +

+

+

+ R

Circular charged ring

Q

E2

E1

E

θ α

+Q E

F – Q E


Chapter 1 15

c(2) Unit and dimensional formula : S. I. unit – volt

CoulombJoule

= C.G.S. unit – Stat volt (e.s.u.); 1 volt

3001

= Stat volt Dimension – ][][ 132 −−= ATMLV

(3) Types of electric potential : According to the nature of charge potential is of two types

(i) Positive potential : Due to positive charge. (ii) Negative potential : Due to negative charge.

(4) Potential of a system of point charges : Consider P is a point at which net electric potential is to be determined due to several charges. So net potential at P

( )

...4

4

3

3

2

2

1

1 +−

+++=rQ

krQ

krQ

krQ

kV In general ∑=

=X

i i

i

rkQ

V1

Note : ≅ At the centre of two equal and opposite charge V = 0 but 0≠E

≅ At the centre of the line joining two equal and similar charge 0,0 =≠ EV

(5) Electric potential due to a continuous charge distribution : The potential due to a continuous charge distribution is the sum of potentials of all the infinitesimal charge elements in which the distribution may

be divided i.e., ∫∫ ==r0

dQdVV

πε4,

(6) Graphical representation of potential : When we move from a positive charge towards an equal negative charge along the line joining the two then initially potential decreases in magnitude and at centre become zero, but this potential is throughout positive because when we are nearer to positive charge, overall potential must be positive. When we move from centre towards the negative charge then though potential remain always negative but increases in magnitude fig. (A). As one move from one charge to other when both charges are like, the potential first decreases, at centre become minimum and then increases Fig. (B).

(7) Potential difference : In an electric field potential difference between two points A and B is defined as equal to the amount of work done (by external agent) in moving a unit positive charge from point A to point B.

r1

r2

r3

P

– Q1 r4

– Q2 + Q3

– Q4

+ q – q

V

x

Y

X O

(A)

v

x

Y

X O

(B)

+ q + q


Chapter 1 16

ci.e.,

0qW

VV AB =− in general VQW ∆= . ; =∆V Potential difference through which charge Q moves.

Electric Field and Potential Due to Various Charge Distribution.

(1) Point charge : Electric field and potential at point P due to a point charge Q is

rrQ

kE ˆor2

==

2rQ

kE

=

041πε

k , rQ

kV =

Note : ≅ Electric field intensity and electric potential due to a point charge q, at a distance t1 + t2

where t1 is thickness of medium of dielectric constant K1 and t2 is thickness of medium of dielectric constant K2 are :

2

2210 )(41

KtKt

QE

1 +=

πε ;

)221041

KtK(t

QV

1 +=

πε

(2) Line charge

(i) Straight conductor : Electric field and potential due to a charged straight conducting wire of length l and charge density λ

(a) Electric field : )sin(sin βαλ+=

rk

Ex and )cos(cos αβλ−=

rk

Ey

If α = β; αλsin

2rk

Ex = and Ey = 0

If l → ∞ i.e. α = β = 2π

; rk

Exλ2

= and Ey = 0 so r

Enet02πε

λ=

If α = 0, 2π

β = ; r

kEE yx

λ== |||| so

rk

EEE yxnetλ222 =+=

(b) Potential :

++

−+=

1

1log

2 22

22

0 lr

lrV eπε

λ for infinitely long conductor crV e +

−= log

2 0πελ

r

Q P

+ +

+ +

l + β

α P

r

Ey

Ex


Chapter 1 17

c(ii) Charged circular ring : Suppose we have a charged circular ring of radius R and charge Q. On it’s

axis electric field and potential is to be determined, at a point ‘x’ away from the centre of the ring.

(a) Electric field : Consider an element carrying charge dQ . It’s electric field ( )22 xRKdQ

dE+

= directed

as shown. It’s component along the axis is θcosdE and perpendicular to the axis is θsindE . By

symmetry ∫ = 0sinθdE , hence 212222 )(

.)(

cosxRx

xRkdQ

dEE++

== ∫∫ θ

( ) 2322 xR

kQxE

+= directed away from the centre if Q is positive

(b) Potential : 22

0

.4

1

Rx

QV

+=

πε

Note : At centre x = 0 so Ecentre= 0 and R

kQVcentre =

At a point on the axis such that x >> R 2x

kQE = and

xkQ

V =

At a point on the axis if 2

Rx ±= ,

20

max36 a

QE

πε=

(3) Surface charge :

(i) Infinite sheet of charge : Electric field and potential at a point P as shown

)(2 0

orEE ∝=εσ

and Cr

V +−=02ε

σ

(ii) Electric field due to two parallel plane sheet of charge : Consider two large, uniformly charged parallel. Plates A and B, having surface charge densities are Aσ and Bσ respectively. Suppose net electric

field at points P, Q and R is to be calculated.

At P, )(21

)(0

BABAP EEE σσε

+=+=

At Q, )(21

)(0

BABAQ EEE σσε

−=−= ; At R, )(21

)(0

BABAR EEE σσε

+−=+−=

x

+

O

2

R−

2

R

x

E

+ + + + + + +

+ + +

+ +

+ + +

+ r

P

+ + + + + + +

+ + +

+ +

+ +

+

+ + •

A

• • R P

EB

EA

EB

EA EA EB

B

Q + + + + + + +

+ + +

+ +

+ +

+

+ +

+

+

+

+

+

+ +

+ +

+

+

+

+ +

dE sin θ

dE cos θ

Ed

θ θ

R P

x

dQ


Chapter 2 18

cNote : If σσ +=A and σσ −=B then 0,,0

0

=== RQp EEEεσ

. Thus in case of two infinite

plane sheets of charges having equal and opposite surface charge densities, the field is non-zero only in the space between the two sheets and is independent of the distance between them i.e., field is uniform in this region. It should be noted that this result will hold good for finite plane sheet also, if they are held at a distance much smaller then the dimensions of sheets i.e., parallel plate capacitor.

(iii) Conducting sheet of charge :

0εσ

=E

Cr

V +−=0ε

σ

(iv) Charged conducting sphere : If charge on a conducting sphere of radius R is Q as shown in figure then electric field and potential in different situation are –

(a) Out side the sphere : P is a point outside the sphere at a distance r from the centre at which electric field and potential is to be determined.

Electric field at P

2

0

2

20

.4

1r

RrQ

Eout εσ

πε== and

rR

rQ

Vout0

2

0

.4

1εσ

πε==

×=×=

24 RAQπσ

σ

(b) At the surface of sphere : At surface Rr =

So, 0

20

.4

1εσ

πε==

R

QEs and

00

.4

1εσ

πεR

RQ

Vs ==

(c) Inside the sphere : Inside the conducting charge sphere electric field is zero and potential remains constant every where and equals to the potential at the surface.

0=inE and inV = constant sV=

Note : ≅ Graphical variation of electric field and potential of a charged spherical conductor with distance

+ +

+

+

+ +

+ + +

+

r P

+Q

R

+ + +

+

+

+ + +

+ +

+

+ +

R

+Q + +

+

+

+

+ + +

+ +

+

+ +

Hollow Solid

E=σ/ε0

+ +

+ + + +

– – –

–

– –


Chapter 2 19

c

(4) Volume charge (charged non-conducting sphere) :

Charge given to a non conducting spheres spreads uniformly throughout it’s volume.

(i) Outside the sphere at P

2

0

.4

1

r

QEout πε

= and rQ

Vout .4

1

0πε= by using

3

34

R

Q

πρ =

2

0

3

3 r

REout ε

ρ= and

rR

Vout0

3

3ερ

=

(ii) At the surface of sphere : At surface Rr =

0

20 3

.4

1ερ

πεR

R

QEs == and

0

2

0 3.

41

ερ

πεR

RQ

Vs ==

(iii) Inside the sphere : At a distance r from the centre

3

0

.4

1

R

QrEin πε

= 03ε

ρr= rEin ∝ and

0

22

3

22

0 6)3(

2

]3[4

1ε

ρπε

rRR

rRQVin

−=

−=

Note : ≅ At centre 0=r So, sVRQ

V23

.4

123

0centre =×=

πε i.e.,

outsurfacecentre VVV >>

≅ Graphical variation of electric field and potential with distance

+Q

r

P R +

+ +

+

+ +

+

+

+ +

+ +

+

+ +

+ + + +

+

+

+

+

+

+

+ +

+ + + +

O

O R

r

routV

1∝

VS

r =R

V - r graph

O Ein=0

E

O R

r

2

1

rEout ∝

E - r graph

E-r graph

O

E

r

2

1

rEout ∝

r = R

Ein ∝ r

R O

+

+

+ + + +

+ + +

+

+ +

+ + + V-r graph

R O

+ +

+ + +

+ + + +

+

+ +

+ +

rVout

1∝

VC

VS

r = R r

+


Chapter 2 20

c(5) Electric field and potential in some other cases

(i) Uniformly charged semicircular ring : lengthcharge

=λ

At centre :

2

022

2R

QRK

Eεπ

λ==

R

QR

KQV

04πε==

(iii) Charged cylinder of infinite length

(a) Conducting (b) Non-conducting

For both type of cylindrical charge distribution r

Eout02πε

λ= , and

REsuface

02πελ

= but for conducting

0=inE and for non-conducting 2

0in

2 Rr

Eπελ

= . (we can also write formulae in form of ρ i.e., r

RE

0

2

out 2ερ

= etc.)

(ii) Hemispherical charged body :

At centre O, 04ε

σ=E

02ε

σRV =

(iv) Uniformly charged disc

At a distance x from centre O on it’s axis

+−=

220

12 Rx

xE

εσ

−+= xRxV 22

02εσ

Note : ≅ Total charge on disc Q = σπR2

≅ If x → 0, 02

–~εσ

E i.e. for points situated near the disc, it behaves as an infinite sheet of charge.

σ

R

x O

• + +

+ +

+ + +

+ +

+ + +

+ + + +

+ + + + +

+

O

+Q

+

R

+

+ +

+ + + +

+ +

+

r P

R +

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

r P


Chapter 2 21

cConcepts

No point charge produces electric field at it’s own position.

Since charge given to a conductor resides on it’s surface hence electric field inside it is zero.

The electric field on the surface of a conductor is directly proportional to the surface charge density at that point i.e, σ∝E

Two charged spheres having radii 1r and 2r charge densities 1σ and 2σ respectively, then the ratio of electric field on their

surfaces will be 2

1

22

2

1

2

1

r

r

EE

==σσ

=24 r

Q

πσ

In air if intensity of electric field exceeds the value CN/103 6× air ionizes.

A small ball is suspended in a uniform electric field with the help of an insulated thread. If a high energy x–ray bean falls on the ball, x-rays knock out electrons from the ball so the ball is positively charged and therefore the ball is deflected in the direction of electric field.

Electric field is always directed from higher potential to lower potential.

A positive charge if left free in electric field always moves from higher potential to lower potential while a negative charge moves from lower potential to higher potential.

The practical zero of electric potential is taken as the potential of earth and theoretical zero is taken at infinity.

An electric potential exists at a point in a region where the electric field is zero and it’s vice versa.

A point charge +Q lying inside a closed conducting shell does not exert force another point charge q placed outside the shell as shown in figure

Actually the point charge +Q is unable to exert force on the charge +q because it can not produce electric field at the position of +q. All the field lines emerging from the point charge +Q terminate inside as these lines cannot penetrate the conducting medium (properties of lines of force).

The charge q however experiences a force not because of charge +Q but due to charge induced on the outer surface of the shell.

+ +

+

+ + +

+

+ +

+

+ +

+

+ + +

E = 0

+

+

+ +

+ + +

+

+

+ + +

+

+ +

E = 0

+

F= QE

X–Ray

E

+q +Q

+

+Q

+ +

+

+

+ +

+ + +

+

+

+

+

– –

– – –

–

–


Chapter 2 22

c

Example: 21 A half ring of radius R has a charge of λ per unit length. The electric field at the centre is

=

041πε

k

[CPMT 2000; CBSE PMT 2000; REE 1999]

(a) Zero (b) Rkλ

(c) Rkλ2

(d) R

kπλ

Solution: (c) θRddl =

Charge on .θλRddl =

Field at C due to dERRd

kdl == 2.θλ

We need to consider only the component θcosdE , as the component θsindE will cancel out because of the field at C due to the symmetrical element dl′,

The total field at C is ∫=2

0cos2

πθdE

Rkd

Rk λθθλ π

2cos22

0== ∫

=

202 R

Q

πε

Example: 22 What is the magnitude of a point charge due to which the electric field 30 cm away has the magnitude 2 newton/coulomb ]1094/1[ 29

0 Nm×=πε [MP PMT 1996]

(a) coulomb11102 −× (b) coulomb11103 −× (c) coulomb11105 −× (d) coulomb11109 −×

Solution: (a) By using 20

.4

1rQ

Eπε

= ; ( )22

9

10301092

−×××=

Q⇒ CQ 11102 −×=

Example: 23 Two point charges Q and – 3Q are placed at some distance apart. If the electric field at the location of Q is E, then at the locality of – 3Q, it is

(a) E− (b) E/3 (c) E3− (d) – E/3

Solution: (b) Let the charge Q and – 3Q be placed respectively at A and B at a distance x Now we will determine the magnitude and direction to the field produced by charge – 3Q at A, this is E

as mentioned in the Example.

∴2

3xQ

E = (along AB directed towards negative charge)

Now field at location of – 3Q i.e. field at B due to charge Q will be 32E

xQ

E' == (along AB directed

away from positive charge)

Example: 24 Two charged spheres of radius 1R and 2R respectively are charged and joined by a wire. The ratio of electric field of the spheres is [CPMT 1999 Similar to CBSE 2003]

(a) 2

1

RR

(b) 1

2

RR

(c) 22

21

R

R (d)

21

22

R

R

Solution: (b) After connection their potential becomes equal i.e., 2

2

1

1 ..

RQk

RQ

k = ; ⇒ 2

1

2

1

RR

QQ

=

– e + e E ≠ 0

V = 0 Examples based on electric field and electric potential

θ θ

θ

dθ

dl

dl′ dE

C

A B

x

Q – 3 Q


Chapter 2 23

cRatio of electric field .

1

22

1

2

2

1

2

1

RR

RR

QQ

EE

=

×=

Example: 25 The number of electrons to be put on a spherical conductor of radius 0.1m to produce an electric field of 0.036 N/C just above its surface is [MNR 1994]

(a) 5107.2 × (b) 5106.2 × (c) 5105.2 × (d) 5104.2 ×

Solution: (c) By using 2R

QkE = , where R = radius of sphere so 0.036 =

( )29

1.0109

ne×× ⇒ 5105.2 ×=n

Example: 26 Eight equal charges each +Q are kept at the corners of a cube. Net electric field at the centre will be

=

041πε

k

(a) 2r

kQ (b)

2

8

r

kQ (c)

2

2

r

kQ (d) Zero

Solution: (d) Due to the symmetry of charge. Net Electric field at centre is zero.

Note :

Example: 27 q, 2q, 3q and 4q charges are placed at the four corners A, B, C and D of a square. The field at the centre O of the square has the direction along.

[CPMT 1989]

(a) AB (b) CB (c) AC (d) BD

Solution: (b) By making the direction of electric field due to all charges at centre. Net electric field has the direction along CB

Example: 28 Equal charges Q are placed at the vertices A and B of an equilateral triangle ABC of side a. The magnitude of electric field at the point A is

(a) 204 a

Qπε

(b) 204

2aQ

πε (c) 2

043

aQ

πε (d) 2

02 aQ

πε

Solution: (c) As shown in figure Net electric field at A

+q

a

a

a

+q

+q

E=0

Equilateral triangle

a a

a

a

q

q q

q

E = 0

Square

3q

2q q

4q

O

A B

C D


Chapter 2 24

c 60cos222

CBCB EEEEE ++=

2

0.

41

aQ

EE CB πε==

So, 204

3aQ

Eπε

=

Example: 29 Four charges are placed on corners of a square as shown in figure having side of 5 cm. If Q is one micro coulomb, then electric field intensity at centre will be [RPET 1999]

(a) CN /1002.1 7× upwards

(b) CN /1004.2 7× downwards

(c) CN /1004.2 7× upwards

(d) CN /1002.1 7× downwards

Solution: (a) |||| AC EE > so resultant of AC EE & is ACCA EEE −= directed toward Q

Also |||| DB EE > so resultant of BE and DE i.e.

DBBD EEE −= directed toward – 2Q charge hence Net electric field at centre is

( ) ( )22BDCA EEE += .… (i)

By proper calculations CNEA /1072.0

102

5

10109|| 7

22

69 ×=

×

××=−

−

CNEB /1044.1

102

5

102109|| 7

22

69 ×=

×

×××=

−

−; CNEC /1044.1

102

5

102109|| 7

22

69 ×=

×

×××=

−

−

CNED /1072.0

102

5

10109|| 7

22

69 ×=

×

××=−

−; So, N/C.||E||E||E ACCA

710720 ×=−=

and ./1072.0|||||| 7 CNEEE DBBD ×=−= Hence from equation – (i) CNE /1002.1 7×= upwards

Example: 30 Infinite charges are lying at x = 1, 2, 4, 8…meter on X-axis and the value of each charge is Q. The value of intensity of electric field and potential at point x = 0 due to these charges will be respectively

(a) Q91012× N/C, 1.8 × 104 V (b) Zero, 1.2 × 104V

(c) Q9106× N/C, 9 × 103 V (d) Q9104 × N/C , 6 × 103 V

Solution: (a) By the superposition, Net electric field at origin

∞++++= ...

8

1

4

1

2

1

1

12222

kQE x = 0 x = 1 x = 2 x = 4 x = 8

6

A

B C +

+

FB FC E

a

a

a

Q – 2Q

+ 2Q – Q

– 2Q Q

+2Q – Q

O

A B

C D

ECA

EC

EB EA

ED

EBD Enet


Chapter 2 25

c

∞++++= ...

641

161

41

1kQE

∞++++ ...641

161

41

1 is an infinite geometrical progression it’s sum can be obtained by using the

formula r

aS

−=∞ 1

; Where a = First term, r = Common ratio.

Here 1=a and 41

=r so, 34

4/111

.....641

161

41

1 =−

=∞++++ .

Hence CNQQE /101234

109 99 ×=×××=

Electric potential at origin

∞+

×+

×+

×+

×=

−−−−

.......8101

4101

2101

1101

41 6666

0πεV

−×=

∞++++××= −

21

1

1109............

81

41

21

110109 369 volt4108.1 ×=

Note : ≅ In the arrangement shown in figure +Q and – Q are alternatively and

equally spaced from each other, the net potential at the origin O is x

QV e 2log

.4

1

0πε=

Example: 31 Potential at a point x-distance from the centre inside the conducting sphere of radius R and charged with charge Q is [MP PMT 2001]

(a) RQ

(b) xQ

(c) 2x

Q (d) xQ

Solution: (a) Potential inside the conductor is constant.

Example: 32 The electric potential at the surface of an atomic nucleus (Z = 50) of radius V5109 × is

(a) 80 V (b) V6108 × (c) 9 V (d) V5109 ×

Solution: (b) Vr

neV 6

15

1999 108

109

106.150109109 ×=

×

××××=××=

−

−

Example: 33 Eight charges having the valves as shown are arranged symmetrically on a circle of radius 0.4m in air. Potential at centre O will be

+Q – Q +Q – Q

x 2x 3x 4x O

+5 µC

+11 µC

– 5 µC

+7 µC

– 5 µC +7 µC

+15 µC

– 7 µC

O


Chapter 2 26

c(a) volt41063 × (b) volt101063 × (c) volt61063 × (d) Zero

Solution: (a) Due to the principle of superposition potential at O

voltV 46

96

01063

4.01028

1094.01028

41

×=×

××=×

×=−−

πε

Example: 34 As shown in the figure, charges +q and –q are placed at the vertices B and C of an isosceles triangle. The potential at the vertex A is [MP PET 2000]

(a) 220

2.

41

ba

q

+πε (b)

220.

41

ba

q

+πε (c)

220

)(.

41

ba

q

+

−πε

(d) Zero

Solution: (d) Potential at A = Potential due to (+q) charge + Potential due to (– q) charge

0)(

41

.4

1220220=

+

−+

+=

ba

q

ba

qπεπε

Example: 35 A conducting sphere of radius R is given a charge Q. consider three points B at the surface, A at centre and C at a distance R/2 from the centre. The electric potential at these points are such that (a) VA = VB = VC (b) VA = VB ≠ VC (c) VA ≠ VB ≠ VC (d) VA ≠ VB = VC

Solution: (a) Potential inside a conductor is always constant and equal to the potential at the surface.

Example: 36 Equal charges of 9103

10 −× coulomb are lying on the corners of a square of side 8 cm. The electric

potential at the point of intersection of the diagonals will be

(a) 900 V (b) V2900 (c) V2150 (d) V21500

Solution: (d) Potential at the centre O

2/

.4

14

0 a

QV

πε×= given CQ 910

310 −×= ⇒ mcma 21088 −×==

2

108

103

10

10952

9

9−

−

×

××××=V volt21500=

A point charge Q is placed outside a hollow spherical conductor of radius R, at a distance (r > R) from its

centre C. The field at C due to the induced charges on the conductor is

=

041πε

K

(a) Zero (b) ( )2Rr

QK

−

(c) 2r

QK directed towards Q (d)

2r

QK directed away from Q

Solution: (c) A according to the figure shown below. The total field at C must be zero. The field at C due to the

point charge is2r

QKE = towards left. The field at C due to the induced charges must be

2r

KQ

towards right i.e. directed towards Q.

+

– –

–

– –

–

+ +

+

+ +

+

r

C

R

+ Q

Tricky example: 3

A

B C –q +q

a

b b

O

Q Q

Q Q a

2a


Chapter 2 27

c

A point charge q is placed at a distance of r from the centre of an uncharged conducting sphere of

radius R (< r). The potential at any point on the sphere is

(a) Zero (b) rq

.4

1

0πε (c)

20

.4

1

r

qRπε

(d) R

qr 2

0.

41πε

Solution: (c) Since, potential V is same for all points of the sphere. Therefore, we can calculate its value at the centre of the sphere.

∴ '.4

1

0V

rq

V +=πε

; where V′ = potential at centre due to induced charge = 0 (because net

induced charge will be zero) ∴ rq

V .4

1

0πε= .

Potential Due to Concentric Spheres.

To find potential at a point due to concentric sphere following guideline are to be considered

Guideline 1: Identity the point (P) at which potential is to be determined.

Guideline 2: Start from inner most sphere, you should know where point (P) lies w.r.t. concerning sphere/shell (i.e. outside, at surface or inside)

Guideline 3: Then find the potential at the point (P) due to inner most sphere and then due to next and so on.

Guideline 4: Using the principle of superposition find net potential at required shell/sphere.

Standard cases

Case (i) : If two concentric conducting shells of radii r1 and r2(r2 > r1) carrying uniformly distributed charges Q1 and Q2 respectively. What will be the potential of each shell

To find the solution following guidelines are to be taken.

Here after following the above guideline potential at the surface of inner shell is

2

2

01

1

01 .

41

.4

1rQ

rQ

Vπεπε

+=

and potential at the surface of outer shell

2

2

02

1

02 .

41

.4

1rQ

rQ

Vπεπε

+=

Q1

r1

r2

Q2

Tricky example: 4


Chapter 2 28

cCase (ii) : The figure shows three conducting concentric shell of radii a, b and c (a < b < c) having charges Qa, Qb and Qc respectively what will be the potential of each shell

After following the guidelines discussed above

Potential at A;

++=

cQ

bQ

aQ

V cbaA

041πε

Potential at B;

++=

cQ

bQ

bQ

V cbaB

041πε

Potential at C;

++=

cQ

cQ

cQ

V cbaC

041πε

Case (iii) : The figure shows two concentric spheres having radii r1 and r2 respectively (r2 > r1). If charge on inner sphere is +Q and outer sphere is earthed then determine.

(a) The charge on the outer sphere (b) Potential of the inner sphere

(i) Potential at the surface of outer sphere 0.4

1.

41

20202 =+=

rQ'

rQ

Vπεπε

⇒ QQ' −=

(ii) Potential of the inner sphere 2010

1)(

41

.4

1rQ

rQ

V−

+=πεπε

−=

210

114 rr

Qπε

Case (iv) : In the case III if outer sphere is given a charge +Q and inner sphere is earthed then

(a) What will be the charge on the inner sphere (b) What will be the potential of the outer sphere (i) In this case potential at the surface of inner sphere is zero, so if Q' is the charge induced on inner sphere

then 04

1

2101 =

+=

rQ

rQ'

Vπε

i.e., Qrr

Q'2

1−=

(Charge on inner sphere is less than that of the outer sphere.) (ii) Potential at the surface of outer sphere

2020

2 .4

1.

41

rQ

rQ'

Vπεπε

+=

+−= Q

rr

Qr

V2

1

202 4

1πε

−=

2

1

201

4 rr

rQπε

Example: 37 A hollow metal sphere of radius 5 cm is charged such that the potential on its surface is 10 volts. The potential at the centre of the sphere is

(a) Zero (b) 10 V

(c) Same as at a point 5 cm away from the surface (d) Same as at a point 25 cm away from the surface

Solution: (b) Inside the conductors potential remains same and it is equal to the potential of surface, so here potential at the centre of sphere will be 10 V

Example: 38 A sphere of 4 cm radius is suspended within a hollow sphere of 6 cm radius. The inner sphere is charged to a potential 3 e.s.u. When the outer sphere is earthed. The charge on the inner sphere is

Q′ +Q

r1

r2

Qc

Qb Qa

c

b a B A

r2

r1

+Q

Examples based on concentric spheres


Chapter 2 29

c (a) 54 e.s.u. (b)

41

e.s.u. (c) 30 e.s.u. (d) 36 e.s.u.

Solution: (d) Let charge on inner sphere be +Q. charge induced on the inner surface of outer sphere will be –Q.

So potential at the surface of inner sphere (in CGS)

64

3QQ

−=

⇒ 36=Q e.s.u.

Example: 39 A charge Q is distributed over two concentric hollow spheres of radii r and )( rR > such that the surface densities are equal. The potential at the common centre is [IIT-JEE 1981]

(a) )(4)(

0

22

rRrRQ+

+πε

(b) rR

Q+

(c) Zero (d) )(4

)(22

0 rRrRQ+

+πε

Solution: (d) If 1q and 2q are the charges on spheres of radius r and R respectively, in accordance with conservation of charge

21 qqQ += ….(i)

and according to the given problem 21 σσ =

i.e., 2

22

1

44 Rq

rq

ππ= ⇒

2

2

2

1

Rr

qq

= …. (ii)

So equation (i) and (ii) gives )( 22

2

1rR

Qrq

+= and

)( 22

2

2rR

QRq

+=

Potential at common centre

+=

Rq

rq

V 21

041πε )(

)(.

41

)()(41

220

22220 rR

rRQrR

QRrR

Qr++

=

++

+=

πεπε

Example: 40 A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of – 3Q, the new potential difference between the two surfaces is

(a) V (b) 2V (c) 4V (d) –2V

Solution: (a) If a and b are radii of spheres and spherical shell respectively, potential at their surfaces will be

aQ

V .4

1

0sphere πε

= and bQ

V .4

1

0shell πε

=

and so according to the given problem.

shellsphere VVV −=

−=

baQ 11

4 0πε …. (i)

Now when the shell is given a charge –3Q the potential at its surface and also inside will change by

−=

bQ

V3

41

00 πε

So that now

−=

bQ

aQ

V3

41

0sphere πε

and

−=

bQ

bQ

V3

41

0shell πε

hence

Vba

QVV =

−=−

114 0

shellsphere πε

Example: 41 Three concentric metallic spheres A, B and C have radii a, b and c )( cba << and surface charge densities on them are σσ −, and σ respectively. The valves of AV and BV will be

4cm

6cm

+Q

q1

q2

r R

+ + + +

+ +

+ + +

+ +

+ +

Q

a

b

Sphere


Chapter 2 30

c (a)

+−−− cb

ba

cba2

00),(εσ

εσ

(b) ca

cba2

),( −−

(c)

+−−− cb

ca

cba2

00 ),(σε

σε

(d)

+− c

cb

ca 22

0εσ

and )(0

cba +−εσ

Solution: (a) Suppose charges on A, B and C are ba qq , and cq

Respectively, so 22 4

4aq

aq

aa

A πσπ

σσ ×=⇒== , 22 4

4bq

b

qb

bB πσ

πσσ ×−=⇒=−=

and 22 4

4cq

cq

cc

C πσπ

σσ ×=⇒==

Potential at the surface of A

ininsurface )()()( CBAA VVVV ++=

++=

cq

bq

aq cba

041πε

×+

×−+

×=

cc

bb

aa 222

0

44)(44

1 πσπσπσπε

[ ]]0

cbaVA −−=εσ

Potential at the surface of B

++=++=

cq

bq

bq

VVVV cbaCBAB

0insurfaceout 4

1)()()(

πε

×+

×−

×=

cc

bb

ba 222

0

4444

1 πσπσπσπε

+−= cb

ba2

0εσ

Electric Lines of Force.

(1) Definition : The electric field in a region is represented by continuous lines (also called lines of force). Field line is an imaginary line along which a positive test charge will move if left free.

Electric lines of force due to an isolated positive charge, isolated negative charge and due to a pair of charge are shown below

(Radially outward)

+

(Radially inward)

– + – • N

+ +

σ σ σ

A

B

C

a

b c


Chapter 2 31

c(2) Properties of electric lines of force

(i) Electric field lines come out of positive charge and go into the negative charge.

(ii) Tangent to the field line at any point gives the direction of the field at that point.

(iii) Field lines never cross each other.

(iv) Field lines are always normal to conducting surface.

(v) Field lines do not exist inside a conductor.

(vi) The electric field lines never form closed loops. (While magnetic lines of forces form closed loop)

(vii) The number of lines originating or terminating on a charge is proportional to the magnitude of charge. In the following figure electric lines of force are originating from A and terminating at B hence QA is positive while QB is negative, also number of electric lines at force linked with QA are more than those linked with QB hence |Q||Q| BA >

(viii) Number of lines of force per unit area normal to the area at a point represents magnitude of intensity (concept of electric flux i.e., EA=φ )

(ix) If the lines of forces are equidistant and parallel straight lines the field is uniform and if either lines of force are not equidistant or straight line or both the field will be non uniform, also the density of field lines is proportional to the strength of the electric field. For example see the following figures.

A

B

EA EB

(A)

+ +

+

+

+ +

– –

–

–

–

– (B)

A B

QA QB

+ –

N S


Chapter 2 32

c

(3) Electrostatic shielding : Electrostatic shielding/screening is the phenomenon of protecting a certain region of space from external electric field. Sensitive instruments and appliances are affected seriously with strong external electrostatic fields. Their working suffers and they may start misbehaving under the effect of unwanted fields.

The electrostatic shielding can be achieved by protecting and enclosing the sensitive instruments inside a hollow conductor because inside hollow conductors, electric fields is zero.

(i) It is for this reason that it is safer to sit in a car or a bus during lightening rather than to stand under a tree or on the open ground.

(ii) A high voltage generator is usually enclosed in such a cage which is earthen. This would prevent the electrostatic field of the generator from spreading out of the cage.

(iii) An earthed conductor also acts as a screen against the electric field. When conductor is not earthed field of the charged body C due to electrostatic induction continues beyond AB. If AB is earthed, induced positive charge neutralizes and the field in the region beyond AB disappears.

Equipotential Surface or Lines.

If every point of a surface is at same potential, then it is said to be an equipotential surface

or

for a given charge distribution, locus of all points having same potential is called “equipotential surface” regarding equipotential surface following points should keep in mind :

(1) The density of the equipotential lines gives an idea about the magnitude of electric field. Higher the density larger the field strength.

(2) The direction of electric field is perpendicular to the equipotential surfaces or lines.

(3) The equipotential surfaces produced by a point charge or a spherically charge distribution are a family of concentric spheres.

E

+ +

+ +

+ + + +

– – – –

– – – –

E

E

E

E

E = 0 q = 0

Hollow space

Conductor

C

– A B

+ + + + +

– – – – –

C

A B – – – – –

• X • Y

(A)

EX = EY

• X • Y

(B)

EX > EY


Chapter 2 33

c

(4) For a uniform electric field, the equipotential surfaces are a family of plane perpendicular to the field lines.

(5) A metallic surface of any shape is an equipotential surface e.g. When a charge is given to a metallic surface, it distributes itself in a manner such that its every point comes at same potential even if the object is of irregular shape and has sharp points on it.

If it is not so, that is say if the sharp points are at higher potential then due to potential difference between these points connected through metallic portion, charge will flow from points of higher potential to points of lower potential until the potential of all points become same.

(6) Equipotential surfaces can never cross each other

(7) Equipotential surface for pair of charges

Concepts

Unit field i.e. 1N/C is defined arbitrarily as corresponding to unit density of lines of force.

– + + +

Pair of two equal and opposite charges Pair of two equal and similar charges

+

+ +

+

+ +

+

+ +

+

+ +

+

+ +

+

• V = const.

O

Metallic charged sphere

+

+ +

+ + + +

+

+ +

+ +

+ +

+ +

V = const.

Charged metallic body of irregular shape

V = V2

V = V1

V1 V2 V3 V4 V5

Equipotential surface

V1 > V2 > V3 > V4 > V5


Chapter 2 34

c Number of lines originating from a unit charge is

0

1ε

It is a common misconception that the path traced by a positive test charge is a field line but actually the path traced by a unit positive test charge represents a field full line only when it moves along a straight line.

Both the equipotential surfaces and the lines of force can be used to depict electric field in a certain region of space. The advantage of using equipotential surfaces over the lines of force is that they give a visual picture of both the magnitude and direction of the electric field.

Example: 42 Three positive charges of equal value q are placed at the vertices of an equilateral triangle. The resulting lines of force should be sketched as in

(a) (b) (c) (d)

Solution (c) Option (a) shows lines of force starting from one positive charge and terminating at another. Option (b) has one line of force making closed loop. Option (d) shows all lines making closed loops. All these are not correct. Hence option (c) is correct

Example: 43 A metallic sphere is placed in a uniform electric field. The lines of force follow the path (s) shown in the figure as

(a) 1 (b) 2 (c) 3 (d) 4

Solution: (d) The field is zero inside a conductor and hence lines of force cannot exist inside it. Also, due to induced charges on its surface the field is distorted close to its surface and a line of force must deviate near the surface outside the sphere.

Example: 44 The figure shows some of the electric field lines corresponding to an electric field. The figure suggests

[MP PMT 1999]

•

•

•

•

• •

+q

+q +q

1

2

3

4

1

2

3

4

A B C

Examples based on electric lines of force


Chapter 2 35

c

(a) CBA EEE >> (b) CBA EEE == (c) BCA EEE >= (d) BCA EEE <=

Solution: (c)

Example: 45 The lines of force of the electric field due to two charges q and Q are sketched in the figure. State if

(a) Q is positive and qQ >

(b) Q is negative and qQ >

(c) q is positive and qQ <

(d) q is negative and qQ <

Solution: (c) q is +ve because lines of force emerge from it and qQ < because more lines emerge from q and less

lines terminate at Q.

Example: 46 The figure shows the lines of constant potential in a region in which an electric field is present. The magnitude of electric field is maximum at

(a) A (b) B (c) C (d) Equal at A, B and C

Solution: (b) Since lines of force are denser at B hence electric field is maximum at B

Example: 47 Some equipotential surface are shown in the figure. The magnitude and direction of the electric field is

(a) 100 V/m making angle 120o with the x-axis (b) 100 V/m making angle 60o with the x-axis

(c) 200 V/m making angle 120o with the x-axis (d) None of the above

Solution: (c) By using θcosdrEdV = suppose we consider line 1 and line 2 then

(30 – 20) = E cos 60o (20 – 10) × 10–2 So mvoltE /200= making in angle 120o with x-axis

Q q

30 V

20 V

• C

• B

40 V

50 V

A •

10 20 30

θ

20 V 30 V 40 V

θ θ θ

cm

θ = 30o

y

x

10 20 30 40 30o

20 V 30 V 40 V E

dr

1 2

120o

x

y


Chapter 2 36

c Which of the following maps cannot represent an electric field

(a) (b) (c) (d)

Solution: (a) If we consider a rectangular closed path, two parallel sides of it considering with lines of force as shown, then we find that work done along the closed path abcd is abE1 – cdE2 ≠ 0. Hence the field cannot represent a conservative field. But electric field is a conservative field. Hence a field represented by these lines cannot be an electric field.

A charge Q is fixed at a distance d in front of an infinite metal plate. The lines of force are represented

by

(a) (b) (c) (d)

Solution: (a) Metal plate acts as an equipotential surface, therefore the field lines should act normal to the surface of the metal plate.

Relation Between Electric Field and Potential.

In an electric field rate of change of potential with distance is known as potential gradient. It is a vector quantity and it’s direction is opposite to that of electric field. Potential gradient relates with electric field according to the following

relation ;drdV

E −= This relation gives another unit of electric field is metervolt

.

In the above relation negative sign indicates that in the direction of electric field potential decreases.

In space around a charge distribution we can also write

kEjEiEE zyxˆˆˆ ++=

where ,dxdV

Ex −= dydV

Ey −= and dzdV

Ez −=

+Q A B

dr

E

Tricky example: 5

• +Q

+Q +Q +Q +Q

c d

a b

Tricky example: 6


Chapter 2 37

cWith the help of formula ,

drdV

E −= potential difference between any two points in an electric field can be

determined by knowing the boundary conditions ∫∫ −=−=2

1

2

1

cos..r

r

r

rdrEdrEdV θ .

For example: Suppose A, B and C are three points in an uniform electric field as shown in figure.

(i) Potential difference between point A and B is

∫ ⋅−=−B

AAB drEVV

Since displacement is in the direction of electric field, hence θ = 0o

So, ∫∫ −=⋅−=⋅−=−B

A

B

AAB EddrEdrEVV 0cos

In general we can say that in an uniform electric field dV

E −= or dV

E =||

Another example dV

E =

(ii) Potential difference between points A and C is :

)(cos)(cos ABEACEdrEVVC

AAC −=−=−=− ∫ θθ = – Ed

Above relation proves that potential difference between A and B is equal to the potential difference between A and C i.e. points B and C are at same potential.

Concept

Negative of the slope of the V-r graph denotes intensity of electric field i.e. ErV

−==θtan

Example: 48 The electric field, at a distance of 20 cm from the centre of a dielectric sphere of radius 10 cm is 100 V/m. The ‘E’ at 3 cm distance from the centre of sphere is [RPMT 2001]

(a) 100 V/m (b) 125 V/m (c) 120 V/m (d) Zero

Solution: (c) For dielectric sphere i.e. for non-conducting sphere 2

.

r

qkEout = and

3R

kqrEin =

22 )1020(100

−×=

KQEout ⇒ KQ = 100 × (0.2)2 so

32

222

)1010(

)103()2.0(100−

−

×

×××=inE = 120 V/m

• •

• C

B A

d

θ

+ +

+

+

+ +

– –

–

–

–

–

E

V1 = V V2 = 0 d

+ –

E = V/d Example based on E = – dV/dr


Chapter 2 38

cExample: 49 In x-y co-ordinate system if potential at a point P(x, y) is given by axyV = ; where a is a constant, if r is

the distance of point P from origin then electric field at P is proportional to [RPMT 2000]

(a) r (b) r–1 (c) r—2 (d) r2

Solution: (a) By using drdV

E −= aydxdV

Ex −=−= , axdydV

Ey −=−=

Electric field at point P aryxaEEE yx =+=+= 2222 i.e., E ∝ r

Example: 50 The electric potential V at any point x, y, z (all in metres) in space is given by V = 4x2 volt. The electric field at the point (1m, 0, 2m) in volt/metre is [MP PMT 2001; IIT-JEE 1992; RPET 1999]

(a) 8 along negative X-axis (b) 8 along positive X-axis

(c) 16 along negative X-axis (d) 16 along positive Z-axis

Solution: (a) By using dxdV

E −= ⇒ xxdxd

E 8)4( 2 −=−= . Hence at point (1m, 0, 2m). E = – 8 volt/m i.e. 8 along – ve

x-axis.

Example: 51 The electric potential V is given as a function of distance x (metre) by V = (5x2 + 10x – 9) volt. Value of electric field at x = 1m is [MP PET 1999]

(a) – 20 V/m (b) 6 V/m (c) 11 V/m (d) – 23 V/m

Solution: (a) By using dxdV

E −= ; )1010()9105( 2 +=−+−= xxxdxd

E ,

at x = 1m mVE /20−=

Example: 52 A uniform electric field having a magnitude E0 and direction along the positive X-axis exists. If the electric potential V, is zero at X = 0, then, its value at X = +x will be [MP PMT 1987]

(a) V(x)= +xE0 (b) V(x)= – xE0 (c) V(x)= x2E0 (d) V(x)= – x2E0

Solution: (b) By using )()(

12

12

rrVV

rV

E−−

−=∆∆

−= ; 0

0)(0 −

−−=

xxV

E ⇒ V(x) = – xE0

Example: 53 If the potential function is given by V = 4x + 3y, then the magnitude of electric field intensity at the point (2, 1) will be [MP PMT 1999]

(a) 11 (b) 5 (c) 7 (d) 1

Solution: (b) By using i.e., 22yx EEE += ; 4)34( −=+−=−= yx

dxd

dxdV

Ex

and 3)34( −=+−=−= yxdyd

dydV

Ey

∴ CNE /5)3()4( 22 =−+−=


Chapter 2 39

c The variation of potential with distance R from a fixed point is as shown below. The electric field at

mR 5= is [NCERT 1975]

(a) 2.5 volt/m

(b) – 2.5 volt/m

(c) mvolt/52

(d) mvolt /52

−

Solution: (a) Intensity at 5 m is same as at any point between B and C because the slope of BC is same throughout (i.e. electric field between B and C is uniform). Therefore electric field at R = 5m is equal to the slope of

line BC hence by drdV

E−

= ;

mV

E 5.246)50(=

−−

−=

Note : ≅ At R = 1m , mV

E 5.2)02()05(

−=−−

−=

and at mR 3= potential is constant so E = 0.

Work Done in Displacing a Charge.

(1) Definition : If a charge Q displaced from one point to another point in electric field then work done in this process is VQW ∆×= where ∆V = Potential difference between the two position of charge Q.

( θcos. rErEV ∆=∆=∆ where θ is the angle between direction of electric field and direction of motion of

charge).

(2) Work done in terms of rectangular component of E and r : If charge Q is given a

displacement )ˆˆ( 321 krjrirr ++= in an electric field ).ˆˆˆ( 321 kEjEiEE ++= The work done is

)().( 332211 rErErEQrEQW ++== .

Conservation of Electric Field.

As electric field is conservation, work done and hence potential difference between two point is path independent and depends only on the position of points between. Which the charge is moved.

5 4 3 2 1

1 2 3 4 5 6 0

Pi

l i

Distance R in

5

4

3

2

1

1 2 3 4 5 6 O

A B

C

Distance R in metres

Pot

entia

l in

volts

A B

I

II

III

WI = WII = WIII

Tricky example: 7


Chapter 2 40

cConcept

No work is done in moving a charge on an equipotential surface.

Example: 54 A charge (– q) and another charge (+Q) are kept at two points A and B respectively. Keeping the charge (+Q) fixed at B, the charge (– q) at A is moved to another point C such that ABC forms an equilateral triangle of side l. The network done in moving the charge (– q) is

(a) l

Qq

041πε

(b) 2

041

l

Qqπε

(c) Qql04

1πε

(d) Zero

Solution: (d) Since l

kQVV CA ==

so 0)( =−= AC VVqW

Example: 55 The work done in bringing a 20 coulomb charge from point A to point B for distance 0.2 m is 2 Joule. The potential difference between the two points will be (in volt) [RPET 1999 Similar to MP PET 1999]

(a) 0.2 (b) 8 (c) 0.1 (d) 0.4

Solution: (c) VQW ∆= . ⇒ 2 = 20 × ∆V ⇒ ∆V = 0.1 volt

Example: 56 A charge +q is revolving around a stationary +Q in a circle of radius r. If the force between charges is F then the work done of this motion will be

[CPMT 1975, 90, 91, 97; NCERT 1980, 83; EAMCET 1994; MP PET 1993, 95;

MNR 1998; AIIMS 1997; DCE 1995; RPET 1998]

(a) F × r (b) rF π2× (c) r

Fπ2

(d) 0

Solution: (d) Since +q charge is moving on an equipotential surface so work done is zero.

Example: 57 Four equal charge Q are placed at the four corners of a body of side ‘a’ each. Work done in removing a charge – Q from its centre to infinity is [AIIMS 1995]

(a) 0 (b) a

Q

0

2

42πε

(c) a

Q

0

22πε

(d) a

Q

0

2

2πε

Solution: (c) We know that work done in moving a charge is W = Q∆V

Here )( 0 ∞−= VVQW 0=∞V ∴ W = Q × V0

Also aQ

aQ

a

QV

0000

24

24

2/.

41

4πεπεπε

==×=

WI = WII = WIII

I II

III A B

+ Q + q

Examples based on work done

A

B C

l l

l + Q

– q

Q Q

–Q

A B

C D Q Q

O

a

a

a a


Chapter 2 41

cSo,

aQ

W0

22πε

=

Example: 58 Two point charge 100 µC and 5 µC are placed at point A and B respectively with AB = 40 cm. The work done by external force in displacing the charge 5 µC from B to C, where BC = 30 cm, angle

229

0/109

41

and2

CNmABC ×==πε

π [MP PMT 1997]

(a) 9 J (b) J2081

(c) J259

(d) J49

−

Solution: (d) Potential at B due to +100 µC charge is

voltVB6

2

69 10

49

1040

10100109 ×=

×

×××=

−

−

Potential at C due to +100 µC charge is

voltVC6

2

69 10

59

1050

10100109 ×=

×

×××=

−

−

Hence work done in moving charge +5µC from B to C )(105 6

BC VVW −×= −

×−××= +− 666 10

49

1059

105W J49

−=

Example: 59 There is an electric field E in x-direction. If the work done in moving a charge 0.2 C through a distance of 2 metres along a line making an angle 60o with the x-axis is 4J, what is the value of E [CBSE 1995]

(a) 4 N/C (b) 8 N/C (c) CN/3 (d) 20 N/C

Solution: (d) By using VqW ∆×= and θcosrEV ∆=∆

So, θcosrqEW ∆=

60cos22.04 ×××== EjW

⇒ E = 20 N/C

Example: 60 An electric charge of 20 µC is situated at the origin of X-Y co-ordinate system. The potential difference between the points. (5a, 0) and (– 3a, 4a) will be

(a) a (b) 2a (c) Zero (d) 2

a

Solution: (c) a

kQVA 5

= and a

kQVB 5

=

∴ 0=− BA VV

Example: 61 Two identical thin rings each of radius R, are coaxially placed a distance R apart. If Q1 and Q2 are

respectively the charges uniformly spread on the two rings, the work done in moving a charge q from the centre of one ring to that of the other is

(a) Zero (b) 24

)12)((

0

21

R

QQq

πε−−

(c) R

QQq

0

21

42)(

πε+

(d) 24

)12(

0

2

1

R

QQ

q

πε

−

O x 60o

2m

0.2C A

Q

5a

5a

B (–3a, 4a)

A (5a, 0)

A

C

50 cm

30 cm

40 c

m

+ 100 µC

+ 50 µC π / 2

B


Chapter 2 42

cSolution: (b) Potential at the centre of first ring

220

2

0

1

44 RR

QR

QVA

++=

πεπε

Potential at the centre of second ring 22

0

1

0

2

44 RR

QR

QVB

++=

πεπε

Potential difference between the two centres 24

))(12(

0

21

R

QQVV BA

πε−−

=−

∴ Work done 24

))(12(

0

21

R

QQqW

πε−−

=

A point charge q moves from point A to point D along the path ABCD in a uniform electric field. If the

co-ordinates of the points A, B, C and D are (a, b, 0), (2a, 0, 0), (a, – b, 0) and (0, 0, 0) then the work done by the electric field in this process will be [IIT-JEE 1989]

(a) – qEa

(b) Zero

(c) 2E (a + b)q

(d) b

qEa2

Solution: (a) As electric field is a conservative field

Hence the work done does not depend on path

∴ AODABCD WW = ODAO WW +=

= Fb cos 90o + Fa cos 180o = 0 + qEa (– 1)= – qEa

Equilibrium of Charge.

(1) Definition : A charge is said to be in equilibrium, if net force acting on it is zero. A system of charges is said to be in equilibrium if each charge is separately in equilibrium.

(2) Type of equilibrium : Equilibrium can be divided in following type:

(i) Stable equilibrium : After displacing a charged particle from it's equilibrium position, if it returns back then it is said to be in stable equilibrium. If U is the potential energy then in case of stable equilibrium

2

2

dxUd

is positive i.e., U is minimum.

(ii) Unstable equilibrium : After displacing a charged particle from it's equilibrium position, if it never

returns back then it is said to be in unstable equilibrium and in unstable equilibrium 2

2

dxUd

is negative i.e., U is

maximum.

R

Q2 Q1

R R

A B 1 2

Y

X

C

B D

A E

X

Y

C

B D

E

A (a,b,0)

√a2 + b2

√a2 + b2

O θ θ

θ a

b

b

√a2 + b2

Tricky example: 8


Chapter 2 43

c(iii) Neutral equilibrium : After displacing a charged particle from it's equilibrium position if it neither

comes back, nor moves away but remains in the position in which it was kept it is said to be in neutral

equilibrium and in neutral equilibrium 2

2

dxUd

is zero i.e., U is constant

(3) Guidelines to check the equilibrium

(i) Identify the charge for which equilibrium is to be analysed.

(ii) Check, how many forces acting on that particular charge.

(iii) There should be atleast two forces acts oppositely on that charge.

(iv) If magnitude of these forces are equal then charge is said to be in equilibrium then identify the nature of equilibrium.

(v) If all the charges of system are in equilibrium then system is said to be in equilibrium

(4) Different cases of equilibrium of charge

Case – 1 : Suppose three similar charge qQ ,1 and

2Q are placed along a straight line as shown below

Charge q will be in equilibrium if |||| 21 FF =

i.e., 2

2

1

2

1

=

xx

QQ

; This is the condition of

equilibrium of charge q. After following the guidelines we can say that charge q is in stable equilibrium and this system is not in equilibrium

Note : 12

11 /QQ

xx

+=

and 21

21 /QQ

xx

+=

e.g. if two charges +4µC and +16 µC are separated by a distance of 30 cm from each other then for equilibrium a third charge should be placed between them at a distance

cmx 104/161

301 =

+= or cmx 202 =

Case – 2 : Two similar charge 1Q and 2Q are

placed along a straight line at a distance x from each other and a third dissimilar charge q is placed in between them as shown below

Charge q will be in equilibrium if |||| 21 FF =

i.e., 2

2

1

2

1

=

xx

QQ

.

Note : Same short trick can be used here to

find the position of charge q as we discussed in Case–1 i.e.,

12

11 /QQ

xx

+= and

212

1 /QQ

xx

+=

It is very important to know that magnitude of charge q can be determined if one of the extreme charge (either 1Q or )2Q is in equilibrium i.e. if 2Q

is in equilibrium then 2

21||

=

xx

Qq and if 1Q is

in equilibrium then 2

12||

=

xx

Qq (It should be

remember that sign of q is opposite to that of )or( 21 QQ )


Chapter 2 44

cCase – 3 : Two dissimilar charge 1Q and 2Q are placed along a straight line at a distance x from each

other, a third charge q should be placed out side the line joining 1Q and 2Q for it to

experience zero net force.

(Let |Q2| < |Q1|)

Short Trick :

For it's equilibrium. Charge q lies on the side of chare which is smallest in magnitude and

121 −=

/QQ

xd

(5) Equilibrium of suspended charge in an electric field

(i) Freely suspended charged particle : To suspend a charged a particle freely in air under the influence of electric field it’s downward weight should be balanced by upward electric force for example if a positive charge is suspended freely in an electric field as shown then

In equilibrium mgQE = Q

mgE =⇒

Note : ≅ In the above case if direction of electric field is suddenly reversed in any figure then acceleration

of charge particle at that instant will be a = 2g. (ii) Charged particle suspended by a massless insulated string (like simple pendulum) : Consider

a charged particle (like Bob) of mass m, having charge Q is suspended in an electric field as shown under the influence of electric field. It turned through an angle (say θ) and comes in equilibrium.

So, in the position of equilibrium (O′ position)

QET =θsin ….(i)

mgT =θcos ….(ii)

By squaring and adding equation (i) and (ii) ( ) ( )22 mgQET +=

Dividing equation (i) by (ii) mgQE

=θtan ⇒ mgQE

θ 1tan−=

(iii) Equilibrium of suspended point charge system : Suppose two small balls having charge +Q on each are suspended by two strings of equal length l. Then for equilibrium position as shown in figure.

x

x1 x2

q

Q2

A B O F1 F2 Q1

x

x1 x2

q

Q1 Q2

A B O F2 F1

x d

Q1 – Q2 q

+Q

F = QE

mg

E

E

+Q

F = QE

mg

+ + + + + + + + +

– – – – – – – – – F = QE

mg

+Q d V

Vmgd

Emg

neQ ===

or or

QE

E

θ

θ

l

mg

T sin θ

T cos θ

T

O

O′


Chapter 2 45

c

eFT =θsin ….(i)

mgT =θcos ….(ii)

( ) ( )222 mgFT e +=

and mgFe=θtan ; here

2

2

041

xQ

Fe πε= and θsin

2l

x=

(iv) Equilibrium of suspended point charge system in a liquid : In the previous discussion if point charge system is taken into a liquid of density ρ such that θ remain same then

In equilibrium θsin'TFe' = and θρ cos)( 'TgVmg =−

∴ 2

0

2

)(4)(tan

xgVmgK

QgVmg

Fe'

ρπερθ

−=

−=

When this system was in air 2

0

2

4tan

mgx

QmgFe

πεθ ==

∴ So equating these two gives us

−

=−

=⇒−

=ρρρ

mVVm

mK

Vmkm1

1)(

11

If σ is the density of material of ball then

−

=

θθ

K1

1

Example: 62 A charge q is placed at the centre of the line joining two equal charges Q. The system of the three charges will be in equilibrium. If q is equal to

[CPMT 1999; MP PET 1999, MP PMT 1999; CBSE 1995; Bihar MEE 1995; IIT 1987]

(a) 2Q

− (b) 4Q

− (c) 4Q

+ (d) 2Q

+

Solution: (b) By using Tricky formula 22/

=

xx

Qq

⇒ 4Q

q = since q should be negative so 4Q

q −= .

T sin θ

T cos θ T

Fe

θ

mg

+Q

l l θ θ

x +Q

T′ sin θ

T′ cos θ T′

F′e

θ

(mg – Vρg)

+Q

l l θ θ

x +Q

QE

mg

Q Examples based on equilibrium of charge


Chapter 2 46

Example: 63 Two point charges +4q and +q are placed at a distance L apart. A third charge Q is so placed that all the three charges are in equilibrium. Then location and magnitude of third charge will be

(a) At a distance3L

from +4q charge,9

4q

(b) At a distance 3L

from +4q charge, 9

4q−

(c) At a distance 3

2L from +4q charge,

94q

−

(d) At a distance 3

2Lfrom +q charge,

94q

+

Solution: (c) Let third charge be placed at a distance 1x from +4q charge as shown

Now

qq

Lx

41

1

+

=3

2L= ⇒

32L

x =

For equilibrium of q, 9

43/4

2 qL

LqQ =

+= ⇒

94q

Q −= .

Example: 64 A drop of 610− kg water carries 610− C charge. What electric field should be applied to balance it’s weight (assume g = 10 m/sec2)

(a) ,/10 mV Upward (b) ,/10 mV Downward (c) 0.1 V/m Downward (d) ,/1.0 mV

Upward

Solution: (a) In equilibrium QE = mg

Qmg

E = =6

6

10

1010−

− × = 10 V/m; Since charge is positive so electric field will be upward.

Example: 65 A charged water drop of radii 0.1 mµ is under equilibrium in some electric field. The charge on the drop

is equivalent to electronic charge. The intensity of electric field is [RPET 1997]

(a) CN /61.1 (b) CN /2.25 (c) CN /262 (d) CN /1610

Solution: (c) In equilibrium QE = mg ; Q

gr

Qmg

E.

34 3

==ρπ

19

336

106.1

1010)101.0()14.3(34

−

−

×

××××= = 262 N/C

Example: 66 The bob of a pendulum of mass 8 gµ carries an electric charge of 10102.39 −× coulomb in an electric

field of metervolt /1020 3× and it is at rest. The angle made by the pendulum with the vertical will be

(a) 27o (b) 45o (c) 87o (d) 127o

Solution: (b) qET =θsin , mgT =θcos

∴ mgqE

=θtan

L

x1 x2

Q +4q +q

E

T cos

qE θ

mg

T sin

T

θ


Chapter 3 47

18.9108

1020102.39tan

6

310

=××

×××=

−

−

θ

⇒ o45=θ

Example: 67 Two small spherical balls each carrying a charge Q = 10 Cµ (10 micro-coulomb) are suspended by two

insulating threads of equal lengths 1 m each, from a point fixed in the ceiling. It is found that in equilibrium threads are separated by an angle 60o between them, as shown in the figure. What is the

tension in the threads. (Given : ( )29

0/109

41

CNm×=πε

)

(a) 18 N

(b) 1.8 N

(c) 0.18 N

(d) None of these

Solution: (b) From the geometry of figure

r = 1m

In the condition of equilibrium eo FT =30sin

2

269

1

)1010(.109

21 −×

×=×T

⇒ T= 1.8 N

Example: 68 Two similar balloons filled with helium gas are tied to L m long strings. A body of mass m is tied to another ends of the strings. The balloons float on air at distance r. If the amount of charge on the balloons is same then the magnitude of charge on each balloon will be

(a) 2/12

tan2

θ

kmgr

(b) 2/1

2tan

2

θ

mgr

k

(c) 2/1

cot2

θ

kmgr

(d) 2/1

tan2

θ

mgrk

Solution: (a) In equilibrium

mgR =2 …. (i) θsinTFe = …. (ii) θcosTR = …. (iii)

From equation (i) and (iii)

mgT =θcos2 …. (iv)

Dividing equation (ii) by equation (iv)

mgFe=θtan

21

⇒ mgr

Qk

2

2

tan21

=θ ⇒2/12

tan2

= θθ

kmgr

T cos

m

m

L L θ θ

T sin

F

R R

r

T

60o

Q Q

+10 µC Fe

1m 1m

30o

r T sin 30o

T cos 30o T

mg

30o

+10 µC

30o

m

L L θ θ

r

Q Q


Chapter 3 48

Time Period of Oscillation of a Charged Body.

(1) Simple pendulum based : If a simple pendulum having length l and mass of bob m oscillates about

it's mean position than it's time period of oscillation gl

T π2=

Case – 1 : If some charge say +Q is given to bob and an electric field E is applied in the direction as shown in figure then equilibrium position of charged bob (point charge) changes from O to O′.

On displacing the bob from it’s equilibrium position 0′. It will oscillate under the effective acceleration g′, where

( ) ( )22 QEmgmg' +=

( )22 / mQEgg' +=⇒

Hence the new time period is g'l

T π21 =

( )( )21

221 2

QE/mg

lT

+

= π

Since g' >g, hence T1 < T

i.e. time period of pendulum will decrease.

Case – 2 : If electric field is applied in the downward direction then.

Effective acceleration

mQEgg' /+=

So new time period

( )QE/mgl

T+

= π22

T2 < T

Case – 3 : In case 2 if electric field is applied in upward direction then, effective acceleration.

mQEgg' /−=

So new time period

( )QE/mgl

T−

= π23

T3 > T

Case – 4 : In the case 3,

if 23T

T = i.e., ( )mQEgl

/2

−π

glπ2

21

= ⇒ QE = 3 mg

i.e., effective vertical force (gravity + electric) on the bob = mg – 3 mg = – 2 mg, hence the equilibrium position O′′ of the bob will be above the point of suspension and bob will oscillate under on effective acceleration 2g directed upward.

Hence new time period gl

T2

24 π= , T4 < T

l

O

QE

mg mg′

O′

E

θ

d

l

O

mg + QE

θ T l

E

mg

θ l

QE

E

mg

QE O′′

E


Chapter 3 49

(2) Charged circular ring : A thin stationary ring of radius R has a positive charge +Q unit. If a negative charge – q (mass m) is placed at a small distance x from the centre. Then motion of the particle will be simple harmonic motion.

Electric field at the location of – q charge

( )23

220

.4

1

Rx

QxE

+=

πε

Since x<< R, So 2x neglected hence 3

0

.4

1

R

QxE

πε=

Force experienced by charge – q is3

0

.4

1

R

QxqF

πε−=

xF −∝⇒ hence motion is simple harmonic

Having time period qQmRπε

πT 034

2=

(3) Spring mass system : A block of mass m containing a negative charge – Q is placed on a frictionless horizontal table and is connected to a wall through an unstretched spring of spring constant k as shown. If electric field E applied as shown in figure the block experiences an electric force, hence spring compress and block comes in new position. This is called the equilibrium position of block under the influence of electric field. If block compressed further or stretched, it execute oscillation having time

period km

πT 2= . Maximum compression in the spring due to electric field = k

QE

Neutral Point.

A neutral point is a point where resultant electrical field is zero. It is obtained where two electrical field are equal and opposite. Thus neutral points can be obtained only at those points where the resultant field is subtractive. Thus it can be obtained.

(1) At an internal point along the line joining two like charges (Due to a system of two like point charge) : Suppose two like charges. 1Q and 2Q are separated by a distance x from each other along a line as shown in following figure.

If N is the neutral point at a distance 1x from 1Q and at a

distance ( )12 xxx −= from 2Q then –

At N ||| 21 toduetodue QE.F.QE.F.| =

i.e., 21

1

0.

41

x

Qπε

=22

2

0.

41

x

Qπε

2

2

1

2

1

=⇒

xx

QQ

Short τrick : 12

11 /QQ

xx

+= and

212

1 /QQ

xx

+=

Note : ≅ In the above formula if 21 QQ = , neutral point lies at the centre so remember that resultant field at the midpoint of two equal and like charges is zero.

(2) At an external point along the line joining two like charges (Due to a system of two unlike point charge) : Suppose two unlike charge 1Q and 2Q separated by a distance x from each other. Here neutral point lies outside the line joining two unlike charges and also it lies nearer to charge which is smaller in magnitude.

If 21 QQ < then neutral point will be obtained on the side of

1Q , suppose it is at a distance l from 1Q

Hence at neutral point ; ( )2

22

1

lx

kQ

l

kQ

+=

2

2

1

+=⇒

lxl

QQ

Short τrick : ( )112 −=

/QQ

xl

Note : ≅ In the above discussion if |||| 21 QQ = neutral

point will be at infinity.

x

x1 x2

Q1 N

Q2

l

N Q1

Q2

x

x O

R – q

+Q + +

+

+

+ +

+ +

+ +

+ + + + + +

+ + +

m, – Q

E

k


Chapter 3 50

Zero Potential Due to a System of Two Point Charge.

If both charges are like then resultant potential is not zero at any finite point because potentials due to like charges will have same sign and can therefore never add up to zero. Such a point can be therefore obtained only at infinity.

If the charges are unequal and unlike then all such points where resultant potential is zero lies on a closed curve, but we are interested only in those points where potential is zero along the line joining the two charges.

Two such points exist, one lies inside and one lies outside the charges on the line joining the charges. Both the above points lie nearer the smaller charge, as potential created by the charge larger in magnitude will become equal to the potential created by smaller charge at the desired point at larger distance from it.

I. For internal point :

(It is assumed that 21 QQ < ).

( )1

2

1

1

xxQ

xQ

−= ⇒ ( )112

1 +=

/QQx

x

II. For External point :

( )1

2

1

1

xxQ

xQ

+= ⇒ ( )112

1 −=

/QQx

x

Example: 69 Two similar charges of +Q as shown in figure are placed at points A and B. – q charge is placed at point C midway between A and B. – q charge will oscillate if

(a) It is moved towards A

(b) It is moved towards B

(c) It is moved along CD

(d) Distance between A and B is reduced

Solution: (c) When – q charge displaced along CD, a restoring force act on it which causes oscillation.

Example: 70 Two point charges (+Q) and (– 2Q) are fixed on the X-axis at positions a and 2a from origin respectively. At what position on the axis, the resultant electric field is zero

(a) Only ax 2= (b) Only ax 2−= (c) Both ax 2±= (d) 23a

x = only

Solution: (b) Let the electric field is zero at a point P distance d from the charge +Q so at P. 0)(

)2(.22=

+

−+

da

Qk

d

Qk

⇒ 22 )(

21

dad += ⇒

)12( −=

ad

x

x1 x2

Q1 P

Q2

x1

Q1 Q2

x

P

D

C – q

A B

+Q +Q

P + Q – 2Q

x

d

a 2a

N Examples based on oscillation of charge and neutral


Chapter 3 51

Since d > a i.e. point P must lies on negative x-axis as shown at a distance x from origin hence

aaa

adx 212

=−−

=−= . Actually P lies on negative x-axis so ax 2−= .

Example: 71 Two charges 9e and 3e are placed at a distance r. The distance of the point where the electric field intensity will be zero is [MP PMT 1989]

(a) ( )13 +

r from 9e charge (b)

311+

r from 9e charge

(c) ( )31−

r from 3e charge (d)

311+

r from 3e charge

Solution: (b) Suppose neutral point is obtained at a distance 1x from charge 9e and 2x from charge 3e

By using

1

21

1QQ

xx

+

= =

ee

r

93

1+

+

=

3

11

r

Example: 72 Two point charges – Q and 2Q are separated by a distance R, neutral point will be obtained at

(a) A distance of )12( −

R from – Q charge and lies between the charges.

(b) A distance of)12( −

R from – Q charge on the left side of it

(c) A distance of )12( −

Rfrom 2Q charge on the right side of it

(d) A point on the line which passes perpendicularly through the centre of the line joining – Q and 2Q charge.

Solution: (b) As already we discussed neutral point will be obtained on the side of charge which is smaller in magnitude i.e. it will obtained on the left side of – Q charge and at a distance.

12

−

=

QQ

Rl ⇒

)12( −=

Rl

Example: 73 A charge of + 4µC is kept at a distance of 50 cm from a charge of – 6µC. Find the two points where the potential is zero

(a) Internal point lies at a distance of 20 cm from 4µC charge and external point lies at a distance of 100 cm from 4µC charge.

(b) Internal point lies at a distance of 30 cm from 4µC charge and external point lies at a distance of 100 cm from 4µC charge

(c) Potential is zero only at 20 cm from 4µC charge between the two charges

(d) Potential is zero only at 20 cm from – 6µC charge between the two charges

Solution: (a) For internal point X, cm

QQ

xx 20

14650

11

21 =

+=

+

= and for external point Y,

cm

QQ

xx 100

14650

11

21 =

−=

−

=

r

x1 x2

N 9e 3e

50cm 100cm

4µC X Y

– 6µC

20cm


Chapter 3 52

Two equal negative charges – q are fixed at points (0, a) and (0, – a) on the y-axis. A positive charge Q is released from rest at the point (2a, 0) on the x-axis. The charge Q will

[IIT-JEE 1984, Bihar MEE 1995, MP PMT 1996]

(a) Execute simple harmonic motion about the origin

(b) Move to the origin and remains at rest

(c) Move to infinity

(d) Execute oscillatory but not simple harmonic motion.

Solution: (d) By symmetry of problem the components of force on Q due to charges at A and B along y-axis will cancel each other while along x-axis will add up and will be along CO. Under the action of this force charge Q will move towards O. If at any time charge Q is at a distance x from O.

2122220 )()(4

12cos2

xa

x

xa

qQFF

+×

+

−=⇒

πεθ

i.e., ( ) 23220

2.

41

xa

qQxF

+−=

πε

As the restoring force F is not linear, motion will be oscillatory (with amplitude 2a) but not simple harmonic.

Electric Potential Energy.

(1) Potential energy of a charge : Work done in bringing the given charge from infinity to a point in the electric field is known as potential energy of the charge. Potential can also be written as potential energy

per unit charge. i.e. QU

QW

V == .

(2) Potential energy of a system of two charges : Since work done in bringing charge Q2 from ∞ to

point B is ,2 BVQW = where VB is potential of point B due to charge Q1 i.e. r

QVB

1

041πε

=

So, rQQ

UW 21

02 .

41πε

==

This is the potential energy of charge Q2, similarly potential energy of charge Q1 will be rQQ

U 21

01 .

41πε

=

Hence potential energy of Q1 = Potential energy of Q2 = potential energy of system rQQ

kU 21= (in C.G.S.

rQQ

U 21= )

Note : ≅ Electric potential energy is a scalar quantity so in the above formula take sign of Q1 and Q2.

2a

x

a

a

O C

θ

A

B

– q

Q

– q

Q1

r

Q2

A B

Tricky example: 9


Chapter 3 53

(3) Potential energy of a system of n charges : In a system of n charges electric potential energy is calculated for each pair and then all energies so obtained are added algebraically. i.e.

++= .........

41

23

32

12

21

0 rQQ

rQQ

Uπε

and in case of continuous distribution of charge. As VdQdU .= ⇒

∫= dQVU

e.g. Electric potential energy for a system of three charges

Potential energy

++=

31

13

23

32

12

21

041

rQQ

rQQ

rQQ

πε

While potential energy of any of the charge say Q1 is

+

31

13

12

21

041

rQQ

rQQ

πε

Note : ≅ For the expression of total potential energy of a system of n charges consider 2

)1( −nn number of

pair of charges.

(4) Electron volt (eV) : It is the smallest practical unit of energy used in atomic and nuclear physics. As electron volt is defined as “the energy acquired by a particle having one quantum of charge 1e when

accelerated by 1volt” i.e. CJ

CeV1

106.11 19 ××= − J19106.1 −×= = 1.6 × 10–12 erg

Energy acquired by a charged particle in eV when it is accelerated by V volt is E = (charge in quanta) × (p.d. in volt)

Commonly asked examples :

S.No. Charge Accelerated by p.d.

Gain in K.E.

(i) Proton 5 × 104 V K = e × 5 × 104 V = 5 × 104 eV = 8 × 10–15 J [JIPMER 1999]

(ii) Electron 100 V K = e × 100 V = 100 eV = 1.6 × 10–17 J [MP PMT 2000; AFMC 1999]

(iii) Proton 1 V K = e × 1 V = 1 eV = 1.6 × 10–19 J [CBSE 1999]

(iv) 0.5 C 2000 V K = 0.5 × 2000 = 1000 J [JIPMER 2002]

(v) α-particle 106 V K = (2e) × 106 V = 2 MeV [MP PET/PMT 1998]

(5) Electric potential energy of a uniformly charged sphere : Consider a uniformly charged sphere of radius R having a total charge Q. The electric potential energy of this sphere is equal to the work done in bringing the charges from infinity to assemble the sphere.

R

QU

0

2

203πε

=

(6) Electric potential energy of a uniformly charged thin spherical shell :

R

QU

0

2

8πε=

Q1

Q2 Q3

r31

r23

r12


Chapter 3 54

(7) Energy density : The energy stored per unit volume around a point in an electric field is given by

202

1Volume

EU

Ue ε== . If in place of vacuum some medium is present then 202

1EU re εε=

Concepts

Electric potential energy is not localised but is distributed all over the field

If a charge moves from one position to another position in an electric field so it’s potential energy change and work done in this changing is if UUW −=

If two similar charge comes closer potential energy of system increases while if two dissimilar charge comes closer potential energy of system decreases.

Example: 74 If the distance of separation between two charges is increased, the electrical potential energy of the system [AMU 1998]

(a) May increases or decrease (b) Decreases

(c) Increase (d) Remain the same

Solution: (a) Since we know potential energy rQQ

kU 21.=

As r increases, U decreases in magnitude. However depending upon the fact whether both charges are similar or disimilar, U may increase or decrease.

Example: 75 Three particles, each having a charge of 10µC are placed at the corners of an equilateral triangle of side

10cm. The electrostatic potential energy of the system is (Given )/1094

1 229

0CmN −×=

πε

[AMU 1998]

(a) Zero (b) Infinite (c) 27 J (d) 100 J

Solution: (c) Potential energy of the system,

×

××=

−

31.0

)1010(109

269U = 27 J

Example: 76 Three charges Q, +q and +q are placed at the vertices of a right-angled isosceles triangle as shown. The net electrostatic energy of the configuration is zero if Q is equal to [IIT (Screening) 2000]

(a) 21+

−q

(b) 21

2

+

− q

(c) – 2 q

10µC 10µC

10µC

10 cm

10 cm

10 cm

Examples based on electric potential energy

+q +q

Q

a

∞ Q q

U = 0


Chapter 3 55

(d) +q

Solution: (b) Potential energy of the configuration 02

..

.2

=++=a

Qqk

aqk

aQq

kU ⇒ 12

2

+

−=

qQ

Example: 77 A charge 10 e.s.u. is placed at a distance of 2 cm from a charge 40 e.s.u. and 4 cm from another charge of 20 e.s.u. The potential energy of the charge 10 e.s.u. is (in ergs)

(a) 87.5 (b) 112.5 (c) 150 (d) 250

Solution: (d) Potential energy of 10 e.s.u. charge is

.2504

20102

4010ergU =

×+

×=

Example: 78 In figure are shown charges q1 = + 2 × 10–8 C and q2 = – 0.4 × 10–8 C. A charge q3 = 0.2 × 10–8 C in moved along the arc of a circle from C to D. The potential energy of q3 [CPMT 1986]

(a) Will increase approximately by 76%

(b) Will decreases approximately by 76%

(c) Will remain same

(d) Will increases approximately by 12%

Solution: (b) Initial potential energy of q3 93231 109

18.0××

+=

qqqqUi

Final potential energy of q3 93231 109

2.08.0××

+=

qqqqU f

Change in potential energy = Uf – Ui

Now percentage change in potential energy 100×−

=i

if

u

UU

+

×

−

=

18.0

10012.0

1

213

32

qqq

qq On putting the values %76–~−

Three charged particles are initially in position 1. They are free to move and they come in position 2 after some time. Let U1 and U2 be the electrostatics potential energies in position 1 and 2. Then

(a) U1 > U2 (b) U2 > U1

(c) U1 = U2 (d) U2 ≥ U1

Solution: (a) Particles move in a direction where potential energy of the system is decreased.

40 esu

10 esu

20 esu

4 cm 2 cm

C

q1

q3

80 c

m

D 60 cm

80 cm

A B

q2

Tricky example: 10

q3

C

A B

q1 q2

80 c

m

D 60 cm


Chapter 3 56

Motion of Charged Particle in an Electric Field.

(1) When charged particle initially at rest is placed in the uniform field :

Let a charge particle of mass m and charge Q be initially at rest in an electric field of strength E

(i) Force and acceleration : The force experienced by the charged particle is QEF = . Positive charge

experiences force in the direction of electric field while negative charge experiences force in the direction opposite to the field. [Fig. (A)]

Acceleration produced by this force is m

QEmF

a ==

Since the field E in constant the acceleration is constant, thus motion of the particle is uniformly accelerated.

(ii) Velocity : Suppose at point A particle is at rest and in time t, it reaches the point B [Fig. (B)]

V = Potential difference between A and B; S = Separation between A and B

(a) By using atuv += , tmE

Qv += 0 , ⇒ m

QEtv =

(b) By using asuv 222 += , sm

QEv ××+= 202

mQV

v22 =

=

sV

E ⇒ mQV

v2

=

(iii) Momentum : Momentum p = mv, QEtm

QEtmp =×= or mQV

mQV

mp 22

=×=

(iv) Kinetic energy : Kinetic energy gained by the particle in time t is

mtEQ

mQEt

mmvK2

)(21

21 2222

2 === or QVmQV

mK =×=2

21

(2) When a charged particle enters with an initial velocity at right angle to the uniform field :

When charged particle enters perpendicularly in an electric field, it describe a parabolic path as shown

(i) Equation of trajectory : Throughout the motion particle has uniform velocity along x-axis and horizontal displacement (x) is given by the equation x = ut

Since the motion of the particle is accelerated along y–axis, we will use equation of motion for uniform

acceleration to determine displacement y. From 2

21

atutS +=

Fig. (A)

E

A B S

Fig. (B)

+Q

– Q F=QE

F = QE E


Chapter 3 57

We have 0=u (along y-axis) so 2

21

aty =

i.e., displacement along y-axis will increase rapidly with time (since )2ty ∝

From displacement along x-axis ux

t =

So 2

21

=

ux

mQE

y ; this is the equation of parabola which shows 2xy ∝

(ii) Velocity at any instant : At any instant t, uvx = and m

QEtvy =

So 2

222222||

mtEQ

uvvvv yx +=+==

If β is the angle made by v with x-axis than muQEt

v

v

x

y ==βtan .

Concepts

An electric field is completely characterized by two physical quantities Potential and Intensity. Force characteristic of the field is intensity and work characteristic of the field is potential.

If a charge particle (say positive) is left free in an electric field, it experiences a force )( QEF = in the direction of electric field and

moves in the direction of electric field (which is desired by electric field), so its kinetic energy increases, potential energy decreases, then work is done by the electric field and it is negative.

Example: 79 An electron (mass = kg31101.9 −× and charge = .)106.1 19 coul−× is sent in an electric field of

intensity ./101 6 mV× How long would it take for the electron, starting from rest, to attain one–tenth the

velocity of light

(a) sec12107.1 −× (b) sec6107.1 −× (c) sec8107.1 −× (d) sec10107.1 −×

Solution: (b) By using m

QEtv = ⇒

31

6198

101.9

10)106.1(103

101

−

−

×

×××=××

t ⇒ .107.1 10 sect −×=

Example: 80 Two protons are placed m1010− apart. If they are repelled, what will be the kinetic energy of each proton at very large distance

(a) J191023 −× (b) J19105.11 −× (c) J191056.2 −× (d) J281056.2 −×

Solution: (d) Potential energy of the system when protons are separated by a distance of 1010− m is

JU 1910

2199

102310

)106.1(109 −−

−

×=×××

=

Q E

Examples based on motion of

vx

vy

β

v

P+ p+

Y

X

E

u

P(x, y)


Chapter 3 58

According to law of conservation of energy at very larger distance, this energy is equally distributed in

both the protons as their kinetic energy hence K.E. of each proton will be .105.11 19 J−×

Example: 81 A particle A has a charge +q and particle B has charge +4q with each of them having the same mass m. When allowed to fall from rest through the same electrical potential difference, the ratio of their speeds

B

A

vv

will becomes [BHU 1995; MNR 1991]

(a) 2 : 1 (b) 1 : 2 (c) 1 : 4 (d) 4 : 1

Solution: (b) We know that kinetic energy QVmvK == 2

21

. Since, m and V are same so, Qv ∝2 ⇒

.21

4===

qq

QQ

vv

B

A

B

A

Example: 82 How much kinetic energy will be gained by an −α particle in going from a point at 70 V to another point

at 50 V [RPET 1996]

(a) 40 eV (b) 40 keV (c) 40 MeV (d) 0 eV

Solution: (a) Kinetic energy VQK ∆= ⇒ VeK )5070()2( −= eV40=

Example: 83 A particle of mass 2g and charge Cµ1 is held at a distance of 1 metre from a fixed charge of .1mC If the

particle is released it will be repelled. The speed of the particle when it is at a distance of 10 metres from the fixed charge is [CPMT 1989]

(a) 100 m/s (b) 90 m/s (c) 60 m/s (d) 45 m/s

Solution: (b) According to conservation of energy

Energy of moving charge at =A Energy of moving charge at B

2363

963

9 )102(21

101010

1091

1010109 v−

−−−−

××+×

××=×

××

⇒ 81002 =v ⇒ m/sec90=v

A mass of 1g carrying charge q falls through a potential difference V. The kinetic energy acquired by it is E. When a mass of 2g carrying the charge q falls through a potential difference V. What will be the kinetic energy acquired by it

(a) 0.25 E (b) 0.50 E (c) 0.75 E (d) E

Solution: (d) In electric field kinetic energy gain by the charged particle K = qV. Which depends charge and potential difference applied but not on the mass of the charged particle.

Tricky example: 11

1m

10m

A B

1 mC 1 µC

Moving charge

Fixed charge


Chapter 3 59

Force on a Charged Conductor.

To find force on a charged conductor (due to repulsion of like charges) imagine a small part XY to be cut and just separated from the rest of the conductor MLN. The field in the cavity due to the rest of the conductor is E2, while field due to small part is E1. Then

Inside the conductor 021 =−= EEE or 21 EE =

Outside the conductor 0

21 εσ

=+= EEE

Thus 0

21 2εσ

== EE

To find force, imagine charged part XY (having charge dAσ placed in the cavity MN having field E2).

Thus force 2)( EdAdF σ= or dAdF0

2

2εσ

= . The force per unit area or electric pressure is 0

2

2εσ

=dAdF

The force is always outwards as 2)( σ± is positive i.e., whether charged positively or negatively, this force will try to expand the charged body.

A soap bubble or rubber balloon expands on given charge to it (charge of any kind + or –).

Equilibrium of Charged Soap Bubble.

For a charged soap bubble of radius R and surface tension T and charge density .σ The pressure due to

surface tension RT

4 and atmospheric pressure outP act radially inwards and the electrical pressure )( elP acts

radially outward.

The total pressure inside the soap bubble 0

2

outin 24

εσ

−+=RT

PP

Excess pressure inside the charged soap bubble 0

2

excessoutin 24

εσ

−==−RT

PPP . If air pressure inside and

outside are assumed equal then outin PP = i.e., 0excess =P . So, 0

2

24

εσ

=RT

This result give us the following formulae

(1) Radius of bubble 208

σε T

R =

(2) Surface tension 0

2

8εσ R

T =

(3) Total charge on the bubble TRRQ 028 επ=

(4) Electric field intensity at the surface of the bubble R

kTRT

Eπ

ε328

0

==

(A)

+ + +

+ +

+ +

+ + + + + + + +

+ +

+ + + + M

N

Y

X

E2 E1

E1 E2

Inside E = 0 L

+

+

+ + +

+

+ +

+

+ +

+ +

E2

(B)

air

Pout

Pin

PT

air

Uncharged

Pout

Pin

PT

+

+ +

+ Pelec

Charged


Chapter 3 60

(5) Electric potential at the surface 0

83

επ RTRTkV ==


Chapter 3 61

c Electric Dipole.

(1) General information : System of two equal and opposite charges separated by a small fixed distance is called a dipole.

(i) Dipole axis : Line joining negative charge to positive charge of a dipole is called its axis. It may also be termed as its longitudinal axis.

(ii) Equatorial axis : Perpendicular bisector of the dipole is called its equatorial or transverse axis as it is perpendicular to length.

(iii) Dipole length : The distance between two charges is known as dipole length (L = 2l)

(iv) Dipole moment : It is a quantity which gives information about the strength of dipole. It is a vector quantity and is directed from negative charge to positive charge along the axis. It is denoted as p

and is

defined as the product of the magnitude of either of the charge and the dipole length.

i.e. )2( lqp

=

Its S.I. unit is coulomb-metre or Debye (1 Debye = 3.3 × 10–30 C × m) and its dimensions are M0L1T1A1.

Note : ≅ A region surrounding a stationary electric dipole has electric field only.

≅ When a dielectric is placed in an electric field, its atoms or molecules are considered as tiny dipoles.

≅ Water (H2O), Chloroform (CHCl3), Ammonia (NH3), HCl, CO molecules are some example of permanent electric dipole.

(2) Electric field and potential due to an electric dipole : It is better to understand electric dipole with magnetic dipole.

A B

+q 2l

Equ

ator

ial a

xis

Dipole axis – q

+

+

– – O2–

H+

H+

+ – +


Chapter 4 62

c

S.No. Electric dipole Magnetic dipole

(i) System of two equal and opposite charges separated by a small fixed distance.

System of two equal and opposite magnetic poles (Bar magnet) separated by a small fixed distance.

(ii) Electric dipole moment : )2( lqp

= , directed from

q− to +q. It’s S.I. unit is coulomb × meter or

Debye.

Magnetic dipole moment : )2( lmM

= , directed

from S to N. It’s S.I. unit is ampere × meter2.

(iii) Intensity of electric field

If a, e and g are three points on axial, equatorial and general position at a distance r from the centre of dipole

on axial point 3

0

2.

41

r

pEa πε

= (directed from – q to +q)

on equatorial point 3

0.

41

r

pEe πε

= (directed from +q to –q)

on general point )1cos3(.4

1 23

0+= θ

πε r

pEa

Angle between – aE

and p

is 0o, eE

and p

is 180o,

E

and p

is (θ + α) (where θα tan21

tan = )

Electric Potential – At a 2

0.

41

r

pVa πε

= , At e 0=V

At g 2

0

cos.

41

r

pV

θπε

=

Intensity of magnetic field

If a, e and g are three points on axial, equatorial and general position at a distance r from the centre of dipole

on axial point 3

0 2.

4 r

MBa π

µ= (directed from S to N)

on equatorial point 3

0 .4 r

MBe π

µ= (directed from N to S)

on general point )1cos3(.4

23

0 += θπµ

r

MBa

Angle between – aB

and M

is 0o, eB

and M

is

180o, B

and M

is (θ + α) (where θα tan21

tan = )

(3) Dipole (electric/magnetic) in uniform field (electric/magnetic)

(i) Torque : If a dipole is placed in an uniform field such that dipole (i.e. p

or M

) makes an angle θ

with direction of field then two equal and opposite force acting on dipole constitute a couple whose tendency is to rotate the dipole hence a torque is developed in it and dipole tries to align it self in the direction of field.

A

–q +q 2l

B

p

a +q

Equatorial line

Axial line

θ + α

aE

g

E

eE

e

θ

2l

p

– q a

Equatorial line

Axial line

θ

e

θ + α

aB

eB

B

M

g

2l

N S

N S – m + m

M


Chapter 4 63

cConsider an electric dipole in placed in an uniform

electric field such that dipole (i.e. p ) makes an angle θ

with the direction of electric field as shown

(a) Net force on electric dipole 0=netF

(b) Produced torque τ = pE sinθ )( EP ×=τ

A magnetic dipole of magnetic moment M is placed in uniform magnetic field B by making an angle θ as shown

(a) Net force on magnetic dipole 0=netF

(b) torque τ = MB sinθ )( BM ×=τ

(ii) Work : From the above discussion it is clear that in an uniform electric/magnetic field dipole tries to align itself in the direction of electric field (i.e. equilibrium position). To change it’s angular position some work has to be done.

Suppose an electric/magnetic dipole is kept in an uniform electric/magnetic field by making an angle θ1 with the field, if it is again turn so that it makes an angle θ2 with the field, work done in this process is given by the formula

)cos(cos 21 θθ −= pEW

If θ1 = 0o and θ2 = θ i.e. initially dipole is kept along the field then it turn through θ so work done

)cos1( θ−= pEW

)cos(cos 21 θθ −= MBW

If θ1 = 0o and θ2 = θ then W = MB(1 – cosθ)

(iii) Potential energy : In case of a dipole (in a uniform field), potential energy of dipole is defined as work done in rotating a dipole from a direction perpendicular to the field to the given direction i.e. if θ1 = 90o and θ2 = θ then –

)cos90(cos θ−== pEUW ⇒ U = – pE cosθ

)cos90(cos θ−== MBUW ⇒ U = – MB cosθ

+q +q

–q –q

θ1 θ2

E

P

P θ

E M

M θ

B

B

θ1 θ2

M

F = qE

F = qE

θ

B

A

–q

+q

θ

N

S

F = mB

F = mB


Chapter 4 64

c(iv) Equilibrium of dipole : We know that, for any equilibrium net torque and net force on a particle

(or system) should be zero.

We already discussed when a dipole is placed in an uniform electric/magnetic field net force on dipole is always zero. But net torque will be zero only when θ = 0o or 180o

When θ = 0o i.e. dipole is placed along the electric field it is said to be in stable equilibrium, because after turning it through a small angle, dipole tries to align itself again in the direction of electric field.

When θ = 180o i.e. dipole is placed opposite to electric field, it is said to be in unstable equilibrium.

θ = 0o θ = 90O θ = 180o

Stable equilibrium Unstable equilibrium

τ = 0 τmax = pE τ = 0

W = 0 W = pE Wmax = 2pE

Umin = – pE U = 0 Umax = pE

θ = 0o θ = 90O θ = 180o

Stable equilibrium Unstable equilibrium

τ = 0 τmax = MB τ = 0

W = 0 W = MB Wmax = 2MB

Umin = – MB U = 0 Umax = MB

(v) Angular SHM : In a uniform electric/magnetic field (intensity E/B) if a dipole (electric/magnetic) is slightly displaced from it’s stable equilibrium position it executes angular SHM having period of oscillation. If I = moment of inertia of dipole about the axis passing through it’s centre and perpendicular to it’s length.

For electric dipole : pEI

T π2= and For Magnetic dipole : MB

IT π2=

(vi) Dipole-point charge interaction : If a point charge/isolated magnetic pole is placed in dipole field at a distance r from the mid point of dipole then force experienced by point charge/pole varies

according to the relation 3

1r

F ∝

(vii) Dipole-dipole interaction : When two dipoles placed closed to each other, they experiences a force due to each other. If suppose two dipoles (1) and (2) are placed as shown in figure then

Both the dipoles are placed in the field of one another hence potential energy dipole (2) is

31

0212122

2.

41

0cosr

ppEpEpU

πε×−=−=−=

then by using drdU

F −= , Force on dipole (2) is dr

dUF 2

2 −=

⇒

−=3

21

02

2.

41

r

ppdrd

Fπε

4

21

0

6.

41

r

ppπε

−=

Similarly force experienced by dipole (1) 4

21

01

6.

41

r

ppF

πε−= so

421

021

6.

41

r

ppFF

πε−==

+q +q –q – q

O O E2 E1

P1 P2 r

1 2

E

p

E

E

p M

B

M

M

B

B


Chapter 4 65

cNegative sign indicates that force is attractive.

421

0

6.

41

||r

ppF

πε= and

4

1r

F ∝

S. No. Relative position of dipole Force Potential energy

(i)

421

0

6.

41

r

ppπε

(attractive) 3

21

0

2.

41

r

ppπε

(ii)

421

0

3.

41

r

ppπε

(repulsive) 3

21

0.

41

rpp

πε

(iii)

421

0

3.

41

r

ppπε

(perpendicular to r ) 0

Note : ≅ Same result can also be obtained for magnetic dipole.

(4) Electric dipole in non-uniform electric field : When an electric dipole is placed in a non-uniform field, the two charges of dipole experiences unequal forces, therefore the net force on the dipole is not equal to zero. The magnitude of the force is given by the negative derivative of the potential energy w.r.t.

distance along the axis of the dipole i.e. drEd

pdrdU

F .−=−= .

Due to two unequal forces, a torque is produced which rotate the dipole so as to align it in the direction of field. When the dipole gets aligned with the field, the torque becomes zero and then the unbalanced force acts on the dipole and the dipole then moves linearly along the direction of field from weaker portion of the field to the stronger portion of the field. So in non-uniform electric field

(i) Motion of the dipole is translatory and rotatory

(ii) Torque on it may be zero.

Concepts

For a short dipole, electric field intensity at a point on the axial line is double than at a point on the equatorial line on electric dipole i.e. Eaxial = 2Eequatorial

It is intresting to note that dipole field 31

rE ∝ decreases much rapidly as compared to the field of a point charge .

12

∝

rE

1P

+q – q

2P

+q – q

+q +q

–q –q

2P

1P

r

+q

–q

1P

r

2P

+q – q

+q

–

E′


Chapter 4 66

c

Example: 84 If the magnitude of intensity of electric field at a distance x on axial line and at a distance y on equatorial line on a given dipole are equal, then x : y is [EAMCET 1994]

(a) 1 : 1 (b) 2:1 (c) 1 : 2 (d) 1:23

Solution: (d) According to the question 3

03

0.

412

.4

1

y

p

x

pπεπε

= ⇒ 1:)2( 3/1=yx

Example: 85 Three charges of (+2q), (– q) and (– q) are placed at the corners A, B and C of an equilateral triangle of side a as shown in the adjoining figure. Then the dipole moment of this combination is [MP PMT 1994; CPMT 1994] (a) qa

(b) Zero

(c) 3aq

(d) qa3

2

Solution: (c) The charge +2q can be broken in +q, +q. Now as shown in figure we have two equal dipoles inclined at an angle of 60o. Therefore resultant dipole moment will be

60cos222 pppppnet ++=

p3=

qa3=

Example: 86 An electric dipole is placed along the x-axis at the origin O. A point P is at a distance of 20 cm from this

origin such that OP makes an angle 3π

with the x-axis. If the electric field at P makes an angle θ with x-

axis, the value of θ would be [MP PMT 1997]

(a) 3π

(b)

+ −

23

tan3

1π (c)

32π

(d)

−

23

tan 1

Solution: (b) According to question we can draw following figure.

As we have discussed earlier in theory απθ +=3

3

tan21

tanπα = ⇒

23

tan 1−=α

So, 23

tan3

1−+=πθ

Example: 87 An electric dipole in a uniform electric field experiences [RPET 2000]

(a) Force and torque both (b) Force but no torque (c) Torque but no force (d) No force and no torque

Solution: (c) In uniform electric field Fnet = 0, τnet ≠ 0

Example: 89 Two opposite and equal charges 4 × 10–8 coulomb when placed 2 × 10–2 cm away, form a dipole. If this dipole is placed in an external electric field 4 × 108 newton/coulomb, the value of maximum torque and the work done in rotating it through 180o will be [MP PET 1996 Similar to MP PMT 1987]

(a) 64 × 10–4 Nm and 64 × 10–4 J (b) 32 × 10–4 Nm and 32 × 10–4 J

Examples based on electric dipole – q +q

60O

P P

A

B C

+2q

– q – q a

a a

O

y

x

P

E

π/3 θ

p

α


Chapter 4 67

c(c) 64 × 10–4 Nm and 32 × 10–4 J (d) 32 × 10–4 Nm and 64 × 10–4 J

Solution: (d) τmax = pE and Wmax = 2pE p = Q × 2l = 4 × 10–8 × 2 × 10–2 × 10–2 = 8 × 10–12 C-m

So, τmax = 8 × 10–12 × 4 × 108 = 32 × 10–4 N-m and Wmax = 2 × 32 × 10–4 = 64 × 10–4 J

Example: 90 A point charge placed at any point on the axis of an electric dipole at some large distance experiences a force F. The force acting on the point charge when it’s distance from the dipole is doubled is

[CPMT 1991; MNR 1986]

(a) F (b) 2F

(c) 4F

(d) 8F

Solution: (d) Force acting on a point charge in dipole field varies as 3

1

rF ∝ where r is the distance of point charge

from the centre of dipole. Hence if r makes double so new force 8

'F

F = .

Example: 91 A point particle of mass M is attached to one end of a massless rigid non-conducting rod of length L. Another point particle of the same mass is attached to other end of the rod. The two particles carry charges +q and – q respectively. This arrangement is held in a region of a uniform electric field E such that the rod makes a small angle θ (say of about 5 degrees) with the field direction (see figure). Will be minimum time, needed for the rod to become parallel to the field after it is set free

(a) pE

mLt

22π= (b)

qEmL

t22

π= (c)

pEmL

t22

3π= (d)

qEmL

t2π=

Solution: (b) In the given situation system oscillate in electric field with maximum angular displacement θ. It’s time period of oscillation (similar to dipole)

pEI

T π2= where I = moment of inertia of the system and qLp =

Hence the minimum time needed for the rod becomes parallel to the field is pEIT

t24π

==

Here 222

222 MLLM

LMI =

+

= ⇒

qEML

EqLML

t2222

2 ππ=

××=

An electric dipole is placed at the origin O and is directed along the x-axis. At a point P, far away from the dipole, the electric field is parallel to y-axis. OP makes an angle θ with the x-axis then

(a) 3tan =θ (b) 2tan =θ (c) θ = 45o (d) 2

1tan =θ

Solution: (b) As we know that in this case electric field makes an angle θ +α with the direction of dipole

Where θα tan21

tan =

Here θ +α = 90o ⇒ θα −= 90

Hence θθ tan21

)90tan( =− ⇒ θθ tan21

cot =

Tricky example: 12

O

Y

X

E

θ

P

P

+q

–q

θ E


Chapter 4 68

c⇒ 2tan2 =θ ⇒ 2tan =θ

Electric Flux.

(1) Area vector : In many cases, it is convenient to treat area of a surface as a vector. The length of the vector represents the magnitude of the area and its direction is along the outward drawn normal to the area.

(2) Electric flux : The electric flux linked with any surface in an electric field is basically a measure of total number of lines of forces passing normally through the surface. or

Electric flux through an elementary area ds is defined as the scalar product of area of field i.e.

θφ cosdsEdsEd =⋅=

Hence flux from complete area (S) ∫= θφ cosdsE = ES cosθ

If θ = 0o, i.e. surface area is perpendicular to the electric field, so flux linked with it will be max.

i.e. φmax = E ds and if θ = 90o, φmin = 0

(3) Unit and Dimensional Formula

S.I. unit – (volt × m) or 2m

CN −

It’s Dimensional formula – (ML3T–3A– 1)

(4) Types : For a closed body outward flux is taken to be positive, while inward flux is to be negative

Gauss’s Law.

(1) Definition : According to this law, total electric flux through a closed surface enclosing a charge is

0

1ε

times the magnitude of the charge enclosed i.e. )(1

.0

encQε

φ =

(2) Gaussian Surface : Gauss’s law is valid for symmetrical charge distribution. Gauss’s law is very helpful in calculating electric field in those cases where electric field is symmetrical around the source producing it. Electric field can be calculated very easily by the clever choice of a closed surface that encloses the source charges. Such a surface is called “Gaussian surface”. This surface should pass through the point where electric field is to be calculated and must have a shape according to the symmetry of source.

Area ds

sd

d

θ

sd

E

(B)

Body

Negative-flux (A)

Positive – flux

Body

n E


Chapter 4 69

ce.g. If suppose a charge Q is placed at the centre of a hemisphere, then to calculate the flux through this

body, to encloses the first charge we will have to imagine a Gaussian surface. This imaginary Gaussian surface will be a hemisphere as shown.

Net flux through this closed body 0ε

φ Q=

Hence flux coming out from given hemisphere is .2 0ε

φ Q=

(3) Zero flux : The value of flux is zero in the following circumstances

(i) If a dipole is enclosed by a surface

0;0 == encQφ

(ii) If the magnitude of positive and negative charges are equal inside a closed surface

,0=encQ

so, φ = 0

(iii) If a closed body (not enclosing any charge) is placed in an electric field (either uniform or non-uniform) total flux linked with it will be zero

0=Tφ 2aEutoin == φφ

(4) Flux emergence : Flux linked with a closed body is independent of the shape and size of the body and position of charge inside it

Sphere

0=Tφ

• Q

+q –q

+q

+4q

–5q

sd

sd

a

y

E

a

x sd

0ε

φQ

T =

•Q

0ε

φQ

T = 0ε

φQ

T =

0ε

φQ

T =

• Q •Q

•Q

ERin2πφ −= ERout

2πφ += 0=T

φ

E


Chapter 4 70

c

(i) If a hemispherical body is placed in uniform electric field then flux linked with the curved surface

ERcurved2πφ +=

(ii) If a hemispherical body is placed in non-uniform electric field as shown below. then flux linked with the curved surface.

ERcurved22πφ =

(v) If charge is kept at the centre of cube

).(1

0

Qtotal εφ =

06ε

φ Qface =

08ε

φ Qcorner =

012εφ Q

edge =

(iv) If charge is kept at the centre of a face

First we should enclosed the charge by assuming a Gaussian surface (an identical imaginary cube)

0ε

φ Qtotal =

02ε

φ Qcube = (i.e. from 5 face only)

00 1025

1εε

φ QQface =

= .

Concept

In C.G.S. π

ε41

0 = . Hence if 1C charge is enclosed by a closed surface so flux through the surface will be πφ 4= .

Example: 91 Electric charge is uniformly distributed along a long straight wire of radius 1 mm. The charge per cm length of the wire is Q coulomb. Another cylindrical surface of radius 50 cm and length 1 m symmetrically encloses the wire as shown in the figure. The total electric flux passing through the cylindrical surface is

[MP PET 2001]

(a) 0ε

Q

(b) 0

100ε

Q

(c) )(

10

0πεQ

E

n

R

n

R

Example based on electric flux and Gauss’s

50 cm

1 m

+ +

+

+ +

+

Q

Q


Chapter 4 71

c(d)

)(100

0πεQ

Solution: (b) Given that charge per cm length of the wire is Q. Since 100 cm length of the wire is enclosed so QQenc 100=

⇒ Electric flux emerging through cylindrical surface 0

100ε

φQ

= .

Example: 92 A charge Q is situated at the corner A of a cube, the electric flux through the one face of the cube is [CPMT 2000]

(a) 06ε

Q (b)

08εQ

(c) 024ε

Q (d)

02εQ

Solution: (c) For the charge at the corner, we require eight cube to symmetrically enclose it in a Gaussian surface. The

total flux 0ε

φ QT = . Therefore the flux through one cube will be .

8 0εφ Q

cube = The cube has six faces and

flux linked with three faces (through A) is zero, so flux linked with remaining three faces will .8 0εφ

Now

as the remaining three are identical so flux linked with each of the three faces will be

00 241

81

31

εεQQ

=

×= .

Example: 93 A square of side 20 cm is enclosed by a surface of sphere of 80 cm radius. Square and sphere have the same centre. Four charges + 2 × 10–6 C, – 5 × 10– 6 C, – 3 × 10– 6 C, +6 × 10– 6 C are located at the four corners of a square, then out going total flux from spherical surface in N–m2/C will be

(a) Zero (b) (16 π) × 10– 6 (c) (8π) × 10–6 (d) 36π × 10–6

Solution: (a) Since charge enclosed by Gaussian surface is

0)106103105102( 6666. =×+×−×−×= −−−−

encφ so 0=φ

Example: 94 In a region of space, the electric field is in the x-direction and proportional to x, i.e., ixEE ˆ0= . Consider

an imaginary cubical volume of edge a, with its edges parallel to the axes of coordinates. The charge inside this cube is

(a) Zero (b) 300 aEε (c) 3

00

1aE

ε (d) 2

0061

aEε

Solution: (b) The field at the face ABCD = .00 ixE

∴ Flux over the face ABCD = – (E0x0)a2

The negative sign arises as the field is directed into the cube.

The field at the face EFGH = .)( 00 iaxE +

∴ Flux over the face EFGH = 200 )( aaxE +

The flux over the other four faces is zero as the field is parallel to the surfaces.

Q

D

C G

a a

X

H

B

A E

F

a x0

Y

Z


Chapter 4 72

c∴ Total flux over the cube qaE

212

0 ==

where q is the total charge inside the cube. ∴ .300 aEq ε=

In the electric field due to a point charge + Q a spherical closed surface is drawn as shown by the dotted circle. The electric flux through the surface drawn is zero by Gauss’s law. A conducting sphere is inserted intersecting the previously drawn Gaussian surface. The electric flux through the surface

(a) Still remains zero

(b) Non zero but positive

(c) Non-zero but negative

(d) Becomes infinite

Solution: (b) Due to induction some positive charge will lie within the Gaussian surface drawn and hence flux becomes something positive.

Application of Gauss’s Law.

Gauss’s law is a powerful tool for calculating electric field in case of symmetrical charge distribution by

choosing a Gaussian surface in such away that E is either parallel or perpendicular to it’s various faces.

e.g. Electric field due to infinitely long line of charge : Let us consider a uniformly charged wire of infinite length having a constant linear charge density is

.lengthcharge

=λλ Let P be a point distant r from the wire at which the electric field is to be

calculated.

Draw a cylinder (Gaussian surface) of radius r and length l around the line charge which encloses the charge Q ( lQ .λ= ). Cylindrical Gaussian surface has three surfaces;

two circular and one curved for surfaces (1) and (2) angle between electric field and normal to the surface is

90o i.e., .90o=θ

So flux linked with these surfaces will be zero. Hence total flux will pass through curved surface and it is

∫= θφ cosdsE …. (i)

According to Gauss’s law

0ε

φ Q= …. (ii)

Equating equation (i) and (ii) ∫ =0ε

QdsE

⇒ ∫ =⇒=00

2ε

πε

QrlEx

QdsE

⇒ rk

rrlQ

Eλ

πελ

πε2

22 00

===

=04

1πε

K

+ + + + + + +

P l

1

2

a

n

n

Tricky example: 13

+Q


Chapter 4 73

c Capacitance.

(1) Definition : We know that charge given to a conductor increases it’s potential i.e., VQ ∝ ⇒ CVQ =

Where C is a proportionality constant, called capacity or capacitance of conductor. Hence capacitance is the ability of conductor to hold the charge.

(2) Unit and dimensional formula : S.I. unit is Farad (F)Volt

Coulomb=

Smaller S.I. units are mF, µF, nF and pF ( FmF 3101 −= , FF 6101 −=µ , FnF 9101 −= ,

FFpF 121011 −== µµ )

C.G.S. unit is Stat Farad FaradStatF 111091 ×= . Dimension : ][][ 2421 ATLMC −−= .

(3) Capacity of an isolated spherical conductor : When charge Q is given to a spherical conductor of radius R, then potential at the surface of sphere is

RQ

kV .=

=04

1πε

k

Hence it’s capacity RVQ

C 04πε== ⇒ .RRπεC90

1091

4×

==

in C.G.S. RC =

Note : ≅ If earth is assumed to be spherical having radius .6400 kmR = It’s theortical capacitance

39

1064001091

×××

=C Fµ711= . But for all practical purpose capacitance of earth is taken

infinity.

(4) Energy of a charged conductor : When a conductor is charged it’s potential increases from 0 to V as shown in the graph; and work is done against repulsion, between charge stored in the conductor and charge coming from the source (battery). This work is stored as “electrostatic potential energy”

From graph : Work done = Area of graph QV21

=

Hence potential energy QVU21

= ; By using ,CVQ = we can write

C

QCVQVU

221

21 2

2 ===

(5) Combination of drops : Suppose we have n identical drops each having – Radius – r, Capacitance – c, Charge – q, Potential – v and Energy – u.

If these drops are combined to form a big drop of – Radius – R, Capacitance – C, Charge – Q, Potential – V and Energy – U then –

(i) Charge on big drop : nqQ =

+ + +

+ + +

+ + +

+ + +

+ + + +

+ + Q

O

R

V

Q

Electrostatic Potential and Capacitance (Electrostatics Part 5)

Chapter 5 74

c(ii) Radius of big drop : Volume of big drop = n × volume of a single drop i.e., 33

34

34

rnR ππ ×= , rnR /31=

(iii) Capacitance of big drop : cnC /31=

(iv) Potential of big drop : cn

nqCQ

V3/1

== vnV /32=

(v) Energy of big drop : 23/23/12 )()(21

21

vncnCVU == unU /35=

(6) Sharing of charge : When two conductors joined together through a conducting wire, charge begins to flow from one conductor to another till both have the same potential, due to flow of charge, loss of energy also takes place in the form of heat.

Suppose there are two spherical conductors of radii 1r and ,2r having charge 1Q and ,2Q potential 1V and

,2V energies 1U and 2U and capacitance 1C and 2C respectively, as shown in figure. If these two spheres are

connected through a conducting wire, then alteration of charge, potential and energy takes place.

(i) New charge : According to the conservation of charge QQQQQ =+=+ '2

'121 (say), also

20

10

2

1'2

'1

44

rr

VCVC

Q

Qπεπε

== , 2

1'2

'1

rr

Q

Q= ⇒

2

1'2

'1 11

rr

Q

Q+=+ ⇒

2

21'2

'2

'1

rrr

Q

QQ +=

+

⇒

+

=21

22 rr

rQQ' and similarly

+

=21

11 rr

rQQ'

(ii) Common potential : Common potential capacityTotal chargeTotal

)( =V 21

'2

'1

21

21

CCQQ

CCQQ

++

=++

= ⇒ 21

2211

CCVCVC

V++

=

(iii) Energy loss : As electrical energy stored in the system before and after connecting the spheres is

222

211 2

121

VCVCUi += and 2

21

221121

221 )(

21

.)(21

++

+=+=CC

VCVCCCVCCU f

so energy loss 221

21

21

2)V(V

)C(CCC

UUΔU fi −+

=−=

Concept

Capacity of a conductor is a constant term, it does not depend upon the charge Q, and potential (V) and nature of the material of the conductor.

r1

Q1

C1

V1

U1

Q1=C1V1

r2

Q2=C2V2

Q2

C2

V2

U2

(A) (B)

r1

Q1′ C1

V

U1′

Q1′=C1V1

r2

Q2′=C2V2

Q2′ C2

V

U2′


Chapter 5 75

c

Example: 95 Eight drops of mercury of same radius and having same charge coalesce to form a big drop. Capacitance of big drop relative to that of small drop will be [MP PMT 2002, 1990; MNR 1999, 87; DCE 1998]

(a) 16 times (b) 8 times (c) 4 times (d) 2 times

Solution: (d) By using relation cnC .3/1= ⇒ ccC 2.)8( 3/1 ==

Example: 96 Two spheres A and B of radius 4 cm and 6 cm are given charges of Cµ80 and Cµ40 respectively. If they are connected by a fine wire, the amount of charge flowing from one to the other is [MP PET 1991]

(a) Cµ20 from A to B (b) Cµ16 from A to B (c) Cµ32 from B to A (d) Cµ32 from A to B

Solution: (d) Total charge CQ µ1204080 =+= . By using the formula

+

=21

11'

rrr

QQ . New charge on sphere A is

+

=BA

AA rr

rQQ ' Cµ48

644

120 =

+= . Initially it was ,80 Cµ i.e., Cµ32 charge flows from A to B.

Example: 97 Two insulated metallic spheres of Fµ3 and Fµ5 capacitances are charged to V300 and V500 respectively.

The energy loss, when they are connected by a wire, is [Pb PMT 1999; CPMT 1999; KCET (Engg.) 2000]

(a) J012.0 (b) J0218.0 (c) J0375.0 (d) J75.3

Solution: (c) By using 221

21

21 )()(2

VVCC

CCU −

+=Δ ; JU 375.0=Δ

Example: 98 64 small drops of mercury, each of radius r and charge q coalesce to form a big drop. The ratio of the

surface density of charge of each small drop with that of the big drop is [KCET 2002]

(a) 1 : 64 (b) 64 : 1 (c) 4 : 1 (d) 1 : 4

Solution: (d) ;4/

4/ 2

2

2

==

rR

Qq

RQ

rq

Big

Small

ππ

σσ

since R = n1/3r and Q = nq

So 3/1

1

nBig

Small =σσ

⇒ 41

=Big

Small

σσ

Two hollow spheres are charged positively. The smaller one is at 50 V and the bigger one is at 100 V. How should they be arranged so that the charge flows from the smaller to the bigger sphere when they are connected by a wire [Kerala PET 2002]

(a) By placing them close to each other

(b) By placing them at very large distance from each other

(c) By placing the smaller sphere inside the bigger one

(d) Information is insufficient

Solution: (c) By placing the smaller sphere inside the bigger one. The potential of the smaller one will now be 150 V. So on connecting it with the bigger one charge will flow from the smaller one to the bigger one.

+ + +

+ + + +

+ +

+ +

+ +

+ + Examples based on sharing of charge, drops and general concept of capacity

Tricky example: 14


Chapter 5 76

c Capacitor.

(1) Definition : A capacitor is a device that stores electric energy. It is also named condenser.

or A capacitor is a pair of two conductors of any shape, which are close to each other and have equal and opposite charge.

(2) Symbol : The symbol of capacitor are shown below

or variable capacitor

(3) Capacitance : The capacitance of a capacitor is defined as the magnitude of the charge Q on the positive

plate divided by the magnitude of the potential difference V between the plates i.e., VQ

C =

(4) Charging : A capacitor get’s charged when a battery is connected across the plates. The plate attached to the positive terminal of the battery get’s positively charged and the one joined to the negative terminal get’s negatively charged. Once capacitor get’s fully charged, flow of charge carriers stops in the circuit and in this condition potential difference across the plates of capacitor is same as the potential difference across the terminals of battery (say V).

(5) Charge on capacitor : Net charge on a capacitor is always zero, but when we speaks of the charge Q on a capacitor, we are referring to the magnitude of the charge on each plate. Charge distribution in making of parallel plate capacitor can easily be understand by reading carefully the following sequence of figures –

+ +

+

+ + +

+Q – –

–

–

–

–

–Q

+Q

2Q+

2Q+

⇒ ⇒

X X

⇒

2Q+

2Q+

2Q−

X

2Q+

Y

+ Q – Q

X Y

⇒

–Q +Q

V + –

+ –

+ –

C

V


Chapter 5 77

c(6) Energy stored : When a capacitor is charged by a voltage source (say battery) it stores the electric

energy. If C = Capacitance of capacitor; Q = Charge on capacitor and V = Potential difference across

capacitor then energy stored in capacitor C

QQVCVU

221

21 2

2 ===

Note : ≅ In charging capacitor by battery half the energy supplied is stored in the capacitor and remaining half energy (1/2 QV) is lost in the form of heat.

(7) Types of capacitors : Capacitors are of mainly three types as described in given table

Parallel Plate Capacitor Spherical Capacitor Cylindrical Capacitor

It consists of two parallel metallic plates (may be circular, rectangular, square) separated by a small distance

A = Effective overlaping area of each plate

d = Separation between the plates

Q = Magnitude of charge on the inner side of each plate

σ = Surface density of charge of

each plate

=

AQ

V = Potential difference across the plates

E = Electric field between the plates

=

0εσ

Capacitance : dAε

C 0=

in C.G.S. : πdA

C4

=

If a dielectric medium of dielectric constant K is filled completely between the plates then capacitance increases by K times KCC'=

It consists of two concentric conducting spheres of radii a and b (a < b). Inner sphere is given charge +Q, while outer sphere is earthed

Capacitance ab

ab.πεC

−= 04

in C.G.S. ab

abC

−= . In the presence of

dielectric medium (dielectric constant K)

between the spheres ab

abKC

−= 04' πε

Special Case : If outer sphere is given a charge +Q while inner sphere is earthed

Induced charge on the inner sphere

Qba

Q .' −= , ab

b.πεC'

−=

2

04

This arrangement is not a capacitor. But it’s capacitance is equivalent to the sum of capacitance of spherical capacitor and spherical conductor i.e.

bab

abab

b00

2

0 44.4 πεπεπε +−

=−

It consists of two concentric cylinders of radii a and b (a < b), inner cylinder is given charge +Q while outer cylinder is earthed. Common length of the cylinders is l then

Capacitance

=

ab

log

lC

e

02πε

In the presence of dielectric medium (dielectric constant K) capacitance increases by K times and

=

abKl

C

elog

2' 0πε

+

+

+

+ +

+ +

+ –

–

–

– –

–

– –

– –

– –

– –

– –

+Q –Q

a b

+ +

+

+

+ +

– –

–

–

–

– d

A

+Q – Q

air

– – a

b

Q′ +Q – –

– –

– –

– – –

a b q – q

l


Chapter 5 78

c

Concepts

It is a very common misconception that a capacitor stores charge but actually a capacitor stores electric energy in the electrostatic field between the plates.

Two plates of unequal area can also form a capacitor because effective overlapping area is considered.

If two plates are placed side by side then three capacitors are formed. One between distant earthed bodies and the first face of the first plate, the second between the two plates and the third between the second face of the second plate and distant earthed objects. However the capacitances of the first and third capacitors are negligibly small in comparision to that between the plates which is the main capacitance.

Capacitance of a parallel plate capacitor depends upon the effective overlapping area of plates )( AC ∝ , separation between the

plates

∝

dC

1 and dielectric medium filled between the plates. While it is independent of charge given, potential raised or

nature of metals and thickness of plates.

The distance between the plates is kept small to avoid fringing or edge effect (non-uniformity of the field) at the bounderies of the plates.

Spherical conductor is equivalent to a spherical capacitor with it’s outer sphere of infinite radius. A spherical capacitor behaves as a parallel plate capacitor if it’s spherical surfaces have large radii and are close to each other. The intensity of electric field between the plates of a parallel plate capacitor (E = σ/ε0) does not depends upon the distance

between them. The plates of a parallel plate capacitor are being moved away with some velocity. If the plate separation at any instant of time is

‘d’ then the rate of change of capacitance with time is proportional to 2

1

d.

Radial and non-uniform electric field exists between the spherical surfaces of spherical capacitor.

Two large conducting plates X and Y kept close to each other. The plate X is given a charge 1Q while plate Y is given a charge

)( 212 QQQ > , the distribution of charge on the four faces a, b, c, d will be as shown in the following figure.

A

d

+ +

+

+

+ +

– –

–

–

–

–

⇒

X X Y

a c

b d

Q1 Q2

+

2

21 QQ

−

2

21 QQ

−

2

21 QQ

−−

2

21 QQ


Chapter 5 79

c

Example: 99 The capacity of pure capacitor is 1 farad. In D.C. circuit, its effective resistance will be [MP PMT 2000]

(a) Zero (b) Infinite (c) 1 ohm (d) ohm21

Solution: (b) Capacitor does not work in D.C. for D.C. it’s effective resistance is infinite i.e. it blocks the current to flow in the circuit.

Example: 100 A light bulb, a capacitor and a battery are connected together as shown here, with switch S initially open. When the switch S is closed, which one of the following is true [MP PMT 1995]

(a) The bulb will light up for an instant when the capacitor starts charging

(b) The bulb will light up when the capacitor is fully charged

(c) The bulb will not light up at all

(d) The bulb will light up and go off at regular intervals

Solution: (a) Current through the circuit can flow only for the small time of charging, once capacitor get’s charged it blocks the current through the circuit and bulb will go off.

Example: 101 Capacity of a parallel plate condenser is 10µF when the distance between its plates is cm8 . If the distance

between the plates is reduced to 4cm, its capacity will be

[CBSE 2001; Similar to CPMT 1997; AFMC 2000]

(a) 10µF (b) 15µF (c) 20µF (d) 40µF

Solution: (c) dd

AC

10 ∝=ε

∴ 1

2

2

1

dd

CC

= or FCdd

C µ201048

12

12 =×=×=

Example: 102 What is the area of the plates of a F3 parallel plate capacitor, if the separation between the plates is mm5

[BHU 2002; AIIMS 1998]

(a) 2910694.1 m× (b) 2910529.4 m× (c) 2910281.9 m× (d) 2910981.12 m×

Solution: (a) By using the relation dA

C 0ε= ⇒ .10694.11085.8

1053 2912

3

0m

CdA ×=

×

××==

−

−

ε

Example: 103 If potential difference of a condenser )6( Fµ is changed from 10 V to 20 V then increase in energy is

[CPMT 1997, 87]

(a) J4102 −× (b) J4104 −× (c) J4103 −× (d) J4109 −×

Solution: (d) Initial energy 212

1CVUi = ; Final energy 2

221

CVU f =

∴ Increase in energy )1020(10621

)(21 2262

122 −××=−=−=∆ −VVCUUU if .109 4 J−×=

– +

+ +

–

– Example based on simple concepts of capacitor

s + –


Chapter 5 80

cExample: 104 A spherical capacitor consists of two concentric spherical conductors. The inner one of radius 1R maintained

at potential 1V and the outer conductor of radius 2R at potential 2V . The potential at a point P at a distance

x from the centre (where 12 RxR >> ) is [MP PMT 1997]

(a) )( 112

21 RxRRVV

−−−

(b) xRR

RxRVxRRV)(

)()(

12

122211

−−+−

(c) )( 12

21 RR

xVV

−+ (d) x

RRVV

)()(

21

21

++

Solution: (b) Let 1Q and 2Q be the charges on the inner and the outer sphere respectively. Now 1V is the total potential

on the sphere of radius R1,

So, 2

2

1

11 R

QRQ

V += …….. (i)

and 2V is the total potential on the surface of sphere of radius 2R ,

So, 2

1

2

22 R

QRQ

V += …….. (ii)

If V be the potential at point P which lies at a distance x from the common centre then

1

11

1

2

21

RQ

Vx

QRQ

xQ

V −+=+= 11

111

11

)(11V

xRxRQ

VRx

Q +−

=+

−= ……..(iii)

Substracting (ii) from (i)

2

2

1

121 R

QRQ

VV −=− ⇒ 11122121 )( QRQRRRVV −=− ⇒ 12

21211

)(RR

RRVVQ

−−

=

Now substituting it in equation (iii), we have

1121

21211

)()()(

VRRxR

RRVVxRV +

−−−

= ⇒ )(

)()(

12

122211

RRxRxRVxRRV

V−

−+−=

Example: 105 The diameter of each plate of an air capacitor is 4 cm. To make the capacity of this plate capacitor equal to that of 20 cm diameter sphere, the distance between the plates will be [MP PET 1996]

(a) m3104 −× (b) m3101 −× (c) 1 cm (d) cm3101 −×

Solution: (b) According to the question RdA

00 4πε

ε= ⇒ .101

10104

)102(4

32

22

mR

Ad −

−

−

×=××

×==

ππ

π

Example: 106 A spherical condenser has inner and outer spheres of radii a and b respectively. The space between the two is filled with air. The difference between the capacities of two condensers formed when outer sphere is earthed and when inner sphere is earthed will be [MP PET 1996]

(a) Zero (b) a04πε (c) b04πε (d)

− abb

a04πε

Solution: (c) Capacitance when outer sphere is earthed ab

abC

−= .4 01 πε and capacitance when inner sphere is earthed

abb

C−

=2

02 .4πε . Hence bCC .4 012 πε=−

Example: 107 After charging a capacitor of capacitance Fµ4 upto a potential 400 V, its plates are connected with a resistance of Ωk1 . The heat produced in the resistance will be [CBSE PMT 1994]

(a) 0.16 J (b) 1.28 J (c) 0.64 J (d) 0.32 J

Solution: (d) This is the discharging condition of capacitor and in this condition energy released


Chapter 5 81

c 2

21

CVU = J32.0)400(10421 26 =×××= − .32.0 J=

Example: 108 The amount of work done in increasing the voltage across the plates of a capacitor from V5 to V10 is W. The work done in increasing it from V10 to V15 will be (a) 0.6 W (b) W (c) 1.25 W (d) 1.67 W

Solution: (d) As we know that work done initialfinal UU −= )(21 2

122 VVC −=

When potential difference increases from V5 to V10 then

)510(21 22 −= CW ……..(i)

When potential difference increases from V10 to V15 then

)1015(21

' 22 −= CW ……..(ii)

On solving equation (i) and (ii) we get .67.1' WW =

In an isolated parallel plate capacitor of capacitance C, the four surface have charges 1Q , 2Q , 3Q and

4Q as shown. The potential difference between the plates is [UPSEAT 2003; IIT-JEE 1999]

(a) C

QQQQ2

4321 +++ (b)

CQQ

232 +

(c) C

QQ2

32 − (d)

CQQ

241 +

Solution: (c) Plane conducting surfaces facing each other must have equal and opposite charge densities. Here as the plate areas are equal, 32 QQ −= .

The charge on a capacitor means the charge on the inner surface of the positive plate (here it is 2Q )

Potential difference between the plates ecapacitanc

charge=

CQ

CQ

22 22 ==

.22

)( 3222

CQQ

CQQ −

=−−

=

Dielectric.

Dielectrics are insulating (non-conducting) materials which transmits electric effect without conducting we know that in every atom, there is a positively charged nucleus and a negatively charged electron cloud surrounding it. The two oppositely charged regions have their own centres of charge. The centre of positive charge is the centre of mass of positively charged protons in the nucleus. The centre of negative charge is the centre of mass of negatively charged electrons in the atoms/molecules.

Q1

Q2

Q3

Q4

Tricky example: 15


Chapter 5 82

c(1) Type of Dielectrics : Dielectrics are of two types –

(i) Polar dielectrics : Like water, Alcohol, 2CO , 3NH ,

HCl etc. are made of polar atoms/molecules.

In polar molecules when no electric field is applied centre of positive charges does not coincide with the centre of negative charges.

A polar molecule has permanent electric dipole moment )(p

in the absence of electric field also. But a

polar dielectric has net dipole moment is zero in the absence of electric field because polar molecules are randomly oriented as shown in figure.

In the presence of electric field polar molecules tends to line up in the direction of electric field, and the substance has finite dipole moment.

(ii) Non polar dielectric : Like 2N , 2O , Benzene,

Methane etc. are made of non-polar atoms/molecules. In non-polar molecules, when no electric field is applied the centre of positive charge coincides with the centre of negative charge in the molecule. Each molecule has zero dipole moment in its normal state.

When electric field is applied, positive charge experiences a force in the direction of electric field and negative charge experiences a force in the direction opposite to the field i.e., molecules becomes induced electric dipole.

In general, any non-conducting, material can be called as a dielectric but broadly non conducting material having non polar molecules referred to as dielectric because induced dipole moment is created in the non polar molecule.

(2) Polarization of a dielectric slab : It is the process of inducing equal and opposite charges on the two faces of the dielectric on the application of electric field.

Suppose a dielectric slab is inserted between the plates of a capacitor. As shown in the figure.

Induced electric field inside the dielectric is iE , hence this induced

electric field decreases the main field E to iEE − i.e., New electric field

between the plates will be iEEE −=' .

(3) Dielectric constant : After placing a dielectric slab in an electric field. The net field is decreased in that region hence

If =E Original electric field and ='E Reduced electric field. Then KEE

='

where K is called dielectric constant

K is also known as relative permittivity )( rε of the material or SIC (Specific Inductive Capacitance)

The value of K is always greater than one. For vacuum there is no polarization and hence 'EE = and 1=K

105o

H+ H+

O– –

P

+ + – – 0=P

+ +

E

–

–

+ – +

– +

– –

+ – +

+ –

P

– + – + – +

– + – + – +

+ + + + + + + + + +

– – – – – – – – – –

E

Ei + – + –

+ – + –

+ – + –

+ – + –

+ – + +


Chapter 5 83

c(4) Dielectric breakdown and dielectric strength : If a very high electric field is created in a dielectric,

the outer electrons may get detached from their parent atoms. The dielectric then behaves like a conductor. This phenomenon is known as dielectric breakdown.

The maximum value of electric field (or potential gradient) that a dielectric material can tolerate without it’s electric breakdown is called it’s dielectric strength.

S.I. unit of dielectric strength of a material is mV

but practical unit is mmkV

.

Variation of Different Variables (Q, C, V, E and U) of Parallel Plate Capacitor.

Suppose we have an air filled charged parallel plate capacitor having variables are as follows :

Charge – Q, Surface charge density – AQ

=σ , Capacitance – dA

C 0ε=

Potential difference across the plates – dEV .=

Electric field between the plates – 00 εε

σAQ

E ==

Energy stored – QVC

QCVU

21

221 2

2 ===

(1) When dielectric is completely filled between plates : If a dielectric slab is fills completely the gap

between the plates, capacitance increases by K times i.e., d

AKC 0'

ε= ⇒ KCC ='

The effect of dielectric on other variables such as charge. Potential difference field and energy associated with a capacitor depends on the fact that whether the charged capacitor is disconnected from the battery or battery is still connected.

Quantity Battery is Removed

Battery Remains connected

Capacity C′ = KC C′ = KC

Charge Q′ = Q (Charge is conserved) Q′ = KQ

Potential V′ = V/K V′ = V (Since Battery maintains the potential difference)

Intensity E′ = E/K E′ = E

Energy U′ = U/K U′ = U/K

Note : ≅ If nothing is said it is to be assumed that battery is disconnected.

d

A K K

d

V

A

+ +

+

+

+ +

– –

–

–

–

–

+Q – Q

Air

d

V

A


Chapter 5 84

c(2) When dielectric is partially filled between the plates : If a dielectric slab of thickness )( dtt < is

inserted between the plates as shown below, then =E Main electric field between the plates, =iE Induced electric

field in dielectric. =−= )(' iEEE The reduced value of electric field in the dielectric. Potential difference between the

two plates of capacitor is given by

tKE

tdEtEtdEV .)(')(' +−=+−=

⇒

+−=

+−=

+−=

Kt

tdAQ

Kt

tdKt

tdEV00

'εε

σ

Now capacitance of the capacitor

'

'VQ

C = ⇒

Kt

td

AC

+−= 0'

ε

+++++++−

=................)(

'

3

3

2

2

1

1321

0

Kt

Kt

Kt

tttd

AC

ε

+++

=

4

4

3

3

2

2

1

1

0'

Kt

Kt

Kt

Kt

AC

ε

(3) When a metallic slab is inserted between the plates :

Capacitance )(

' 0

tdA

C−

=ε

∞='C (In this case capacitor is said to be short circuited)

(4) When separation between the plates is changing : If separation between the plates changes then it’s

capacitance also changes according to d

C1

∝ . The effect on other variables depends on the fact that whether the

charged capacitor is disconnected from the battery or battery is still connected.

+ +

+

+

+ +

– –

–

–

–

–

d

A

+ +

+

+

+ +

– –

–

–

–

–

Ei

E

d

A

t

K=∞

d

A K=∞

d

A or

A K1 K2 K3

t1 t2 t2

d

A

d

K1 K2 K3 K4

t1 t2 t3 t4


Chapter 5 85

c(i) Separation is increasing

Quantity Battery is removed

Battery remains connected

Capacity Decreases because

dC

1∝ i.e., CC <'

Decreases i.e., CC <'

Charge Remains constant because a battery is not present

i.e., QQ ='

Decreases because battery is present i.e., QQ <'

Remaining charge )'( QQ − goes back to the

battery.

Potential difference

Increases because CQ

V = ⇒ C

V1

∝ i.e., VV >' VV =′ (Since Battery maintains the potential

difference)

Electric field

Remains constant because 00 εε

σAQ

E ==

i.e., EE ='

Decrease because 0εA

QE = ⇒ QE ∝

i.e., EE <'

Energy Increases because

CQ

U2

2

= ⇒ C

U1

∝

i.e., UU >'

Decreases because 2

21

CVU = ⇒ CU ∝

i.e., UU <'

(ii) Separation is decreasing

Quantity Battery is removed Battery remains connected

Capacity Increases because

dC

1∝ i.e., CC >' Increases i.e., CC >'

Charge Remains constant because battery is not present

i.e., QQ ='

Increases because battery is present i.e., QQ >'

Remaining charge )'( QQ − supplied from the

battery.

Potential difference

Decreases because CQ

V = ⇒C

V1

∝ i.e., VV <' VV =′ (Since Battery maintains the potential

difference)

Electric field

Remains constant because 00 εε

σAQ

E ==

i.e., EE ='

Increases because 0εA

QE = ⇒ QE ∝

i.e., EE >'

Energy Decreases because

CQ

U2

2

= ⇒ C

U1

∝

i.e., UU <'

Increases because 2

21

CVU = ⇒ CU ∝

i.e., UU >'

A

d′

A

d′

V


Chapter 5 86

c Force Between the Plates of a Parallel Plate Capacitor.

Field due to charge on one plate on the other is 02ε

σ=E , hence the force QEF =

×−=

02εσσAF A

0

2

2εσ

−=

⇒ A

QAF

0

2

0

2

22||

εεσ

==

Energy Density Between the Plates of a Parallel Plate Capacitor.

The energy stored in a capacitor is not localised on the charges or the plates but is distributed in the field. And as in case of a parallel plate capacitor field is only between the plates i.e. in a volume (A× d), the so called energy density.

Hence Energy density AdV

dA

Ad

CV 20

2

212

1

VolumeEnergy

===ε

.21

21 2

0

2

0 EdV εε =

=

Concepts

In the expression of capacitance of parallel plate capacitor filled partially with dielectric term

+−

Kt

td is known as effective

air separation between the plates.

When dielectric is partially filled between the plates of a parallel plate capacitor then it’s capacitance increases but potential

difference decreases. To maintain the capacitance and potential difference of capacitor as before (i.e., dA

c 0ε= , dV0εσ

= )

separation between the plates has to be increased. Suppose separation is increased by 'd so in this case

dA

Kt

tdd

A 00

'

εε=

+−+

which gives us d't

tK

−=

Example: 109 The mean electric energy density between the plates of a charged capacitor is (here Q = Charge on the capacitor and A = Area of the capacitor plate) [MP PET 2002]

(a) 2

0

2

2 A

Q

ε (b)

202 A

Q

ε (c)

AQ

0

2

2ε (d) None of these

Solution: (a) Energy density 2

0

22

00

20

221

21

A

QAQ

Eεε

εε =

== .

Example: 110 Plate separation of a Fµ15 capacitor is 2 mm. A dielectric slab (K = 2) of thickness 1 mm is inserted between the plates. Then new capacitance is given by [BHU 1994, Similar to BHU 2000]

(a) Fµ15 (b) Fµ20 (c) Fµ30 (d) Fµ25

Solution: (b) Given FdA

C µε

150 == ……..(i)

d

A

+ +

+

+

+ +

– –

–

–

–

–

Air

Example based on capacitor with dielectric


Chapter 5 87

c Then by using

210

10102' 3

33

00−

−− +−×=

+−=

A

Kt

td

AC

εε 30 10

32

××= Aε ; From equation (i) .20' FC µ=

Example: 111 There is an air filled pF1 parallel plate capacitor. When the plate separation is doubled and the space is filled

with wax, the capacitance increases to .2 pF The dielectric constant of wax is [MNR 1998]

(a) 2 (b) 4 (c) 6 (d) 8

Solution: (b) Given that capacitance pFC 1=

After doubling the separation between the plates 2

'C

C =

and when dielectric medium of dielectric constant k filled between the plates then 2

'CK

C =

According to the question, 22

' ==CK

C ⇒ .4=K

Example: 112 If a slab of insulating material m5104 −× thick is introduced between the plate of a parallel plate capacitor, the

distance between the plates has to be increased by m5105.3 −× to restore the capacity to original value. Then

the dielectric constant of the material of slab is [AMU 1999]

(a) 10 (b) 12 (c) 6 (d) 8

Solution: (d) By using 'dt

tK

−= ; here mt 5104 −×= ; md 5105.3' −×= ⇒

55

5

105.3104

104−−

−

×−×

×=K = 8

Example: 113 The force between the plates of a parallel plate capacitor of capacitance C and distance of separation of the plates d with a potential difference V between the plates, is [MP PMT 1999]

(a) d

CV2

2

(b) 2

22

2d

VC (c)

2

22

d

VC (d)

CdV 2

Solution: (a) Since A

QF

0

2

2ε= ⇒ .

22

2

0

22

dCV

AVC

F ==ε

Example: 114 A capacitor when filled with a dielectric 3=K has charge 0Q , voltage 0V and field 0E . If the dielectric is

replaced with another one having K = 9, the new values of charge, voltage and field will be respectively

(a) 000 3,3,3 EVQ (b) 000 3,3, EVQ (c) 00

0 3,3

, EV

Q (d) 3

,3

, 000

EVQ

Solution: (d) Suppose, charge, potential difference and electric field for capacitor without dielectric medium are Q, V and E respectively

With dielectric medium of 3=K With dielectric medium of 9=K

Charge QQ =0 Charge 0' QQQ ==

Potential difference 30V

V = Potential difference 39

' 0VVV ==

Electric field 30E

E = Electric field .39

' 0EEE ==


Chapter 5 88

cExample: 115 A slab of copper of thickness b is inserted in between the plates of parallel plate capacitor as shown in the

figure. The separation between the plates is d. If 2d

b = then the ratio of capacities of the capacitor after and

before inserting the slab will be [IIT-JEE 1976; Similar to Orissa JEE 2002, KCET 2001, MP PMT 1994]

(a) 1:2 (b) 2 : 1 (c) 1 : 1 (d) 2:1

Solution: (b) Capacitance before inserting the slab dA

C 0ε= and capacitance after inserting the slab td

AC

−= 0'ε .

Where 2d

bt == so d

AC 02

'ε

= hence, .12'

=CC

Example: 116 The capacity of a parallel plate condenser is 0C . If a dielectric of relative permitivity rε and thickness equal to

one fourth the plate separation is placed between the plates, then its capacity becomes C. The value of 0C

C

will be

(a) 14

5+r

r

εε

(b) 13

4+r

r

εε

(c) 12

3+r

r

εε

(d) 1

2+r

r

εε

Solution: (b) Initially capacitance dA

C 00

ε= ……..(i) Finally capacitance

r

ddd

AC

ε

ε4/

4

0

+−= ……..(ii)

By dividing equation (ii) by equation (i) 13

4

0 +=

r

r

CC

εε

An air capacitor of capacity FC µ10= is connected to a constant voltage battery of 12 V. Now the

space between the plates is filled with a liquid of dielectric constant 5. The charge that flows now from battery to the capacitor is [MP PMT 1997]

(a) Cµ120 (b) Cµ600 (c) Cµ480 (d) Cµ24

Solution: (c) Initially charge on the capacitor CQi µ1201210 =×=

When dielectric medium is filled, so capacitance becomes K times, i.e. new capacitance CC µ50105' =×=

Final charge on the capacitor CQf µ6001250 =×=

Hence additional charge supplied by the battery .480 CQQ if µ=−=

Cu b d

A = Area

Tricky example: 16


Chapter 5 89

c Grouping of Capacitors.

Series grouping Parallel grouping

(1) Charge on each capacitor remains same and equals to the main charge supplied by the battery

V = V1 + V2 + V3

(2) Equivalent capacitance

321

1111CCCCeq

++= or 113

12

11 )( −−−− ++= CCCCeq

(3) In series combination potential difference and energy distribution in the reverse ratio of capacitance i.e.,

CV

1∝ and

CU

1∝ .

(4) If two capacitors having capacitances C1 and C2 are

connected in series then Addition

tionMultiplica=

+=

21

21CC

CCCeq

VCC

CV .

21

11

+

= and VCC

CV .

21

22

+

=

(5) If n identical capacitors each having capacitances C are connected in series with supply voltage V then

Equivalent capacitance nC

Ceq = and Potential

difference across each capacitor nV

V =' .

(1) Potential difference across each capacitor remains same and equal to the applied potential difference

Q = Q1 + Q2 + Q3

(2) Ceq = C1 + C2 + C3

(3) In parallel combination charge and energy distributes in the ratio of capacitance i.e. Q ∝ C and U ∝ C

(4) If two capacitors having capacitance C1 and C2 respectively are connected in parallel then

21 CCCeq +=

QCC

CQ .

21

11

+

= and QCC

CQ .

21

22

+

=

(5) If n identical capacitors are connected in parallel

Equivalent capacitance nCCeq = and Charge on each

capacitor nQ

Q ='

Redistribution of Charge Between Two Capacitors.

When a charged capacitor is connected across an uncharged capacitor, then redistribution of charge occur to equalize the potential difference across each capacitor. Some energy is also wasted in the form of heat.

Suppose we have two charged capacitors 1C and 2C after disconnecting these two from their respective batteries. These two capacitors are connected to each other as shown below (positive plate of one capacitor is connected to positive plate of other while negative plate of one is connected to negative plate of other)

Charge on capacitors redistributed and new charge on them will be

+

=21

1'1 CC

CQQ ,

+

=21

2'2 CC

CQQ

The common potential 21

2211

21

21

CCVCVC

CCQQ

V++

=++

= and loss of energy 221

21

21 )()(2

VVCC

CCU −

+=∆

Note : ≅ Two capacitors of capacitances C1 and C2 are charged to potential of V1 and V2 respectively. After disconnecting from batteries they are again connected to each other with reverse polarity i.e., positive plate

of a capacitor connected to negative plate of other. So common potential 21

2211

21

21

CCVCVC

CCQQ

V+−

=+−

= .

+ + + +

– – – –

+ + + +

+ + + +

– – – –

– – – –

+Q –Q +Q –Q +Q –Q C1 C2 C3

Q V1 V2 V3

+ –

V

+ + + +

– – – –

+Q2 –Q2

V

+ + + +

– – – –

+Q1 –Q1

+ + + +

– – – –

+Q3 –Q3 Q Q3

Q2

Q1

+ + + + + +

– – – – – – C2

Q2

V2

C1

Q1

V1


Chapter 5 90

c Circuit With Resistors and Capacitors.

(1) A resistor may be connected either in series or in parallel with the capacitor as shown below

Series RC Circuit Parallel RC Circuit

In this combination capacitor takes longer time to charge. Resistor has no effect on the charging of capacitor.

The charging current is maximum in the beginning; it decreases with time and becomes zero after a long time.

Resistor provides an alternative path for the electric current.

(2) Three states of RC circuits

(i) Initial state : i.e., just after closing the switch or just after opening the switch.

(ii) Transient state : or instantaneous state i.e., any time after closing or opening the switch.

(iii) Steady state : i.e., a long time after closing or opening the switch. In the steady state condition, the capacitor is charged or discharged.

(3) Charging and discharging of capacitor in series RC circuit : As shown in the following figure (i) when switch S is closed, capacitor start charging. In this transient state potential difference appears across capacitor as well as resistor. When capacitor gets fully charged the entire potential difference appeared across the capacitor and nothing is left for the resistor. [shown in figure (ii)]

(i) Charging : In transient state of charging charge on the capacitor at any instant

−=

−RC

t

eQQ 10 and

potential difference across the capacitor at any instant

−=

−RC

t

eVV 10

V0

VC

O t

VC = V0(1 – e–t/RC)

Q0

Q

O t

Q = Q0(1 – e–t/RC)

(i)

V0 S

C + –

R

+ –

i V V′

Transient state

C

V0 S

+ – R

+ –

V0

Steady state

charge)Max (00

C

QV =

(ii)

V0 S

C R

V0 S

C

R


Chapter 6 91

c(ii) Discharging : After the completion of charging, if battery is removed capacitor starts discharging. In

transient state charge on the capacitor at any instant RCteQQ /0

−= and potential difference cross the capacitor at

any instant CRteVV /0

−= .

(iii) Time constant (τ) : The quantity RC is called the time constant i.e., RC=τ .

In charging : It is defined as the time during which charge on the capacitor rises to 0.63 times (63%) the

maximum value. That is when RCt == τ , 01

0 639.0)1( QeQQ =−= − or

In discharging : It is defined as the time during which charge on a capacitor falls to 0.37 times (37%) of the

initial charge on the capacitor that is when RCt == τ , 01

0 37.0)( QeQQ == −

(iv) Mixed RC circuit : In a mixed RC circuit as shown below, when switch S is closed current flows through the branch containing resistor as well as through the branch contains capacitor and resistor (because capacitor is in the process of charging)

When capacitor gets fully charged (steady state), no current flows through the line in which capacitor is

connected. Therefore the current through resistor 1R is ( )rRV+1

0 , hence potential difference across resistance will be

equal to ( ) 11

0 RrR

V+

. The same potential difference will appear across the capacitor, hence charge on capacitor in

steady state ( )rRRCV

Q+

=1

10

Q0

Q

t

Q= Q0 e–t/RC

O

V0

VC

t

VC = V0 e–t/RC

O

S

R1

i

V0

C R2

R

Transient state

S

R1

i

V0

C R2

R

i

Steady state

No current


Chapter 6 92

cConcepts

In series combination equivalent capacitance is always lesser than that of either of the individual capacitors e.g. If two capacitors having capacitances 3µF and 6µF respectively are connected in series then equivalent capacitance

FCeq µ26363=

+×

= Which is lesser than the smallest capacitance (3µF) of the network.

In parallel combination, equivalent capacitance is always greater than the maximum capacitance of either capacitor in network.

If n identical plates arranged as shown below, constitutes (n – 1) capacitors in series. If each capacitors having capacitance dA0ε

then dn

ACeq )1(

0

−=

ε

In this situation except two extreme plates each plate is common to adjacent capacitors.

If n identical plates are arranged such that even no. of plates are connected together and odd number of plates are connected together, then (n – 1) capacitors will be formed and they will be in parallel grouping.

Equivalent capacitance CnC )1(' −= Where C = capacitance of a capacitor dA0ε=

If n identical capacitors are connected in parallel which are charged to a potential V. If these are separated and connected in series then potential difference of combination will be nV.

Network Solving.

To solve capacitive network for equivalent capacitance following guidelines should be followed.

Guideline 1. Identify the two points across which the equivalent capacitance is to be calculated.

Guideline 2. Connect (Imagine) a battery between these points.

Guideline 3. Solve the network from the point (reference point) which is farthest from the points between which we have to calculate the equivalent capacitance. (The point is likely to be not a node)

(1) Simple circuits : Suppose equivalent capacitance is to be determined in the following networks between points A and B

+ + + +

– – – –

+ + + +

– – – –

+ + + +

– – – –

+ + + +

– – – –

– +

1

2

3

4

5

6

7


Chapter 6 93

cSuppose equivalent capacitance is to be determined in the following networks between points A and B

(i)

(ii)

(iii)

(2) Circuits with extra wire : If there is no capacitor in any branch of a network then every point of this branch will be at same potential. Suppose equivalent capacitance is to be determine in following cases

(i)

CAB = 3C

(ii)

CAB = 2C

3µF 6µF

3µF A B

⇒ ⇒ A B 3µF

2µF FABC µ532 =+=

3µF 6µF

3µF A B

Fµ263

=×

2µF

6µF 6µF

A B

6µF

3µF

3 + 3 = 6µF

2µF

6µF

B

3µF

A

3µF

⇒ ⇒

Fµ32

6=

2µF

6µF

B

6µF

A

μFABC 523 =+=

⇒

2µF

6µF 6µF

A B

6µF

3µF

Fµ326=

A

B

9µF 9µF 9µF

9µF 9µF 9µF

6µF 9µF 6µF

Series

Fµ33

9=

+ –

By similar process CAB = 3µF

⇒

A

B

9µF 9µF 9µF

9µF 9µF 9µF

6µF 9µF 6µF

+ –

A

B

9µF 9µF

9µF 9µF

9µF 6µF

+ –

A

B

9µF 9µF

9µF 9µF

6µF 3µF 6µF

Parallel 6 + 3 = 9µF

⇓

⇐

C

C

C B A

+ –

A B

C C C ⇒ A B

C C C

A A

A A

B

B B

B ⇒

No p.d. across vertical

branch so it is removed

⇒ ⇒

B C

C

A A B C

C

C

A B

A A

A A

B

B

C

C

C

⇒


Chapter 6 94

c(iii)

CAB = 3C

(iv)

(v) Since there is no capacitor in the path APB, the points A, P and B are electrically same i.e., the input and output points are directly connected (short circuited).

Thus, entire charge will prefer to flow along path APB. It means that the capacitors connected in the circuit will not receive any charge for storing. Thus equivalent capacitance of this circuit is zero.

(3) Wheatstone bride based circuit : If in a network five capacitors are arranged as shown in following

figure, the network is called wheatstone bridge type circuit. If it is balanced then 4

3

2

1

CC

CC

= hence C5 is removed

and equivalent capacitance between A and B

(i) (ii) (iii)

43

43

21

21

CCCC

CCCC

CAB ++

+=

+ –

C

C C

C

B A Parallel C + C = 2C

⇓

+ –

C

2C

C

B A

Series

3

2

2

2 C

CC

CC=

+

×

Parallel

35

3

2 CC

C=+

+ –

2C/3

B A C

⇐

A C5

C2

C4

D

E

C1

C3

B

A

C2

C4

C1

C3 B

E D

C5

A B C5

C2

C4

D

E

C1

C3

C5

C

C

C B A

+ –

A B

C

C

C A B

C

C

C

A

A

B

B

A B C

C C

C

⇒ ⇒

A B C

C C

C

A

A

B

B A

Hence equivalent capacitance between

A and B is 3

5C

A B

C

C C

P

⇒


Chapter 6 95

c(4) Extended wheatstone bridge : The given figure consists of two wheatstone bridge connected

together. One bridge is connected between points AEGHFA and the other is connected between points EGBHFE.

This problem is known as extended wheatstone bridge problem, it has two branches EF and GH to the left and right of which symmetry in the ratio of capacities can be seen.

It can be seen that ratio of capacitances in branches AE and EG is same as that between the capacitances of the branches AF and FH. Thus, in the bridge AEGHFA; the branch EF can be removed. Similarly in the bridge EGBHFE branch GH can be removed

3C

CAB2

=

(5) Infinite chain of capacitors : In the following figure equivalent capacitance between A and B

(i)

Suppose the effective capacitance between A and B is CR. Since the network is infinite, even if we remove one pair of capacitors from the chain, remaining network would still have infinite pair of capacitors, i.e., effective capacitance between X and Y would also be CR

Hence equivalent capacitance between A and B

RR

RAB C

CCCCCC

C =+++

=21

21 )( ⇒

−

+= 141

2 2

12

CCC

CAB

A

B

C1 C1

C2 C2 C2

C1

⇒ ∞

⇒ A

C

C

C C

C C

B C C

E G

H F

A

C

C

C C

C C

B

E G

H F

A

B

X

C2

C1

CR

Y

CR

Parallel (C2 + CR)

A

B

C1

CR

Series

R

R

CCC

CCC

++

+

21

21 )(

(C2 + CR) ⇒


Chapter 6 96

c

(ii) For what value of C0 in the circuit shown below will the net effective capacitance between A and B be independent of the number of sections in the chain

Suppose there are n sections between A and B

and the network is terminated by C0 with equivalent capacitance CR. Now if we add one more sections to the network between D and C (as shown in the following figure), the equivalent capacitance of the network CR will be independent of number of sections if the capacitance between D and C still remains C0 i.e.,

Hence 021

0210

)(CCCCCC

C+++×

= ⇒ 021022

0 =−+ CCCCC

On simplification

−

+= 141

2

120 C

CCC

2

(6) Network with more than one cell :

(i)

Potential difference across C1 is )( 2121

2 EECC

C−

+

and potential difference across C2 is

)( 2121

1 EECC

C−

+

(ii) Potential difference between the ends of this arrangement is given V.

(7) Advance case of compound dielectrics : If several dielectric medium filled between the plates of a parallel plate capacitor in different ways as shown.

(i) The system can be assumed to be made up of two capacitors C1 and C2 which may be said to connected in series

2

011 d

AKC

ε= ,

2

022 d

AKC

ε= and

21

111CCCeq

+= ⇒ dA

KKKK

Ceq0

21

21 .2 ε

+

=

Also 21

212KKKK

Keq +=

C2 C1

E2

E1

C2 C1

E1 – E2

⇒

K1 K2 A

d/2 d/2

2 1

C1 C2 E A B

+ – C1 C2 E

A B

+ – V

⇒ ⇒

(E – V)

C1 C2

A

B

C1 C1

C2 C2 C2

C1

C2 C0

C1 C

D

Parallel C2 + C0

A

B

C2 C0 C0 CR

C1 ⇓ C

D


Chapter 6 97

c(ii) In this case these two capacitors are in parallel and

,2

011 d

AKC

ε=

dAK

C2

022

ε=

Hence, 21 CCCeq += ⇒ dAKK

Ceq021 .

2ε

+=

Also 2

21 KKKeq

+=

(iii) In this case C1 and C2 are in series while this combination is in parallel with C3

d

AKd

AK

C 0101

1

2

2 εε== ,

dAK

d

AK

C 0201

2

2

2 εε== and

dAK

d

AK

C2

2

2 0303

3εε

==

Hence, 321

21 CCC

CCCeq +

+=

dAk

dAk

dAk

dAk

dAk

203

0201

0201

εεε

εε

++

×= ⇒

dAk

kkkk

Ceq03

21

21 .2

ε

+

+=

Also

+

+=21

213

2 kkkkk

keq

1

2

K2

K1

d

A/2

A/2

2

K3 A/2

A/2 K2 K1

d/2 d/2

C1 C2

C3


Chapter 6 98

Example: 117 Three capacitors of 2µ f, 3µ f and 6µ f are joined in series and the combination is charged by means of a 24 volt battery. The potential difference between the plates of the 6µ f capacitor is

[MP PMT 2002 Similar to MP PMT 1996]

(a) 4 volts (b) 6 volts (c) 8 volts (d) 10 volts

Solution: (a) Equivalent capacitance of the network is 61

31

211

++=eqC

FCeq µ1=

Charge supplied by battery Q = Ceq.V ⇒ 1 × 24 = 24 µC

Hence potential difference across 6µ F capacitor 6

24= .4 volt=

Example: 118 Two capacitors each of 1 µ f capacitance are connected in parallel and are then charged by 200 V D.C. supply. The total energy of their charges in joules is [MP PMT 2002]

(a) 0.01 (b) 0.02 (c) 0.04 (d) 0.06

Solution: (c) By using formula 2

21

VCU eq=

Here FCeq µ2=

∴ 26 )200(10221

×××= −U

J04.0=

Example: 119 Five capacitors are connected as shown in the figure. The equivalent capacitance between the point A and B is

[MP PMT 2002; SCRA 1996; Pantnagar 1987]

(a) 1 µ f (b) 2 µ f (c) 3 µ f (d) 4 µ f

Solution: (b)

Hence equivalent capacitance between A and B is 2µF.

Example based on series and parallel grouping of capacitors

A

B 2µf

1µf 1µf 1µf


⇒

A

B

2µf

2µf

2µf 1µf 1µf

+ –

Series

Fµ12

2=

⇓

A

B

1µf 1µf


A

B

2µf 1µf

2µf

Series

Fµ12

2= ⇐

A

B

2µf

2µf

2µf 1µf 1µf

2µF

+ – 24V

3µF 6µF

Q

+ – 200V

1µF

1µF


Chapter 7 99

Example: 120 In the following network potential difference across capacitance of 4.5 µF is [RPET 2001; MP PET 1992]

(a) 8 V (b) 4 V (c) 2 V (d) 6 V

Solution: (a) Equivalent capacitance FCeq µ35.495.49=

+×

=

Charge supplied by battery Q = Ceq × V = 3 × 12 = 36 µC

Hence potential difference across 4.5 µF 5.4

36= = 8V.

Example: 121 A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constants K1, K2 and K3 as shown in fig. If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constant K is given by [IIT Screening 2000]

(a) 321 2

1111KKKK

++= (b) 321 2

111KKKK

++

=

(c) 321

21 2KKK

KKK +

+= (d) 321 2KKKK ++=

Solution: (b) The effective capacitance is given by

+

+=)(

12

11

2130 KKKAd

Ceq ε

The capacitance of capacitor with single dielectric of dielectric constant K is d

AKC 0ε=

According to question CCeq = i.e., d

AK

KKKd

A 0

213

0

12

1

εε=

+

+

⇒ .1

211

213 KKKK ++=

3µF

6µF

4.5µF

12 V

+ –


3µF

6µF

4.5µF

12 V

+ –

A/2

d

K1 K2 d/2

A A = Area of plates

K3

A/2


Chapter 7 100

Example: 122 Two capacitors C1 = 2µF and C2 = 6µF in series, are connected in parallel to a third capacitor C3 = 4µF. This arrangement is then connected to a battery of e.m.f. = 2 V, as shown in the fig. How much energy is lost by the battery in charging the capacitors ? [MP PET 2001]

(a) J61022 −× (b) J61011 −× (c) J610332 −×

(d) J610

316 −×

Solution: (b) Equivalent capacitance FCCC

CCCeq µ5.54

862

321

21 =+×

=++

=

∴ JVCU eq622 1011)2(5.5

21

.21 −×=××==

Example: 123 In the circuit shown in the figure, each capacitor has a capacity of 3µF. The equivalent capacity between A and B is [MP PMT 2000]

(a) Fµ43

(b) 3 µF (c) 6 µF (d) 5 µF

Solution: (d)

Hence equivalent capacitance .536363

FCeq µ=++×

=

Example: 124 Given a number of capacitors labelled as 8µF, 250 V. Find the minimum number of capacitors needed to get an arrangement equivalent to 16 µF, 1000 V [AIIMS 2000]

(a) 4 (b) 16 (c) 32 (d) 64

Solution: (c) Let C = 8 µF, C′ = 16 µF and V = 250 volt, V′ = 1000 V

Suppose m rows of given capacitors are connected in parallel which each row contains n capacitor then

Potential difference across each capacitors nV

V'

= and equivalent capacitance of network n

mCC =' .

On putting the values, we get n = 4 and m = 8. Hence total capacitors = m × n = 8 × 4 = 32.

Short Trick : For such type of problem number of capacitors 2''

×=

VV

CC

n . Here 32250

10008

16 2

=

=n

Example: 125 Ten capacitors are joined in parallel and charged with a battery up to a potential V. They are then disconnected from battery and joined again in series then the potential of this combination will be [RPET 2000]

(a) V (b) 10 V (c) 5 V (d) 2 V

Solution: (b) By using the formula nVV =' ⇒ .10' VV =

A B 3µF

3µF

3µF

3µF Parallel 3 + 3 = 6µF

A B

A B

A B

B

⇒

A B 3µF

3µF

6µF

⇒

C1 C2

C3

+ –

2 V

A B


Chapter 7 101

Example: 126 For the circuit shown, which of the following statements is true [IIT-JEE 1999]

(a) With S1 closed, V1 = 15 V, V2 = 20 V (b) With S3 closed, V1 = V2 = 25 V

(c) With S1 and S2 closed V1 = V2 = 0 (d) With S1 and S3 closed V1 = 30 V, V2 = 20 V

Solution: (d) When S3 is closed, due to attraction with opposite charge, no flow of charge takes place through S3. Therefore, potential difference across capacitor plates remains unchanged or V1 = 30 V and V2 = 20 V.

Alternate Solution

Charges on the capacitors are – pCq 60)2()30(1 == , pCq 60)3()20(2 == or )say(21 qqq ==

The situation is similar as the two capacitors in series are first charged with a battery of emf 50 V and then disconnected.

When S3 is closed, V1 = 30 V and V2 = 20 V.

Example: 127 A finite ladder is constructed by connecting several sections of FF µµ 4,2 capacitor combinations as shown in the figure. It is terminated by a capacitor of capacitance C.

What value should be chosen for C, such that the equivalent capacitance of the ladder between the points A and B becomes independent of the number of sections in between

(a) Fµ4 (b) Fµ2 (c) Fµ18 (d) Fµ6

Solution: (a) By using formula

+

+= 141

2 2

12

CCC

C ; FCFC

µµ

24

2

1

==

We get .4 FC µ=

Example: 128 Figure shows two capacitors connected in series and joined to a battery. The graph shows the variation in potential as one moves from left to right on the branch containing the capacitors. [MP PMT 1999]

(a) 21 CC >

(b) 21 CC =

(c) 21 CC <

(d) The information is insufficient to decide the relation between 1C and 2C

Solution: (c) According to graph we can say that potential difference across the capacitor 1C is more than that across 2C .

Since charge Q is same i.e., 2211 VCVCQ == ⇒ 1

2

2

1

VV

CC

= ⇒ 21 CC < ).( 21 VV >

q q + – + –

2PF 3PF

+ –

50V

⇒ + – + –

q = 60pC q = 60pC

V1 = 30V V2 = 20V

+ – + – V1 = 30V V2 = 20V

C1 = 2pF C2 = 3pF

S1 S2

S3

4µF 4µF 4µF

2µF 2µF 2µF C

A

B

C1

+ –

C2

V

X


Chapter 7 102

Example: 129 Two condensers of capacity C and 2C are connected in parallel and these are charged upto V volt. If the battery is removed and dielectric medium of constant K is put between the plates of first condenser, then the potential at each condenser is [RPET 1998; IIT-JEE 1988]

(a) 2+K

V (b)

VK

32 + (c)

22+KV

(d) 2

3+KV

Solution: (d) Initially

Equivalent capacitance of the system CCeq 3=

Total charge VCQ )3(=

Finally Equivalent capacitance of the system CKCCeq 2+=

Hence common potential .2

3)2(

3)2( +

=+

=+

=K

VCK

CVCKC

QV

Example: 130 Condenser A has a capacity of 15 µF when it is filled with a medium of dielectric constant 15. Another condenser B has a capacity 1 µF with air between the plates. Both are charged separately by a battery of 100V. after charging, both are connected in parallel without the battery and the dielectric material being removed. The common potential now is [MNR 1994]

(a) 400V (b) 800V (c) 1200V (d) 1600V

Solution: (b) Charge on capacitor A is given by CQ 461 10151001015 −− ×=××=

Charge on capacitor B is given by CQ 462 10100101 −− =××=

Capacity of capacitor A after removing dielectric Fµ1151015 6

=×

=−

Now when both capacitors are connected in parallel their equivalent capacitance will be Ceq Fµ211 =+=

So common potential .800102

)101()1015(6

44

V=×

×+×=

−

−−

Example: 131 A capacitor of Fµ20 is charged upto 500V is connected in parallel with another capacitor of Fµ10 which is

charged upto 200V. The common potential is [CBSE 2000; CPMT 1999; BHU 1997]

(a) 500V (b) 400V (c) 300V (d) 200V

Solution: (b) By using 21

2211

VCVCVC

V+

= ; C1 = 20 µF, V1 = 500 V, C2 = 10 µF and V2 = 200 V

.4001020

2001050020VV =

+×+×

=

Example: 132 In the circuit shown [DCE 1995]

+ – V

C

2C

Q

2C

KC

C1 = 4µF C2 = 8µF

12 V

6 V

+ –

+ –


Chapter 7 103

(a) The charge on C2 is greater than that of C1 (b) The charge on C2 is smaller than that of C1

(c) The potential drop across C1 is smaller than C2 (d) The potential drop across C1 is greater than C2

Solution: (d) Given circuit can be redrawn as follows

FCeq µ38

1284=

×=

So CQ µ16638

=×=

Hence potential difference voltV 44

161 == and voltV 2

816

2 == i.e. V1 > V2

Example: 133 As shown in the figure two identical capacitors are connected to a battery of V volts in parallel. When capacitors are fully charged, their stored energy is 1U . If the key K is opened and a material of dielectric

constant 3=K is inserted in each capacitor, their stored energy is now 2U . 2

1

UU will be [IIT 1983]

(a) 53 (b)

35 (c) 3 (d)

31

Solution: (a) Initially potential difference across both the capacitor is same hence energy of the system is

2221 2

121

CVCVCVU =+= ……..(i)

In the second case when key K is opened and dielectric medium is filled between the plates, capacitance of both the capacitors becomes 3C, while potential difference across A is V and potential difference across B is

3V hence energy of the system now is

2

22 3

)3(21

)3(21

+=

VCVCU 2

610

CV= …….(ii)

So, 53

2

1 =UU

Example: 134 In the following figure the resultant capacitance between A and B is Fµ1 . The capacitance C is [IIT 1977]

(a) Fµ1132

(b) Fµ3211

(c) Fµ3223

(d) Fµ2332

C1 = 4µF C2 = 8µF

12 – 6 = 6 V + –

V1 V2

K

C B

C A

V

+

–

A C 1µF

8µF 6µF 4µF

12µF

2µF 2µF

B


Chapter 7 104

Solution: (d) Given network can be simplified as follows

Given that equivalent capacitance between A and B i.e., CAB Fµ1=

But

932932

+

×=

C

CCAB hence .

2332

1

932932

FCC

Cµ=⇒=

+

×

Example: 135 A Fµ1 capacitor and a Fµ2 capacitor are connected in parallel across a 1200 volts line. The charged

capacitors are then disconnected from the line and from each other. These two capacitors are now connected to each other in parallel with terminals of unlike signs together. The charges on the capacitors will now be

(a) Cµ1800 each (b) Cµ400 and Cµ800 (c) Cµ800 and Cµ400 (d) Cµ800 and Cµ800

Solution: (b) Initially charge on capacitors can be calculated as follows

Q1 = 1 × 1200 = 1200 µC and Q2 = 2 × 1200 = 2400 µC

Finally when battery is disconnected and unlike plates are

connected together then common potential 21

12'CCQQ

V+−

=

2112002400

+−

= V400=

Hence, New charge on 1C is Cµ4004001 =×

And New charge on 2C is .8004002 Cµ=×

Example: 136 The two condensers of capacitances Fµ2 and Fµ3 are in series. The outer plate of the first condenser is at

1000 volts and the outer plate of the second condenser is earthed. The potential of the inner plate of each condenser is

(a) 300 volts (b) 500 volts (c) 600 volts (d) 400 volts

1Q + –

+ – 2µF

1µF

+ – 1200 V

2Q

1C

2C

2µ

1µ C

2C

+ – + + – –

V

– + – – + +


A

C 1µF

8µF 6µF 4µF

12µF

2µF 2µF

B

Series

Fµ4126

126=

+

×

A

C 1µF

8µF 4µF 4µF

4µF

B Series

Fµ38

48

48=

+

×

Parallel Fµ844 =+

⇒

A

C

Fµ3

8

B

Parallel

Fµ932

98

38

=+

Fµ9

8

⇓

⇐

A

C 1µF

Fµ3

8

B

Series

Fµ98

8181=

+×

8µF

⇓

C B A

Fµ932


Chapter 7 105

Solution: (d) Here, potential difference across the combination is VVV BA 1000=−

Equivalent capacitance FCeq µ56

3232=

+×

=

Hence, charge on each capacitor will be )( BAeq VVCQ −×= Cµ1200100056

=×=

So potential difference between A and C, VVV CA 6002

1200==− ⇒ 6001000 =− CV ⇒ VVc 400=

Example: 137 Four identical capacitors are connected in series with a 10V battery as shown in the figure. The point N is earthed. The potentials of points A and B are

(a) VV 0,10 (b) VV 5.25.7 − (c) VV 55 − (d) VV 5.2,5.7

Solution: (b) Potential difference across each capacitor will be V5.24

10=

Hence potential difference between A & N i.e., VVV NA 5.75.25.25.2 =++=−

⇒ VVV AA 5.70 ==− While 5.2=− BN VV ⇒ 5.20 =− BV ⇒ VVB 5.2−=

Example: 138 In the figure below, what is the potential difference between the points A and B and between B and C respectively in steady state [IIT-JEE 1979]

(a) 100 volts both (b) 75=ABV volts, 25=BCV volts

(c) 25=ABV volts, 75=BCV volts (d) 50=ABV volts 50=BCV volts

Solution: (c) In steady state No current flows in the given circuit hence resistances can be eliminated

2µF 3µF

A B 1000 V +

0 V

C

⇒

3µF

3µF 1µF

1µF

1µF

100 V

B

A C

B B

A

A

C

C

Parallel 1+1= 2µF

Parallel 3+3= 6µF

6µF 2µF

1µF

100 V

B

A C

A C

V1=VAB V2=VBC

+ –

Line (1)

Line (2)

+ –

10V

A B C

N

C C C

3µF

3µF 1µF

1µF

1µF

100 V

B

A C


Chapter 7 106

By using the formula to find potential difference in series combination of two capacitor

+=

+

= VCC

CVV

CCC

V22

12

21

21 and.

VVV AB 2510062

21 =×

+== ; .75100

626

2 VVV BC =×

+==

Example: 139 A capacitor of capacitance Fµ5 is connected as shown in the figure. The internal resistance of the cell is 0.5Ω. The amount of charge on the capacitor plate is [MP PET 1997]

(a) Cµ0 (b) Cµ5 (c) Cµ10 (d) Cµ25

Solution: (c) In steady state current drawn from the battery Ai 1)5.011(

5.2=

++=

In steady state capacitor is fully charged hence No current will flow through line (2)

Hence potential difference across line (1) is voltV 221 =×= , the same potential difference appears across the capacitor, so charge on capacitor CQ µ1025 =×=

Example: 140 When the key K is pressed at time 0=t . Which of the following statements about the current i in the resistor AB of the adjoining circuit is true [CBSE 1995]

(a) mAi 2= at all t (b) i oscillates between 1mA and 2mA

(c) i = 1mA at all t (d) At t = 0, i = 2mA and with time it goes to 1mA

Solution: (d) At 0=t whole current passes through capacitance; so effective resistance of circuit is Ω1000 and current

mAAi 21021000

2 3 =×== − . After sufficient time, steady state is reached; then there is no current in

capacitor branch; so effective resistance of circuit is Ω=+ 200010001000 and current

mAAi 11012000

2 3 =×== − i.e., current is mA2 at 0=t and with time it goes to mA1 .

Example: 141 The plates of a capacitor are charged to a potential difference of 320 volts and are then connected across a resistor. The potential difference across the capacitor decays exponentially with time. After 1 second the potential difference between the plates of the capacitor is 240 volts, then after 2 and 3 seconds the potential difference between the plates will be [MP PET 1998]

(a) 200 and 180 volts (b) 180 and 135 volts (c) 160 and 80 volts (d) 140 and 20 volts

Solution: (b) During discharging potential difference across the capacitor falls exponentially as teVV λ−= 0 (λ = 1/RC)

1 Ω 1 Ω

2 Ω 5µF Line (1)

Line (2)

i 0.5 Ω

1 Ω 1 Ω

2 Ω 5µF

+ – 2.5 V

K A B

2 V

C=

1µF

1000

Ω

1000 Ω


Chapter 7 107

Where V = Instantaneous P.D. and =0V max. P.D. across capacitor

After 1 second V1 = 320 (e–λ) ⇒ 240 = 320 (e–λ) ⇒ 43

=−λe

After 2 seconds V2 = 320 (e–λ)2 ⇒ volt18043

3202

=

×

After 3 seconds V3 = 320 (e–λ)3 = volt13543

3203

=

×

Example: 142 Five similar condenser plates, each of area A. are placed at equal distance d apart and are connected to a source of e.m.f E as shown in the following diagram. The charge on the plates 1 and 4 will be

(a) d

AdA 00 2

,εε −

(b) d

AVdAV 00 2

,εε −

(c) d

AVdAV 00 3

,εε −

(d) d

AVdAV 00 4

,εε −

Solution: (b) Here five plates are given, even number of plates are connected together while odd number of plates are connected together so, four capacitors are formed and they are in parallel combination, hence redrawing the figure as shown below.

Capacitance of each

Capacitor is dA

C 0ε=

Potential difference across each capacitor is V

So charge on each capacitor VdA

Q 0ε=

Charge on plate (1) is dAV0ε+

While charge on plate 4 is 20 ×−dAVε

.2 0

dAVε

−=

Example: 143 Four plates are arranged as shown in the diagram. If area of each plate is A and the distance between two neighbouring parallel plates is d, then the capacitance of this system between A and B will be

(a) d

A04ε (b)

dA03ε

(c) d

A02ε (d)

dA0ε

1 2 3 4 5 –

+ V

V + –

1 2

3 2

3 4

5 4

+ + +

– – –

+ + +

– – –

+ + +

– – –

+ + +

– – –

d

d

B A


Chapter 7 108

Solution: (c) To solve such type of problem following guidelines should be follows

Guideline 1. Mark the number (1,2,3……..) on the plates

Guideline 2. Rearrange the diagram as shown below

Guideline 3. Since middle capacitor having plates 2, 3 is short circuited so it should be eliminated from the circuit

Hence equivalent capacitance between A and B dA

CAB02

ε=

A capacitor of capacitance C1 = 1µF can withstand maximum voltage V1 = 6 KV (kilo-volt) and another capacitor of capacitance C2 = 3µF can withstand maximum voltage V2 = 4KV. When the two capacitors are connected in series, the combined system can withstand a maximum voltage of

[MP PET 2001]

(a) 4 KV (b) 6 KV (c) 8 KV (d) 10 KV

Solution: (c) We know Q = CV

Hence (Q1)max = 6 mC while (Q2)max = 12 mC

However in series charge is same so maximum charge on C2 will also be 6 mC (and not 12 mC) and hence

potential difference across C2 will be KVF

mCV 2

36

2 ==µ

and as in series V = V1 + V2

So KVKVKVV 826max =+=

Tricky example: 17

1

A B

2

3

4

1 2 B A

3 4

1 2

3

B A

4

2 3


Chapter 7 109

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