electrostatics theory

Electrostatics

AIM IIT 011 – 45769447 [email protected] 1

Content

1. Introduction of Electrostatics

2. Properties of charges

3. Coulomb’s Law

4. Electric Field

5. Electric Filed lines and its properties

6. Electric potential energy

Equipotential Curve

7. Gauss’s Law and its application

8. Properties of conductor

9. Electric Dipole

10. Capacitor

11. Van de Graaff Generator

Electrostatics


1. Introduction of Electrostatics

Electrostatics is the branch of Physics, which deals with static electric charges or charges at rest or slow-moving. Electrostatics: Properties of Stationary Charge

Magnetism: Properties of Moving charge

2. Properties of charge

“Object is charged to indicate that it has a charge imbalance.”

Electric Charge is an intrinsic property of matter (like mass) which causes it to experience a force when near other

electrically charged matter. There are two types of charge, positive and negative.

Electric Charge

SI Symbol Q or q

Si units Coulomb

Other units e

Derivation from other quantities ItQ

Electron e Negative C19106.1

Proton p Positive C19106.1

Neutron n Neutral Zero

When a glass rod is rubbed with silk, the transfer of electrons takes place. Glass rod loses some electrons

and hence it becomes positively charged and silk gains those electrons and it becomes negatively charged.

e

Electron Transfer

+ _

Thus, there are two types of charges namely positive charge and negative charge.

Note: Charge produces Electric field (Stationary Charge) and Magnetic field (Moving Charge), and radiates energy.

“Charge can be transferred from one part of the system to another system, but net charge will have a constant

value” or “charge can never be created nor be destroyed”

Properties of electric charge:

1. Like charges repel each other whereas unlike charges attract each other.

2. Law of conservation of charge: The algebraic sum of total charges on a system is always constant.

3. It is electron which is responsible for charging of a body and not the proton.

4. The charge on a body cannot be a fraction of electron charge. (Charge is quantized)

neq

5. Charge is always associated with mass. In Charging, the mass of a body changes. If electrons are removed from

the body, the mass of the body will decreases and the body will becomes positively charged. On the other hand, if

electrons are added to a body, the mass of the body will increase and the body will acquire a net negative.

6. Accelerated charge radiates energy.

http://en.wikipedia.org/wiki/Force

Electrostatics


Problem 1: Which one charge is not possible on a body?

A. C20108 B. C19104.2 C. C191016 D. C19102.3

Solution: (A). 2

1

106.1

108.019

19

e

qn ; which must be an integer value, Charge is always transferring in the

integral multiple of the charge of electron.

(B). 2

3

106.1

104.219

19

e

qn

(C). 10106.1

101619

19

e

qn

(D). 2106.1

102.319

19

e

qn

Hence, A, B are not possible and C, D is possible.

For the point of view of electrostatic, there are two type of materials; Conductors and Insulators

In conductor electric charges free to move, the outer electrons of each atom or molecule are only weakly

bound to it. These electrons are almost free to move throughout the body of the material and called free electrons

(conduction electrons). When such a materials is placed in an electric field, the free electrons move in a direction

opposite to the field. Such materials are called conductors. (Example: Metal, human body and Earth)

Insulator, in which all the electrons are tightly, bound their respective atoms or molecule. There are no free

electrons. When such material is placed in an electric field, the electrons may slightly shift opposite to the field but

they cannot leave their parent atoms and hence can’t move through long distance. Such materials are called

dielectrics. (Example: Glass, rubber and plastic)

Methods of charging conductors

i. By rubbing

ii. By conduction (By contact)

iii. By Induction (From a distance)

1. By rubbing (Frictional electricity): When a glass rod is rubbed with a silk cloth the glass rod acquires some positive

charge and the silk cloth acquire negative charge by the same amount.

Comb is passed through dry hair. Cloud also becomes charged by friction.

2. By conduction (Charging by contact): After conduction charge becomes equally Distributed between A & B, if A

and B are exactly same in shape, size material and finishing, Otherwise unequally.

+

+ + + + +

+

+

+

+

Charged

A

B

q q/2

A

q/2

+ + +

+ A

+ +

+ +

Gets Charged + +

+ + +

+

Uncharged

Electrostatics


3. By Induction (From a distance): It is also possible to charge a conductor in a way that does not involve contact. A

positively charged rod brought close to metal sphere. In the sphere free electrons close to the rod move close to this

side.

Earth-

Insulated Rod +

+ + + +

+

+ - - +

+

+ +

-

- - +

+ + +

+

+ - - +

+

+ +

-

- -

- - - -

-

- -

-

- -

Grounding wire

Problem 2: Three small identical balls have charges C12103 , C12108 and C12104 respectively. They are brought in contact and then separated. Calculate (i) charge on each ball (ii) number of electrons in excess or deficit on each ball after contact. Solution: (i) The charge on each ball

Cqqq

q 44321 103103

483

3

(ii) Since the charge is positive, there is a shortage of electrons on each ball.

7

19

4

10875.1106.1

103

e

qn

∴ Number of electrons = 710875.1 .

Problem 3: Two spherical conductors B and C having equal radii and carrying equal charges on them repel each

other with a force F when kept apart at some distance. A third spherical conductor having same radii as that of B but

uncharged is brought in contact with B, then brought in contact with C and finally removed away from both. Find the

new force of repulsion between B and C.

Solution: Initially, Force between B and C is 2

2

R

KQF

B C Q Q A Q = 0

R

A third uncharged sphere ‘A’, is brought in contact with B, (Charge transfer will take place by conduction)

C

Q/2

Q A

Q/2

R

B

Amount of transferred Charge is

depend upon the shape of

conductor, material.

Then third charged sphere with charge Q/2, is brought in contact with C, (Again charge transfer take place till equal

charge)

C Q/2 Q A

Q/2

R

B

3Q/4 3Q/4 A

Q/2

R

B C

Electrostatics


Here, Total Charge is conserved. Total Charge on C and A is 2/32/ QQQ , which equally distributed on the both

sphere, Charge on each sphere is 4/3Q .

Finally;

3Q/4 Q/2

R

B C

Finally Force between B and C

FR

KQ

R

QQKF

8

3

8

34/32/'

2

2

2

3. Coulomb’s Law:

q2

r q1

“The force between two point charges is directly proportional to product of their magnitude 1q &

2q , and inversely

proportional to the square of the distance ‘r’ between them. The direction of forces is along the line joining the two

point charges.”

21qqF ……………………. (i)

2

1

rF …………………… (ii)

Combined two equations

2

21

r

qqkF

Where k is a constant 0

229

4

1/109

CNmk and 0 = permittivity of the free space.

Vector Form: )(^

2

21 rr

qqkF

)(3

21

rr

qqkF

Important Points Regarding Coulomb’s law

1. Charges are assumed to be rest.

2. Charges are assumed to be point particle.

3. Magnitude of the force

2

21

r

qqkF

4. Direction of the force between two charges:

Q Q Q

Force between two

positive charges

Q Q Q

Force between two

opposite charges

Force between two

negative charges

Repulsive Force Repulsive Force Attractive Force

F F

F

F F

F

(i). “Same charges push each other in the opposite direction (Force is repulsive) and opposite charges pull each

other in the direction of towards each other (Force is attractive) and this force always along the straight line joined

both charges.”

Electrostatics


(ii). If force is negative force is attractive

If force is positive force is repulsive

(iii). Force on the both charge is equal in magnitude

(iv). There is no any force exerted on any point charge by self.

5. Effect of the medium on the force between two charges: 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 tow charge decreases by a factor of K known as dielectric constant or specific inductive capacity (SIC) of the

medium, K is also called relative permittivity of the medium (relative means with respect to free space).

Hence in the presence of the medium: 2

21

04

1

r

qqF

r

Kr = Dielectric constant/ relative permittivity of the medium.

1q

In Medium

F

2q 1q

F/K

2q In Vacuum

Relative Permittivity of the medium: Dielectric constant is the ratio of the force of attraction or repulsion between

the two similar point charges in the air to the ratio of the force of attraction or repulsion between point charges

separated by same distance in the medium.

Medium

air

F

FK

6. Principle of Superposition of electric force: Total force on a given charges is the vector sum of all the individual

force exerted on it by all other charges, each individual force being calculated by Coulomb’s law.

......514131211

FFFFF

q1

q2

q3

q4

q5 q

7. Force on a test charge due to continuous distribution of charges.

r

qo

q

dq

^

2

0

0 .4

rr

dqqF

Electrostatics


8. Charge densities:

Linear Charge Density: Charge per unit length,

l

Q

m

C

Surface Charge density: Charge per unit area

A

Q

2m

C

Volume Charge Density: Charge per unit volume

V

Q

3m

C

Problem 4: Two point charges A and B separated by a distance R attract each other with a force of N31012 . Find

the force between A and B when the charges on them are doubled and distance is halved.

Solution: Given that NR

KQF 3

2

2

1012

Q Q

Force between two

opposite charges

Attractive Force

F F

Then, the charges on them are doubled and distance is halved,

Q2 Q2

Force between two

opposite charges

F ’ F ’

New Force is 2

2

216

)2/(

)2)(2('

R

KQ

R

QQKF

Magnitude of new force is NF 192.010121616 3 .

Problem 5: If a charge q is placed at the centre of the line joining two equal charges Q such that the system is in

equilibrium then find the value of q.

Solution: System is equilibrium means that net force on each particle is zero.

Q Q

q

Force on middle charge is always zero,

Force on charge located at left hand side or right hand side,

0)2()( 2

2

2

r

KQ

r

KQqFFF Qqnet (For equilibrium)

By solving this 4

Qq

Electrostatics


Problem 6: Two small identical balls P and Q, each of mass gm10

3, carry identical charges and are suspended by

threads of equal lengths. At equilibrium, they position themselves as shown in figure. What charge on each ball?

600

600

Solution: Free body diagram

600

600

mg

Fe

T

Tsin600

Tcos600

mgT 060sin

2

2060cos

l

kqT

With the help of these two equations

Ck

mglq

kq

mgl 7

0

2

2

20 10

60tan60tan

For tension in the string

2

22)(

l

kqmgT

Problem 7: Point charges having values C1.0 , C2.0 , C3.0 and C2.0 are placed at the corners A, B, C

and D respectively of square of side one meter. Calculate the magnitude of the force on a charge of C1 placed at

centre of the square.

Solution:

1 m

C2.0

D

211222 BCABAC

2AC

DOBOCOmmAO 2

12

2

1

C1.0

C3.0 C2.0

C

B A

C1

O

BF

CF DF

AF

F

Where NFA 0018.0

2

1

)101)(101.0(109

2

669

(along OC)

Electrostatics


NFC 0054.0

2

1

)101)(103.0(109

2

669

(Along OC)

Where NFB 0036.0

2

1

)101)(102.0(109

2

669

(Along OD)

NFD 0036.0

2

1

)101)(102.0(109

2

669

(Along OD)

Force on charge at O due to A (Repulsive) and due to C is (attractive) along the AC

NFFF CA 0072.00054.00018.01

Similarly force on charge due to B and D is along OD

NFFF DB 0072.00036.00036.01

There are two forces equal in magnitude and acting perpendicular

Net force NFFFFFF 01018.00072.0414.1290cos2 0

21

2

2

2

1

Alternate Method: Use vector form.

Problem 8: Three charges 321, qandqq are placed as shown in the figure. The x-component of the force on

1q is proportional to

‒q3

a Ѳ

y

b

X

+q2 ‒q1

A. sin2

3

2

2

a

q

b

q B. cos

2

3

2

2

a

q

b

q

C. sin2

3

2

2

a

q

b

q D. cos

2

3

2

2

a

q

b

q

Solution: Answer is C

‒q3

a

Ѳ

2

31

2a

qkqF

b

2

211

b

qkqF

+q2

‒q1

cos2F

sin2F

X component of the force

sin21 FFFx

sin2

31

2

21

a

qkq

b

qkqFx

sin2

3

2

2

a

q

b

qFx

Electrostatics


Problem Set 1 1. What is the Coulomb’s law of electrostatics? Explain the dependency of the medium. Write law in the vector

form.

2. Define the dielectric constant of the medium?

3. Two point charges of C2 and C6 repel each other with a force of 12 N. If each is given an additional

charge of C4 , what will be the new force?

4. Explain: Two identical metallic spheres of exactly equal masses are taken. One is given a positive charge q

coulombs and the other an equal negative charge. Are their masses after charging equal?

5. Two point charges each of q C separated by 1m distance experience a force of F. How much force is experienced

by them if they are immersed in water, keeping distance of separation between them same. (Dielectric constant for

water K = 80).

6. Force of attraction between two point charges placed at a distance d is F. What distance apart should they be

kept in the same medium so that force between them is F/3?

7. Does Coulomb’s law of electric force obey the Newton’s law?

8. How does the force between two point charges change if the dielectric constant of the medium in which they are

kept increases?

9. What is quantization of electric charge?

10. Two charged spherical conductors, each of radius R, are distant d ( Rd 2 ). They carry charges + q and – q. Will

the force of attraction between them be exactly 2

0

2

4 d

q

?

11. Two identical metallic spheres, having unequal opposite charges are placed at a distance 0.9 m apart in air. After

bringing them in contact with each other they are again placed at the same distance apart. Now the force of

repulsion between them is 0.025 N. Calculate the finale charge on each of them.

12. In what ways does a charge differs from mass?

13. Two similar and equal charged identical metal spheres A and B repel each other with a force of N5102 . A

third identical uncharged sphere C is touched with A and then placed at the midpoint between A and B. Calculate

the net electric force on C.

14. Three point charges of CC 3,2 and C3 are kept at the vertices A, B, C respectively of an equilateral

triangle of side 20 cm. What should be the sign and the magnitude of the charge to be placed at the midpoint (M) of

side BC so that the charge at A remains in equilibrium?

Electrostatics


15. Two point charges of values q and 2q are kept at a distance d apart from each other in air. A third charge Q is to

be kept along the same line in such a way that net force acting on q and 2q is zero. Calculate the position of charge Q

in terms of q and d.

4. Electric Field

A field of force surrounding a charged particle.

Electric field strength: - The strength of electric field at a point is defined as the force experienced by one coulomb

of positive charge placed at that point. It is a vector quantity and its direction is the direction in which the force acts

on positive charge placed at that point.

0

00 q

FLimitE

q

Where, F is the force on charge 0q due to electric field due to charge Q.

Hence, electric field due to a point charge Q is given by

2

04

1

r

QE

Electric filed due to system of n charges (use vector addition)

......514131211

EEEEE

Electric field due to continuous distribution of charge (use integration)

dEE

r

P

q

dq

^

2

0

.4

1r

r

dqE

Problem 9: An infinite number of charges, each q coulomb, are placed along x – axis at ,......9,3,1 mmmx

Calculate the electric field at the point 0x due to these charges, if (i). All the charges are same sign (ii) consecutive

charges have opposite signs.

Solution: (i). ..........931 EEEE

x = 0

q ∞

q q

x = 1 x = 3 x = 9

.........931 222

kqkqkq

E

9

11

1.......

9

1

3

1

1

1222

kqkqE

(Sum of infinite G.P.; r

aS

1, Where a is first term and r is common ratio)

1

0 8

9

4

NCq

E

Electrostatics


(ii). ..........931 EEEE

x = 0

q ∞

q -q

x = 1 x = 3 x = 9

.........931 222

kqkqkqE

9

11

1.......

9

1

3

1

1

1222

kqkqE

1

0 10

9

4

NCq

E

Problem 10: Point charges having values CC 5,1 and C2 are placed at the corners A, B and C respectively

of an equilateral triangle of side 2 m in free space. Determine the magnitude of intensity at the point D midway

between A and C.

Solution:

A

1 µC

‒5 µC

EA 2 µC D

C

EC

B

EB

The intensity AE at D due to charge at A is given by 2

69

1

10109

AE

mmACAD 12

2

1

2

1

CNEA /109 3 ; Along DC

The intensity CE at D due to charge at C is given by

CNEC /1

102109

2

69

CNEC /1018 3 (Long DA)

The magnitude of the resultant of CE and DE is given by

CNCNEE AC /109/)1091018( 333

The intensity BE at D due to charge at B is given by

CNCNEB /1015/3

105109 3

69

Along DB

In right angled triangle mBDBD

32

30cos 0

If E is the magnitude of the intensity, then CNE /10749.1)1015()109( 42323

Electrostatics


Problem 11: Find the electric field at a perpendicular distance r from the midpoint of the straight wire of length

l and has charge Q .

Solution: Charge density is l

Q

Let’s take a small line element length of dx and has charge dq at a distance x from the centre of the straight wire.

Electric field at the point due to small charge element R is given by

2)(RP

kdqE (Along the RP); where OPr

And 2222 rxOPORRP

22 rx

kdqE

Electric field at point P due to the similar point charge at S

22 rx

kdqE

Vertical component due to both point charge has cancelled and resultant electric field due to the point charges is

given by cos2EdE

S

l

x

dx

P Ө

Ө

R

sinE

sinE

cos2E

dx

x E

E

O

So, net electric field due to straight wire

cos2

2/

0

22

l

rx

kdqE

Where dxl

Qdq and

22cos

rx

x

2/

0

2/322

2/

0

2/322

2/

0

22 )(

2

)(2cos2

lll

rx

dx

l

kQrdx

rxl

kQr

rx

kdqE

22

2/

0

22

2/

0

2224

12

2/

2/1212

rl

l

rl

kQ

rl

l

rl

kQ

rx

x

rl

kQrE

ll

22 4

12

rlr

kQE

2/322 )( rx

dxI

Let tanrx 2secrdx

2222233

2 1sincos

sec

sec

rx

x

rrr

d

r

drI

Electrostatics


Important Point: Electric field due to infinite length wire

2

2

02

2

0220

4

12

4

12

4

12

l

rr

kLimit

l

rrl

kQLimit

rlr

kQLimitE

lll

r

kE

2

(Alternate method: use Gauss’s Law)

5. ELECTRIC FIELD LINES/ ELECTRIC FORCE LINES: Graphical Representation

Electric filed in a region can be graphical represented by drawing certain curves known as lines of electric force or

electric field lines. Electric field provides a means for visualizing the direction and magnitude of electric field.

Electric

field lines

+Q

The following points should be remembered:

1. Lines of force diverge out from a positive charge and converge at a negative charge or Electric field lines

extend away from positive charge (where they originate) and towards negative charge (where they terminate)

2. The tangent to a line of force at any point gives the direction of electric field at that point.

3. There is no physical existence of electric field lines.

4. Lines must lie along the radii.

5. Lines of force never intersect.

6. In a uniform field, the filed lines are straight parallel and uniformly spaced.

7. Electric field line always flow from higher potential to lower potential.

8. Filed lines never exist inside a conductor and start or end normally from the surface of a conductor.

9. Electric field lines also give the indication of equipotential surfaces.

10. The number of lines of force originating from or ending on a charge is proportional to the magnitude of

the charge.

Important figure:

Positive Charge

─

Negative Charge

+

─ +

Electrostatics


Positive Charge Positive Charge

─ +

+

- Q

A.

B. B

Electric field is zero everywhere

inside a metal (conductor).

Field lines do not enter a metal

plus these are perpendicular to

metal surface. BA EE

FORCE ON A CHARGE q IN ELECTRIC FIELD

EqF

EqF

Force on positive charge always in the direction of electric field, and force on negative charge opposite to electric

field direction.

Problem 12: A uniform electric field E is created between two parallel, charged plates as shown in figure (29-W9).

An electron enters the field symmetrically between the plates with of speed u0.The length of each plate isl . Find the

angle of deviation of the path of the electron as it comes out of the field.

v0

+ + + + + + + + +

‒ ‒ ‒ ‒ ‒ ‒ ‒ ‒ ‒

l

E

Ө

Solution: The acceleration of the electron is m

eEa in the upward direction. The horizontal velocity remain u0 as

there is no acceleration in this direction. Thus, the time taken in crossing the field is

0u

lt … (1)

The upward component of the velocity of the electron as it emerges from the field region is

0mu

eElatuy

The horizontal component of the velocity remains

.0uux

Electrostatics


The angle ɵ made by the resultant velocity with the original direction is given by

2

0

tanmu

eEl

u

u

x

y

Thus, the electron deviates by an angle

2

0

1tanmu

eEl

6. Electric potential energy:

The electrostatic potential energy of a system of point charges is the work required to assemble this system

of charges by bringing them in form an infinite distance.

Electrostatic Potential energy of the system of two particles:

O

P

r1

R

r2

Q dr

r S

Assumptions: First, we assume that the test charge q is so small that it does not disturb the original configuration, namely the charge Q at the origin (or else, we keep Q fixed at the origin by some unspecified force). Second, in bringing the

charge q from R to P, we apply an external force extF just enough to counter the repulsive electric force EF (i.e,

Eext FF ). This means there is no net force on or acceleration of the charge q when it is brought from R to P, i.e.,

it is brought with infinitesimally slow constant speed. Work done by the external force is the negative of the work done by the electric force, and gets fully stored in the form of potential energy of the charge q. Thus, work done by external forces in moving a charge q from R to P is

P

R

E

P

R

extRP drFdrFW

This work done is against electrostatic repulsive force and gets stored as potential energy.

21

22

11111

2

1

2rr

kQqr

kQqdrr

kQqdrr

kQqW

r

r

r

r

P

R

RP

If point R at the infinite then 2r

1r

kQqU

Electrostatic Potential:

Electric potential is a property of an electric field, regardless of whether a charged object has been placed in

that field; it is measured in joules per coulomb, or volts.

The Electric potential at a point in an electric field is the energy required (Work to be done by an external

force) per unit charge to bring a small positive charge from infinity to that point without any acceleration to it.”

Electric potential due to a point charge

r

kQ

q

WVP

P

r

Q

Note: i. Most important is potential difference.

Electrostatics


ii. The potential at a point due to a positive source charge is positive and that due to negative source charge

is negative.

Relationship between E and V:

1. If electric field is given

f

i

dlEV .

P

dlEV .

Where

kEjEiEE zyx

kdzjdyidxdl

2. If potential is given then

k

dz

dVj

dy

dVi

dx

dVE

3. Potential due to a group of charges

.............321 VVVV

4. Potential due to continuous charge distribution

dVV

Electric potential energy: - “Potential Energy of a system of point charges is defined as the amount of work

done to assemble this system by bringing them in from an infinite distance.”

Potential Energy of two point charges

r

qqU 21

04

1

Change in potential energy of a point charge in moving it from one point A to another B in fixed uniform field:

WVVqU AB )(

Equipotential Surface:

Electric filed lines

Equipotential Surface

- -----------

Equipotential Surface

Properties of Equipotential Surface:

i. There is no electric field in any direction lying on or along the equipotential surface. (Because there is no

potential difference)

ii. Electric field and hence lines of force are always perpendicular to the Equipotential surface.

iii. No work is required to be done in moving a charge from one point to another on an equipotential

surface.

Electrostatics


Problem 13: Calculate (i) the potential at a point due a charge of C7104 located at 0.09m away (ii) work done in

bringing a charge of C9102 from infinity to the point.

Solution: (i) The potential due to the charge q1 at a point is r

qV 1

04

1

47

9 10409.0

104109

V Volt

(ii) Work done in bringing a charge q2 from infinity to the point is W = q2 V = 2 × 10−9 × 4 × 104 W = 8 × 10−5 J Problem 14: Calculate the electric potential at a point P, located at the centre of the square of point charges shown in the figure.

nCq 121

d =1.3m

nCq 174 nCq 313

nCq 242

P

Solution: Potential at a point P is

r

q

r

q

r

q

r

qV 4321

04

1

The distance md

r 919.02

3.1

2

Total charge = 4321 qqqq

= C910)17312412(

= C91036

∴ 919.0

1036109

99 C

V

= 352.6 V

Problem 15: Three charges – 2 × 10−9C, +3 × 10−9C, – 4 × 10−9C are placed at the vertices of an equilateral triangle ABC of side 20 cm. Calculate the work done in shifting the charges A, B and C to P, Q and R respectively which are the mid points of the sides of the triangle.

nCq 21

nCq 32 nCq 43

A

P

Q

R

C B

Solution: The potential energy of the system of charges,

r

qq

r

qq

r

qqU 133221

04

1

Work done in displacing the charges from A, B and C to P, Q and R respectively

if UUW

Electrostatics


iU and

fU are the initial and final potential energy of the system.

JUi 7105.402.0

)102(104

02.0

)104(103

02.0

103102109

9999999

JUi 710901.0

)102(104

02.0

)104(103

01.0

103102109

9999999

∴ Work done = J7105.4 Problem 16: An alpha particle with kinetic energy 10 Me V is heading towards a stationary tin nucleus of atomic 50. Calculate the distance of closest approach.

Solution: In approaching the nucleus, kinetic energy of alpha particle is converted into electrical potential energy, so

if r is the distance of closest approach:

.).(4

1

4

1

2

1 21

0

21

0

2

EK

qqr

r

qqmv

)(

)50)(2(

4

1

0 KM

eer

)106.11010(

)50106.1)(106.12()109(

96

19199

r

fmmr 4.14104.14 15

Passage for Q 17 to 19: Two fixed charges – 2Q and Q are located at the points with coordinates ( - 3a, 0) and ( + 3a,

0) respectively in the x-y plane.

Given that all points in the x- y plane where the electric potential due to the two charges is zero, lie in the circle.

17. Radius of the given circle

A. 3a B. 4a C. 5a D. 6a

18. The expression of potential )(xV at a general point on the x axis is

A.

xaxa 3

2

3

1 B.

xaxa 3

2

3

1 C.

xaxa 2

3

2

1 D.

xaxa 2

3

2

1

19. If a particle of charge +q starts form rest at the centre of the circle, by a short quantitative argument that the

particle eventually crosses the circle, its speed when it does so

A.

2/1

0 24

1

ma

Qq

B.

2/1

04

1

ma

Qq

C.

2/1

0

2

4

1

ma

Qq

D.

2/1

0 44

1

ma

Qq

Solution: Let ),( yxP be general point on x – y plane. Electric potential at point P would be

QQ VVV 2

220

220 )3(

2

4

1

)3(4

1

yax

Q

yax

QV

……………………………….(1)

Given that potential is zero

Electrostatics


2222222 )4()5()3()3(4 ayaxyxayxa

)0,3( a

Q -2Q

P(x, y)

)0,3( a

This equation of a circle of radius a4 and centre at )0,5( a

)0,(a

V=0

0

P(x, y)

)0,5( a )0,9( a

(b). On x –axis potential will be undefined (or say )at ax 3 and ax 3 , because charge Q and Q2 are

placed at these two points. So, between axa 33 we can find potential by putting 0y in equation (1).

Therefore,

xaxa

QV

3

2

3

1

4 0 for axa 33

0V at ax

V at ax 3

And V at ax 3

For ax 3 , there is again a point where potential will become zero so for ax 3 , we can write:

xaax

QV

3

2

3

1

4 0 for ax 3

For ax 3 , we can write

axax

QV

3

2

3

1

4 0 for ax 3

Potential at centre at ax 5 will be,

a

Q

aa

QV

00 48

2

2

1

4

= positive

Potential on the circle is zero.

Since, potential > potential on the circumference on it, the particle will cross the circle because positive charge

moves from higher potential to lower potential. Speed of the particle, while crossing the circle would be,

ma

qQ

m

Vqv

08

2

Here, V is potential difference between the centre and circumference of the circle.

Electrostatics


Problem Set 2 1. In an electric field an electron is kept freely. If the electron is replaced by a proton, what will be the relationship

between the force experienced by them?

2. Two point charges of C3 each are 100 cm apart. At what point on the line joining the charges will the electric

intensity be zero?

3. What is the signs of 1q and

2q ?

q2

q1 4. Draw lines of force to represent a uniform electric field.

5. Sketch the line of force due to two equal positive point charges placed near each other?

6. Consider three charged rods A, B and C. A and B repel each other while A and C attract each other. What will be

the nature of force between B and C?

7. Two point electric charges of unknown magnitude and sign are placed a distance‘d’ apart. The electric field

intensity is zero at a point, not between the charges but on the line joining them. Write two essential conditions for

this to happen.

8. Two point charges of C19105 and C191020 are separated by a distance 2 m. Find the point on the line

joining them at which electric field intensity is zero.

9. What is an equipotential surface? Write down any tow properties of the equipotential surface.

10. What is the work done in moving a change of 10 nC between two points on an equipotential surface?

11. Draw an equipotential surface in a uniform in a uniform electric filed.

12. Draw an equipotential surface for a point change .0Q

13. A positive change is moved in an electrostatic filed from a point at high potential to a point at low potential. How

does its kinetic energy and potential energy change?

14. Three concentric metal spheres C and BA, have radii 321 R and R,R respectively and have

change 321 Q and Q,Q . What is the potential and intensity of electric field at a point P between the spheres

B and C at a distance r from the centre O of the spheres?

15. The work done in moving a charge of 3 C between two points is 6 .J what is the potential difference between the

points?

16. Explain what is meant by an electric line of force? Give its 4 important properties? Explain with sketches.

Electrostatics


7. Gauss’ law

Electric Flux of an electric field through a surface: The electric flux through a Gaussian surface is proportional to

the net number of electric field lines passing through that surface.

Or

Electric Flux: - The electric flux linked with the surface are the product of the surface area and the component of

electric intensity taken perpendicular to the surface of the field.

dSE .

Direction of surface is perpendicular to the surface pointing outwards.

The flux of the net electric field through a closed surface equals the net charge enclosed by the surface divided

by0 .

0

.

inqdSE

Problem 20: The electric field in a region is given by^^

jqipE

, where p & q are constants. Find the net flux

passing through a circle area of radius of r parallel to (i). x – y plane (ii). y – z plane (iii). z – x plane.

Solution: (i). ^^

jqipE

, area vector parallel to x – y plane is ^

2 krS

Flux 0)(.^

2^^

krjqipdSE

(ii). ^^

jqipE

, area vector parallel to z – y plane is ^

2 irS

Flux 2^

2^^

)(. rpirjqipdSE

(iii). ^^

jqipE

, area vector parallel to z – x plane is ^

2 jrS

Flux 2^

2^^

)(. rqjrjqipdSE

Applications of Gauss’s law:

1. Charged Conductor: An electric conductor has a large number of free electrons and when placed in an electric

field, these electrons redistribute themselves to make the field zero at all the points inside the conductor.

If a charge is injected anywhere in the conductor, it must come over to the surface of the conductor so that

the interior is always charge free. Also, if the conductor has a cavity, the charge must come over to the surface.

Earthing a conductor

All conductor which are not given any external charge, are also very nearly at the same potential.

The potential of the earth is often taken to be zero.

If a conductor is connected to the earth, the potential of the conductor becomes equal to that of the earth, i.e. zero.

If the conductor was at some other potential, charge will flow from it to earth or from earth to it to br ing its

potential to zero.

Electrostatics


C

+

+

+

R

2+

Conductor

¯ ¯ ¯

¯

¯ ¯

¯

¯ ¯

¯

+

+

+

+

+ .+q

.+q -q

+

R

Conductor

¯ ¯ ¯

¯

¯ ¯

¯

¯ ¯

¯

.+q

-q

The potential due to the charge at the centre r

kqV

The potential due to the charge at the inner surface r

kqV

The potential due to the charge at the outer surface r

kqV

Net potential = r

kqV

After earthing, the charge at outer surface flows to earth and potential of the sphere becomes zero.

2. Electric field due to a uniformly charged Sphere: Here is a spherical geometry, where the charges are evenly

distributed throughout the volume. If the total charge in the sphere is Q, and the sphere has a radius R, then the

volume charge density is

3

3C/m

34 R

Q

By symmetry, the E field is everywhere radial from the center of the sphere.

(i). If point is inside the Sphere: Use a spherical Gaussian surface,

which is perpendicular to

E everywhere. The area vector

S is parallel to

E , and the total area is 24 r so when the Gaussian surface radius is

Rr , then

0

.

inqdSE

32

003

44 rrEAdE

rE03

3

003 4

1

3

3

4 R

Qrr

R

QE

E S

R

r

Electrostatics


(ii). If point is outside the Sphere: Use a spherical Gaussian surface,

which is perpendicular to

E everywhere. The area vector

S is parallel

to

E , and the total area is 24 r so when the Gaussian surface radius

is Rr , then

0

.

inqdSE

32

003

44 RrEAdE

2

0

3

3 r

RE

2

04

1

r

QE

E S

R

r

(iii). Electric field on the surface:

Put Rr , in any result

2

04

1

R

QE

Potential due to Solid sphere

Potential at an outside point: r

kqV

Potential on the surface: R

kqV

Potential at the inside point: 3R

kqrE Potential:

2

2

22

3

R

r

R

kqV

3. Field due to an infinite long straight charged wire: Consider a long charge with linear charge density . Calculate the electric field at a point P which is at a distance x from the line charge.

The electric flux ( ) through curved surface = cosEds

0

)2(

lrlEEds

(∵ 00 and the surface area of the curved part is rl2 )

Electrostatics


r

kE

2

Note: The line charge is assumed infinitely long because this alone would ensure the direction of electric field

radially outward at all points around it.

4. Electric field due to a plane Sheet of Charge: Consider a large plane sheet of charge with surface charge density

(Charge per unit area) .

02

E

Does not depend upon the distance from the charge sheet. (Class Notes)

5. Electric field due to a thick charged conducting plate:

0

E

Electric field inside the conductor 0E

6. The electric field outside a spherical shell of charge with radius R and total charge q is directly radially and has

magnitude

Outside point 2r

kqE Rr Potential:

r

kqV

Inside Point E = 0 r < R Potential at inside and surface R

kqV

Exercise: Derive an expression for electric field and potential due to

(i). Point charge (ii). Linear charge distribution (iii). Ring (iv). Disc

(v). Shell (vi). Sphere (vii). Charge sheet (viii). Bulk sheet

Use Gauss’s law or Coulomb’s law for it.

8. Properties of conductor

1. Excess charge on a conductor resides on its outer surface: The interior of a conductor can have no excess charge

in the static situation.

+ + + + +

+

+ +

+

+

+

00

Eq

dsE in

S

Because 0inq

Gaussian surface

+

+

+

+ +

+

+

+

Conductor

E = 0

2. Inside a conductor, electric field is zero.

3. Electrostatic potential is constant throughout the volume of the conductor and has same value (as inside) on its

surface.

4. Electric field at the surface of charged conductor:

Electrostatics


Electric field is given^

0

nE

, where the surface charge density is and ^

n is unit vector normal to the

surface in the outward direction.

It is perpendicular to the surface and it is not constant value through total surface it depends upon surface

charge density.

For a non uniform conductor charge density varies inversely as the radius of curvature R of that part of

the conductor,

R

1

+ + + + +

+

+ +

+

+

+

21 RR 21

21 EE

+

+

+

+ +

+

+

+

2

1

5. Electrostatic shielding: Consider a conductor with cavity, with no charge insi de the cavity. A remarkable result is

that the electric field inside the cavity is zero, whatever is the size and shape of the cavity and whatever is the

charge on the conductor and the external fields in which it might be placed. This is known as electrostatic shielding.

0inq

So, electric field inside

the cavity is zero. +q -

-

- -

-

-

-

Problem 21: A point charge causes an electric flux of CNm /106 23 to pass through a spherical Gaussian surface of 10 cm radius centred on the charge. (i) If the radius of the Gaussian surface is doubled, how much flux will pass through the surface? (ii) What is the value of charge? Solution: (i) If the radius of the Gaussian surface is doubled, the electric flux through the new surface will be the same, as it depends only on the net charge enclosed within and it is independent of the radius.

CNm /106 23

(ii). 0

q

Cq 8123 1031.51085.8106

Electrostatics


Problem Set 3 1. Define electric Flux. What are the SI unit of electric flux? Is electric flux a vector or a scalar?

2. A change q is placed at a corner of a cube of side ‘a’. What is the electric flux associated with the cube?

3. 21 sands Are hollow concentric spheres enclosing charge QandQ 2 respectively as shown in

(i) What is the ratio of the electric flux through 21 SandS ?

(ii) How will the electric flux through the sphere 1S change, if a medium of dielectric constant 5 is introduced

in the space inside 1S in place of air?

4. State Gauss’s theorem. Using it derive an expression for the electric filed intensity due to on infinitely long

straight wire of linear charge density ./ mC

5. State Gauss’s theorem in electrostatics. Derive an expression for the electric filed intensity due to an infinite plane

thin sheet of electric charge. Or

State Gauss’s theorem in electrostatics and express it mathematically. Using it, derive an expression for electric field

intensity at a point near a thin infinite plane sheet of electric charge.

6. An early model for an atom considered it to have a positively charged point nucleus of charge Ze surrounded by

a uniform density of negative charge up to a radius .R The atom as a whole is neutral. For this model, what is the

electric filed at a distance r from the nucleus when (i) ,Rr (ii) andRr (iii) Rr ? Use Gauss’s theorem.

Electrostatics


9. Electric Dipole

“When two equal & opposite electric charge, +q and – q (small in magnitude) are separated by a small distance 2a,

then this system is known as electric dipole”

-q 2 a

q

Electric dipole Moment:

aqp 2

Direction of

p from negative to the positive charge. (Along the axis of dipole).

Electric potential and Electric field intensity due to dipole:

-q 2a q

P

r

θ

Potential 2

cos

r

kpV

Electric field 1cos3 2

3

r

kpE

Important points should be remembered

1. When 00 (P is on the axis of dipole)

2r

kpV and

3

2

r

kpE (direction is parallel to p)

Point P is called an “end point on position”.

2. When 090 (P is on the equator of dipole )

0V and 3r

kpE (direction is antiparallel to p)

Point P is called a “broad side-on position”.

3. Electric dipole in uniform electric field

-q

2a

q P

θ

E

E

O

F

F

Torque on an electric dipole placed in an electric field

)(2

EaqEp

4. Work done in rotating an electric dipole in a uniform electric filed

]cos[cossin fipEdpEW

f

i

If dipole is placed along the direction of the electric field initially, then 0i and f , in such case

Electrostatics


]cos1[ fpEW

Work done by external force, if the dipole is rotated from an angle 1 to

2 ,

)cos(cos 21 pEWWforceelctricforceexternal

5. Potential energy of a dipole placed in a uniform electric field

f

i

dpEUUdU fi

sin)()(

6. Equilibrium of dipole

When an electric dipole is placed in a uniform electric field net force on it is zero for any position of the dipole in the

electric field. But torque acting on it is zero only at 00 and 0180 . Thus, we can say that at these two positions of

the dipole, net force or torque on it is zero or the dipole is in equilibrium. Of this 00 is the stable equilibrium

position of the dipole because potential energy in this position is minimum pEU and when displaced from this

position a torque starts acting on it which is restoring in nature and which has a tendency to bring the dipole back in

its equilibrium position. On, the other hand, at 0180 , the potential energy of the dipole is maximum pEU

and when it displaced from this position, the torque has tendency to rotate it in other direction. This torque is not

restoring in nature. So, this equilibrium is known as unstable equilibrium position.

Problem 22: Find the electric field due to an electric dipole (i). at a point on its axial line (point on the axis) (ii). For points on the equatorial plane. Solution: (i). Let the point P be at distance r from the centre of the dipole on the side of the charge q, as shown in Fig.

Electric field due to – q charge

^

2

0 )(4p

ar

qE q

…………………………………(1)

Similarly, electric field due to + q charge

^

2

0 )(4p

ar

qEq

…………………………………(1)

The net electric filed at the point P

qqp EEE

^

2

0

^

2

0 )(4)(4p

ar

qp

ar

qE p

^

222

0 )(

4

4

1p

ar

qaE p

For ar

^

3

0

4

4p

r

qaqE p

Direction of this electric field along the direction of dipole moment.

Electrostatics


(ii). Electric field on the perpendicular bisector

The magnitudes of the electric fields due to the two charges +q and –q are given by

)(4 22

0 ar

qE q

)(4 22

0 ar

qE q

and are equal.

The directions of qE and qE are as shown in Fig. Clearly, the components normal to the dipole axis cancel away.

The components along the dipole axis add up. The total electric field is opposite to^

p .

We have ^

cos)( pEEE qq

^

2/322

0 )(4

2p

ar

qaE

At large distance ar , this reduces to

^

3

04

2p

r

qaE

)2( aqp

At a point on the dipole axis

3

04

2

r

pE

At a point on the equatorial plane

3

04

2

r

pE

Notice the important point that the dipole field at large distances falls off not as 2

1

r but as

3

1

r.

Electrostatics


10. Capacitor

Capacitor: Storehouse for potential energy”

A combination of two conductors placed close to each other is called capacitor one of the conductor is given a

positive charge and the other is given an equal negative charge.

The charge on a positive plate is called the charge on a capacitor and the potential difference between them is called

the potential difference between them is called the potential of the capacitor.

V

+Q

‒Q

V

Charge on a capacitor does not mean that the total charge given to capacitor. Total charge = 0QQ

“Capacitor stores the electrostatic energy in the form of electric charge.”

Even a single conductor can be used as a capacitor by assuming the other at infinity.

Any conductor can store some amount of charge; this capacity to store charge is called capacitance of conductor.

For example a spherical conductor of radius R, has maximum potential at surface

R

kQV

Where Q is charge given to conductor

Capacitor and capacitance: - The charge Q given to an isolated conductor is proportional to its electrical potential V,

CVQorVQ

or V

QC coulomb/volt or farad.

Where ‘C’ is the constant of proportionality and known as capacity or capacitance of a conductor.

So, capacity of spherical capacitance Rk

R

V

QC 04

Capacitance depends upon, Medium and geometrical parameter.

In case of Earth, R = 6400km, FC 711 , so we have to take a conductor of size of earth’s size to make a

capacitance of F711 .

The unit of capacitance is farad = coulomb/volt.

One microfarad faradF 610 and

One micro-micro farad .10)( 12 faradfaradpicoF

We cannot give infinite charge to any conductor, it has limit. In another word, capacity of storing charge

is called capacitance of conductor. At this amount of charge it has maximum potential.

V

QC , this does not mean that capacitance will depend upon charge given and potential, capacitance

will depend on the geometry of capacitor and medium.

Electrostatics


Method to find capacitance

Step 1: Give a charge Q to the conductor

Step 2: Find the potential on it due to charge Q

Step 3: Then find the capacitance using the formula

VQC /

Parallel plate Capacitor

- σ

d

A

V

qC

A

dqdEdV 0

00

σ

000 22

E

d

When the medium between two plates is an air, then the capacity of this capacitor is given by

d

AC 0

Effect of dielectric

d

AKCKC 0'

- q

V q

-

- KEE /'

-

-

-

+

+

+

+

+

K = ∞ (Conductor)

K = 1 (Vacuum)

If vacuumC be capacity of a capacitor with vacuum or air between its plates and dielectricC

vacuum

dielectric

C

CK

Dielectric constant is also known as specific inductivity capacity of the dielectric.

Effect of dielectrics: by placing a solid dielectric plate between the plates of capacitor we can have

1. It solves the problem of maintaining two large metal sheets at a very small separation without actual

contact.

2. It increases the maximum possible difference which can be applied between the plates of the capacitor

without dielectric breakdown. Many dielectric materials can tolerate stronger electric field without break down than

can air.

3. It increases the capacity of the capacitor.

Electrostatics


Amount of charge induced on the dielectric slab placed in a capacitor

-qi +qi ‒q +q

+

+ +

+

+ +

+

+ +

+

+ +

+

‒

‒

‒ ‒

‒

‒ ‒

‒

‒ ‒

‒

‒

‒

0E

iE

netE

inet EEE 0

iEEK

E 0

0 Where K

EEnet

0

KE

K

EEEi

110

0

0

K

qqi 11

00

q

K

Kqi

1

When area of slab SA is not equal to area of plate than

QK

K

A

Aq s

i

1

In case of conductor K = ∞, then

QA

Aq s

i

Spherical capacitor

The spherical capacitor consists of two concentric spherical conductors of radii a and b respectively (a<b). The space

between the two conductors is filled by a dielectric of dielectric constant K. A charge q is given to the capacitance of

spherical condenser is given by

ab

abKC 04

Cylindrical capacitor

The cylindrical capacitor consists of two coaxial cylindrical conductors of radii a and b respectively. The

space between the two is filled with a dielectric of dielectric constant K. The inner cylinder is given a positive charge

q per unit length and the outer cylinder is earthed. The capacitance capacitor is give by

)/(log

2 0

ab

KlC

e

Where l is the length of cylindrical capacitor?

Electrostatics


Combination of capacitors

Capacitors in series: When a number of capacitors having capacities etcCCC ,,, 321 are connected in series,

then the resultant capacitance C is given by

....1111

321

CCCC

Note: Charge flow through all the capacitors is always constant.

Capacitors in parallel: When a number of capacitors having capacities etcCCC ,,, 321 are connected in

parallel, then the resultant capacitance C is given by

....321 CCCC

Note: Potential at all the capacitors is always equal.

Note:

C1 1K

t1

2K

t2

3K

t3

4K

t4

C2 C3 C4

4403302201104321 /

1

/

1

/

1

/

111111

tAKtAKtAKtAKCCCCC

44

0

33

0

22

0

11

0

//// Kt

A

Kt

A

Kt

A

Kt

AC

When a slab of thickness t is introduced, then the capacity is given by

)/(

0

Kttd

AC

Energy stored in a capacitor

It the capacitor is given a charge q coulomb so that its potential is raised by V volts, then the energy W joules

stored is given by

qVW2

1

CqCV /2

1

2

1 22

Where C is in faraday.

Loss of energy: - When two capacitors are connected together, some potential energy is dissipated as heat. Hence

there is a loss of energy which is given by

22

21

21

21

2

1)(

)(2

1VCVV

CC

CCeq

Electrostatics


This loss is minimum; when21 VV , since there is no flow of electric charge.

Kirchhof’s law:

First law: This is basically the law of conservation of charge. Following points are important regarding the

first law.

(i). In case of a battery, both terminals of the battery supply equal amount of charge.

(ii). In an isolated system (not connected to either of the terminals of a battery or to the earth) net charge

remains constant.

Second law: In a capacitor potential drops when one moves from positive plate to the negative plate by q/C

and in a battery it drops by an amount equal to the emf of the battery.

Problem 23: Find the capacitance of a spherical conductor enclosed by an earthed concentric spherical shell.

Solution: Charge Q is given to inner sphere

Q

‒Q

r2

r1

2

1

Potential at point 1 21

1r

kQ

r

kQV

Potential at point 2 is 02 V (Because of earthing)

Potential difference 21

21r

kQ

r

kQVVV

Hence capacitance is given by

12

2104

rr

rr

V

QC

If 12 rr ; 10

212

210 4

/14 r

rrr

rrC

This corresponds to that of an isolated sphere.

Electrostatics


Problem 24: Three capacitors of capacities 5 FFandF 34, are connected with the first and second in series

and the third in parallel with them. Calculate the capacity of the combination.

Solution: If sC is the combined capacity of series combination of ,45 FFand Then

20

9

4

1

5

11

sC

FCs 20

9

F3

F4 F5 F9/20

F3

Now, consider the parallel combination of two capacitors of capacitances F20

9and F3

Capacity of combination FF 22.5320

9

Problem 25: You are given three capacitors, each of capacitance .9 F In how many ways can they be combined?

What will be the effective capacitance in each case?

Solution: Arrangement 1. Connect all the three capacitors in parallel (Figure 1)

Combined capacity F27999

Figure 1 Figure 2

Arrangement 2: Connect all the three capacitors in series (figure 2)

FCOrC

33

1

9

1

9

1

9

11

Figure 3 Figure 4

Arrangement 3: Cannot the series combination of two capacitors in parallel with the third capacitor (Figure

3)

If C’ is the capacity of the series combination, then

FCOrC

2

9'

9

2

9

1

9

1

'

1

Capacity of combination, FFC 5.1392

9

Electrostatics


Arrangement 4: Cannot a parallel combination of two capacitors in series with a third capacitor (figure 4)

Capacity of parallel combination F1899

If C is the total capacity, then FCOrC

69

1

18

11

Problem 26: In Fig 4.24, find the equivalent capacitance of the combination.

Given: FandCFCFC 00.4,00.5,00.10 321

1C

2C

3C V

Solution: Capacity of series combination, 12C

FF 3

10

510

510

Total capacity, 312 CCC

FFF 33.73

22

3

10

Problem 27: Each of the uncharged capacitors in Fig. has a capacitance of .25 F What charge Shall Flow through

the meter M when the switch S is closed?

V4200

S

M

C C C

Solution: Total capacity = 3C

FF 75253

mCVF 315420075

Problem 28: An isolated air capacitance C is charged to a potential .V Now the charging battery is disconnected and

a dielectric slab of dielectric constant 3 is inserted between its plates, completely filling the space between the

plates, then how do the following change:

(i) Capacitance, (ii) potential difference,

(ii) field between the plates (iv) energy stored by the capacitor.

Solution: (i) When a dielectric of dielectric constant 3K is inserted fully then .3' CC

Electrostatics


(ii) New potential difference 33'

'V

C

Q

C

QV

iii) New field between the plates .,33

'' strengthfieldoriginalEwhere

E

d

V

d

VE

(iv) New energy stored UwhereUCVV

VCU ,3

1

32

1

3)3(

2

1''

2

1'

22

2

is original value of energy stored.

Problem 29: A parallel plate capacitor has plates of area 200 cm2 and separation between the plates 1 mm. Calculate (i) the potential difference between the plates if 1n C charge is given to the capacitor (ii) with the same charge (1n C) if the plate separation is increased to 2 mm, what is the new potential difference and (iii) electric field between the plates. Solution: The capacitance of the capacitor

nFFd

AC 177.010177.0

101

102001085.8 9

3

4120

(i). The potential difference between the plates

VC

QV 65.5

10177.0

1019

9

(ii). If the plate separation is increased from 1 mm to 2 mm, the capacitance is decreased by 2, the potential difference increases by the factor 2 ∴ New potential difference is 5.65 × 2 = 11. 3 V

(iii). Electric field is, 56501085.810200

101124

9

00

A

QE N/C

Problem 30: A parallel plate capacitor with air between the plates has a capacitance of 8 pF. What will be the capacitance, if the distance between the plates be reduced to half and the space between them is filled with a substance of dielectric constant 6.

Solution: pFd

AC 80

0

When the distance is reduced to half and dielectric medium fills the gap, the new capacitance will be

pFpFCd

A

d

AC rrr 9686222

2/0

00

Problem 31: The plates of a parallel plate capacitor have an area of 90 cm2 each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply. How much electrostatic energy is stored by the capacitor? Solution: Capacitance of a parallel plate capacitor

Fd

AC 11

3

4120 10186.3

105.2

10901085.8

Energy of the capacitor JCV 62112 1055.240010186.32

1

2

1

Electrostatics


Problem 32: A parallel-plate capacitor is connected to a battery. The space between the two plates is empty. If

the separation between the capacitor plates is tripled while the capacitor remains connected to the battery, what is the ratio of the final stored energy to the initial stored energy?

Example 33: A dielectric is inserted between the plates of a parallel-plate capacitor, completely filling the

region between the plates. Air initially filled the region between the two plates. The capacitor was connected to a

battery during the entire process. True or false: (a) The capacitance value of the capacitor increases as the dielectric is inserted between the plates.

(b) The charge on the capacitor plates decreases as the dielectric is inserted between the plates. (c) The electric field between the plates does not change as the dielectric is inserted between the plates.

(d) The energy storage of the capacitor decreases as the dielectric is inserted between the plates.

Electrostatics


Example 34: (a) Two identical capacitors are connected in parallel. This combination is then connected across

the terminals of a battery. How does the total energy stored in the parallel combination of th e two capacitors compare to the total energy stored if just one of the capacitors were connected across the terminals of the

same battery? (b) Two identical capacitors that have been discharged are connected in series. This combination

is then connected across the terminals of a battery. How does the total energy stored in the series combination of the two capacitors compare to the total energy stored if just one of the capacitors were connected across

the terminals of the same battery?

Electrostatics


Example 35: (a) The potential difference between the plates of a 3.00-μF capacitor is 100 V. How much energy

is stored in the capacitor? (b) How much additional energy is required to increase the potential difference between the plates from 100 V to 200 V?

Example 36: An air-gap parallel-plate capacitor that has a plate area of 2.00 m2

and a separation of 1.00 mm is charged to 100 V. (a) What is the electric field between the plates? (b) What is the electric energy density

between the plates? (c) Find the total energy by multiplying your answer from Part (b) by the volume between

the plates. (d) Determine the capacitance of this arrangement. (e) Calculate the total energy from 2

2

1CVU ,

and compare your answer with your result from Part (c).

Electrostatics


Example 37: A spherical capacitor consists of a thin spherical shell that has a radius R

1 and a thin, concentric

spherical shell that has a radius R2, where R

2 > R

1. (a) Show that the capacitance is given by

12

1204

RR

RRC

.

(b) Show that when the radii of the shells are nearly equal, the capacitance approximately is given by the

expression for the capacitance of a parallel-plate capacitor, C = ∈0A/d , where A is the area of the sphere

and12 RRd .

Electrostatics


Example 37: A 100-pF capacitor and a 400-pF capacitor are both charged to 2.00 kV. They are then disconnected from the voltage source and are connected together, positive plate to negative plate and negative

plate to positive plate. (a) Find the resulting potential difference across each capacitor. (b) Find the energy dissipated when the

connections are made.

Electrostatics


Example 38: A parallel combination of two identical 2.00-μF parallel-plate capacitors (no dielectric is in the

space between the plates) is connected to a 100-V battery. The battery is then removed and the separation between the plates of one of the capacitors is doubled. Find the charge on the positively charged plate of each of

the capacitors.

Electrostatics


11. Van de Graaff Generator The working of Van de Graaff generator is based on the principle of electrostatic induction and action of points. Suppose we have a large spherical conducting shell of radius R, on which we place a charge Q. This charge spreads itself uniformly all over the sphere. The field outside the sphere is just that of a point charge Q at the centre; while the field inside the sphere vanishes. So the potential outside is that of a point charge; and inside it is constant, namely the value at the radius R. We thus have: Potential inside conducting spherical shell of radius R carrying charge Q = constant

R

QV

04

1

………………………………….(1)

Now, as shown in Figure, let us suppose that in some way we introduce a small sphere of radius r, carrying some charge q, into the large one, and place it at the centre. The potential due to this new charge clearly has the following values at the radii indicated:

Electrostatics


Potential due to small sphere of radius r carrying charge q

r

Q

04

1

(at the surface of the small sphere)

R

Q

04

1

(at the large shell of radius R)

Taking both charges q and Q into account we have for the total potential V and the potential difference the values

R

q

R

QRV

04

1)(

r

q

R

QrV

04

1)(

And

Rr

qRVrV

11

4)()(

0 ……………………………………(2)

Assume now that q is positive. We see that, independent of the amount of charge Q that may have accumulated on the larger sphere and even if it is positive, the inner sphere is always at a higher potential: the difference V(r )–V(R) is positive. The potential due to Q is constant upto radius R and so cancels out in the difference. This means that if we now connect the smaller and larger sphere by a wire, the charge q on the former will immediately flow onto the matter, even though the charge Q may be quite large. The natural tendency is for positive charge to move from higher to lower potential. Thus, provided we are somehow able to introduce the small charged sphere into the larger one, we can in this way keep piling up larger and larger amount of charge on the latter. The potential (Eq. 1) at the outer sphere would also keep rising, at least until we reach the breakdown field of air. A hollow metallic sphere A is mounted on insulating pillars as shown in the Fig. A pulley B is mounted at the centre of the sphere and another pulley C is mounted near the bottom. A belt made of silk moves over the pulleys. The pulley C is driven continuously by an electric motor. Two comb−shaped conductors D and E having number of needles are mounted near the pulleys. The comb D is maintained at a positive potential of the order of 104 volt by a power supply. The upper comb E is connected to the inner side of the hollow metal sphere.

Electrostatics


Because of the high electric field near the comb D, the air gets ionised due to action of points, the negative charges in air move towards the needles and positive charges are repelled on towards the belt. These positive charges stick to the belt, moves up and reaches near the comb E. As a result of electrostatic induction, the comb E acquires negative charge and the sphere acquires positive charge. The acquired positive charge is distributed on the outer surface of the sphere. The high electric field at the comb E ionises the air. Hence, negative charges are repelled to the belt, neutralises the positive charge on the belt before the belt passes over the pulley. Hence the descending belt will be left uncharged. Thus the machine continuously transfers the positive charge to the sphere. As a result, the potential of the sphere keeps increasing till it attains a limiting value (maximum). After this stage no more charge can be placed on the sphere, it starts leaking to the surrounding due to ionisation of the air. The leakage of charge from the sphere can be reduced by enclosing it in a gas filled steel chamber at a very high pressure. The high voltage produced in this generator can be used to accelerate positive ions (protons, deuterons) for the purpose of nuclear disintegration.

Electrostatics


Problem Set 4

1. For what position of an electric dipole in a uniform electric filed its potential energy is

(i) Minimum and (ii) maximum?

2. On what factors does the capacitance of a parallel plate capacitor depend?

Or

Name three factors on which capacitance of a parallel plate capacitor depends.

3. The distance between the plates of a parallel plate capacitor id d. A metal plate of thickness 2

dis placed between

the plates, without touching either of the plates. What will the new capacity?

4. Find the equivalent capacity when two capacitors of FandF 2010 , respectively are joined in

(i) parallel, (ii) series.

5. Three capacitors 321, CandCC are connected in parallel. A Charge Q is given to the arrangement. How is the

charge shared by the capacitors?

6. Derive an expression for the energy stored in a charged parallel plate capacitor with air as the medi um between

its plates.

7. When two capacitors of capacitance 21 CandC are connected in series, the net capacitance is ;3 F when

connected in parallel its value is .6 F Calculate value of 21 CandC .

8. A parallel plate capacitor with air between the plates has a capacitance of .8pF The separation between the

plates is now reduced by half and the space them is filled with a medium of dielectric constant 5. Calculate the value

of capacitance in the second case.

10. YandX are two parallel plate capacitors having the same area of plates and same separation between the

plates. X has air between the plates and Y contains a dielectric medium of .5r

(i) Calculate the potential difference between the plates of X and Y .

(ii) What is the ratio of electrostatic energy stored in YandX ?

Electrostatics


Assignment 1 1. Two equal unlike charges placed 3 cm apart in air attract each other with a force of 40 N. The magnitude of each

charge in micro coulomb is

A. 0.2 B. 2 C. 20 D. 200

2. Five point charge each of charge + q Coulomb are placed on five vertices of a regular hexagon of side h as shown

in figure.

+ q +q

+q

+q +q

B

F ─ q

A

C O

D E A. The force on – q at O due to charge + q at A and D are balanced

B. The force on – q due to charges at B and E are balanced

C. The resultant force on – q at O is 2

2

04

1

h

q

along OE

D. The resultant force on – q at O is 2

2

04

1

h

q

along OC

3. 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:

A. 2

Q B.

4

Q C.

4

Q D.

2

Q

4. Four charges are arranged at the corners of a square ABCD as shown in figure. The force on the charge at the

centre kept at the centre O is

A

C +q

+q +2q

-2q D

B

O

A. zero B. along the diagonal AC

C. along the diagonal BD D. perpendicular to side AB

5. Two equal metal balls are charge to 10 and – 20 units of charge. Then they are brought in contact with each other

and again separated to original distance. The ratio of the magnitudes of force between the two balls before and

after contact is

A. 8 : 1 B. 1 : 8 C. 1 : 2 D. 2 : 1

6. Which one charge is not possible on the body?

A. C20106.1 B. C19104.2 C. C19102.3 D. C19108.4

7. A charge Q is placed at each of two opposite corners of a square. A charge q is placed at each of the other two

corners. If the net electric force on Q is zero, then Q/q is

A. - 1 B. 1 C. 2

1 D. 22

Electrostatics


Assignment 2

1. 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:

2. On moving a charge of 20 coulombs by 2 cm, 2 J of work is done, then the potential difference between the points

is

A. 0.1 V B. 8 V C. 2 V D. 0.5 V

3. Three charges Q, +q and +q are placed at the right angle (isosceles triangle) as shown. The net electrostatic energy

of the configuration is zero if Q is equal to

+q

Q

+q

A. 21

q B.

22

2

q C. – 2 q D. + q

4. Figure shows some electric field lines corresponding to an electric field. The figure suggests that ( E = electric field

and V = potential)

A. BA EE B. BA EE

C. BA VV D. BA VV A

B

5. A metallic shell has a point charge q kept inside its cavity. Which one of the following diagram correctly represents

the electric lines of forces?

A. B. C. D.

6. Three infinitely long charges sheets are placed as shown in figure. The electric field at point P is:

A. ^

0

2k

B.

^

0

2k

C. ^

0

4k

D.

^

0

4k

z

z = 3a

z = - a

P

2 z = 0

x

Electrostatics


7. A charge is placed at the centre of cube, the flux emitted through its one face is

A. 0

q B.

02

q C.

06

q D.

012

q

8. In the case of a charged metallic sphere, potential (V) changes with respect to distance (s) the centre as:

V

A. S

V

S B.

V

S C.

V

S D.

9. A hollow metal sphere of radius 5 cm is charged such that the potential on its surface is 10 V. 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

10. Three concentric conducting spherical shells have radii r, 2r and 3r and charges 21,qq and 3q respectively.

Innermost and outermost shells are earthed as shown in figure. Select the correct alternative(s)

q

q3 q2

q1

A. 231 qqq B. 4

21

qq C. 3

1

3 q

q D.

3

1

2

3 q

q

11. Two concentric shells of radii R and 2R have given charges q and - 2q as shown in figure. In region Rr :

-2q

2R R

q

r

A. 0E B. 0E

C. 0V D. 0V

12. A large non-conducting sheet M is given a uniform charge density. Two uncharged small metal rods A and B are

placed near the sheet as shown in figure

A B

M

A. M attracts A B. M attracts B C. A attracts B D. B attracts A

Electrostatics


Assignment 3 1. If a spherical conductor comes out from the closed surface of the sphere, then total flux emitted from the surf ace

will be

A. 0

1( the charge enclosed by surface) B. 0

( charge enclosed by surface)

C. 04

1

( charge enclosed by surface) D. 0

2. The work done in placing a charge of 18108 coulomb on a condenser of capacity 100 micro farad is:

A. 321032 B. 321016 C. 26101.3 D. 10104

3. A parallel plate capacitor of plate area A, separation is filled with dielectric as shown in the figure. The dielectric

constants are 1K and

2K . Net capacitance is

d 1K

1K

A. )( 21

0 KKd

A

B.

21

210

KK

KK

d

A C.

21

2102

KK

KK

d

A D.

21

2102

KK

KK

d

A

4. Find the value of C if the equivalent capacitance between points A and B is F1 :

A

C

A. F4.1 B. F8.1

C. F9.1 D. F6.1

F2 F2

F2

F4 F8

F6

F1

B

5. Seven capacitors each of capacitance F2 are connected in a configuration to obtain an effective

capacitance F11

10. Which of the following combination will achieve the desired results?

A. B.

Electrostatics


C. D.

6. Which of the following options is correct for the isolated conductor show in the figure having potential

CBA VVV ,, and DV at the points A, B, C and D respectively?

A. VA = VB > VC >VD B. VA = VB < VC < VD

C. VA < VB < VC < VD D. VA = VB = VC = VD

+ +

+

+ + +

+ +

+ +

+ + + D C

B

A

7. The separation between the plates of parallel plate capacitor is d and area of each plate is A, when a slab of

material of dielectric constant K and thickness t is introduced between the plates, the capacitance becomes:

A.

Ktd

A

11

0 B.

Ktd

A

11

0 C.

Ktd

A

11

0 D.

Ktd

A

11

0

8. A charge ‘Q’ is distributed over two concentric hollow spheres of radii ‘r’ and ‘R’ (>r) such that surface densities

are equal, then potential at the common centre

A. R

KQ B.

r

KQ C.

)(

)(22 rR

rRKQ

D. zero

9. The figure shows two identical parallel plate capacitors connected to a battery with the switch S closed. The

switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric

constant (or relative permittivity) 3. Then, the ratio of the total electrostatic energy stored in both capacitors before

and after the introduction of the dielectric

A B

S A.3/5 B. 4/3 C. 4/5 D. None

C C V

10. To obtain F3 capacity from three capacitors of F2 each, they will be arranged

A. All the three in series B. All the three in parallel

C. Two capacitors in series and third in parallel with the combination of first two

D. Two capacitors in parallel and the third in series with the combination of first two

11. In bringing an electron towards another electron, electrostatic potential energy of the system:

A. decreases B. Increases C. remain unchanged D. become zero

Electrostatics


12. The equivalent capacitance between points A and B of circuit shown in the figure is:

A

F3

F3

A. F9 B. F1

C. F5.4 D. F6

B

F3

13. Two conductors of radius 1R and

2R capacitances 1C and

2C have charges 1q and

2q respectively when they

are joined together by a conducting wire, charge redistributes in these conductors 1'q and

2'q respectively, then

2

1

'

'

q

q

A. 2

1

R

R B.

1

2

R

R C.

2

1

R

R D.

2

2

1

R

R

14. Consider a parallel plate capacitor with plate area A. A charge positive q is given to one plate and negative

charge q to other plate. Then force with which both plates attract each other is (A = area of the plate)

A. 0

2

A

q B.

0

22

A

q C. zero D.

0

2

2 A

q

Electrostatics


Assignment 4

1. A glass rod rubbed with silk acquires a charge of C12108 . The number of electrons it has gained or lost

A. 7105 (gained) B. 7105 (lost) C. 8102 (lost) D. 12108 (lost) 2. The electrostatic force between two point charges kept at a distance d apart, in a medium εr = 6, is 0.3 N. The force between them at the same separation in vacuum is A. 20 N B. 0.5 N C. 1.8 N D. 2 N 3. Electric field intensity is 400 V/m at a distance of 2 m from a point charge. It will be 100 V/ m at a distance? A. 50 cm B. 4 cm C. 4 m D. 1.5 m 4. Two point charges +4q and +q are placed 30 cm apart. At what point on the line joining them the electric field is zero? A. 15 cm from the charge q B. 7.5 cm from the charge q C. 20 cm from the charge 4q D. 5 cm from the charge q 5 A dipole is placed in a uniform electric field with its axis parallel to the field. It experiences A. only a net force B. only a torque C. both a net force and torque D. Neither a net force nor a torque 6. If a point lies at a distance x from the midpoint of the dipole, the electric potential at this point is proportional to

A. 2

1

x B.

3

1

x C.

4

1

x D.

2/3

1

x

7. Four charges +q, +q, −q and –q respectively are placed at the corners A, B, C and D of a square of side a. The electric potential at the centre O of the square is

A. a

q

04

1

B.

a

q2

4

1

0 C.

a

q4

4

1

0 D. zero

8. Electric potential energy (U) of two point charges is

A. 2

0

21

4 r

qq

B.

r

qq

0

21

4 C. cospE D. sinpE

9. The work done in moving 500 μC charge between two points on equipotential surface is A. zero B. finite positive C. finite negative D. infinite 10 Which of the following quantities is scalar? A. dipole moment B. electric force C. electric field D. electric potential 11 The unit of permittivity is

A. 212 mNC B. 22 CNm C. mH / D. 22 mNC 12 The number of electric lines of force originating from a charge of 1 C is A. 1.129 × 1011 B. 1.6 × 10−19 C. 6.25 × 1018 D. 8.85 × 1012 13 The electric field outside the plates of two oppositely charged plane sheets of charge density σ is

A. 02

B.

02

C.

0

D. zero

Electrostatics


14. The capacitance of a parallel plate capacitor increases from 5 μf to 60 μf when a dielectric is filled between the plates. The dielectric constant of the dielectric is A. 65 B. 55 C. 12 D. 10 15. A hollow metal ball carrying an electric charge produces no electric field at points A. outside the sphere B. on its surface C. inside the sphere D. at a distance more than twice 16. State Coulomb’s law in electrostatics and represent it in vector form. 17. What is permittivity and relative permittivity? How are they related? 18. Explain the principle of superposition. 19. Define electric field at a point. Give its unit and obtain an expression for the electric field at a point due to a point charge. 20. Write the properties of lines of forces. 21. What is an electric dipole? Define electric dipole moment? 22. Derive an expression for the torque acting on the electric dipole when placed in a uniform field. 23. What does an electric dipole experience when kept in a uniform electric field and non−uniform electric field? 24. Derive an expression for electric field due to an electric dipole (a) at a point on its axial line (b) at a point along the equatorial line. 25. Define electric potential at a point. Is it a scalar or a vector quantity? Obtain an expression for electric potential due to a point charge. 26. Distinguish between electric potential and potential difference. 27. What is an equipotential surface? 28. What is electrostatic potential energy of a system of two point charges? Deduce an expression for it. 29. Derive an expression for electric potential due to an electric dipole. 30. Define electric flux. Give its unit. 31. State Gauss’s law. Applying this, calculate electric field due to (i) an infinitely long straight charge with uniform charge density (ii) an infinite plane sheet of charge of q. 32. What is a capacitor? Define its capacitance. 33. Explain the principle of capacitor. Deduce an expression for the capacitance of the parallel plate capacitor. 34. What is dielectric? Explain the effect of introducing a dielectric slab between the plates of parallel plate capacitor.

Electrostatics


35. A parallel plate capacitor is connected to a battery. If the dielectric slab of thickness equal to half the plate separation is inserted between the plates what happens to (i) capacitance of the capacitor (ii) electric field between the plates (iii) potential difference between the plates. 36. Deduce an expression for the equivalent capacitance of capacitors connected in series and parallel.

37. Prove that the energy stored in a parallel plate capacitor is C

QCV

22

1 22

38. What is meant by dielectric polarisation? 39. State the principle and explain the construction and working of Van de Graaff generator. 41. The sum of two point charges is 6 μ C. They attract each other with a force of 0.9 N, when kept 40 cm apart in vacuum. Calculate the charges. 42. Two small charged spheres repel each other with a force of 2 × 10−3 N. The charge on one sphere is twice that on the other. When one of the charges is moved 10 cm away from the other, the force is 5 × 10−4 N. Calculate the charges and the initial distance between them. 43. Four charges +q, +2q, +q and –q are placed at the corners of a square. Calculate the electric field at the intersection of the diagonals of the square of side10 cm if q = 5/3 × 10−9C. 44. Two charges 10 × 10−9 C and 20 × 10−9C are placed at a distance of 0.3 m apart. Find the potential and intensity at a point mid−way between them. 45. An electric dipole of charges 2 × 10−10C and –2 × 10−10C separated by a distance 5 mm, is placed at an angle of 60o to a uniform field of 10Vm−1. Find the (i) magnitude and direction of the force acting on each charge. (ii) Torque exerted by the field 46. An electric dipole of charges 2 × 10−6 C, −2 × 10−6 C are separated by a distance 1 cm. Calculate the electric field due to dipole at a point on its. (i) axial line 1 m from its centre (ii) equatorial line 1 m from its centre. 47. Two charges +q and –3q are separated by a distance of 1 m. At what point in between the charges on its axis is the potential zero? 48. Three charges +1μC, +3μC and –5μ C are kept at the vertices of an equilateral triangle of sides 60 cm. Find the electrostatic potential energy of the system of charges. 49. Two positive charges of 12 μC and 8 μC respectively are 10 cm apart. Find the work done in bringing them 4 cm closer, so that, they are 6 cm apart. 50. Find the electric flux through each face of a hollow cube of side 10 cm, if a charge of 8.85 μC is placed at the centre. 51. A spherical conductor of radius 0.12 m has a charge of 1.6 × 10−7C distributed uniformly on its surface. What is the electric field (i) inside the sphere (ii) on the sphere (iii) at a point 0.18 m from the centre of the sphere? 52. The area of each plate of a parallel plate capacitor is 4 × 10−2 sq m. If the thickness of the dielectric medium between the plates is 10−3 m and the relative permittivity of the dielectric is 7. Find the capacitance of the capacitor.

Electrostatics


53. Two capacitors of unknown capacitances are connected in series and parallel. If the net capacitances in the two combinations are 6μF and 25μF respectively, find their capacitances. 54. Two capacitances 0.5 μF and 0.75 μF are connected in parallel and the combination to a 110 V battery. Calculate the charge from the source and charge on each capacitor. 55. Three capacitors are connected in parallel to a 100 V battery as shown in figure. What is the total energy stored in the combination of capacitor?

56. A parallel plate capacitor is maintained at some potential difference. A 3 mm thick slab is introduced between the plates. To maintain the plates at the same potential difference, the distance between the plates is increased by 2.4 mm. Find the dielectric constant of the slab. 57. A dielectric of dielectric constant 3 fills three fourth of the space between the plates of a parallel plate capacitor. What percentage of the energy is stored in the dielectric? 58. Find the charges on the capacitor shown in figure and the potential difference across them.

59. Three capacitors each of capacitance 9 pF are connected in series (i) What is the total capacitance of the combination? (ii) What is the potential difference across each capacitor, if the combination is connected to 120 V supply?

Electrostatics


Answer Sheet Problem Set 1

1. Class Notes 2. Class Notes

3. – 4 N (attractive) 4. No their masses not equal. 5. F/80 6. d3 7. Yes 8. Force decreases

9. neq 10. No 11. 1. 5 µ C 12. Class Notes 13. N5102 14. N393.3 along AM

15. )21(

dfrom charge q or

21

2

dfrom charge 2q

Problem Set 2

1. Class Notes 2. At midpoint 50 cm 3. q1 is negative and q2 is positive 4. Class Notes 5. Class

Notes 6. Attractive 7. Class Notes 8. 2/3 m 9. Class Notes 10. Zero 11. Class Notes

12. Class Notes 13. Potential Energy decreases and kinetic energy increases

14.

3

321

04

1

R

Q

r

Q

r

QVp

2

21

04

1

r

Q

r

QE

15.2 Volt 16. Class Notes

Problem Set 3

1. v – m or N m2C -1, Scalar 2. 02/ q 3. (i). 1/3 (ii). 1/5 th 4, 5, 6: See Class Notes

Assignment 1

1. B 2. A, B, D 3. B 4. C 5. A 6. A, B 7. D

Assignment 2

1. C 2. A 3. B 4. A, D 5. C 6. B 7. C 8. B 9. B 10. A, B, C 11. A, C

12. A, B, C, D

Assignment 3 1. A 2. A 3. C 4. A 5. A 6. B 7. D 8. C 9. A 10. C 11. B 12. A 13. A 14. D Assignment 4 1 (b) 2 (c) 3 (c) 4 (c) 5 (d) 6 (a) 7 (d) 8 (b) 9 (a) 10 (d) 11 (a) 12 (a) 13 (d) 14 (c) 15 (c) 35 (i) increases (ii) remains the same (iii) remains the same 41 q1 = 8 × 10−6C , q2 = –2 × 10−6 C 42 q1 = 33.33 × 10−9C, q2 = 66.66 ×10−9C, x = 0.1 m 43 0.9 × 104 Vm–1 44 V = 1800 V, E = 4000 Vm−1 45 2 × 10−9N, along the field, τ = 0.866 × 10−11 Nm 46 360 N/C, 180 N C–1 47 x = 0.25 m from +q 48 –0.255 J 49 5.70 J 50 1.67 × 105 Nm2C−1 51 zero, 105 N C–1, 4.44 × 104 N C−1 52 2.478 × 10−9F 53 C1 = 15 μF, C2 = 10μF 54 q = 137.5 μC, q1 = 55 μC, q2 = 82.5 μC 55 0.3 J 56 εr = 5 57 50% 58 q1 = 144 × 10−6C, q2 = 96 × 10−6C, q3 = 48 × 10−6C V1 = 72 V, V2 = 48 V 59 3 pF, each one is 40 V

Electrostatics


Introduction of Advance level Problem: A thin half ring of radius cmR 20 is uniformly charged with total charge Q = 0.70 nC . Find the magnitude of

the electric field strength at the curvature centre of this half ring. Reference: IE Irodove

Solution: Let P and Q are the two small elements whose azimuthal angle is with respect to C.

Q

dq C

cosE

E

d

P

Electric Field due to small elements P has two components Vertical and horizontal; Similar Elements Q has two

components, Horizontal components sinE will cancel. Hence, only Resultant electric field on the center will due to

vertical component cosE .

cosEdE If ]90,90[ 00

cos2EdE If ]90,0[ 0

And E = is the electric field due to small element Q has charge dq

2R

KdqE

Where;

dQ

dq

SO total electric field due to this ring

90

0

2cos2

d

Q

R

KE

2

0

2

90

0

2 4

2)01(

2cos

2

R

Q

R

KQd

R

KQE

0

222

R

QE

Put the values and get

mkVE /10.0

. A thin non conducting ring of radius R has linear charge density cos0 , where 0 is a constant, is the

azimuthal angle. Find the magnitude of the electric field strength

(a). at the centre of the ring;

(b). on the axis of the ring as a function of the distance x form its centre. Investigate the obtained function at

Rx

Electrostatics


Advance Assignment 1:

All questions have only one answer

1. Two vertical metallic plates carrying equal and opposite charges are parallel to each other. A small spherical

metallic ball is suspended by a long insulating thread such that it hangs freely in the centre of the two metallic

plates. The ball, which is unchanged, is taken slowly towards the positively charged plate and is made to touch the

plate. The ball will

A. Stick to the positively charged plate

B. Come back to the original position and will remain there

C. Oscillation between the two plates touching each plate in turn

D. Oscillate between the plates without touching them

2. An electron of mass em initially at rest, moves through a certain distance in a uniform electric field in time 1t . A

proton of mass pm , also, initially at rest, takes time 2t to move through an equal distance in this uniform electric

field. Neglecting the effect of gravity, the ratio 12 / tt is nearly equal to:

A. 1 B. 2/1/ ep mm C. 2/1

/ pe mm D. 1836

3. A charge + q is fixed at each of the points ,.....5,3, 000 xxxxxx on the x –axis and charge – q is fixed at

each of the points ,.....6,4,2 000 xxxxxx . Here, 0x is a positive constant. Take the electric potential at a

point due to a charge Q at a distance r from it to be rQ 04/ . Then the potential at the origin due to the above

system of charge is:

A. 0 B. 2ln8 0

q C. D.

004

2ln

x

q

4. Force between two dipoles varies with distance between them

A. Inversely with fourth power of the distance

B. Inversely with third power of the distance

C. Inversely with second power of the distance

D. None

5. Four charges equal to – Q are placed at the four corners of a square and a charge q is at its centre . If the system is

in equilibrium the value of q is:

A. 2214

Q

B. 2214

Q

C. 2212

Q

D. 2212

Q

6. If the electric flux entering and leaving an enclosed surface respectively is 21 and , the electric charge inside

the surface will be:

A. 021 B. 012 C. 021 / D. 012 /

Electrostatics


7. At any point on the right bisector of the line joining two equal and opposite charges

A. The electric field is zero B. The electric potential is zero

C. The electric potential decreases with increasing distance from the centre

D. The electric field is perpendicular to the line joining the charge

8. A sheet of aluminum foil of negligible thickness is introduced between the plates of a capacitor. The capacitance

of the capacitor:

A. Increases B. Decreases C. Remain uncharged D. Becomes infinite

9. A non-conducting ring of radius 0.5 m carries a total charges of C101011.1 distributed non-uniformly on its

circumference producing an electric field E every where in space. The value of the integral

0l

l

Edl (l =0 being

centre of the ring) in volts is

A. + 2 B. – 1 C. – 2 D. zero

10. A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric

materials having constants 21, KK and 3K as shows. 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:

A/2

d

A

d/2

K1 K2

K3

A/2 A.

321 2

1111

KKKK B.

321 2

111

KKKK

C. 3

21

21 21

KKK

KK

K

D.

32

32

31

311

KK

KK

KK

KK

K

Where A = area of plates

11. A uniform electric field pointing in positive x-direction exists in a region. Let A be the origin, B be the point on the

axis at x =1 cm and C be the point on the y-axis at y =1 cm. Then the potentials at the point A, B and C satisfy:

A. BA VV B. BA VV C. CA VV D. CA VV

12. Consider the charge configuration and a spherical Gaussian surface as shown in the figure. When calculating the

flux of the electric field over the spherical surface, the electric field will be due to

-q1

+q1

q2

A. 2q B. only the positive charges C. all the charges D. 1q and 1q

Electrostatics


13. Two fixed, equal, positive charges, each of magnitude 5105 coulomb are located at points A and B separated

by a distance of 6 m. An equal and opposite charge moves towards them along the line COD, the perpendicular

bisector of the line AB. The moving charge, when it reaches the point C at a distance of 4 m from O, has a kinetic

energy of 4 joule. Then, the distance of the farthest point D which the negati ve charge will reach before returning

towards C

-q O

A. 4.48 m B. 2.24 m C. 8.81 D. None

D

B

C

A + q

14. A parallel plate capacitor is connected to a battery. The plates are pulled apart with a uniform speed. If x is the

separation between the plates, the time rate of change of electrostatic energy of a capacitor is proportional to:

A. 2x B. x C. 1x D. 2x

15. In the circuit as shown in the figure, the effective capacitance between A and B is

A. F8 B. F4

C. F2 D. F3

F4

B

A

F4

F4

F2 F2

16. A cube of metal is given a positive charge Q. For the above system, which of the following statements is true?

A. Electric potential at the surface of the cube is zero

B. Electric potential within the cube is zero

C. Electric field is normal to the surface of the cube

D. Electric field varies within the cube

17. A charge (- q) and another charge (+Q) are kept 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 net

work done in moving the charge (- q) is

A. l

Qq

04

1

B.

2

04

1

l

Qq

C. Qql

04

1

D. zero

18. Two conductors of radius 1R and 2R capacitances 1C and 2C have charges 1q and 2q respectively when they

are joined together by a conducting wire, charge redistributes in these conductors 1'q and 2'q respectively, then

1'q =

A. )( 21

21

2 qqRR

R

B. )( 21

21

1 qqRR

R

C. )( 21

21

2 qqCC

C

D. )( 21

21

1 qqCC

C

Electrostatics


19. An isolated conductor initially free from charge is charged by repeated contacts with a plate which after each

contact is replenished to a charge Q. If q is charge on the conductor after first operation, then maximum charge

which can be given to the conductor in this way is

A. Q B. Q + q C. zero D. qQ

Qq

20. A capacitor is given a charge q. The distance between the plates of the capacitors is d. One of the plates is fixed

and other plate is moved from the other till the distance between them 2d. Then work done by the external force is

A. zero B. 0

2

A

dq C.

0

2

2 A

dq D.

0

22

A

dq

21. A point charge q is placed on the top of a cone of semi vertex angle . The electric flux through the base of the

cone

A. zero B. 02

q C.

0

q D.

02

)cos1(

q

22. Incorrect statement is

A. Electric field is in the direction in which the potential steepest.

B. Magnitude of Electric field is given by the change in the magnitude of potential per unit displacement

normal to the equipotential surface at the point.

C. A charge is placed inside a cavity in conductor, then electric field is zero in cavity.

D. Electric field at the surface of a charged conductor is normal to the surface and does not depend upon

surface charge density

Electrostatics


Advance Assignment 2

Section I: Q. No. 1 to 8 has only one correct answer

1. Consider a neutral conducting sphere. A positive point charge is placed outside the sphere. The net charge on the

sphere is then

A. Negative and distributed uniformly over the surface of the sphere

B. Negative and appears only at the point on the sphere closest to the point charge

C. Negative and distributed non-uniformly over the entire surface of the sphere

D. zero

2. A thin spherical conducting shell of radius R has a charge q. Another charge Q is placed at the centre of the shell.

The electrostatic potential at a point P at distance 2

R from the centre of the shell is:

A. 24

)(

0

RQq

B.

R

Q

04

2

C. R

q

R

Q

00 4

2

4

2

D.

R

q

R

Q

00 44

2

3. A charge Q is distributed over two concentric hollow spheres of radii r and R(R>r) such that their surface densities

are equal. Find the potential at the common centre 1

04

k

A. rR

kQ

B.

22

)(

rR

rRkQ

C.

r

kQ D.

R

kQ

4. Two identical thin rings, each of radiuses R, are coaxially placed a distance R apart. If 21 QandQ 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. RQQq 021 42/12)(

C. RQQq 021 4/)(2 D. RQQq 021 4212)/(

5. Consider the situation shown in the figure. The capacitor A has a charge q on it whereas B is uncharged. The

charge appearing on the capacitor B a long time after the switch is closed is:

A. zero B. q/2

C.q D. 2q

-

-

- +

+

+

+

-

- +

q

S

B A

Electrostatics


6. Two identical capacitors have the same capacitance C. One of them is charged to potential 1V and the other to

2V .

The negative ends are also connected, the decrease in energy of the combined system is:

A. 2

2

2

14

1VVC B. 2

2

2

14

1VVC C. 221

4

1VVC D. 221

4

1VVC

7. The capacitance of parallel plate capacitor with plate area A and separation d is C. The space between the plates is

filled with two wedges of dielectric constants 1K and

2K , respectively. The capacitance of resulting capacitor is:

d

A A.

1

2

21

021 log)( K

K

KKd

AKK

B.

2

1

21

021 log)( K

K

KKd

AKK

C. 1

2

21

021 log)( K

K

KKd

AKK

D.

2

1

21

021 log)( K

K

KKd

AKK

K1 K2

8. Positive and negative point charges of equal magnitude are kept at

2,0,0a

and

2,0,0

a, respectively. The

work done by the electric field when another positive charge is moved from )0,0,( a to )0,,0( a is positive

A. positive B. negative

C. zero D. depends upon the path connecting the initial points

Section II: Q. No. 9 to 12 have one or more correct answer

9. A conducting sphere 1S and radius r is attached to an insulating handle. Another conducting sphere 2S of radius R

is mounted on an insulating stand. 2S is initially uncharged. 1S is given a charge Q. brought into contact with 2S

and removed. 1S is recharged such that the charge on it is again Q; and it again brought into contact with 2S and

removed. This procedure repeated n times

A. The electrostatic energy of 2S after n such contacts with 1S is

2

0

14

1

rR

R

r

QR

R

B. The electrostatic energy of 2S after n such contacts with 1S is

2

0

18

1

rR

R

r

QR

R

C. The limiting value of this energy as n is 2

0

2

4 r

RQ

D. The limiting value of this energy as n is 2

0

2

8 r

RQ

Electrostatics


10. A parallel-plate capacitor is connected to a battery. A metal sheet of negligible thickness is placed between the

plates. The sheet remains parallel to the plates of the capacitor.

A. The battery will supply more charge B. The capacitance will increase

C. The potential difference between the plates will increase

D. Equal and opposite charges will appear on the two faces of the metal plate.

11. The electric field in a region is directed outward and is proportional to the distance r from the origin. Taking the

electric potential at the origin to be zero,

A. it is uniform in the region B. it is proportional to r

C. it is proportional to 2r D. it increases as one goes away from the origin

12. An electric dipole is placed at the centre of a sphere. Mark the correct options:

A. The flux of the electric field through the sphere is zero

B. The electric field is zero at every point of the sphere

C. The electric field is not zero anywhere on the sphere

D. The electric field is zero on a circle on the sphere

Section III: Passage for Q. 13 to Q. 18

Passage for from Q. 13 to Q. 15

The nuclear charge (Ze) is non-uniformly. Distributed within a nucleus of radius R. The charge density )(r {charge

per unit volume} is dependent only on the radial distance r from the centre of the nucleus as shown in figure. The

electric field is along the radial direction

)(r

d

a R

r

13. The electric field at r = R is

A. independent of a B. directly proportional to a

` C. directly proportional to 2a D. inversely proportional to a

14. For a = 0, the value of d (maximum value of as shown in the figure) is

A. 34

3

R

Ze

B.

3

3

R

Ze

C.

33

4

R

Ze

D.

33 R

Ze

15. The electric field within the nucleus is generally observed to be linearly dependent on r. This implies

A. a = 0 B. 2

Ra C. a = R D.

3

2Ra

Electrostatics


Passage for Q 16 to 18

Two points A and B are 2 cm apart and a uniform electric field E acts along the straight line AB directed from A to B

with E = 200N/C. A particle of charge C6101 is taken from A to B along AB.

16. The force on the charge

A. Zero B. N6101 C. N6102 D. N4102

17. The potential difference BA VV

A. Zero B. 2 volt C. 3 volt D. 4 volt

18. The work done on the charge by electric field

A. Zero B. J6102 C. J6104 D. J6101

Section IV: Q. No. 13 to 14 are statement – 1 (Assertion) and Statement – 2 (Reason) type questions

and carries 3 marks each for correct answer and –1 mark for each wrong answer. The following six

problems consists two statements. You have to examine these two statements carefully and decide if

the Assertion and Reason are individually true and if so, whether the Reason is the correct

explanation of the Assertion. Select your answer to these problems using the code given below and

mark answer accordingly:

A. Statement -1 is True, Statement – 2 is True; Statement -2 is correct explanation for statement -

1.

B. Statement-1 is True, Statement –2 is True; Statement-2 is not correct explanation for

statement-1

C. Statement -1 is True, Statement – 2 is False

D. Statement -1 is False, Statement – 2 is True

19. STATEMENT-1

For practical purposes, the earth is used as a reference at zero potential in electrical circuits.

STATEMENT-2

The electrical potential of a sphere of radius R with charge Q uniformly distributed on the surface is given by

R

Q

04 .

Electrostatics


20. Statement 1: A charge q is placed at the centre of a metallic shell as shown in figure. Electric field at point P on

the shell due to charge q is zero.

Statement 2: Net electric field in a conductor under electrostatic conditions is zero.

P

q

A. A B. B C. C D. D E. E

Section V: Match the column

21.

Column I Column II

A. Equipotential surface is always p. Vector

B. Electric potential gradient is q. normal to electric field

C. In series combination of capacitors, what is same on each capacitors is r. potential difference

D. In parallel combination of capacitors, what is same for each capacitors is s. charge

22. Two spherical shells are shown in figure, suppose r is the distance of a point from their common centre. Then,

q2

q1

R2

R1 = radius of inner shell

Column I Column II

A. Electric field for 1Rr p. is constant for q2 and vary for q1

B. Electric potential for 1Rr q. is zero for q2 and vary for q1

C. Electric potential for 21 RrR r. is constant

D. Electric field for 21 RrR s. zero

Electrostatics


Advance Assignment 3

Section I: Q. No. 1 to 8 has only one correct answer

1. A long, hollow conducting cylinder is kept coaxially inside another long, hollow conducting cylinder of larger

radius. Both the cylinder is initially electrically neutral.

A. A potential difference appears between the two cylinders when a charge density is given to the inner

cylinder

B. A potential difference appears between the two cylinders when a charge density is given to the outer

cylinder

C. No potential difference between the two cylinders when a uniform line charge is kept along the axis of

the cylinders

D. No potential difference appears between the two cylinders when same charge density is given to both the

cylinders.

2. Six charges, three positive and three negative of equal magnitude are to be placed at the vertices of a regular

hexagon such that the electric field at O is double the electric field when only one positive charge of same

magnitude is placed at R. Which of the following arrangements of charge is possible for, P, Q, R, S, T and U

respectively?

P Q

O R

S T

U

A. ,,,,, B. ,,,,, C. ,,,,, D. ,,,,,

3. A spherical portion has been removed from solid sphere having a charge uniformly distributed in its volume as

shown in the figure. The electric field inside the emptied space is

B

A. zero everywhere B. non-zero and uniform C. non-uniform D. Zero at only its centre

4. The electrostatic potential V at any point (x,y,z) in space is given by 24xV .

A. They y – and z- components of the electrostatic field at any point are zero

B. The x-component at any point is given by

ix8

C. The x-component at a point (1, 0, 2) is

i8

D. They y- and z-components of the field are constant in magnitude.

5. The magnitude of electric field

E in the annular region of a charged cylindrical capacitor:

A. is same through

Electrostatics


B. is higher near the outer cylinder than near the inner cylinder

C. varies as 1/r where r is the distance from the axis

D. varies as 2/1 r where r is the distance from the axis

6. Two isolated metallic spheres of radii R and 2R are charged such that both of these have same charge density .

The spheres are located far away from each other, and connected by a thin conducting wire. Then, new charge

density on the bigger sphere

A. 6

B.

6

7 C.

6

5 D. None

7. Two uniformly charged plane sheets 1S and

2S having charge densities 1 and

2 (21 ) are placed at a

distance d parallel to each other. A charge 0q is moved along a line of length a (a < d) at an angle 045 with the

normal to 1S . Then the work done by an electric field

A. 22

)( 210 aq B.

22

)( 210 aq C.

2

)( 210 aq D. None

8. Two equal point charges are fixed at axandax on the x-axis. Another point charge Q is placed at the

origin. The change in the electrical potential energy of Q, when it is displaced by a small distance x along the x -axis, is

approximately proportional to:

A. x B. 2x C. 3x D. 1/x

Section II: Q. No. 9 to 12 have one or more correct answer

9. A spherical symmetric charge system is centered at origin. Given Electric potential

0

0

0

00

4

4

Rrr

Q

RrR

Q

V

R0

r

V

A. within 02Rr total enclosed net charge is Q B. Electric field is discontinued at 0Rr

C. Change is only present at 0Rr D. Electrostatic energy is zero for 0Rr

Electrostatics


10. A charge + Q is fixed at the origin of the coordinate system while a small electric dipole of dipole moment

P pointing away from the charge along the x axis is set free from a point far away from the origin.

A. The kinetic energy of the dipole when it reaches to a point (d, 0) is 2

04

1

d

pQ

B. The kinetic energy of the dipole when it reaches to a point (d, 0) is 2

0

2

4

1

d

pQ

C. The force on the charge + Q at this moment is 3

0

2

4

1

d

pQ

D. The force on the charge + Q at this moment is 3

04

1

d

pQ

11. Four point charges +8mC, - 1 mC, - 1mC, and + 8 mC are fixed at the points mmm2

3,

2

3,

2

27 and

m2

27respectively on the y axis. A particle of mass kg4106 and charge C1.0 moves along the – x direction.

Its speed at x is 0v . (Assume that space is gravity free)

A. The least value of 0v for which the particle cross the origin is 3 m/sec

B. The least value of 0v for which the particle cross the origin is 4 m/sec

C. The kinetic energy of the particle at the origin is 4103 J

D. The kinetic energy of the particle at the origin is 4103.5 J

12. The separation between the plates of a charged parallel-plate capacitor is increased. Which of the following

quantities will change?

A. charge on the capacitor B. potential difference across the capacitor

C. energy of the capacitor D. energy density between the plates

Section III: Passage for Q. 13 to Q. 18


A, B, C, D are four thin, similar metallic parallel plates, equally separated by distance d, and connected to a cell of

p.d.(V), as shown in figure.

D B

1

V

2 3

A C

Electrostatics


13. The potential of plate D

A. zero B. V C. V/2 D. None

14. If B and C be connected by a wire, then what will be the potential of plate D

A. zero B. V C. V/2 D. None

15. How will the electric field will change in the spacing between the plates B and C

A. zero B. Increase C. decrease D. None


In an isolated system (neither connected to the terminal of a battery nor to any other source of charge e.g. earth)

net charge remains constant. From the two terminals of a battery or from two plates of capacitor equal and

opposite charges enter or leave.

Two capacitors of capacity F6 and F3 are charged to 100 V and 50 Volt separately and connected as shown in

figure. Now all the three switches 21, SS and 3S are closed

50V S1 100V

200V

F6 F3 1 2 3 4

S2 S3

16. Which plate (s) form an isolated system:

A. Plate 1 and plate 4 separately B. Plate 2 and plate 3 separately

C. Plates 1 and 4 jointly D. Plates 2 and 3 jointly

17. Charge on both the capacitor in steady state will be (on F6 first):

A. C400 , C400 B. C700 , C250 C. C800 , C350 D. C300 , C450

18. suppose 1q , 2q and 3q be the magnitude of charges flow through switches 1S , 2S and 3S after they are closed.

Then:

A. 31 qq and 02 q B. 2

231

qqq

C. 231 2qqq D. 321 qqq

Electrostatics


Section IV: Q. No. 13 to 14 are statement – 1 (Assertion) and Statement – 2 (Reason) type questions

and carries 3 marks each for correct answer and –1 mark for each wrong answer. The following six

problems consists two statements. You have to examine these two statements carefully and decide if

the Assertion and Reason are individually true and if so, whether the PReason is the correct

explanation of the Assertion. Select your answer to these problems using the code given below and

mark answer accordingly:

A. Statement -1 is True, Statement – 2 is True; Statement -2 is correct explanation for statement -

1.

B. Statement-1 is True, Statement –2 is True; Statement-2 is not correct explanation for

statement-1

C. Statement -1 is True, Statement – 2 is False

D. Statement -1 is False, Statement – 2 is True

19. Statement 1: The dielectric constant of a conductor is infinite.

Statement 2: Dielectric constant depends on the availability of free electrons in the substance.

A. B. C. D.P

20. Statement 1: A parallel plate capacitor is connected across a battery through a key. A dielectric slab of constant k

is introduced between the plates. The energy stored in the capacitor becomes k times.

Statement 2: The surface density of charge on the plates remains uncharged.

A. B. C. D.

Section V: Match the column 21.

Column I Column II

A. Electric field is a vector whose dimensions are p. 133 ATML

B. Electric flux is a scalar whose dimensions are q. 132 ATML

C. Dimension of electric dipole moment r. 13 AMLT

D. Dimension of Electric potential is s. LTAM 0

22. In the figure shown P is a point on the surface of an imaginary sphere, Match the following

Electrostatics


P q1

q2

Column I Column II

A. Electric field at point P p. due to q1 only

B. Electric flux through a small area at P q. due to q2 only

C. Electric flux through whole system r. due to both charges

Advance Assignment 1 1. D 2. B 3. D 4. A 5. B 6. B 7. B 8. C 9. A 10. D 11. B 12. C 13. A 14. A

15. B 16. C 17. D 18. B, D 19. D 20. C 21. D 22. D

Advance Assignment2 1. D 2. D 3. B 4. B 5. A 6. C 7. B 8. C 9. B, D 10. D 11. C 12. A, C 13. A

14. B 15. C 16. D 17. D 18. C 19. B 20. D 21. A – q, B – p, C – s, D – r 22. A – s, B – r, C – p, D – q

Advance Assignment 3 1. A 2. D 3. B 4. A,B,C 5. C 6. C 7. A 8. D 9. A,B,D 10. A, C

11. A, C 12. B,C 13. A 14. A 15. A 16.D 17.B 18. D 19. B 20. C

21. A – r, B – p, C – s, D – q 22. A – r , B – r, C - p

electrostatics theory

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