(cbse) electrostatics chapter-1 electrostatics

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(CBSE) Electrostatics

MC SQUARE ACADEMY, 1/107 Old Rajinder nagar New Delhi – 110060, Ph: 8130030691, 9899662803

1.1 INTRODUCTIONElectrostatics deals with the study of forces, fields and potentials arising from static charges.

1.2 ELECTRIC CHARGECharge is an inherent property of matter that is responsible for electrical and magnetic effects.The subject of the electrical effect of charges at rest is called electrostatics. When both electricaland magnetic effect are present, the interaction between charges is referred to as electromagnetic.

There exist two types of charges in nature: positive and negative. Like charges repel, andunlike charges attract.

The SI unit of charge is the coulomb (C). It is defined in terms of electric current. One coulombis the charge transferred in one second across the section of wire carrying a current of oneampere.

q = It

1C = (1A) (1s)

The c.g.s. unit of charge is 1 electrostatic unit (e.s.u.) of charge or stat coulomb.

1 coulomb = 3 × 109 stat coulomb.

Another unit of charge is one electromagnetic unit (e.m.u.)

1 e.m.u of charge = 3 × 1010 e.s.u. of charge = 10 coulomb.

Table 1.1

Objects acquiring two kinds of charges on rubbing

Positive Charge Negative Charge

1. Glass rod Silk cloth

2. Fur or wool Ebonite

3. Dry hair Plastics

1.2.1 Quantization of Charge

Electric charges appear only in discrete amounts, it is said to be quantized. The quantumof charge, first directly measured in 1909 by R. A. Millikan, is approximately

e = 1.602 × 10–19 C

A charge q must be an integer multiple of this basic unit. That is q = 0, ±e, ±2e, ±3e etc.Although the mass of the proton is about 1800 times greater than that of the electron,their charges have the same magnitude.

qe = –e ; qp = +e

Note that the electron itself is not the charge: Charge is a property, like mass, of elementaryparticles, such as the electron.

Chapter-1ELECTROSTATICS

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Electrostatics (CBSE)


Example: 1.1Estimate the number of electrons in 100 g of water. How much is the total negative charge on these

electrons ?Solution :

A molecule of water (H2O) is made up of two hydrogen atoms, each containing one electron andone oxygen atom containing 8 electrons. Thus each molecule of water has 10 electrons.

Molecular weight of water = 2 + 16 = 18As number of molecules in 18 g of water = 6.023 × 1023

Number of molecules in 100 g of water = 236.023 10 100

18

As each molecule has 10 electrons.

Total number of electrons in 100 g of water = 23

256.023 10 10010 3.35 1018

As charge on each electron = –1.6 × 10–19 C Total negative charge on these electronsq = 1.6 × 10–19 × 3.35 × 1025 Cq = 5.36 × 106 C

1.2.2 Conservation of Charge

For an isolated system, the total charge remains constant, charge is neither created nordestroyed, it is transferred from one body to the other.

By 'isolated' we mean that there are no paths, such as wires or damp air, by which chargescan leave or enter the system. To apply the law of conservation of charges, we add upthe number of elementary charges before the interaction and then again after it. Forexample,

In a chemical reaction

Na+ + Cl– NaCl

(+e) + (–e) = (0)

In a radioactive decay

n p + e– + (antineutrino)

(0) = (+e) + (–e) + (0)

1.2.3 Invariance of ChargeThe numerical value of an elementary charge is independent of velocity. It is providedby the fact that an atom is neutral. The difference in masses of an electron and a protonsuggests that electrons move much faster in an atom than protons. If the charges weredependent on velocity, the neutrality of atoms would be violated.

1.3 CONDUCTORS & INSULATORSSome substances which allow electricity to pass through them easily are called conductors. Theyhave electric charges (electrons) that are comparatively free to move inside the material. Metals,human and animal bodies and earth are conductors. Most of the non-metals like glass, procelein,plastic, nylon, wood offer high resistance to the passage of electricity through them. They arecalled insulators.

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When some charge is transferred to a conductor, it readily gets distributed over the entire surfaceof the conductor. In contrast, if some charge is put on an insulator, it stays at the same place. Theprocess of sharing the charges with the earth is called grounding or earthing.

There is a third category called semiconductors, which offer resistance to the movement of chargeswhich is intermediate between the conductors and insulators.

The human body is a conductor of electricity.

1.4 COULOMB'S LAWThe force of interaction of two stationary point charges in vacuum is directly proportional to theproduct of these charges and inversely proportional to the square of their separation

1 22

kq qFr

....(1)

where k is a constant which depends on the system of units. Its value in SI unit is

k = 9 × 109 Nm2C–2

The constant is often written in the form

k = 0

14

where 0 is called the absolute permitivity of free space or vaccum which is numericallyequal to

0 = 8.85 × 10–12 C2/Nm2

Dimensions of 0 = [M–1L–3T4A2]

The coulomb force acts along the straight line joining the

points of location of the charges. This force is central and spherically symmetric.

The vector form of Coulomb's law isr

q1 q2 12r̂

Fig. 1.1. Two point charges separated by a distance.

12F

= 1 22

ˆkq qr 12r .... (2)

where 1̂2r is the unit vector, directed from q2 to q1 and 12F

is the force on the charge q1 due to

q2. Similarly force on charge q2 due to q1 is 21F

.

21F

= 1 22

ˆkq qr 12-r

1.4.1 Limitations of Coulomb's law

(a) Charges are assumed to be at rest

(b) Coulomb’s law is applicable for point charges.

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1.5 DIELECTRIC CONSTANT OR RELATIVE ELECTRICAL PERMITIVITYWhen the changes are situated in a medium other than free space (vacuum or air), the forcebetwen them is given by

1 22

14m

q qFr

... (1)

where is called absolute electrical permitivity of the intervening medium.

The force between the same two charges held the same distance apart in vacuum is

1 20 2

0

14

q qFr

... (2)

Dividing (2) by (1), we get

0

0

orrm

F KF

where r is called the relative permitivity or dielectric constant of the medium.

Example: 1.2Suppose the spheres in Example 10 have identical sizes. A third sphere of same size but unchanged

is brought in contact with the first, then brought in contact with the second and finally removed from both.What is the new force of repulsion between A and B ?Solution :

Here, Charge on A = 6.5 × 10–7 C;Charge on B = 6.5 × 10–7 C

Their sizes are equal. When third sphere C of same size is brought in contact with A, their chargesare equally shared.

Charge left on A, 7

71

6.5 10 3.25 102

q C

The sphere C carrying 3.25 × 10–7 C of charge is brought in contact with B carrying 6.5 × 10–7

coulomb charge. As their sizes are equal, therefore, charge on each of the spheres B and C becomes7

72

(6.5 3.25) 10 4.875 102

q C C

As 1 2

20

1 ; 0.54

q qF r mr

9 7 7

32

9 10 3.25 10 4.875 10 5.7 10 .(0.5)

F N

1.6 FORCES BETWEEN MULTIPLE CHARGES

The coulomb's law obeys the principle of superposition, which means that the force betweentwo particles is not affected by the presence of other charges. This principle is used to find thenet force exerted on a given particle by other charged particles.

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q1 + +

+

-

q2

q4

q3

12F

13F

14F

Fig. 1.2. Forces acting on q1 to q2, q3 and q4 are shown

The net force 1F

on q1 is simply the vector sum

1F

= 12F

+ 13F

+ 14F

Note that the notation 12F

represents the force on q1 due to q2.

Example: 1.3

Four point positive charges q are located at the corners of a square as shown in the figure. Find thenet force acting on the charge at D.

A

Fig.(4) Four point charges

B C

D q

q q

q

l

l

l

l

Fig. E1.1. Four point charges located at the vertices of a square

Solution :

2 2 2

2 2 2ˆ ˆ ˆ ˆcos45 sin 45o o

DA DB DCkq kq kqF F F F i j i jl l 2l

2 2

2

1 1ˆ ˆ1 12 2 2 22

kq kqF i jl l

2

2

1| | 2 12 2

kqFl

2

2

2 2 12

kqFl

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1.7 ELECTRIC FIELDAn electric charge influences its surroundings by creating an electric field. The second chargedoes not interact directly with the first, rather, it responds to whatever field it encounters.The electric field strength ( E

) at a point is defined as the force per unit charge experienced by

a test charge qt, placed at that point.

E

= tq

F

....(3)

The direction of E is that of the force on a positive charge. In other words, positive charges

experience forces parallel to the field, and negative charges experience forces opposite to thefield Once the field strength is known, the force on any charge q can be found from

F = q E

....(4)From Coulomb's law, the electric field created by a point charges q is given by

E = 2

ˆkqr

r ....(5)

where the unit vector r̂ has it's origin at the source charge q.

r̂ + q E

r

r̂ - q E

r

Fig. 1.3. Electric field due to a positive and negative charge

1.7.1 Electric Field Due to a System of ChargesSince the principle of superposition is valid for Coulomb's law, it is also valid for theelectric field. To calculate the field strength at point due to a system of charges, we firstfind the individual field intensity E1 due to q1, E2 due to q2 and so on.(a) Charge particles :

For N point charges, the resultant field intensity is the vector sum

E

= 1E

+ 2E

+ ............. + NE

= iE

....(6)

+ q1

-

-

P

1E

q3

q2

2E

3E

Fig. 1.4. Electric field at a point is the superposition of individual contribution of each charge.

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(b) Continuous Charge DistributionTo find the electric field due to a continuous distribution of charge, we divide thecharge distribution into infinitesimal elements of charge dq which may be consideredto be point charges. The infinitesimal contribution of the total field produced by suchan element is

d E

= 2ˆk dq

rr ....(7)

where r̂ has its origin at the charge element. The total field is the sum (integral) ofall such contributions over the charge distribution.

E = k 2ˆdq

r r d E

r

P

r̂

dq

Fig. 1.5. Electric field due to continuous charge distribution.

Example 1.4The figure shows a system of two equal (in magnitude) point charges. Charge +q at (0, 1) and –q is

at (1, 0). Find the resultant field intensity at a point P with coordinates (4, 4).

O

P(4, 4) y (m)

q

x (m)

q

(1, 0)

(0, 1)

Fig. E1.2Solution :

1

ˆ ˆ4 3

25 5

i jkqE

2

ˆ ˆ3 4

25 5

i jkqE

1 2ˆ ˆ

25 5kqE E E i j

O

P(4, 4) y (m)

q

x (m)

q

(1, 0)

(0, 1)

E1

E2

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Example: 1.5An electron falls through a distance of 4 cm in a uniform electric field of value 5 × 104 N/C. When

the direction of electric field is reversed, a proton falls through the same distance. Calculate (i) acceleration,(ii) time of fall in each case. Explain why the effect of gravity is negligible in such cases.Solution :

For electron,y1 = 4 cm = 4 × 10–2 mE1 = 5 × 104 N/Cq0 = (–)1.6 × 10–19 coulomb,m1 = 9 × 10–31 kg

Acceleration, 19 4

0 111 31

1 1

1.6 10 5 109 10

q EFam m

= 8.89 × 1015 m/s2

From 21 1 1 1 1

12

y u t a t

21 1 1

102

y a t

or2

11 15

1

2 4 1028.89 10

yta

t1 = 2.99 × 10–9 sec.(b) For proton

Charge q0 = +1.6 × 10–19 C m2 = 1.67 × 10–27 kg

Acceleration, 0 22

22 2

q EFam m

19 4

2 27

1.6 10 5 101.67 10

a

= 4.8 × 1012 m/s2

Similarly, 2

22 12

2

2 2 4 104.48 10

yta

= 1.29 × 10–7 s (y2 = y1)

We observe that accelerationof electron 1015 m/s2 and acceleration of proton 1012 m/s2. Thevalue of g is only 9.8 m/s2, which is negligible.

1.7.2 Electric Field Intensity at any Point on the Axis of a Uniformly Charged RingConsider a circular loop of wire of negligible thickness, radius a and centre O heldperpendicular to the plane of the paper. Let the loop carry a total charge q distributeduniformly over its circumference. We have to determine electric field intensity at any pointP on the axis of the loop, where OP = r, figure below.

Fig. 1.6.

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Consider a small element AB of the loop. Let length of element AB = dl and C be thecentre of the element.Charge on the element AB is

2qdq dl

a

... (1)

Electric field intensity at P due to the charge element AB is 20

1| |4

dqdECP

along PC´

at with the axis.

2 20

1| |4 ( )

dqdEr a

... (2)

dE

can be resolved into two rectangular components.dE cos along PX, the axis of the loop, and dE sin along PY, perpendicular to the axis.For a pair of diametrically opposite elements of the loop, components of electric fieldintensity perpendicular to the axis will cancel, whereas the components along the axis ofthe loop will add. As the loop can be considered to be made up of a large number ofpairs of diametrically opposite elements, therefore,

sin 0dE Hence the resultant electric field intensity E

at P is

| | cosE dE

In OPC, 2 2 1/ 2cos( )

OP rCP r a

2 2 2 2 1/ 20

1| |4 ( ) ( )

dq rEr a r a

Using (1), we get

2 2 3/ 20

1| |4 2 ( )

q dl rEa r a

= 2 2 3/ 204 2 ( )

qr dla r a

= 2 2 3/ 20

(2 )4 2 ( )

qr aa r a

2 2 3/ 20

| |4 ( )

qrEr a

... (3)

The direction of E is along PX, the axis of the loop.

Special Cases(i) When P lies at the centre of the loop.

r = 0, therefore from (3), 0E

(ii) When r >> a (i.e. P lies far off from the loop), neglecting a2 in comparison to r2 in(3), we get

3 20 0

| | ,4 4

qr qEr r

along PX

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This is expression for | |E

at a distance r from a point charge q. Hence a circular loopof charge behaves as a point charge when the observation point (P) is at very largedistance from the loop, compared to the radius of the loop.

1.8 ELECTRIC DIPOLEElectric dipole is a small body (practically a molecule or an atom) in which centre of its positivecharge does not coincide with centre of its negative charge.

Net charge on a dipole is zero. 2 l

–q +q

Dipole moment p of an electric dipole Fig. 1.7.

= One of the charges × distance between the charges p

= q 2l p̂ (Distance between two charges is taken as 2l )

Dipole moment is a vector quantity, where p̂ is a unit vector along dipole axis (from –q to +q).Unit of dipole moment is C m (Coulomb metre).

1.8.1 Electric Field Intensity on Axial Line of Electric Dipole (Side on Position)Consider an electric dipole consisting of two point charges –q and +q separated by a smalldistance 2a. We have to calculate electric intensity E

at a point P on the axial line of thedipole, and at a distance OP = r from the centre O of the dipole, Figure below.

Fig. 1.8

If 1E

is the electric intensity at P due to charge –q, at A, then

1 2 20 0

1 1| |4 4 ( )

q qEAP r a

It is along PASuppose 2E

is the electric intensity at P due to charge +q at B, then

2 2 20 0

1 1| |4 4 ( )

q qEBP r a

It is a along BP produced.As 1E

and 2E

are collinear vectors acting in opposite directions and 2 1| | | |E E

therefore,

the resultant intensity E at P will be difference of two, acting along BP produced.

2 1| | | | | |E E E

= 2 20 0

1 14 ( ) 4 ( )

q qr a r a

2 20

1 14 ( ) ( )

qr a r a

= 2 2

2 2 20

( ) ( )4 ( )

q r a r ar a

2 2 2 2 2 20 0

4 2 24 ( ) 4 ( )

q ar q a rr a r a

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But q × 2a = | |p , the dipole moment

2 2 20

| | 2| | .4 ( )

p rEr a

... (i)

If dipole is short, 2a << r,

4 30 0

| | 2 2 | || |4 4

p r pEr r

... (ii)

The direction of E is along BP produced.

1.8.2 Electric Field Intensity on Equatorial Line of Electrical Dipole (Broad on Position)Consider an electric dipole consisting of two point charges –q and +q separated by a smalldistance AB = 2a with centre at O, Figure below.

Fig. 1.9.We have to find electric intensity E

at a point P on the equatorial line of the dipole, whereOP = r.If 1E

is electric intensity at P due to charge –q at A, then

1 20

1| |4

qEAP

But AP2 = OP2 + OA2 = r2 + a2

1 20

1| |4

qEAP

... (iii)

1E

is represented by PC

.Let PBA =

1E

has two rectangular components :E1 cos along PR || BA and E1 sin along PE.

If 2E

is electric intensity at P, due to charge +q at B, then

2 2 2 20 0

1 1| |4 4 ( )

q qEBP r a

2E

is represented by PD (along BPD). 2E

has two rectangular components,

E2 cos along PR || BA and E2 sin along PF (opposite to PE)From (ii) and (iii), as

1 2| | | |E E

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E1 sin along PE and E2 sin along PF cancel out. Resultant intensity at P is given by

1 1 2| | cos cosE E E

1| | 2 cosE E 1 2| | | |E E

2 20

2| |4 ( )

q OBEr a BP

= 2 2 2 2

0

24 ( )

q ar a r a

= 2 2 3/ 20

24 ( )

q ar a

But q × 2a = | |p , the dipole moment

2 2 3/ 20

| || |4 ( )

pEr a

... (iv)

The direction of E is along ||PR BA

.

In vector form, we can rewrite (iv) as

2 2 3/ 204 ( )

pEr a

Obviously, E is in a direction opposite to the direction of p .

If the dipole is short, 2a << r

30

| || |4

pEr

1.8.3 Electric Field Intensity at any Point due to Short Electric Dipole

In figure below, AB represents a short electric dipole of moment p along .AB

O is the

centre of dipole. We have to calculate electric field intensity E at any point K, where

OK = r, BOK = The dipole moment p can be resolved into two rectangular components :

Fig. 1.10.

(p cos ) along A1B1 and (p sin ) along A2B2 A1B1.Field intensitty at K on the axial line of A1B1

1 30

2 cos| |4pE

r

Let it be represented by KL along OK.

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Field intensity at K on equatorial line of A2B2 is

2 30

sin| |4pE

r

Let it be represented by KM || B2A2 and KL

. Complete the rectangle KLNM. Join KN

.

According to ||gm law, KN

represents resultant intensity ( )E

at K due to the dipole.

As 2 2KN KL KM

2 22 2

1 2 3 30 0

2 cos sin| |4 4p pE E E

r r

=

23

0

3cos 14

pr

Let LKN = In KLN,

30

30

4sintan .4 2 cos

rLN KM pKL KL r p

or1tan tan2

1.8.4 Electic Dipole in a Uniform Electric FieldConsider an electric dipole consisting of two equal and opposite point charges – q and +qseparated by a small distance 2a, having dipole moment | | 2 ,p q a

figure below..

Fig. 1.11.

Let this dipole be held in a uniform external electric field E at an angle with the direction of

E .

Force on charge +q at A = ,qE

along the direction of E .

Force on chare –q at B = ,qE

in a direction opposite to E .

These forces being equal, unlike and parallel form a couple, which rotates the dipole in clock-wise direction, tending to align its axis along the direction of the field.

Draw AC E and ||BC E

perpendicular distance between the forces = arm of couple = AC.As Torque = moment of couple = Force × arm of couple = F × AC = F × AB sin = F × 2a sin = (qE)2a sin = (q × 2a) E sin

= pE sin ( q × 2a = p) . .. (iii)= – pE cos ... (iv)

The vector form is p E

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1.8.5 Potential Energy of a Dipole in a Uniform Electric FieldPotential energy of dipole is the energy possessed by the dipole by virtue of its particularposition in the electric field.Suppose an electric dipole of moment p is oriented at an with the direction of uniformexternal electric field .E

We know, the torque acting on the dipole is = pE sin

Small amount of work done in rotating the dipole through a small angle d against thetorque is

dW = d = p E sin d Total work done in rotating the dipole from orientation 1 to 2 is

2

2

11

sin cosW pE d pE

W = –p E [cos 2 – cos 1] ... (v)P.E. = W = –pE (cos 2 – cos 1) ...(vi)

When the dipole is initially at right angle to E i.e. 1 = 90º, and we have to set it at

angle with E i.e. 2 =

from (vi),W = –pE (cos – cos 90º)W = –pE cos

Potential energy of dipole,U = W = –pE cos

.U p E

Special Case :(a) If = 0º, P.E. = – pE and is minimum(b) If = 90º, P.E. = 0(c) If = 180º, P.E. = + pE and is maximum(d) work done in rotating the dipole from its stable equilibrium position [ = 0º] to unstable

equilibrium position [= 180º] is 2 pE.Note : The self potential energy of the dipole i.e., the energy of system of two charges +q and

–q is 2

04 2q

a

. Thus, the total potential energy of the dipole placed in a uniform electric

field is 2

0

.4 2

qU p Ea

.

1.9 ELECTRIC LINE OF FORCES

A more generalized way of representing an electric field produced by a source is, other than repre-senting it by arrows, in terms of electric lines of force [first put forth by Michael Faraday (1791–1867)]. It is the path along which a unit positive charge would move when kept in an electrostaticfield. In other words, a field line is an imaginary line drawn in such a manner that its direction at anypoint is the same as the direction of the field at that point.

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A line of force is the curve drawn in electric field whose tangent at any point gives the direction ofelectric field intensity at that point.

Fig. 1.12. Direction of electric field

PQ is an electric line of force. The tangent to the line PQ at point A gives the direction of electricfield intensity at A. Similarly, the tangent to the line PQ at B gives the direction of electric fieldintensity at B.

1.9.1 Certain cases of electric lines of forces

(i) The lines of forces due to a single positive point charge are shown below. As positivecharge will repel positive test charge so direction of electric field lines are away frompositive charge.

Fig. 1.13. Lines of force due to positive charge

(ii) The lines of forces due to a single negative point charge are shown below. As nega-tive charge will attract positive test charge so direction of electric field lines are to-wards the negative charge.

Fig. 1.14. lines of force due to negative charge

(iii) Lines of forces due to a pair of equal and opposite charges are shown below.

Fig. 1.15. lines of force due to equal and opposite charges

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(iv) Lines of forces due to two equal positive charges are shown below.

Fig. 1.16. lines of force due to two equal positive charges

(v) Lines of forces due to two unequal positive charges are shown below.

Fig. 1.17. lines of force due to two unequal positive charges

Well it can be seen that when the charges are equal, point M lies at the centre of theline joining the charges. On the other hand when the charges are unequal, the neutralpoint M is closer to the smaller charge.

1.9.2 Properties of Electric Lines of Force

(i) A line of force starts from positive charge and ends at negative charge. If no negativecharge is present, the line goes to infinity. In other words, they are discontinuous.

(ii) Tangent to the line of force at any point gives the direction of the electric intensity atthat point.

(iii) In case of a positively charged body, the electric lines of force are directed away fromthe body. If the body is negatively charged, then the lines of force are directed to-wards the body.

(iv) The electric lines of force are imaginary lines depicting the partial qualitative informa-tion about the field.

(v) Two lines of force always repel each other laterally.

(vi) The number of lines of force crossing per unit area held normally is proportional to theintensity at that point.

(vii) Two lines of force will never intersect each other, because if it happens, there will betwo directions of intensity at that point of intersection which is not feasible.

It can be seen in the figure given below that if two lines intersect at point M. Thenthere are two directions of electric field at a point M given by two tangents MP andMQ, which is not possible.

Fig. 1.18. Electric lines do not intersect

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(viii) The lines of force contract longitudinally on account of the attraction between unlikecharges.

(ix) Electric lines of forces are always normal to the surface, both while starting andending. Thus, there is no component of electric field parallel to the surface of theconductor.

(x) The lines of force exert a lateral pressure on account of repulsion between the likecharges.

1.10 ELECTRIC FLUXThe electric flux through a surface inside an electric field is the total number of electric lines offorce crossing the surface in a direction normal to the surface. It is a property of electric field.Electric flux is a scalar quantity. Electric flux through a surface is denoted by E. But electricflux through a closed surface is denoted by E.

Consider some surface in an electric field E . Let us select a small area element dS

on this

surface. The electric flux of the field over the area element is given by .Ed E dS

or dE = E dS cos or dE = (E cos ) dSor dE = EndS

where En is the component of electric field in the direction of dS

.

(a) (b)Fig. 1.19. (a) and (b) Electric Flux

The electric flux over the whole area is given by .E nS SE dS E dS

The vector for the area is always drawn normal to the surface. For a closed surface, the flux of

the field is given by . .E nS S SE dS E dS E dS

.

Special Cases(i) When > 90º, E is negative(ii) When = 90º, E is zero(iii) When < 90º, E is positiveThe electric flux through close surface depends only upon the charges enclosed by the closedsurface. It does not depend upon the location of the inside charges.

1.11 GAUSS’S LAWIt relates the total flux () of an electric field ( E

) through a closed surface (S) to the net charge

(q) enclosed by that surface. Let the electric field E makes an angle with the positive normal

to the surface S . Then, the quantity

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= ES cos is called the flux of the electric field through the chosen surface. If we draw a vector of magnitudeS along the positive normal, it is called the area vector, S

. Then

= E

.S

Gauss's law states that the total flux of E

linked with a closed surface is (1/0) times thealgebraic sum of the charges enclosed by the closed surface,

E

'S

S

Positive normal: S is outward.Fig. 1.20. Positive normal : S is outward, Negative normal : S is inward

= .s

d

E S = in

o

q

Note that the flux linked with a closed surface is independent of the shape and size of the surface

and position of the charges inside it. While the field E

itself depends on the mutual configuration

of the charges, the flux of E

through an arbitrary closed surface S is determined by the algebraicsum of charges inside S. If the charges insides are displaced, the field E

may change considerably,,

but flux of E remains unchanged.

Note that the electric field E

is the resultant field at the gaussian surface which also includesthe contributions of charges outside the surface.In certain situations Gauss's law can be used to find the electric field without using Coulomb'slaw.

1.11.1 Derivation of Coulomb’s Law from Gauss’s Law

Consider a +q (isolated point charge) at O. Assume a sphere of radius r with O as centre.

Fig. 1.21. Coulomb's law from Gauss's law

From symmetry, it is considered that the electric field intensity E will have the same

magnitude at all points on the surface of the sphere.

Using Gauss's law, 0

. qE ds

E

and ds

and being in the same direction. Also, ˆds n ds

.

P – [19]



0

cos0º qE ds

or 0

qE ds

or 2

0

4 qE r

or 2

04qEr

Now let a charge q0 be placed at the point at which E is calculated. Then force on q0 is

F = q0 E

or 02

04qqF

r

which is Coulomb's law

Thus, Gauss's law is the generalized form of Coulomb's law

1.11.2 Electric Field Due to Line Charge

Let us take a section of an infinite wire having charge density . By symmetry, the electricfield E is radially directed. We have to evaluate an expression for electric field at any pointM at a perpendicular distance r from the wire. Let us choose a Gaussian surface as acylinder of height h, radius r, coaxial with the line. The cylinder is closed at each end normalto the axis. The Gaussian surface consist of: Curved surface A and Ends of the cylinder, Band C.

Fig. 1.22. Electric field due to line charge

The net electric flux . . .A B C

E ds E ds E ds

...(i)

Since at the ends of cylinder, angle between electric field intensity E and n̂ is 90º; there-

fore, there is no contribution to electric flux by them. Hence

. .E ds E dsB C

= 0 ...(ii)

P – [20]



Now, the curved surface of the cylinder is equidistant from the line charge, magnitude of

E is constant everywhere on the curved surface. Electric field intensity E

is normal to thesurface at each point and is in the direction of the outward drawn normal, i.e. E

and theunit vector n̂ normal to curved surface are in the same direction such that = 0º.

Thus, the electric flux through the curved surface of cylinder is

ˆ. (2 )Eds E n ds E ds E hA A A

... (iii)

Using equations (i), (ii) and (iii), we can write

Net electric flux = E(2rh) ...(iv)

Also, charge enclosed in the cylinder = Linear charge density × Length

or q = h ...(v)

By Gauss's law,

0

q

... (vi)

From equations (iv), (v) and (vi)

0

(2 ) hE rh

or electric field intensity 02

Er

Obviously, 1Er

Here, we see E is inversely proportional to the distance r from the line. And the direction ofE is radially outward if the line of charge is positive and inward if it is negative.

1.11.3 Electric Field Intensity due to Charged Shell

Let us study the three cases of evaluating electric field intensity due to charged shell.

(a) At a point outside the shell

Let us take a spherical shell of radius R having uniform charge q distributed on itssurface. The electric lines of force will be directed radially outwards. Let us constructa sphere of radius r as a Gaussian surface. We have to determine E at a point Moutside a sphere such that OM = r (>R).

Fig. 1.23. Electric field due to charged shell at an external point

P – [21]



Applying Gauss's law,

0

ˆ. qE ds E n ds

or we can write,

20 4E r q

or electric field intensity

204

qEr

It is to be noted that the field outside a uniformly charged conducting sphere is thesame as if whole of the charge was concentrated at the centre of the sphere.

(b) At a point on the surface of the shell

Here r = R

Therefore, 204

qER

or 0

,E

where 24qR

is the surface charge density of the shell.

(c) At a point inside the shell

In this case, r(<R) is the distance of the observation point M from the centre O of thesphere. Let us assume a sphere of radius r such that r is less than the radius of thecharged surface.

Fig. 1.24. Electric field due to charged shell at an internal point

As no charge is enclosed by the Gaussian surface, therefore by Gauss's law

0E dS or E = 0

Hence, there is no electric field inside the charged spherical shell.

All the three cases are represented graphically below. It shows the variation of elec-tric field intensity E with distance from the centre of a uniformly charged sphericalshell.

P – [22]



Fig. 1.25.

1.11.4 Electric Field Due to Solid Sphere

Consider a non-conducting solid sphere of radius R with O as the centre has a uniformvolume density of charge . Let we evaluate the electric field intensity at point M at adistance r from its centre O.

(a) At a point outside the solid sphere

Let a point outside the sphere be at r distance from the centre of the sphere.

Let us take a Gaussian surface as a sphere of radius r concentric with the givensphere.

Fig. 1.26. Electric field due to a solid sphere at an external point

By Gauss's law,

0

. qE ds

where q = Total charge on the solid sphere

or 0

cos0º qE ds

or 0

qE ds

or 2

0

4 qE r

or 204

qEr

P – [23]



So, E at any point outside the sphere is such as if whole of the charge is concentrated

at the centre of the sphere. Also, charge 343

q R

(b) At a point on the surface of the solid sphere

In this case, the area of the gaussian surface is equal to the surface area of the sphereof charge, i.e. equal to 4R2.

Applying Gauss's law,

2

0

.4 qE R


qER

(c) At a point inside the solid sphere

Let the observation point from the centre of the sphere be r (<R). A sphere of radius rconcentric with the given sphere be the Gaussian surface.

Fig. 1.27. Electric field due to a solid sphere at an internal point

Let q' be the charge enclosed by the gaussian surface.

Now, 3

3

44 33

qq rR

or 3

3

rq qR

By Gauss's Law

0

. qE ds

Since E and ds

act in the same direction; therefore

or 3

30

cos0º q rE dsR

P – [24]



or 3

30

q rE dsR

or 304

qrER

(for r < R)

Since 34 ,3

q R the above equation becomes

03rE

(for r < R)

So electric field at any point inside the sphere varies directly as the distance of theobservation point from the centre of the sphere.

Fig. 1.28

1.11.5 Electric Field due to Infinite Plane Sheet of Charge

Consider a thin, infinite plane sheet of charge. We have to evaluate electric field intensity atpoint M, distant r from the sheet. Let us take a cylinder having cross-sectional area A andwhose walls are perpendiculars to the sheet of charge.

Fig. 1.29. Electric field due to plane sheet of charge

Assuming this cylinder as a Gaussian surface we know that, by symmetry, the electric fieldon either sides of the sheet should be normal to the plane of sheet, having same magnitudeat all points equidistant from the sheet. Let be the charge per unit area of the sheet. Atthe two cylindrical edges, R and S; and are parallel to each other as shown in the figure.Now, electric flux over these edges

ˆ2 . 2E n ds E ds

P – [25]



The components of electric field E normal to the walls are zero as no lines of force crossthe sidewalls of the cylinder.

Therefore total electric flux over the entire surface of the cylinder = 2 E ds.

Also, the total charge enclosed by the cylinder = ds

Therefore by Gauss' law

0 0

2 . q dsE ds


E

...(i)

It is to be noted that the magnitude of E is independent of the distance from the plate. It isbecause, as we move away from the sheet, more and more charges come into the field ofview and compensate the decrease in the field due to increase in distance.

1.11.6 Electric Field Intensity due to Two Thin Infinite Parallel Sheets of Charge

Let A and B be two thin infinite plane charged sheets held parallel to each other, figurebelow.

Fig. 1.30

Suppose

1 = Uniform surface density of charge on A

2 = Uniform surface density of charge on B

We use equation 02

E

to write the field intensity for each sheet and then apply super--

position principle to calculate the net field intensity in the three regions. As a matter ofconvention, a field pointing from left to right is taken as positive and the one pointing fromright to left is taken as negative.

We assume that

1 > 2 > 0

In region I,

EI = – E1 – E2

1 2

1 20 0 0

1 ( )2 2 2IE

... (ii)

P – [26]



Similarly, in region II,

EII = E1 – E2 = 1 21 2

0 0 0

1 ( )2 2 2

... (iii)

and, in region III,

EIII = E1 + E2 = 1 2

1 20 0 0

1 ( )2 2 2

... (iv)

Special cases : Suppose 1 = , and 2 = – i.e., two thin infinite plane sheets with equaland opposite uniform surface densities of charge are held parallel to each other.

From (ii), EI = 0

From (iii) EIII = 0

From (iv)0 0

2 constant2IIE

Example: 1.6

A point charge +10 C is at a distance of 5 cm directly above the centre of a square of side 10 cmas shown. What is the magnitude of the electric flux through the square ?

Fig. E1.3

Solution :

In figure above, ABCD is a square of side 10 cm. The point of charge + 10 mC is at a distance of 5cm. directly above the centre of ABCD.

We can imagine the square ABCD as one of the six faces of a cube of side 10 cm. The charge +q isat the centre of this cube.

According to Gauss’s theorem, total electric flux through all the six faces of the cube = q/0

Electric flux through the square

6

120

1 1 10 106 6 8.85 10

q

= 1.88 × 105 N m2 C–1

P – [27]



1.12 ELECTRIC POTENTIALElectrostatic potential at a point in an electric field is numerically equal to the amount of work donein bringing unit positive test charge from infinity to that particular point against the electric field.

It is a scalar quantity, its S.I. unit is volt.

1 joule1volt=

1coulomb

Electric potential difference between two points B and A in an electrostatic field as the amount ofwork done in carrying unit positive test charge from A to B (against the electrostatic force of thefield) along any path between the two points.

0

ABB A

WV Vq

1.12.1 Electric Potential due to a Single Charge

Suppose we have to calculate electric potential at any point P due to a single point charge+q at O; where OP = r, figure 1.30.

Fig. 1.31

By definition, electric potential at P is the amount of work done in carrying a unit positivecharge from to P.

At any point A on the line joining OP, where OA = x, the electric intensity is

20

1 ,4

qEx

along OA produced ...(i)

Small amount of work done in moving a unit positive charge from A to B where AB dx

is

. cos180ºdW E dx E dx E dx

... (ii)

Total work done in moving unit + charge from to the point P is

20

14

r r qW E dx dxx

=

2

04

rq x dx

=

0

14

rqx

0 0

1 14 4

q qWr r

... (iii)

This is the potential at P i.e.

04qV W

r

... (iv)

P – [28]



1.12.2 Potential at a Point Due to an Electric Dipole

Let an electric dipole consist of two equal and unlike point charges –q at A and +q at B,separated by a small distance AB = 2a, with centre at O. The dipole moment

| | 2p q a ... (i)

We have to calculate potential at any point P, where OP = r and BOP = q, figure below.

Fig. 1.32

Let AP = r1 and BP = r2

Draw AC PO produced and BD PO. In AOC,

cos OC OCOA a

OC = a cos

Similarly, OD = a cos

Net potential at P du eto the dipole

0 2 0 14 4q qV

r r

0 2 1

1 14

qVr r

Now, 1r AP CP = OP + OC = r + a cos

and r2 = BP DP = OP – OD = r – a cos

2 2 20 0

1 1 1 cos cos4 cos cos 4 cos

q r a r aVr a r a r a

2 2 20

2 cos4 cos

q aVr a

i.e. 2 2 20

cos4 ( cos )

pVr a

... (ii)

P – [29]



Special Cases :

1. When the point P lies on the axial line of the dipole, = 0º

cos = cos 0º = 1

From (ii), 2 204 ( )

pVr a

If a << r, then 204

pVr

Note that due to an electric dipole, potential, 21/V r .

2. When the point P lies on the equitorial line of the dipole, = 90º

cos = cos 90º = 0

i.e. electric potential due to an electric dipole is zero at every point on the equatorialline of the dipole.

Example: 1.7Two charges 3 × 10–8 C and –2 × 10–8 C are located 15 cm apart. At what point on the line joining

the two charges is the electric potential zero ? Take the potential at infinity to be zero.Solution :

Let the potential be zero at a distance x cm from q1 = 3 × 10–8 C r1 = x × 10–2 m;

r2 = (15 – x) × 10–2 m;q2 = –2 × 10–8 C

1 2 1 2

0 1 0 2 0 1 2

14 4 4

q q q qVr r r r

8 89

2 2

3 10 2 100 9 1010 (15 ) 10x x

or8 8

2 2

3 10 2 1010 (15 ) 10x x

2x = 45 – 3xor 5x = 45, x = 9 cm

1.13 EQUIPOTENTIAL SURFACESAs the name suggests, it is a surface in an electric field in which all points are at the same potential.For example, different spherical surfaces around a charged sphere are equipotential surfaces.

According to definition of electric potential, the potential difference between two points P and Q isequal to the work done in moving unit positive test charge from Q to P.

Mathematically, VP – VQ = WQP

Now if the points P and Q lie on an equipotential surface, then VP = VQ. Therefore,

P – [30]



WQP = VP – VQ = 0

Thus, no work is done in carrying the test charge from one point of the equipotential surface to theother.

Since, . cos 0dW E dl E dl

.

Therefore, or cos = 0 or = 90º. It implies E dl

.

In an equipotential surface, the direction of electric field strength and flux density is always at rightangles to the surface.

As electric field intensity is along tangent to the electrostatic lines of force, therefore equipotentialsurfaces are always perpendicular to field lines.

For an isolated point charge, equipotential surfaces are the surfaces of concentric spheres with thecharge at their centre as shown below.

Fig. 1.33. Equipotential surfaces

1.14 ELECTRIC POTENTIAL ENERGYElectric potential energy of a system of point charges as the total amount of work done in bringingthe various charges to their respective positions from infinitely large mutual separations.

1.14.1 Electrical Potential Energy of a System of Two Point Charges

Suppose a point charge q1 is held at a point P1 with position vector 1r in space. Another

point charge q2 is at infinite distance from q1. This is to be brought to the position 2 2( ),P r

where 1 2 12 ,PP r where 1 2 12 ,PP r

figure below..

Fig. 1.34

P – [31]



Now, electric potential at P2 due to charge q1 at P1 is

1

0 12

14

qVr

By definition work done in carrying charge q2 from infinity to P2

W = potential × charge

12

0 124qW q

r

This is stored in the system of two point charges q1 and q2 in the form of electric potentialenergy U. Thus

1 2

0 124q qU W

r

Example: 1.8If one of the two electrons of a hydrogen molecule is removed, we get a hydrogen molecular ion

(H2+). In the ground state of H2

+, the two protons are separated roughly by 1.5 Å and electron is roughly1Å from each proton. Determine the potential energy of the system. Specify your choice of zero of P.E.Solution :

Here q1 = charge on electron (= –1.6 × 10–19 C)q2, q3 = charges on two protons, each = 1.6 × 10–19 Cr12 = distance between q1 and q2 = 1 Å = 10–10 mr23 = distance between q2 and q3 = 1.5 Å = 1.5 × 10–10 mr31 = distance between q3 and q1 = 1 Å = 10–10 m

Taking zero of potential energy at infinity, we have

2 3 3 11 2

0 12 23 31

1. .4

q q q qq qP Er r r

= 19 19

910

( 1.6 10 )( 1.6 10 )9 1010

+

19 2 19 19

10 10

(1.6 10 ) (1.6 10 )( 1.6 10 )1.5 10 1 10

= 9

3810

9 10 1010

×

22(1.6)1.6 1.6 (1.6)

1.5

= –9 × 10–19 × 3.42 = –30.78 × 10–19 J

1.15 RELATION BETWEEN ELECTRIC INTENSITY AND POTENTIALConsider two points A and B in the electrostatic field of a point charge +q placed at O. Let usassume that A and B are so close that the electric field intensity between A and B is uniform and isequal to .E

Fig. 1.35. Relation between E and dV/dr.

P – [32]



In order to move a test charge q0 from A to B, a force 0' 'q E

has to be applied.

Work done = 0 0 0. cos180ºq E dl q Edl q Edl

But, dl = –dr

Work done = –q0E dr

or0

Work done Edrq

Using definition of potential difference, potential difference between A and B, dV = –Edr

ordVEdr

The electric field at a point is equal to the negative gradient of the electric potential.

1.16 CAPACITORSA capacitor essentially constitutes two conducting surfaces separated by an insulating mediumcalled dielectric (which could be air also). Capacitor is used to store electric energy by electrostaticstress in the dielectric (The word condenser is actually a misnomer). The conducting surfaces of acapacitor can be circular, rectangular, spherical or cylindrical in shape.

Fig. 1.36. A capacitor

A capacitor, as shown in the figure, when earthed on plate B holds more charge than in the absenceof it.

1.16.1 Working

When a capacitor is connected across a supply, there is a momentary flow of electronsfrom A to B. As negatively charged electrons are withdrawn from A, it becomes positiveand the electrons collected on B, make it negative. Hence, a potential difference estab-lishes between plates A and B.

1.16.2 Capacitance

The property of a capacitor to store electricity is termed as capacitance. That is, the mea-sure of ability to store charge is the capacitance of conductor. In other words, capacitanceof a capacitor is the amount of charge required to create a unit potential difference be-tween its plates.

P – [33]



Potential difference between two plates is the potential of capacitor. When q coulomb ofcharge is supplied to one of the two plates of a capacitor and if a potential difference of Vvolts is established, then the capacitance will be

Charge

Potential differenceqCV

So, capacitance is the charge required per unit potential difference.

Coulomb =FaradVolt

qCV

1.16.3 Units of Capacitance

So, one farad is the capacitance of a capacitor that requires a charge of one coulomb toestablish a potential difference of one volt. Farad is a very large unit and for general use,smaller units like microfarad (1 F = 10–6 F) nanofarad (1 nF = 10–9 F) micro-micro farador picofarad (1 F = 1 pF = 10–12 F) are employed.

Capacitance depends upon the shape, size of the conductor and the nature of medium inwhich the conductors are placed. Examples of capacitor are mica capacitor, paper capaci-tor, electrolytic capacitor. Varactor diode can also acts as capacitor.

1.16.4 Symbol of Capacitor is Shown Below

Fig. 1.37. Symbol of capacitor

1.16.5 Capacitance of an Isolated Spherical Conductor

Suppose O is the centre and r is radius of an isolated spherical conductor, Figure below. Letcharge +q be given to the sphere. This charge spreads uniformly over the outer surface ofthe sphere whether the sphere is hollow or solid. Therefore, potential at every point on asurface of the sphere is same. As the sphere behaves as if the entire charge were concen-trated at the centre of the sphere, therefore, potential at any point on the surface of thesphere in free space,

Fig. 1.38. Symbol of capacitor

P – [34]



04qV

r

As qCV

04q rC

q

C = 40r

1.16.6 Expression of Capacitance of a Parallel Plate Capacitor

Consider a parallel plate capacitor consisting of two parallel plates of area A square metresseparated by a distance d as shown in the figure.

Fig. 1.39. Parallel plate capacitor

If a charge +q coulomb is supplied to the left plate, then flux passing through the air is = q.

Flux density in the medium q

A A ...(i)

Now electric intensity VEd

...(ii)

And flux density = E ...(iii)

Using equations (i), (ii) and (iii)

q VA d

or, q AV d

or, ACd

If air is the medium between the plates, then capacitance is

0 ACd

If there is uniform dielectric medium of relative permittivity r, present between the plates,then capacitance is

0 ACd

P – [35]



1.17 GROUPING OF CAPACITORSCapacitor can be connected in series combinations and in parallel combinations in various circuits.

1.17.1 Capacitors in Series Combination

When capacitors are connected in the series combination:

Charge on all the capacitors is same.

Potential difference across each is different.

In the figure below, three capacitors are connected in series. We can see that negativeplate of one capacitor is connected to positive plate of next capacitor.

Fig. 1.40. Capacitors in series combination

Let C1, C2 and C3 be the capacitance of three capacitors. Also V1, V2 and V3 be thepotential difference across the three capacitors, V be the applied voltage across combina-tion and C be the combined or equivalent capacitance.

Since, potential difference across each capacitor is different therefore,

V = V1 + V2 + V3

or,1 2 3

1 1 1 1C C C C

1.17.2 Capacitors in Parallel Combination

When capacitors are connected in a parallel combination:

Potential difference across each capacitor is same.

Charge on each is different.

In the figure below, three capacitors are connected in parallel. In this kind of combination,the positive plate of a capacitor is connected with positive plate of other capacitor.

Fig. 1.41. Capacitors in parallel combination

q = q1 + q2 + q3

or, CV = C1V + C2V + C3V

or, C = C1+ C2 + C3

P – [36]



Example: 1.9An electrical technician requires a capacitance of 2 F in a circuit across a potential difference of

1 kV. A large number of 1 F capacitors are available to him, each of which can withstand a potentialdifference of not more than 400 V. Suggest a possible arrangement that requires minimum number ofcapacitors.

Fig. E1.4

Solution :Here, total capacitance

C = 2 FPotential difference, V = 1 kV = 1000 voltCapacity of each capacitor, C1 = 1 FMaximum potential difference across each,

V = 400 voltLet n capacitors of 1 F each be connected in series in a row and m such rows be connected in

parallel as shown in figure.As potential difference across each row = 1000 volt

Potential difference across each capacitor = 1000 400

n

1000 2.5400

n

As n has to be a whole number (not less than 2.5), therefore n = 3.Capacitance of each row of 3 condensers of 1 F each, in series = 1/3.

Total capacitance of m such rows in parallel = 3m

2( ) or 63m F m

Total number of capacitors = n × m = 3 × 6 = 18Hence 1 F capacitors should be connected in six parallel rows, each row containing three capacitors

in series.

1.18 ENERGY STORED IN A CAPACITORWhenever a capacitor is charged, there is an expenditure of energy by the charging supply. Thisenergy gets stored in the electrostatic field set up in the dielectric medium. In the dischargingprocess, the field collapses and the energy is released.

Let at any stage of charging, the potential difference across the plate be V. By definition, thispotential difference is equal to work done in shifting one coulomb of charge from one plate toanother.

P – [37]



If dq is the charge transferred, the work done is

dW = V × dq ...(i)

Since, q = CV

or dq = C dV ...(ii)

Therefore, dW = CV dV

Total work done to set a potential of V units is

2

0 2

V VW C V dV C

or 212

W CV

If C is in farad and V is in volts, then work done will be in joules.

Since the energy stored is the total work done, so energy stored in a capacitor is

212

E CV

Example: 1.10A condenser whose capacitance is 12 F is charged to a potential of 2000 V. Calculate the energy

stored.Solution :

Here, C = 12 F = 12 × 10–6 F,V = 2000 V,E = ?

As 212

E CV

or 6 21 12 10 (2000) 242

E J

1.19 ENERGY DENSITY IN A PARALLEL PLATE CAPACITOREnergy density (u) is defined as the total energy stored per unit volume of the condenser.

i.e., 21

total energy ( ) 2volume ( )

CVUuv Ad

using 0 ACd

and V = Ed,

we get, 2 2

200

1 12 2

A E du Ed Ad

P – [38]



1.20 COMMON POTENTIALWhen two capacitors charged to different potentials are connected by a conducting wire, chargeflows from the one at higher potential to the other at lower potential. This flow continues till theirpotentials become equal. The equal potential of the two capacitors is called common potential.Obviously, no charge is lost in the process of sharing of charges by the two capacitors.Suppose C1, C2 are capacities of two condensers charged to potentials V1 and V2 respectively.The total charge before sharing = C1V1 + C2V2

If V is the common potential on sharing charges, then total charge after sharing= C1V + C2V = (C1 + C2)VAs no charge is lost in the process of sharing, therefore, (C1 + C2)V = C1V1 + C2V2

or 1 1 2 2

1 2

C V C VVC C

1.21 LOSS OF ENERGY ON SHARING CHARGESWhen charges are shared between two bodies, no charge is lost. However, some energy is dissi-pated in the form of heat etc. Hence some energy is lost on sharing charges. To calculate the lossof energy, suppose two capacitors of capacitances C1, C2 charged respectively to potentials V1, V2

are connected together and their charges are shared. The charge flows from one capacitor athigher potential to the other at lower potential till their potentials become equal. The commo poten-tial reached is given by

1 1 2 2

1 2

C V C VVC C

... (i)

Total energy before sharing charges,

2 21 1 1 2 2

1 12 2

E C V C V

Total energy after sharing charges, 22 1 2

1 ( )2

E C C V

Using (i), 2 2

1 1 2 2 1 1 2 22 1 2 2

1 2 1 2

( ) ( )1 ( )2 ( ) 2( )

C V C V C V C VE C CC C C C

Let us calculate : E1 – E2 = 2

2 2 1 1 2 21 1 2 2

1 2

( )1 12 2 2( )

C V C VC V C VC C

= 2 2 2

1 1 1 2 2 2 1 2 1 1 2 2

1 2

( ) ( ) ( )2( )

C V C C C V C C C V C VC C

= 2 2 2 2 2 2 2 2 2 2

1 1 1 2 1 1 2 2 2 2 1 1 2 2 1 2 1 2

1 2

22( )

C V C C V C C V C V C V C V C C V VC C

P – [39]



2 2

1 2 1 2 1 21 2

1 2

( 2 )2( )

C C V V V VE EC C

21 2 1 2

1 21 2

( )2( )

C C V VE EC C

When charges are shared between any two bodies, their potentials become equal, charges acquiredare in the ratio of their capacities. No charge is really lost, but some loss of energy does occur.

1.22 NON POLAR AND POLAR DIELECTRICSNon-polar dielectrics like hydrogen, nitrogen, oxygen, CO2, benzene, methane etc. are made ofnon-polar molecules. In such molecules, the centre of positive charge coincides with the centre ofnegative charge in thee molecule.

Polar dielectrics like water, aocohol, NH3, HCl etc. are made of polar molecules. In such mol-ecules, (under normal conditions with no external field) the centres of positive and negative chargesdo not coincide because of the assymetric shape of the molecules.

1.23 VAN DE GRAFF GENERATORThe Van de Graaff electrostatic generator was designed in the year 1931 by Robert van de Graaff,an MIT physics student who was inspired by large unexplained sparks produced in an industrialprinting press.

1.23.1 Use of Van de Graff Electrostatic Generator

A Van de Graaff device is a mechanical-electrical device, which produces extremely highvoltage of the order of a few million volts at low, safe levels of electric current. We usuallyencounter these devices in our school laboratory, where our physics teachers use them toraise the hair of some lucky student. Yes, we can use a Van de Graaff electrostatic genera-tor to raise your hair, or to jump a large spark to the knuckle of an overly trusting sciencestudent. However, Van de Graaff electrostatic generator devices actually have profes-sional applications. They were originally used as power supplies for the early particle ac-celerators used in research in radioactivity. This was in the days before the invention of thecyclotron and the linear accelerator ring. The early "atom smashers" consisted of a Van deGraaff electrostatic generator connected to a long vacuum tube. Van de Graaff electro-static generator still finds use in particle physics research, and many universities own alarge Van de Graaff electrostatic generator encased in huge pressure chambers filled withinsulating gas. Large Van de Graaff electrostatic generators are also used to power high-energy X-ray machines in order to treat cancer with radiations, obtain X-ray photographsof locomotive engines, or sterilize food with gamma rays. Closer to home are its educa-tional uses. The Van de Graaff generator is an excellent device for studying electrostatics,the science of voltage and electric charge.

1.23.2 Principle of van de Graaff Generator

(a) This generator is based on the following two electrostatic phenomena:

The electric discharge takes place in air or gases readily at pointed conductors. For aspherical conductor of radius r, carrying charge q, the surface density is

2

chargearea 4

qr

It can be seen that for a pointed end, r is very small, therefore surface density is verylarge.

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The particles of air coming in contact with pointed ends similarly charged and arerepelled. In this way, an electric wind is set up which takes away the charged continu-ously from the pointed ends of the conductor. This process of spraying the charge iscalled as corona discharge. That is why conductors used for storing charge are al-ways spheres of large radii.

(b) The charge given to a hollow conductor is transferred to outer surface and is distrib-uted uniformly over it.

1.23.3 Construction

The basic design of Van de Graaff generator is shown in figure given below.

Fig. 1.42 Van de Graaff generator

It consists of a large hollow spherical conducting shell 'S' of radius equal to a few metres,supported by the insulating pillars R1, R2 at a suitable height (of several metres above theground). A long narrow belt of insulating material like, silk, rubber or rayon is wound aroundtwo pulleys P1 and P2. Pulley P1 is at the ground level, whereas pulley P2 is at the centre ofmetallic sphere 'S'. The belt is made to run continuously over the pulleys with the help of amotor. C1 and C2 are two sharply pointed combs fixed as shown in the figure. C1 is knownas the spray comb, held near the lower end of the belt, and C2 is known as the collectingcomb. Positive ions to be accelerated are produced in a discharge tube G and the ionsource lies at the head of the tube inside the spherical shell. The other end of the tubecarrying the target nucleus is earthed. This generator is enclosed in a steel chamber 'C'which is filled with nitrogen or methane at high pressure.

1.23.4 Working

The spray comb is maintained at a positive potential 410 V with respect to the earth byhigh-tension source H.T. As a result of discharging action of sharp points, a positivelycharged electric wind is set up, which sprays positive charge on the belt (i.e. corona dis-charge). As the belt moves, and reaches the sphere, a negative charge is induced on thesharp ends of collecting comb C2 and an equal positive charge is induced on the farther endof C2. This positive charge shifts immediately to the outer surface of conducting sphere 'S'.

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Owing to discharging action of sharp points of comb C2, a negatively charged electric windis set up. This neutralizes the positive charge on the belt. The uncharged belt returns down,collects the positive charge from C1 which in turn is collected by C2. This process is re-peated and thus the positive charge on metallic sphere 'S' goes on accumulating.

The capacity of spherical shell S is expressed as

C = 40R ...(i); where R is radius of the spherical shell.

Now, potential qVC

Using equation (i)

04qV

R

...(ii)

It can be seen from the equation (ii) that the potential V of the spherical shell goes onincreasing with increase in charge q.

The breakdown field of air is nearly 3 × 106 Vm–1. The moment, the potential of conductingshell exceeds this value, air around shell S gets ionized and leakage of charge starts. Toprevent the leakage of charge from the sphere, the generator assembly is enclosed inside asteel chamber filled with nitrogen or methane at high pressures.

If q' be the charge on the ion to be accelerated and V be the potential difference developedacross the ends of the discharge tube then energy acquired by the ions = Vq'. The ions hitthe target with this energy and carry out the artificial transmutation, etc.

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EXERCISEDISCRIPTIVE QUESTIONS1. Derive an expression for dipole field intensity at any point on

(i) at any point

(ii) axis of uniformly charge ring

2. Derive an expression for total work done in rotating an electric dipole through an angle in uniformelectric field.

3. Show that the integral of electric field intensity between any two points depends only on the position ofthese points and is independent of the path followed between these points.

4. Explain the concept of electric potential energy. Derive an expression for potential energy of a systemof two point charges. Generalise the expression for N discrete charges.

5. Using Gauss’s theorem, derive an expression for electric field intensity at a point due to

(i) a line of charge

(ii) a uniformly charged spherical shell

(iii) a charged solid sphere

(iv) an infinite sheet of charge

(v) two parallel sheets of charge

6. Apply Gauss’s theorem to find an expression for the electric intensity at any point due to a point charge.

7. Explain the principle of a capacitor. Derive an expression for the capacitance of a parallel plate capacitor.

8. When two charged conductors having different capacities and different potentials are joined together,show that there is always loss of energy.

9. Deduce the effect of introducing (i) a conducting slab (ii) a dielectric slab in-between the plates of aparallel plate condenser on the capacitance of the condenser.

10. Discuss briefly the principle, construction and working of Van-de-Graff electrostatic generator.

HIGH ORDER THINKING (HOTs)1. Two conductors identical in shape and size, but one of copper and the other of aluminium (which is less

conducting) are both placed in an identical electric field. In which metal more charged will be induced.

2. When is the torque on an electric dipole in a field maximum ?

3. When is an electric dipole in unstable equilibrium in an electric field ?

4. Describe schematically, the equipotential surfaces corresponding to

(i) a constant electric field in the Z direction

(ii) a field that uniformly increases in magnitude but remains in a constant, say z direction

(iii) a single positive charge at the origin

(iv) a uniform grid consisting of long equally spaced parallel charged wires in a plane.

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5. The electrostatic field due to a point charge depends on the distance r as (1/r2), similarly indicate howeach of the following quantities depends on r :

(a) Intensity of light from a point source

(b) Electrostatic potential due to a point charge and

(c) Electrostatic potential at a distance r from the centre of a charged metallic sphere of radius (r < R).

6. A test charge q0 is moved without acceleration from A to C over the path shown in figure below.Calculate potential difference between A and C.

7. (a) A conductor A with cavity as shown below is given a charge Q. Show that the entire charge mustappear on the outer surface of the conductor.

(b) Another conductor B with a charge q is inserted into the cavity keeping B insulated from A. Showthat the total charge on the outside surface of A is (Q + q).

8. The distance between the plates of a parallel plate capacitor is d. A metal plate of thickness d/2 isplaced between the plates, what will be the new capacity ?

9. Given a battery, how would you connect two capacitors, in series or in parallel for them to store thegreater (a) total charge, (b) total energy ?

10. A sphere S1 of radius r1 encloses a total charge Q. If there is another concentric sphere S2 of radiusr2(> r1) and there be no additional charges between S1 and S2, find the ratio of electric flux through S1

and S2.

NUMERICAL ABILITY

1. Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm.What is the mutual force of repulsion if charge on each is 6.5 × 10–7 coulomb. The radii of A and B arenegligible compared to the distance of separation.

(b) What is the force of repulsion if

(i) Each sphere is charged double the above amount and the distance between them is halved ?

(ii) The two spheres are placed in water of dielectric constant 80.

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2. A pendulum bob of mass 80 milligram and carrying a charge of 2 × 10–8 C is at rest in a horizontaluniform electric field of 2 × 104 Vm–1. Find the tension in the thread of the pendulum and the angle itmakes with the vertical.

3. An electric dipole consists of two opposite charges of magnitude q = 1 × 10–6 C separated by 2.0 cm.The dipole is placed in an external field of 1 × 105 NC–1. What maximum torque does the field exert onthe dipole ? How much work must an external agent do to turn the dipole end or end, starting fromposition of alignment ( = 0º) ?

4. An electric dipole consists of charges +2 e and –2 e separated by 0.78 nm. It is in an electric field ofstrength 3.4 × 106 N/C. Calculate the magnitude of the torque on the dipole when the dipole moment is

(a) parallel to, (b) perpendicular to, and (c) antiparallel to the electric field.

5. In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å.

(a) Estimate the potential energy of the system in eV taking the zero of potential energy at infiniteseparation of electron from proton.

(b) What is the minimum work required to free the electron, given that its KE in the orbit is half themagnitude of potential energy obtained in (a) ?

(c) What are the answers to (a) and (b) above, if zero of potential energy is taken at 1.06 Å separation.

6. The electric field in a region is given by 0 03 4ˆ ˆ5 5

E E i E j with E0 = 2.0 × 103 N/C. Find the flux of this

field through a rectangular surface of area 0.2 m2 parallel to the Y-Z plane.

7. A charge q = +1 C is held at O between two points A and B such that AO = 2 m and BO = 1 m.Calculate the value of potential difference (VA – VB). What will be the value of potential difference(VA – VB) if position of B is changed as shown in figure below.

8. Twenty seven charged water droplets each with a diameter of 2 mm and a charge of 10–12 C coalesceto form a single drop. Calculate the potential of the bigger drop.

9. Careful measurement of the electric field at the surface of a black box indicates that the net outwardflux through the surface of the box is 8.0 × 103 NC–1 m2.

(a) What is the net charge inside the box ?

(b) If the net outward flux through the surface of the box were zero, could you conclude that therewere no charges inside the box ? Why or why not ?

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10. The area of a parallel plates of an air capacitor is 0.2 m2 and the distance between them is 0.01 m. Thepotential difference between the plates is 3000 V. When a 0.01 m thick sheet of an insulating material isplaced between the plates, the potential difference decreases to 1000 volt. Determine (i) capacitance ofcapacitor before placing the sheet, (ii) charge on each plate, (iii) dielectric constant of material, (iv)capacitance after placing the insulator, (v) absolute permittivity of the dielectric.

11. Find out the capacitance of a spherical capacitor consisting of two concentric spheres of radii 50 cmand 60 cm. The dielectric constant of the substance filled in space between the two spheres is 6.

12. In the circuit shown below, the emf of each battery is E = 12 volt and the capacitances are C1 = 2.0 Fand C2 = 3.0 F. Find the charges which flow along the paths 1, 2, 3 when key K is pressed.

13. Take the potential of the point B in figure below to be zero. (a) Find the potentials at the points C and D.(b) If a capacitor is connected between C and D, what charge will appear on this capacitor ?

14. The radii of two charged metallic spheres are 5 cm and 10 cm. Each has a charge of 75 C. They areconnected by a conducting wire. Calculate (i) common potential of the spheres after connecting (ii)amount of charge transferred.

15. Two capacitors of 25 F and 100 F are connected in series to a source of 120 V. Keeping their chargesunchanged, they are separated and connected in parallel to each other. Find out (i) potential differencebetween the plates of each capacitor, (ii) energy loss in the process.

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ANSWERSHIGH ORDER THINKING (HOTs)1. In both the metals, induced charge will be equal.

2. When dipole is held at 90º to the electric field.

3. When p is antiparallel to E i.e., q = 180º

4. (i) Equipotential surface are planes parallel to X – Y plane. These are equidistant.(ii) Equipotential surfaces are planes parallel to X – Y plane. AS the field increases unformly, distance

between the planes (differing by fixed potential) decreases.(iii) Equipotential surfaces are concentric spheres with origin at the centre.

(iv) Equipotential surfaces have the shape which changes periodically. At far off distance from the grid,the shape of equipotential surfaces becomes parallel to the grid itself.

5. (a) 2

1Ir

, (b) 1Vr

, (c) V remain constant

6. Ed8. Double9. (a) Cp > Cs, (b) Up > Us

10. 1 : 1

NUMERICAL ABILITY1. (a) 1.521 × 10–2 N, (b) i. 0.24 N, ii. 1.9 × 10–4 N2. 8.801 × 10–4 N, 27º3. 2 × 10–3 N-m, 4 × 10–3 J

4. (a) zero, (b) 8.5 × 10–22 N-m, (c) zero5. (a) –27.16 eV, (b) 13.58 eV, (c) 13.58 eV

6. 240 Nm2/C7. –4500 V, same value8. 81 V

9. (a) 0.07 C,(b) No, any number of charges may be present inside but algebraic sum of these charges is zero.

10. (i) 1.77 × 10–10 F, (ii) 5.31 × 10–9C, (iii) 3, 5.31 × 10–10 F, (iv) 2.66 × 10–11 F m11. 2 × 10–9 F

12. 24 C, –36 C, 12 C

13. (a)503

V , (b) 50 , zero3

V

14. (i) 9 × 106 V, (ii) 25 C15. (i) 38.4 V, (ii) 0.052 J

(cbse) electrostatics chapter-1 electrostatics

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