8. Electrostatics

ELECTROSTATICS

These topics are taken from our Book: ISBN : 9789386320087 Product Name : Electrostatics for JEE Main & Advanced (Study Package for Physics) Product Description : Disha's Physics series by North India's popular faculty for IIT-JEE, Er. D. C. Gupta, have achieved a lot of acclaim by the IIT-JEE teachers and students for its quality and indepth coverage. To make it more accessible for the students Disha now re-launches its complete series in 12 books based on chapters/ units/ themes. These books would provide opportunity to students to pick a particular book in a particular topic. Electrostatics for JEE Main & Advanced (Study Package for Physics) is the 8th book of the 12 book set. • The chapters provide detailed theory which is followed by Important Formulae, Strategy to solve problems and Solved Examples.

• Each chapter covers 5 categories of New Pattern practice exercises for JEE - MCQ 1 correct, MCQ more than 1 correct, Assertion & Reason, Passage and Matching based Questions. • The book provides Previous years’ questions of JEE (Main and Advanced). Past years KVPY questions are also incorporated at their appropriate places. • The present format of the book would be useful for the students preparing for Boards and various competitive exams.

Contents

Contents

1.

Mathematics Used in Physics



Partial differentiation

ii



Uses of differentiation and integration

iii



Area bounded by the curve

iii



Solid angle

iv



Complex number and phasor) v

2. Electrostatics

1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12 1.13 1.14 1.15 1.16 1.17 1.18 1.19 1.20 1.21 1.22

I-VI

1-130

Basics of electrostatics Earthing or grounding Gold leaf electroscope Lightning and lightning conductor Coulomb’s law Electric field Electric field lines (efl) Electric potential energy (epe) Potential difference Equipotential surface Electric flux Gauss’s law or gauss’s theorem Applying gauss’s law : spherical symmetry Applying gauss’s law : cylindrical symmetry Line charge of finite length Applying gauss’s law : planar symmetry Conductor of any shape Mechanical force on the charged conductor Energy density Electric field due to charged ring Charged disc Electric potential energy of system of charges

02 04 04 05 08 14 15 18 19 22 23 26 30 34 35 37 39 39 40 41 44 61

1.23 Electric dipole 64 1.24 A dipole in an electric field 67 1.25 Force between two short dipoles 68 1.26 Quantisation of charge : Millikan oil drop experiment 70 Review of formulae and important points 73 Exercise 1.1 - 1.6 77-106

Hints & solutions 107-130

2.

Capacitance & Capacitors



2.1



131-216

Capacitor or condenser : An introduction

132



2.2 Capacitance or capacity

132



2.3 Charging a capacitor

133



2.4 Calculating capacitance

133



2.5 Capacitance of parallel plate capacitor

135



2.6 Energy stored in a capacitor

137



2.7 Force between the plates of a capacitor 139



2.8

Capacitors in series and parallel

142



2.9

Spherical capacitor

147



2.10 Cylindrical capacitor

149



2.11 Dielectrics

153



2.12 Induced or bound charge

154



2.13 Total energy of the system

159



2.14 Van de graff generator

161



2.15 Kirchhoff’s laws

161



Review of formulae & important points

178



Exercise 2.1-2.6

180-200



Hints & solutions

201-216

Chapter 1

Electrostatics 1.12 GAUSS’S

LAW OR GAUSS’S THEOREM

Gauss's law is the very powerful tool of electrostatics. It can be used to find total flux associated with the closed surface due to charges appearing in the space. By doing this we can calculate electric field at any point of the surface. According to it, the total flux of the net electric field through a closed surface (Gaussian surface) equals the net charge enclosed by the surface divided by Î0 . Thus if qin is the total charge inside the closed surface, then r r qin E.dA = . Î0 Proof of Gauss's law Consider charges q1, q2, ........., qn are inside a closed surface and charges q¢1, q¢2, ......., q¢n are outside the surface. r r r r r r If E1, E2 ,........, En ; E '1, E '2 ,........, E 'n are the fields produced by the respective charges ; inside and outside the closed surface, then resultant electric field at any point P on it r r r r r r r E = ( E1 + E2 + ..... + En ) + ( E '1 + E '2 + ........ + E 'n ) The flux of the resultant electric field through the closed surface r r f= Ñ ò E.dA r r r r r r r r r r r r = éë Ñ ò E1.dA + Ñò E2 .dA + ....... + Ñò En .dAùû + éë Ñò E '1 .dA + Ñò E '2 .dA + ..... + Ñò E 'n .dAùû …(1) r r r r E . dA E ' . dA is the flux q and is the flux to due to charge q which is Here Ñ / Î 1 1 1 0 ò Ñò 1 charge q¢1, which is zero, because it is out off the closed surface. Thus equation (1) can be written as ; q ù é q1 q2 r r + + .... + n ú + [ 0 + 0 + ...... + 0] E × dA = ê Ñò Î0 û ë Î0 Î0

Ñò

r r

Ñò E.dA

or

=

Sqi Î0

=

qin . Î0

Fig. 1.76

…(2)

Note: 1.

The net charge is the algebraic sum of all the enclosed positive and negative charges, and it can be positive, negative, or zero.

2.

The electric field appearing in the Gauss’s law is the net electric field due to all the charges present inside as well as outside the closed surface. On the other hand, the charge qin appearing in the law is only the charge inside the closed surface.

The following observations are useful in calculating total flux 1.

2.

3. 4.

ur If E is perpendicular to a surface area A ( q = 0) at all points, and has the same ur ur magnitude at all points of the surface, then Ñ ò E.d A = Ñò EA cos 0 = EA. For spherical surface EA = E ´ 4 pr 2 = 4pEr 2 . ur If E is parallel to a surface at all points, E^ = 0 or q = 90° and therefore ur ur Ñò E.d A = ò EA cos 90° = 0 . ur If E is zero at all points of a surface, the integral is zero. The surface to which Gauss’s law is applied need not be a real physical surface, such as the surface of a solid body. Indeed, in most applications of this law, one considers an imaginary or geometrical surface that may be in empty space, embedded in a solid body, or partly in space and partly within a body.

Fig. 1.77

ELECTROSTATICS

27

Conductor with excess charge If an excess charge is placed on an isolated conductor, the charge will move entirely over the surface of the conductor. It is because of the fact that like charges repel each other. An internal electric field does appear as the conductor is being charged. However, the excess charge quickly distributes itself such a way that the internal electric field is zero. The movement of charge then ceases, and the net force on each charge is zero; the charges are then in electrostatic equilibrium. ur The fact can be verify by Gauss’s law. It has explained that the electric field E is zero at all points within a conductor when the charges in the conductor are at rest. We may take an ur imaginary surface in the interior of the conductor as shown in figure. Because E is zero everywhere on the surface, ur ur \ Ñò E.d A = 0 By Gauss’s law

ur ur

=

qin Î0

0 =

qin Î0

Ñò E.d A

or which gives

qin = 0.

Cavity in the conductor

Fig. 1.78

Suppose there is a cavity in the charged conductor, and there are no charges within the cavity. ur ur For the surface A, Ñò E.d A = 0 ur Since E is zero in the conductor,, qin = 0. \ Further more, consideration of a surface such as B shows that the net charge on the surface of the cavity must be zero. Thus the entire charge on the conductor lies on its outer surface, not on the cavity wall.

Charge placed inside the cavity 1.

Suppose that there is a conductor inside the cavity but insulated from it and that the inner conductor has charge q. By Gauss's law

Fig. 1.79

ur ur

or

Fig. 1.80

=

qin Î0

0 =

qin Î0

Ñò E.d A

2.

ur Since E is zero inside the conductor, which gives qin = 0. As the charge inside the cavity is q, so there must be a charge on the cavity wall, equal and opposite in sign to the charge q. If outer conductor is initially uncharged before the charge q is inserted, there must be a charge on its outer surface, equal and opposite to the charge on the cavity wall, that is q. Consider a charged conductor having charge q' initially. A charge q is placed in a cavity in the conductor. The charge on the outer surface of the conductor = q' + q.

28

ELECTRICITY & MAGNETISM

3.

Lines of field inside cavity due to charge at its centre Case 1 : Spherical cavity Case 2 : Non-spherical cavity

Fig. 1.81

Conductor in external electric field

Fig. 1.82

Consider an uncharged conductor is placed in an external electric field. The free electrons in the conductor distribute themselves on the surface as shown. Reducing the net electric field inside the conductor to zero and making the net field at the surface perpendicular to the surface.

Potential of earthed conductor is zero Let us consider two concentric conducting shells. Inner shell is given a charge q. The charges are induced at outer shell, as shown in fig. 1.83. The potential of any point on the surface of outer shell V =

1 éq q qù 1 q - + ú= . ê 4p Î0 ë b b b û 4p Î0 b

Now let outer shell is earthed. The charge on outer surface of outer shell neutralise by electrons coming from earth. But negative charge on inner surface of outer shell remain in tact because of attraction of charge of inner shell. Potential at any point on outer shell V =

Ex. 28

1 ìq q ü í - ý=0. 4p Î0 î b b þ

A point charge q is placed on the apex of a cone of semi-

vertex angle q . Show that the electric flux through the base of the cone is

Fig. 1.83

Fig. 1.84 To find A, choose a surface element confined in angle da at an angle a . The area of the element strip

q ( 1 – cos q ) . 2 Î0

dA =

Sol. Consider a Gaussian surface with its centre at the apex of the cone. The flux through the whole sphere is

q , so the flux through the base of the Î0

A q cone f = A0 Î0 where A = area of sphere below the base of the cone and

A0 = area of whole sphere which is 4pR 2 .

Surface area or Thus desired flux

A

( 2pr ) ds

=

2pR sina ( Rd a )

=

2pR 2 sina d a

=

ò0 2pR

q

2

[ r = R sin a]

sin a d a

A =

2pR 2 (1 - cos q)

f =

A q 2pR 2 (1 - cos q ) q = . A0 Î0 4pR 2 Î0

29

ELECTROSTATICS =

As the total lines of force emanating from q1 is equal to the total lines of force terminating to q2, so

q (1 - cos q ) . 2 Î0

Ex. 29

Two charges +q 1 and –q 2 are placed at A and B respectively. A line of force emanates from q1 at an angle a with the line AB. At what angle will it terminate at –q2 ?

or

The number of lines of force emerges is proportional to the amount of charge. The line of force emanating from q1 spread out equally in all

or

Sol.

directions. Hence lines of force per unit solid angle are

q1 2p (1 - cos a ) 4p

=

q2 2 p (1 - cos b ) 4p

q1 a .2sin 2 2 2

=

q2 b .2sin 2 2 2

b 2

=

q1 a sin q2 2

b

=

sin

q1 and the 4p

number of lines of force through cone of half angle a is q1 .2 p (1 - cos a ) . 4p Similarly the number of lines of force terminating on –q2 at angle b is

q2 2 p (1 - cos b ) 4p Fig. 1.85

\

é q aù 2sin -1 ê 1 sin ú 2 úû êë q2

Ans.

Application of Coulomb's law and Gauss's law Coulomb's law is the fundamental law of electrostatics, but for symmetric charge distribution Gauss's law can be used. For electrostatics problems, Gauss's law is equivalent to Coulomb's law. In Gauss's law, the closed surface is of any shape. The

Gaussian surface will often be a sphere, a cylinder, or some other symmetrical form. But it must always be a closed surface, so that a clear distinction can be made between points that are inside the surface, on the surface, and outside the surface.

1.13 APPLYING (i)

GAUSS'S LAW

: SPHERICAL

SYMMETRY

Ch ar ged c on du c t i n g sh el l : Fig. 1.86 shows a charged spherical shell of total charge q and radius R. The charge of the conducting shell will spread uniformly over its outer surface, so charge inside the shell will be zero. For the Gaussian surface, r ³ R, qin = q.

Fig. 1.86

ur ur

Ñò E.d A

=

qin Î0

or

Ñò E.dA cos 0

=

q Î0

or

EÑ ò dA

=

q Î0

or

E ´ 4pr 2

=

q Î0

By Gauss's law

\

ur

æ

1

ö q

E = çè 4 p Î ÷ø 2 0 r

For r < R, the charge inside the closed surface, qin = 0, and so E = 0.

Note: Above obtained results are also applicable to the conducting sphere. Electric potential :

30

ELECTRICITY & MAGNETISM

For the potential at r ³ R, the charge of the shell can be assumed at the centre of the shell, and so electric potential at a distance r from the centre of the shell, V =

æ 1 öq çè 4p Î ÷ø r , 0

r > R.

V =

æ 1 ö q çè 4p Î ÷ø R , 0

r=R

At surface of the shell, r = R, and so

For r < R, the electric field is zero, and so electric potential remain constant from centre to the surface of the shell. \

V =

æ 1 ö q çè 4p Î ÷ø R , 0

r R, the entire charge lies within it. The charge produces an electric field on the Gaussian surface as if the charge were a point charge located at the centre, and so æ 1 ö q , r>R E = ç è 4p Î0 ÷ø r 2

For the Gaussian surface at r < R, charge enclosed in the sphere of radius r, qin

Fig. 1.92

By Gauss's law

or

r r

=

q

4 3 pR 3

´

Ñò E.dA

=

q 'in Î0

2

=

qr 3 / R3 Î0

E ´ 4pr

4 3 qr 3 pr = 3 3 R

\

æ 1 ö qr E = ç è 4p Î0 ÷ø R3

At

r = 0, E = 0.

Electric potential : For the potential at r ³ R, the charge of the sphere can be assumed at its centre, and so electric potential at a distance r from the centre of the sphere, æ 1 öq V = ç è 4p Î0 ÷ø r

,

r>R

At the surface of the shell, r = R, and so æ 1 ö q V = ç è 4p Î0 ÷ø R

Fig. 1.93

,

r =R

Electric potential at r < R : If E1 and E2 are the electric fields outside and inside the sphere, then æ 1 ö q æ 1 ö qr E2 = ç E1 = ç and . ÷ è 4p Î0 ø r 2 è 4p Î0 ÷ø R3

Fig. 1.94

Electric potential at any point is defined as, r ur r V = - ò E .d r ¥

32

ELECTRICITY & MAGNETISM r é R ù = - ê ò E1dr + ò E2 dr ú R ë ¥ û

é Ræ 1 ræ 1 qö qr ö ù = - ê ò¥ ç 4p Î 2 ÷ dr + òR ç 4p Î 3 ÷ dr ú è êë è úû 0 r ø 0 R ø

After simplifying, we get

At the centre of the sphere, r = 0, i.e.,

V =

æ 1 ö éë3R 2 - r 2 ùû çè 4p Î ÷ø q 2R 3 0

V =

æ 1 ö 3q çè 4p Î ÷ø 2R 0

Vcentre

=

Fig. 1.95

3 Vsurface 2

Self energy of the charged sphere Fig. 1.95 shows a sphere of total charge q and radius R. Take an element of thickness dr at a radial distance r all over the sphere. If q1 and q2 are the charges on the element and the sphere inside, then the energy of this system, dU =

1 q1q2 4p Î0 r

where

æ q q1 = ç 4 ç pR 3 è3

and

q2

3q r 2 dr ö 2 4 p = r dr , ÷ R3 ÷ ø

3 æ q ö 4 3 qr = ç r p = ÷ 4 R3 ç pR3 ÷ 3 è3 ø

The total energy of the charged sphere

U

=

=

or

U

=

æ 3q 2 ö æ qr 3 ö r dr ÷ ç ÷ ç æ 1 ö R è R3 ø çè R3 ÷ø R ò0 dU = çè 4p Î0 ÷ø ò0 r 3æ 1 ç 5 è 4p Î0

3 20

0

ö q2 ÷ ø R

q2 R

Fig. 1.96

77

ELECTROSTATICS

MCQ Type 1

Only one option correct 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. There is a point charge +q inside a hollow sphere and a point charge –q just outside its surface. The total flux passing through the sphere (a)

(c) 3.

-q Î0

(b)

2q Î0

(d)

5.

6.

q Î0 7.

zero.

Figure shows three electric field lines. If FA, FB and FC are the forces on a test charge q at the positions A, B and C respectively, then

r A surface has the area vector A = 2i$ + 3 $j m 2 . The flux of an

(

(a)

q1, q2

(b)

q1, q3

(c)

q1, q2, q3

(d)

q1, q2, q3, q4

Figure shows five charged lumps of plastic and an electrically neutral metal coin. The close surface S is indicated in the figure. The net flux through the surface is : (a)

(c)

(c) 4.

FA > FB > FC

(b)

FA > ( FB = FC )

(d)

FA < FB < FC

8.

FA < ( FB = FC )

In the figure the electric lines on the right have twice the separation of those on the left. If a charge particle takes time t to move a distance x in left region, then it will take time to travel the same distance in the right side region is :

(a) (c)

t 2 2t

(b)

t

(d)

2t

)

r V : electric field through it if the field is E = 4i$ m (a) 8 V-m (b) 12 V-m (c) 20 V-m (d) zero In figure a close surface encloses two of the four positively charged particles. Which of the particles contribute to the electric field at point P on the surface :

(b)

(a)

E xercise 1. 1

( q1 - q2 - q3 - q4 + q5 ) Î0

( q1 - q2 - q3 ) Î0

( -q4 + q5 ) Î0

(d) zero The figure shows three situations in which a Gaussian cube sits in an electric field. The arrows and the values indicate the directions of thefieldlinesandthemagnitudes(inN-m2/C) of the flux through the six sides of each cube. The dotted arrows are of the hidden faces. In which situation the cube enclose net positive charge :

(a)

I

(b)

II

(c)

III

(d)

none

A n sw er K ey

1

(d)

2

(b)

3

(c)

4

(c)

Sol. from page 107

5

(a)

6

(d)

7

(b)

8

(b)

ELECTROSTATICS

107

S olutions Exercise 1.1Level -1 1.

(d)

Equal and opposite charges will appear on the sphere. So net charge in the sphere becomes zero. – – – –

2.

(b)

+ + + q + The total flux passing through the sphere

f= 3.

4.

(c)

qin q = Î0 Î0

–q

q

With the increase in the spacing between the field lines, intensity of electric field decreases. So FA > (FB = FC)

1 1 æ Eq ö x = at 2 = ç ÷ t 2 2 2è m ø

(c)

1 æ Eq ö 2 ç ÷ t¢ 2 è 2m ø

and

x=

\

t ¢ = 2 t.

r r f = E. A = 4iˆ.(2iˆ + 3 ˆj ) = 8 V-m Electric field at P is due to all the charges; either inside or outside the close surface.

5.

(a)

6.

(d)

7.

(b)

f=

8.

(b)

We know that f = \

qin q1 - q2 - q3 = . Î0 Î0 qin Î0

q = Î0 .f

For positive charge, f must be positive. In case II, total field lines are (10 + 5 + 3) – (4 + 6 + 3) = 5, are coming out of the cube.

Chapter 2

Capacitance & Capacitors

161

2.15 KIRCHHOFF'S LAWS Kirchhoff's laws are very useful in analysing multiloops circuits.He provided two laws. These are :

Junction rule : The algebraic sum of the charges or currents at any junction is zero. åq = 0

For charge : At junction A as shown in fig. 2.66, q1 + q2 + q3 or q3 For current : Si At A, i1 + i2 + i3 or i3

Loop rule:

= = = = =

0 – (q1 + q2). 0 0 – (i1 + i2).

The algebraic sum of the potential differences across all the circuit elements in a closed loop including with emfs must equal to zero. or å V = 0. (a) Circuit having resistors and cells, then (b)

= 0 å x + å iR Circuit having capacitors and cells, then

(c)

q = 0 C Circuit having resistors, capacitors and cells, then

Fig. 2.65

Fig. 2.66

åx+ å

q + å iR = 0. C The juction rule is an application of the principle of conservation of charge. The loop rule is a consequence of conservation of energy. For circuit analysis, the following procedures should be followed carefully. First of all label all the known and unknown carefully, including an assumed sense of direction for each unknown. Often one does not know in advance the actual direction or sign of an unknown current, emf or charge, but this does not matter. The solution is carried out using the assumed directions, and if the actual direction of a particular quantity is opposite to the assumed direction, the value of the quantity found with negative sign. åx+ å

The following guidelines and sign conventions are useful in analysing circuit problems. 1. 2. 3. 4.

One must remember that, the current in any resistor or charge on any capacitor is the net response of all the sources present in the circuit. Don't think that only nearest one is sending the current or charge. Choose any closed loop in the network, and designate a direction (clockwise or anticlockwise) to traverse the loop for loop rule. Close loop may or may not have any cell, but one of the close loop must include the cell. For a circuit having only one cell, choose a close loop, which include this cell. (a) The p.d. across any resistor can be taken negative when moves along the direction of current and positive for reverse direction of current. (b) The p.d. across capacitor can be taken negative when moves from positive plate of capacitor to negative plate and positive for reverse sense.

Note: See the sign of charge of first coming plate of capacitor only.. (c) The emf of any cell/battery is taken as negative from its positive terminal to negative terminal and positive for reverse sense.

Fig. 2.67

Fig. 2.68

162

ELECTRICITY & MAGNETISM 5. 6. Fig. 2.69

7.

It must be remembered that capacitor in steady state stop the direct current, and so there will be no current in the branch in which capacitor is connected. If capacitors are connected in series with the sources as shown, the charge on each capacitor will be same. It must be remembered that the net charge supply by cell will be zero. Finally, the number of equations obtained must always be equal to the number of unknows.

The R-C series circuit (a)

Fig. 2.70

Charging Consider the circuit as shown in fig. 2.73. If the capacitor is initially has no charge, then the initial p.d. across the capacitor is zero, and the entire emf appears across x . As the capacitor charges, its potential R increases, and p.d. across resistor decreases, corresponding to a decrease in current. After a long time when capacitor becomes fully charged, the entire potential will occur across capacitor and the current becomes zero. Let at any instant the current in the resistor is i, then by loop rule

resistor, causing an initial current i0 =

q +x C

= 0

dq q R- +x dt C

= 0

– iR –

Fig. 2.71

or

-

dq R dt dq qö æ çè x - ÷ø C

\

= x=

q C

dt R

Integrating above expression, we have Fig. 2.72

q

ò 0

or

or

t

dq qö æ çè x - ÷ø C

qö æ lnçx - ÷ è Cø 1 C

=

dt

òR 0

q

=

t R

=

-t RC

0

qö æ l n ç x - ÷ - lnx è Cø

é q ù l n ê1 = - t ú C xû ë RC Substituting, C x = q0 , the maximum or steady state charge on the capacitor and or

RC = t, capacitive time constant \

or

æ qö l n ç1 - ÷ è q0 ø

q

t t

=

-

=

q0 1 - e -t / t

(

)

CAPACITANCE AND CAPACITOR 163 Current :

i

=

dq æ 1ö = q0 ´ e - t / t ´ ç ÷ è tø dt

=

q0 -t / t e RC

=

C x -t / t e RC

xù é ê as i0 = R ú ë û

or

i

= i0 e -t / t

At t = 0,

i

-0 = i0 e = i0

= i0 e -¥ = 0 It means initially capacitor offers no resistance in the circuit. But after charging fully, it offers infinite resistance. Potential of capacitor : and at t = ¥,

i

Potential,

V

=

(

-t / t q q0 1 - e = C C

(

-t / t = V0 1 - e

or

)

)

Energy stored : U

or q = q0 (1 - e

U -t / t

); i = i0 e

-t / t

=

1 CV 2 2

=

2 1 é C êV0 1 - e -t / t ùú û 2 ë

=

1 CV02 1 - e - t / t 2

(

)

(

(

= U0 1 - e-t / t

;V = V0 (1 - e

-t / t

)

)

2

2

); U = U 0 (1 - e

-t / t 2

(a)

)

Capacitive time constant (t) : At t = t,

q

=

(

q0 1 - e -1

)

; 0, 64 q0 = i0e–1 ; 0.37 i0 Therefore time constant is the time in which charge on the capacitor grows to 0.64 times the maximum charge or current decreases to 0.37 times the maximum current.

or at t = t,

(b)

i

(b) Fig. 2.73

Discharging After charging the capacitor, let now source is removed from the circuit. The energy stored in capacitor now used to flow the current. At t = 0, we have q = q0 and Again by loop rule,

U

= U0 =

q02 . 2C

Fig. 2.74

164

ELECTRICITY & MAGNETISM q = 0 C where q is the charge on capacitor at any time t

– iR –

or

-

dq R dt

=

dq = q Integrating above expression, we get or

q

ò

q0

q C -

dt RC t

dq q

=

-

dt

ò RC 0

or

lnq q

=

-

t RC

or

ln

q q0

=

-

t t

q 0

[As RC = t]

or

q

=

q0 e -t / t

Current :

i

=

dq d q0 e - t / t = dt dt

=

æ 1ö q0 e -t / t ´ ç - ÷ è tø

=

C x -t / t e CR

=

x é ù -i0 e -t / t ê As R = i0 ú ë û

=

q q0 - t / t e = C C

Potential :

V

(

= V0 e

(a) Energy :

U

-

)

t t

=

1 CV 2 2

=

1 C V0 e - t / t 2

=

1 CV02 e -2t / t 2

(

)

2

= U 0 e -2t / t .

Finding the time constant (b) Fig. 2.75

If there are many resistors and capacitors are connected in the circuit, then time constant can be obtained by using following steps; (i)

make short circuit at the place of battery.

(ii)

Find equivalent capacitance, if there are many capacitors in the circuit. Assuming equivalent capacitor is Cnet.

(iii) Find net resistance across the capacitor, say it is Rnet . (iv) Finally tnet = Rnet Cnet.

CAPACITANCE AND CAPACITOR 165 Finding the potential difference between any two points in a circuit Let we have to find the potential difference between P and R in above circuit Step 1 : Connect an imaginary battery between the points, let it is VPR. Step 2 : Choose close loop including the imaginary battery. Step 3 : Use loop rule to find VPR. Choose close loop P – R – y – x – P and applying loop rule, we have q + 10 + VPR C

or or

-

= 0

10 + 10 + VPR = 0 1 VPR = 0.

Fig. 2.76

Heat generated It can be obtained by using conservation of total energy of the system. That is Initial energy stored in capacitors + work done by sources = Final energy stored in capacitors + heat generated or Ui + W = Uf + H or H = (Ui – Uf) + W

Note: (a)

Work done will be positive if work is done by the sources and negative if

work is done on the sources. Positive work done by source : If a capacitor C of potential V (V may be zero) is connected to a source of emf x ( x > V)

The work done by source will be

=

x ( Dq )

=

x q f - qi

=

x ( C x - CV )

(

)

= C x (x -V ) . (b)

Negative work done by source : If a capacitor C of potential V is connected to a source of emf x ( x < V) The work done by source will be

= xDq = x (qf – qi) = x (Cx – CV) = – Cx (V – x).

Case 1. If Uf = Ui, the heat generated H = W. Case 2. If capacitors remain the part of same circuit then there always be a positive work.