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Capacitance 1.1

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Section A - Capacitance Calculation(Sphere, Cylinderical, Parallelplate), Sharing of Charge, Com-bination of Capacitor

1. INTRODUCTIONA capacitor can store energy in the form of potentialenergy in an electric field. In this chapter we'lldiscuss the capacity of conductors to hold chargeand energy.

1.1 Capacitance of an isolated conductorWhen a conductor is charged its potential increases.It is found that for an isolated conductor (conductorshould be of finite dimension, so that potential ofinfinity can be assumed to be zero) potential of theconductor is proportional to charge given to it.

q

Isolated conductor

q = charge on conductorV = potential of conductorq V q = CV

Where C is proportionally constant calledcapacitance of the conductor.

1.2 Definition of capacitance :Capacitance of conductor is defined as chargerequired to increase the potential of conductor byone unit.

1.3 Important point about the capacitanceof an isolated conductor:

• It is a scalar quantity.• Unit of capacitance is farad in SI unis and its

dimensional formula is M–1L–2I2T4

• 1 Farad : 1 Farad is the capacitance of a conductor forwhich 1 coulomb charge increases potential by 1 volt.

1 Farad = 1Coulomb

1 Volt

1 F = 10–6 F, 1nF = 10–9 F or 1 pF = 10–12 F• Capacitance of an isolated conductor depends

on following factors :

(a) Shape and size of the conductor:

On increasing the size, capacitance increase.

(b) On surrounding medium :

With increase in dielectric constant K,capacitance increases.

(c) Presence of other conductors:

When a neutral conductor is placed near acharged conductor capacitance of conductorsincreases.

• Capacitance of a conductor does not depend on

(a) Charge on the conductor

(b) Potential of the conductor

(c) Potential energy of the conductor.

1.4 Capacitance of an isolated SpheicalConductor.

EXAMPLE 01

Find out the capacitance of an isolated sphericalconductor of radius R.

Sol. Let there is charge Q on sphere.

Potential KQVR

Hence by formula : Q = CV

CKQQR

C = 40R

1.2 Theory and Exercise Book

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(i) If the medium around the conductor is vacuum orair.:

Cvacuum = 40RR = Radius of spherical conductor. (may be solid orhollow)

(ii) If the medium around the conductor is a dielectricof constant K from surface of sphere to infinitythen

Cmedium = 40KR

(iii)medium

air / vaccum

CC

K = dielectric constant.

2. CAPACITOR :A capacitor or condenser consists of two conductorsseparated by an insulator or dielectric.

(i) When uncharged conductor is brought near to acharged conductor, the charge on conductorsremains same but its potential dcreases resulting inthe increase of capacitance.

(ii) In capacitor two conductors have equal but oppositecharges.

(iii) The conductors are called the plates of the capacitor.The name of the capacitor depends on the shape ofthe capacitor.

(iv) Formulae related with capacitors:(a) Q = CV

A B

A B B A

Q Q QCV V – V V – V

Q = Charge of positive plate of capacitor.V = Potential difference between positive and

negative plates of capacitorC = Capacitance of capacitor.

(v) The capacitor is represented as following :

,

(vi) Based on shape and arrangement of capacitor platesthere are various types of capacitors:(a) Parallel plate capacitor(b) Spherical capacitor.(c) Cylindrical capacitor(v) Capacitance of a capacitor depends on

(a) Area of plates.(b) Distance between the plates.(c) Dielectric medium between the plates.

2.1 Parallel Plate CapacitorTwo metallic parallel plates of any shape but of samesize and separated by small distance constituteparallel plate capacitor. Suppose the area of eachplate is A and the separation between the two platesis d. Also assume that the space between the platescontains vacuum.We put a charge q on one plate and a charge –q onthe other. This can be done either by connectingone plate with the positive terminal and the otherwith negative plate of a battery (as shown in figurea) or by connecting one plate to the earth and bygiving a charge +q to the other plate only. This chargewill induce a charge – q on the earthed plate. Thecharges will appear on the facing surfaces. Thecharges density on each of these surfaces has amagnitude = q/A.

++++++

––––––

+q –q

(a)

or

++++++

––––––

+q –q

(b)

If the plates are large as compard to the separationbetween them, then the electric field between theplates (at point B) is uniform and perpendicular tothe plates except for a small region near the edge.The magnitude of this uniform field E may becalculated by using the fact that both positive andnegative plates produce the electric field in the samedirection (from positive plate towards negative plate)of magnitude /20 and therefore, the net electricfield between the plates will be,

0 0 0

E2 2

Outside the plates (at point A and C) the field dueto positive sheet of charge and negative sheet ofcharge are in opposite directions. Therefore, netfield at these points is zero.The potential difference between the plates is,

0 0

qdV E.d dA

The capacitance of the parallel plate capacitor is,

0q ACV d

or 0AC

d

Capacitance 1.3


2.2 Cylindrical CapacitorCylindrical capacitor consists of two co-axialcylinders of radii a and b and length l. If a charge qis given to the inner cylinder, induced change –qwill reach the inner surface of the outer cylinder.By symmetry, the electric field in region betweenthe cylinders is radially outwards.By Gauss’s theorem, the electric field at a distancer from the axis of the cylinder is given by

0

1 qE2 r

l

The potential difference between the cylinders isgiven by

a a

0 0b b

1 dr q aV E dr q In2 r 2 b

?

l l

– –– –– –– –– –– –– –– –– –– –

+ +

+ +

+ +

+ +

+ +

+ +

+ +

+ +

ab

B

l

A

–

or, 0q 2CaV Inb

l

2.3 Spherical CapacitorA spherical capacitor consists of two concentricspheres of radii a and b as shown. The inner sphereis positively charged to potential V and outer sphereis at zero potential.The inner surface of the outer sphere has an equalnegative charge.The potential difference between the spheres is

0 0

Q QV –4 a 4 b

+–

–

–

–

–

– –

–

–

–

+++

+++++

+

abHence, capacitance

0Q 4 abCV (b a)

EXAMPLE 02Find capacitance of the given system.

2Q Q

d

A=Area

Sol. Arranging charges

0 0 0 0

QE2 2 2A

Now, V = Ed = 0

Qd2A

2Q3

2Q3

2Q

2Q

0Q / 2 ACV d

Electrostatics-1 1.1


Section A - Coloumb's law, ElectrostaticEquillibrium + Properties ofCharge

1. INTRODUCTION :(a) Introduction :

Electromagnetism is a science of the combinatin ofelectrical and magnetic phenomenon.Electromagnetism can be divided into 2 parts :(1) Electrostatics : It deals with the study ofcharges at rest.(2) Electrodynamics : It deals with the study ofcharges in motion (discusses magnetic phenomenon).In this chapter we will be dealing with charges atrest i.e. electrostatics.

(b) Structure of Atom :An atom consists of two parts (i) nucleus (ii) extranuclear part. Nucleus consists of neutrons andprotons and extra nuclear part has electronsrevolving around nucleus.In a neutral atom.number of electrons = number of protons.charge of electrons = charge of protons= 1.602 × 10–19 coulomb.Normally positive charges are positron, proton andpositive ions. In nature practically free existingpositive charge are positive ions and negativecharges are electrons.

(c) Electric ChargeCharge of a material body or particle is the property(acquired or natural) due to which it produces andexperiences electrical and magnetic effects. Someof naturally charged particles are electron, proton,-particle etc.

(d) Types of Charge(i) Positive charge : It is the deficiency ofelectrons compared to protons.(ii) Negative charge : It is the excess of electronscompared to protons.

(e) Units of ChargeCharge is a derived physical quantity. Charge ismeasured in coulomb in S.I. unit. In practice we usemC (10–3C), C (10 –6 C), nC (10–9C) etc.C.G.S unit of charge = electrostatic unit = esu.1 coulomb = 3 × 109 esu of chargeDimensional formula of charge = [M°L°T1I1]

(f) Properties of Charge(I) Charge is a scalar quantity : It adds algebrically

and represents excess, or deficiency of electrons.(II) Charge is transferable : Charging a body implies

transfer of charge (electrons) from one body toanother. Positively charged body means loss ofelectrons, i.e. deficiency of electrons. Negativelycharged body means excess of electrons. This alsoshows that mass of a negatively chargedbody > mass of a positively charged identical body.

(III) Charge is conserved : In an isolated system, totalcharge (sum of positive and negative) remainsconstant whatever change takes place in that system.

(IV) Charge is quantized : Charge on any body alwaysexists in integral multiples of a fundamental unit ofelectric charge. This unit is equal to the magnitudeof charge on electron (1e = 1.6 × 10–19 coulomb). Socharge on anybody Q = ± ne, where n is an integerand e is the charge of the electron. Millikan's oil dropexperiment proved the quantization of charge oratomicity of charge.

1CHAPTER

Electrostatics-I



• Recently, the existence of particles of charge ± 2 e3

and ± 2 e3 has been postulated. These particles are

called quarks but still this is not considered as thequantum of charge because these are unstable(They have very short span of life.)(v) Like point charges repel each other while unlikepoint charges attract each other.(vi) Charge is always associated with mass, i.e.,charge can not exist without mass though mass canexist without charge. The particle such as photonor neutrino which have no (rest) mass can neverhave a charge.(vii) Charge is relativistically invariant : This meansthat charge is independent of frame of reference,i.e., charge on a body does not change whateverbe its speed. This property is worth mentioning asin contrast to charge, the mass of a body dependson its speed and increases with increase in speed.(viii) A charge at rest produces only electric fieldaround itself; a charge having uniform motionproduces electric as well as magnetic field arounditself while a charge having accelerated motionemits electromagnetic radiation.

(g) Conductors and Insulators :Any object can be broadly classified in either of thefollowing two categories :(i) Conductors(ii) Insulators

(i) Conductors : These are the materials that allowflow of charge through them. This category generallycomprises of metals but may sometimes contain non-metals too. (ex. Carbon in form of graphite.)

(ii) Insulators : These are the materials which do notallow movement of charge through them.

(h) Charging of Bodies :An object can be charged by addition or removedof electrons from it. In general an object can eitherbe a conductor or insulator. Thus we are going todiscuss the charging of a conductor and chargingof an insultor in brife.

(i) Charging of Conductors :Conductors can be charged by(a) Rubbing or frictional electricity(b) Conduction & Induction (will be studied in latersections)(c) Thermionic emission (will be study the topic"heat")(d) Photo electric emission (will be studied underthe topic modern physics)

(ii) Charging of Insulators :Since charge cannot flow through insulators, neitherconduction nor induction can be used to charge,insultators, so in order to charge an insulator frictionis used. Whenever an insulator is rubbed against abody exchange of electrons takes place between thetwo. This results in apperance of equal and oppositecharges on the insulator and the other body. Thusthe insulator is charged. For example rubbing ofplastic with fur, silk with glass causes charging ofthese things.

To charge the bodies through friction one of themhas to be an insultator.

2. COULOMB'S LAW :Coulomb, through his experiments found out thatthe two charges 'q1' and 'q2' kept at distance 'r' in amedium as shown in figure-1 exert a force 'F' oneach other. The value of force F is given by

Kq qF

r 1 2

2

F F

+q1 +q2

r

This law gives the net force experienced by q1 andq2 taking in account the medium surrounding them.Where F gives the magnitude of electrostatic force q1and q2 are the magnitudes of the two interacting charges.K is electrostatic constant which depends upon themedium surrounding the two charges.This force F acts along the line joining the twocharges and is repulsive if q1 and q2 are of samesign and is attractive if they are of opposite sign.Let us take some examples on application ofcoulomb's Law.



EXAMPLE 1Charge 5.0 × 10–7 C, –2.5 × 10–7C and 1.0 × 10–7

C are fixed at the corners A, B and C of an equilateraltriangle of side 5.0 cm. Find the electric force onthe charge at C due to the rest two.

9 –7 –7

AC 29 10 5 10 1 10F

(.05)

= 0.18 Nt FAC

C q = 1.0 × 10 C 3 –7

aa = 5 cm

aA Bq = 5 × 10 C 1 –7 q = –2.5 × 10 C 2 –7

FBC

FBC = 9 –7 –7

29 10 –2.5 10 1 10

(.05)

= – .09 NtNet force on C is

Net AC BCF F F

2 2Net AC BC AC BCF (F ) (F ) 2(F )(F )cos

[ = 120°]= 0.15588 Nt

EXAMPLE 2If charge q1 is fixed and q2 is free to move then findout the velocity of q2 when it reaches distance r2after it is release from a distance of r1 from q1 asshown in figure (Assume friction is absent).

r1

q2q1

r2

fixed

Find v of q2 when it reaches distance r2 after it isreleased from rest.

Sol. 1 22

kq qamx

2

1

rv1 2

20 r

kq q dxvdvm x

2

1 2

1 2

kq qv 1 12 m r r

1 2

1 2

2kq q 1 1vm r r

EXAMPLE 3

Ten charged particles are kept fixed on the X axis atpoint x = 10 mm, 20 mm, 30 mm, .......... 100 mm.The first particle has a charge 10–8 C, the second8 × 10–8 C, the third 27 × 10–8C and so on. The tenthparticle has a charge 1000 × 10–8C. Find themagnitude of electric force acting on a 1 C chargeplaced at the origin.

1C q1 q2 q3 q4

Sol. Force of 1C charge = 1

–3 2Kq 1

(10 10 )

+ 2

–3 2Kq 1

(20 10 )

+ 3

–3 2Kq 1 .......

(30 10 )

= –8 3 3 3 3

–4 2 2 2 2K 10 1 2 3 10......

10 1 2 3 10

= 9 × 109 × 10–4 × 55 = 4.95 × 107 Nt[This example explains that the concept ofsuperposition holds in the case of electric forces.Net electric force at the origin is equal to sum ofthe individual electric forces on the 1 C charge]



2CHAPTER

Electrostatics-II

Section A - Questions Based on electric fluxfor various configuration,Question on solid angle concept,Gauss law Based Questions

1. ELECTRIC FLUX :Any group of electric lines of forces passing through agiven surface, we call electric flux and it is denoted by .

• Area as a Vector :Till now we have considered area of a surface as ascalar quantity but for further analysis we treat area ofa surface as a vector quantity whose direction is alongthe normal to the surface. The area vector S

? of asurface which has surface area S can be written as

ˆS Sn?

Where n is the unit vector in the direction alongnormal to the surface.

S ?n

If a surface is three dimensional we consider a smallelemental area dS on this surface and direction ofthis elemental area vector is along the local normalof the surface at the point where elemental area ischosen as shown. Thus

ˆdS dSa

Here a is the unit vector in the direction along thenormal at elemental area dS.

dS?a

• Electric Field Strength in Terms of Electric Flux :Earlier we’ve defined that the density of electric linesgives the magnitude of electric field strength.Mathematically the numerical value of electric fieldstrength at a point in the region of electric field canbe give as the electric flux passing through a unitnormal area at that point.

Flux E.dA

?

, If E?

is constant, E.A ??

In a uniform electric field shown in figure. If bethe flux passing through an area S which is normalto the electric field lines, the value of electric fieldstrength at this surface can be given as

ES

E

area = Sor flux through the surface can be given as

ES

If in an electric field, surface is not normal as shownin figure. Here the are ABCD is inclined at an angle from the normal to electric field. Here we resolvethe area ABCD in two perpendicular componentsas shown in figure. One is S cos , which is areaABCD normal to electric field direction and otheris S sin , which is area CDCDalong the directionof electric field.

area = SC D D'

C'

A B



Here the total flux passing through the given areaABCD is same which is passing through its normalcomponent S cos , thus here the flux through thearea can be given as

EScos [S cos = area of ABCD]If we consider the direction of area vector normalto the area surface, as shown in figure, would bethe angle between S

? and ?E . Thus flux through the

surface ABCD can be given as

E.S ??

AB

E

S

DC

(a) Electric Flux in Non-uniform ElectricFieldIn non-uniform electric field, we can calculate electricflux through a given surface by integrating the aboveexpression for elemental surface area of the givensurface. For this consider the situation shown in figure.If we wish to calculate electric flux through the surfaceM shown in figure. For this we consider an elementalarea dS on the surface M as shown. At this position ifelectric field is E then the electric flux through thiselemental area dS can be given as

d EdScos

E dS

M

Total flux through the surface M can be given as

M

d EdScos

(b) Electric Flux Through a circular Disc :

?

Figure shows a point charge q placed at a distance? from a disc of radius R. Here we wish to find theelectric flux through the disc surface due to the pointcharge q. We know a point charge q originateselectric flux in radially outward direction. The fluxof q which is originated in cone shown in figurepasses through the disc surface.

R

x

E

dS

?q

To calculate this flux, we consider an elemental ringon disc surface of radius x and width dx as shown.Area of this ring (strip) is

dS = 2x dxThe electric field due to q at this elemental ring isgiven as

2 2KqE

(x )

?

If d is the flux passing through this elemental ring,we have

cosEdSd

2 2 2 2

Kq 2 x dx(x ) x

?? ?

= 2 3/ 22 kq xdx

x )2

l(l



Total flux through the disc surface can be given byintegrating this expression over the whole area ofdisc thus total flux can be given as

R

2 2 3/ 20O

q x dxd2 ( x )

?

?

R

2 2 3/ 20 O

q x dx2 ( x )

?

?

R

2 20 O

q 12 x

?

?

2 20

q 1 12 R

?? ?

The above result can be obtained in a much simplerway by using the concept of solid angle and Gauss’sLaw, shortly we’ll discuss it.

(c) Electric Flux Through the LateralSurface of a Cylinder due to a PointCharge :Figure. shows a cylindrical surface of length L andradius R. On its axis at its centre a point charge q isplaced. Here we wish to find the flux coming outfrom the lateral surface of this cylinder due to thepoint charge q. For this we consider an elementalstrip of width dx on the surface of cylinder as shown.The area of this strip is

dS = 2R . dxThe electric field due to the point charge on thestrip can be given as

2 2KqE

(x R )

If d is the electric flux through the strip, we can write cosdSEd

dxx

dS

E

RC

q

L

2 2 2 2

Kq R2 Rdx(x R ) x R

22 2 3/ 2

dx2 KqR(x R )

Total flux through the lateral surface of cylinder canbe given by integrating the above result for thecomplete lateral surface, which can be given as

L / 22

2 2 3/ 20 L / 2

qR dxd2 (x R )

or 2 20

q .4R

?

?

The solution of above integration is left for studentsas exercise. This situation can also be easily handledby using the concepts of Gauss’s Law, we’ll discussin next section.

(d) Electric flux Produced by a Point ChargeThe figure. shows a point charge placed at the centreof a spherical surface of radius R from which electriclines are originated and coming out of the surface ofsphere. For clarity and convenience only lower halfof sphere is drawn in the picture. As the charge q isinside the sphere, whatever flux it originates will comeout from the spherical surface. To find the total flux,we consider an elemental area dS on surface. Theelectric field on the points on surface of sphere canbe given as

2KqER

E

dS



The electric flux coming out from the surface dS is

d E.dS EdS

?

[As = 0 shown in figure]

Thus 2Kqd dSR

Total flux coming out from the spherical surface is

2

Kqd dSR

At every point of spherical surface, magnitude ofelectric field remains same hence we have

2Kq dSR

or 22

Kq 4 RR

[As 2dS 4 R ]

0

q

Thus total flux, the charge q originates is 0

q . Similarly

a charge –q absorbs 0

q electric lines (flux) into it.

Figure shows a charge q enclosed in a closed surfaceS of random shape. Here we can say that the totalelectric flux emerging out from the surface S is thecomplete flux which charge q is originating, henceflux emerging from surface is

S0

q

qS

The above result is independent of the shape ofsurface it only depends on the amount of chargeenclosed by the surface.

(e) Flux Calculation in the Region ofVarying Electric field :In a region electric field depends on x dierection as

E = E0x2

In the cube of edge a shown in figure from frontface electricflux goes in which can be given as

O

y

2a xa

a

a

in out

E = E x0 2

in = E0 (2a)2 . a2 = 4E0a4

From the other surface flux coming out can be given asout = E0(3a)2 . a2 = 9 E0a4

Here out > in for the cubical surface hence netflux =out – in = 5E0a4

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