electrostatics fiitjee

ELECTROSTATICSRSP-PH--1

INTRODUCTION

This study material is intended exclusively for the usage of RADIANCE students preparing for IIT-JEE. Its design is based on our experience with students over the past few years. This materialcovers extensively the fundamental principles and concepts involved, solved problems which high-light the application of these concepts and assignments for practice by the students.Though this study material is written on the basis of the syllabi prescribed by IIT’s for the JointEntrance Examination, it will also prove useful to students who are preparing for other Engineeringexaminations like AIEEE & BITSAT since the content of Physics, Chemistry and Mathematicsremains almost the same for these examinations also.

We would also emphasize that it is the responsibility of the student to inform his/her parents/guardian about his/her performance in the classes and tests.

Wishing you a very successful year at RADIANCE

A Word of Advice

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I think I lost an electron. Are you sure ? Yeah, I’m positive...!!!


oulomb’s law, Electric field, Electric potential , Electrical potential energy of a system of point

charges and of electrical dipoles in a uniform electrostatic field, Electric field lines; Flux of electricfield; Gauss’s law and its application in simple cases, such as, to find field due to infinitely longstraight wire, uniformly charged infinite plane sheet and uniformly charged thin spherical shell.Capacitance; parallel plate capacitor with and without dielectrics; Capacitors in series and paral-lel; Energy stored in a capacitor.

Electric Charge

Electric charge, like mass, is one of the fundamental at 7 tributes of the particle of which thematter is made. Charge is the physical property of certain fundamental particles (like elec-tron, proton) by virtue of which they interact with the other similar fundamental paticles. Todistinguish the nature of interaction, charges are divided into two parts (i) positive (ii) nega-tive. Like charges repel and unlike charges attract. SI unit of charge is coulomb and CGSunit is esu.1C = 3 x 109 esu.Magnitude of the smallest known charges is e = 1.6 x 10-19 C (charge of one electron orproton).

Charging of a body

Basically charging can be done by two methods;1. Conduction 2. InductionOrdinarily, matter contains equal number of protons and electrons. A body can be chargedby the transfer of electrons or redistribution of electrons.The process of charging from an already charged body can happen either by conduction orinduction. Conduction from a charged body, involves transfer of like charges. A positivelycharged body can create more bodies, which are positively charged, but the sum of the totalcharge on all positively charged bodies will be the same as the earlier sum.Induction is a process by which a charged body accomplishes the creation of other chargedbodies, without touching them or losing its own charge.

Properties of Electric ChargeQuantization of chargeCharge exists in discrete packets rather than in continuous amount. i.e. charge on any bodyis the integral multiple of the charge of an electron Q ne , where n = 0, 1, 2,...............

Conservation of chargeCharge is conserved, i.e. total charge on an isolated system is constant. By isolated systemwe mean here a system through the boundary of which no charge is allowed to escape orenter.This does not require that the amount of positive and negative charges are separately con-served only their algebraic sum is conserved.

Charges on a conductorStatic charges reside on te surface of the conductor.

Distribution of chargesThe concentration of the charges is more on a surface with greater curvature.

C


Exercise 1: Two identical metallic spheres of exactly equal masses are taken, one is given a positivecharge q and the other an equal negative charge. These masses after charging are different.Comment on the statement.

Coulomb’s Law

Two point electric charges q1 and q

2 at rest, separated by a distance r exert a force on each

other whose magnitude is given by

F kq q

r 1 2

2 where k is a proportionality constant.

If between the two charges there is fre space then

k = 1

49 10

0

9 2 2

x Nm C , where 0 is the absolute electric permittivity of the free space.

Exercise 2: (i) A negatively charged particle is placed exactly midway between two fixed par-ticles having equal positive charge. What will happen to the charge?(a) If it is displaced at right angle to the line joining the positive charges?(b) Does the Coulomb force that one charge exert on another change if the other chargesbrought nearby?

IIIustration 1:Two particles A and B having charges 8 x 10-6 C and -2 x 10-6 C respectively are heldfixed with a separation of 20 cm. Where should a third charged particle be placed so that itdoes not experience a net electric force ?

Solution:As the net electric force on C should be equal to zero, the force due to A and B must beoppostie in direction. Hence, the particle should be placed on the line AB. As A and B havecharges of opposite signs, C cannot be between A and BAlso A has larger magnitude of charge than B. Hence, C should be placed closer to B thanA. The situation is shown in figure. Suppose BC = x and the charge on C is Q

F

x Q

xiCA

1

4

8 0 10

0 20

6

2

.

.

e jb g and

F

x Q

xiCB

1

4

2 0 10

0

6

2

.e j

bg F F FC CA CB

= 1

4

8 0 10

0 2

2 0 10

0

6

2

6

2

.

.

.x Q

x

x Q

x

L

NMM

O

QPP

e jb g

e jbg i

But FC 0

Hence 1

4

8 0 10

0 2

2 0 10

0

6

2

6

2

.

.

.x Q

x

x Q

x

L

NMM

O

QPP

e jb g

e jbg =0

which gives x = 0.2 m

Principle of superposition

This principle tells us that if charge Q is placed in the vicinity of several charges q1, q

2......q

n,

the the force on Q can be found out by calculating separately the forces F1, F

2......, F

n, exerted

by q1, q

2....q

n respectively on Q and then adding these forces vectorially. Their resultant F is

the total force on Q due to all of charges.


IIIustration 2: It is required to hold equal charges, q in equilibrium at the corners of a square having sidea. What charge when placed at the centre of the square will do this ?

Solution: Let the charge be QAs ABCD is a square of side a

ra a

2

2 2

F

kq

aiBA

2

2,

F

kq

ajBC

2

2,

F

kq

ai jBD

2

2

0 0

245 45

( )cos sin e j

F

kqQ

ai jBQ

( )cos sin

245 45

2

0 0e j

Here i and j have usual meaning. Net force on the charge at B is

F

kq

a

kq

a

kqQ

aiR

FHG

IKJ

2

2

2

2

0

2

0

245

245

( )cos

( )cos

- F

kq

a

kq

a

kqQ

ajR

FHG

IKJ

2

2

2

2

0

2

0

245

245

( )sin

( )sin

For charge , q to be in equilibrium at B, the net force on it must be zero.

F F F FBX BA BD BQ cos cos45 45 00 0

L

NMMM

O

QPPPk

q

a

q

a

Qq

a

2

2

2

2 22

1

2 2

1

20

e j e j. .

Qq

41 2 2e j

Similarly, FBy

= 0, if Qq

4

1 2 2e j.

Electric fieldElectric field due to a point charge is the space surounding it,within which electric force can be experienced by the anothercharge.

Electric field strength or Electric intensity Edi at a point is t

electric force experienced by a unit positive charge at that

point. Mathematically,

EF

q

0

where q0 is positive test charge.

In vector form, electric field at B due to charge Q at A,

E kq

r rr r

1 23 1 2b g

If electric intensity is same at all points in the region, then the field is said to be uniform.Equispaced parallel lines represent uniform electric field, Arrow on the lines gives the di-rection of the electric field.The electric field at a point due to several charges distributed in space is the vector sum of

the fields due to individual charges at the point, E E E En 1 2 .......

The electric field of a continuous charge distribution at some point, E

dq

rr z1

4 02

A

B

X

Y

r1

r2

r r1 2

)A

B

CD

q

450

q

q

q

a

a

r

Q

x

y

FBC

FBD

FBQ

FBA

iej

jej


where dq is the charge on one element of the charge distribution and r is the distance from theelement to the point under consideration and r is the unit vector directed from the position ofelemental charge towards the point where electric field is to be found out.

Exercise 3:(i) Does an electric charge experience a force due to the field that the charge itselfproduces?(ii) Two point charges q and -q are placed at a distance d apart. What are the points at whichresultant electric field is parallel to line joining the two charges ?(iii) Three euqal and similar charges q are placed on each corner of a square of side a. Whatis the electric field at the centre of the square ?

IIIustration 3:What is the electric field at any point on the axis of a uniformly charged rod oflength ‘L’ and linear charge density ' ' ? The point is separated from the nearer end by a.

Solution: Consider an element, dx at a distance, x from the point, P, where we seek to find theelectric field. The elementa charge, dq = dx

Then, dE = kdx

x

2 or E = k

xdx k

xk

a L aa

a L

a

a L

1 1 1 12

z LNMOQP

LNM

OQP

Thus, E =

4

1 1

0 a L a

LNM

OQP [ k =

1

4 0 ]

IIIustration 4:A ring shaped conductor with radius R carries a total charge q uniformly distributedaround it, find the electric field at a point P that lies on the axis of the ring at a distance xfrom its centre.

Solution:Consider a different element of the ring of length

ds. Charge on this element is dq = q

Rds

2

FHGIKJ .

This element sets up a differential electric field dE at

point P.

The resultant field E at P is found by integrating the

effects of all the elements that make up the ring. Fromsymmetry this resultant field must lie along the right

axis. Thus, only the component of dE parallel to this

axis contributes to the final result.

E = d

Ez z E = E

d cos

dE = 1

4

dq

r=

1

4

qds

2 R.

1

R; cos =

x

R02

02 2

FHGIKJ

FHG

IKJ

x x2 2 1 2e j /

To find the total x-component Ex of the field at P, we integrate this expression over all

segment of the ring.

Ex =

dqx

xds

Ecos

1

4 2 R R0 2

z z

2 3 2e j /

The integral is simply the circumference of the ring = 2 R

-----------) x

(R2 + x2)

dE cos

y

O x

R

............

............

.......

dq1/2

(

dE sin

P

dE

a p

----------------------dx

x

aP


E = 1

4 R0 2

qx

x 2 3 2e j /

As q is positive charge, field is directed away from the centre of the ring, along its axis.

Lines of ForceIt has been found quite convenient to visualize the electric field pattern in terms of lines offorce. The electric field pattern vector at a point is related to imaginary lines of force is twoways. The lin of force in an electric field is a curve such that the tangent at any point on itgives the direction of the resultant electric field strength at that point.

(i) Tangent to the line of force at a point gives the direction of E .

(ii) These lins of force are so drawn that their numbr pr unit cross-sectionl area in a region isproportional to intensity of electric field.(iii) Electric line of force can never be closed loops.(iv) Lines of force are imaginary.(v) They emerge from a positive charge and terminate on a negative charge.(vi)Lines of force do not intersect.

Note: When a conductor has a net charge that is at rest, the charge resid entirely on the conductor’ssurface and the electric field is zero, everywhere within the material of the conductor.

Exercise 4:Electric lines of force never cross. Why ?

Gauss’s LawThe net “flow” of electric field through a closed surface depends on the net amount ofelectric charge containd within the surface. This ‘flow’ is described interms of the electricflux through a surface, which is the product of th esurface area and the component of elec-tric field perpendicular to the surface. Gauss’s law sates that the total electric flux througha closed surface is proportional to the total electric charge enclosed with in the surface .This law is useful in calculating fields caused by charge distributions that have varioussymmetry properties.

Mathematically E ds

q.z

0.

z means integral done over a closed surface.

Gauss’s law can be used to evaluate Electric field if te charge distribution is so symmetricthat by proper choice of a Gausian surface we can easily evaluate the above integral..

Exercise 5: Explain whether Gauss’s law is useful in calculating the electric field due to three equalcharges located at the corners of an equilateral triangle ?

IIIustration 5: Figure shows a section of an infinite rod of chargehaving linear charge density which is constant for all pointson the line. Find electric field E at a distance r from the line.

Solution : From symmetry, E due to a uniform linear

charge can only be redially directed, As aGaussian surfae, we can choose a circular cyl-inder of radius r and length I, closed at each

E

S

)

ds

+ + + +++ +++ ++ + +++++

+ + + +++ +++ ++ + +++++

><

r

C

E

c1

c2

l


end by plane caps normal to the axis.

0 0

21

E ds q E ds E ds E ds qin

CCC

in. ; . . .z zzz L

NMM

O

QPP

Cylindrical plane surface, 0 2 90E rI ds ob g E. = I.cos

E =

I =

0 02 2rI r

The direction of E is radially outward for a line of positive charge.

IIIustration 6: Figure shows a spherical symmetric distribution of charge of radius R. Find elec-tric field for points A and B which are lying outside and Inside the charge distribution re-spectively.

Solution:The spherically symmetric distribution of chargemeans that the charge density at any point dependsonly onth distance of the point from the centre andnot on the direction. Secondly, the object can not be aconductor, or else the excess carge will reside on itssurface.Now, apply Gauss’s law to a spherical Gaussian surface of radius r

r > R for point A E 4 r = q 0 02b g e j, .

E ds qenz

E=1

4 02

q

r where q is the total charge

For point B (r < R), 0 024

E ds E r q.z e j

q’ =

q r

R

4

34

3

3

3

= q

r

R

FHGIKJ

3

E = 1

4

1

40

3

20

3

qr

R

r

qr

R

FHGIKJ

Electric PotentialPotential at a point in an electric field, is the amount of work done by an external againstelectric forces in moving a unit positive charge with constant speed from infinitly to thatpoint.As work done by the external agent = - work done by electric force

Hence the required potential V = -

Edq dr

r

q

r

r

z z

4

1

402

0

The potential at a point due to a group of n point charges q1, q

2, q

3..........q

n

V = V1+V

2+...........+V

n(Scalar sum)

1

4 01

q

ri

ii

n

R

B

r

A


The electric potential due to a continuous charge distribution, V = 1

4 0

dq

rzPotential due to a uniformly charged disc

The elemental ring’s area = 2xdx

dV = 1

4

2

02 2

xdx

r x

b g

V = dVxdx

r xR r r

R

z z

LNM

OQP

2 02 2

2 2

0

Potential of a uniformly charged spherical shellIf the charge on the shell = q(i) for r > R

E = kq

r2

zVkq

rdr

kq

r

r

2

(ii) for r = R , V = kq

R

(iii)for r < R , E =kq

r2 (R < r < ) , E = 0 (r < R)

LNMM

OQPP

z zV Edr Edr

R

R

r

=

zkq

rdr

R

2 - 0

R

r

drkq

Rz

Potential of a uniformly charged spherical volumeIf the total charge = Q

(i) for r > R, Volume charge density

Q

R4

33

E (r > R)= kQ

r2 ; V = kQ

rdr

kQ

r

r

2

z

(ii) for r =R, VkQ

r E =

1

4 02

Q

r

(iii) r < R, E (r < R) =

. r

3 0=

Q r

R

.

.4

333

0

E (r < R ) =

Q r

R

.

4 30

= kQ.r

R3

V = - Edr Edr

R r

z zLNMM

OQPP

R

= kQ

rdr

kQr

Rdr =

kQ

2R2

R

3

r

z z LNMM

OQPP

32

2

r

RR

For conducting solid sphere :

(i) (r < R), E =0 , V= kQ

R

(ii) (r = R), E= kQ

R2 , V= kQ

R

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

dx

xR

r

qR

r

p

qR

r

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

R

r

p

Q


(iii) (r > R), E = kQ

R2 , V =kQ

r

Exercise 6: A closed metallic box is charged upto potential V0 what will be the potential at the

centre of the box.

Equipotential surfacesThe locus of points of equal potential is calledan equipotential surface. The electric field isperpendicular to the equipotential at each pointof the surface.

Exercise 7: A solid metallic sphere is placed in auniform electric field. Which of these A, B, Dand D shows the correct path

Relation between field (E) and potential (V)The negative rate of change of potential with distance along a given direction is equal to thecomponent of the field along that direction.

i.e. EdV

drr

or we can say the electric field is along the direction in which the potential decreases at themaximum rate.

Exercise8: (i) If you know E at a given point , can you calculate V at that point ? If not, what

further information do you need.

(ii)If E equals zero at a point , must V equal zero for that point?

IIIustration 7:The electric field in a region is given by E A x i / 3e j .What is the potential in the

region ?

Solution: The electric potential in the region

V(x, y, z) =

z z E dr

A

xdx

A

x

x y z x y z

.

, , , ,b g b g

3 22

IIIustration8:What is the electric potential on the axis of a uniformly charged ring of radius Rcontaining charge q, at a point x away from the centre of the loop ?

Solution : As electric Potential is a scalar quantity . Hence electric potential at P,

Vdq

r rdq z z1

4

1

40 0

Hence V = 1

4 0

q

R x2 2 1 2e j /

IIIustration 9: Given figure shows the lines of con-stant potential in a region in which an electricfield is present. the value of potentials arewritten. At which of the points A, B and C iste magnitude of the electric field is the great-

Equipotential surface

Electric field

>

>

>

>

>

>

HA

B

C

D>

)

dr

P

P’

E

50V40 V

30V 20V 10V

A

C B


est ?

Solution:E

dv

drn

The potential difference betwen any two connective line dv = v1-v

2 = 10 v constant and

hence e will be maxium where the distance dr between the lines minimum.i.e. at B wherethe lines are closest.

IIIustration 10:Some equip-potential surface are shown in fig (1) and (2) what can you say aboutthe magnitude and direction of the electric field

Solution : Electric field is perpendicular to the equipotential surface and in the direction of decreas-ing potential

In Figure 1:The electric field will be at angle making an angle 1200with x-axis.

Magnitude of electric field Ecos 1200 = - 20 10

20 10 10 2

b gb gx

-E .1/2=-10/0.10= E= 200 v/mFigure 2:

Direction of electric field will be redially outward, similar to a point charge kept at the

centre, i.e. V=kq

r when V = 60V =

kq

01.bg kq 6

Hence potential at any distance from the center, V(r) = 6

r

Hence E= -dV

dr=

62r

V mFHGIKJ /

Electrostatic potential energy The electric potential energy of a system of point charges is the amount of work done inbringing the charges from infinity in order to form the system . Two point charges q

1 and q

2

are separatd at a distance r12

, Electric potential energy of the system q1 and q

2

Uq q

r

1

4 0

1 2

12

For three particle system q1, q

2 and q

3

Uq q

r

q q

r

q q

r

FHG

IKJ

1

4 0

1 2

12

1 3

13

2 3

23

We can define electric potential at any point P in an electric field as , VP = U

P/q; where U

P, is

the change in electric potential energy in bringing the test charge q0 from infinity to point p.

IIIustration11.Determine the interaction energy of the point charges of the following setup.

Solution: As you know the interaction energy ofan assembly of charges is given by

20V30V

q2 r

23

1 2

34

a

a

a

a

- q

+q

+q

- q

q1

q1

q2

q3

r12 r

13

r12

)10 20 30 40

300

10V 20V30Vy (cm)

x (cm) 10 cm

20 cm

30 cm

60 V


><

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

+q-q

Equatorial line

1

4 0

q q

r

y j

iji j

n

U= U12

+ U13

+ U14

+ U23

+ U24

+U34

= kq

a

kq

a

kq

a

kq

a

kq

a

kq

a

kq

a

2 2 2 2 2 2 2

2 2

22 2 1

e j e j

Electric DipoleTwo equal and opposite charges separated by a finite distance form electric dipole. It is

characterised by dipole moment vector p .

(1) Charges (+q) and (-q) are called the poles of the dipole.

(2) = the displacement vector from -ve charge to +ve charge .

(3) p = the dipole moment = q

.

(4) The straight line joining the two pols is called axial line.(5) Perpendicular bisector of l is called equatorial line.

Electric field due to a dipole at axial point

Let the charges (-q) and +q are kept at point (-a, 0) & (a, o)respectively in xy plane. Theelectric field at point P (x,0) will be then;

E E E

kq

x ai

kq

x aiaxial q qb g b g b g

2 2

, where i is the unit vector along axis.

= kqx a x a

x ai kq

x a

x ai. .

b g b ge je j

b gbge j

2 2

2 2 2 2 2 2

2 2

= 2

2 2 2

kpx

x ai

e j, as x >> a ,

Ekp

x

23

p = 2aq

Electric field on equatorial line

At P,

E

kq

y ai j

2 2e je jcos sin

E

kq

y ai j

2 2e je jcos sin

E E E

kq

y aip

22 2e j

cos

L

NMM

O

QPP

E

kq

y a

a

y ai

k aq

y aip 2

2

2 2 2 2 1 2 2 2 3 2e j e jb g

e j.

/ /

..................................

......................... ................................................... ..........................

<<

>

>(a

E-

E+

Ep P

2(0, y)

y

x

a

y

- +

> < > >

< >

-q +q Ep

E+E-(-a, 0) O

y

x

(a, 0) (x,0)

x


For a << y , E = kP

y3 or E = -kP

y3

Resultant E is oppositely directed to

P .

Electric field at any point To get electric field at a gneral point due to a dipole,we use the earlier results.We find electric field at A in terns of r and (r >> a).The dipole moment can be vectorially resolved as ;

Electric field at A due to ( cos )P component = 2

3

k P

r

( cos )

Electric field at A due to ( sin )P component = k P

r

( sin )3

E E Ekp

rR P Pcos sin cos sin 2 2

32 24

Resultant , Ekp

rR 3

23 1cos and tantansin

cos

E

E

p

p 2

Dipole in an external electric fieldThe net force on an electric dipole in a uniform external electric field is zero.. However thedipole in the presence of an external electric field experience a torque and has tendency toalign itself along the external electric field.Torque on dipole = force x force arm

= qE q E pE sin sin sin b gbgb g

or p E x

The product of the charge q and separation l is the magnitude of a quantity called the electricdipole moment denoted by p.

The direction of p is along the dipole axis from the negativecharge to the positive charge

asshown in the figure.

As E is a conservative field, work done by an external agent in changing the orientation of

the dipole is stored as potential energy in the system of a dipole present in an externalelectric field.

W= Td pE d z z sin =- pE cos

1

2

2 0 , dipole is perpendicular to the field

We assume, 1090 (as the datum for measuring potential energy can be chosen anywhere)

U = pEcos or U= p E.

IIIustration 12:Two tiny spheres, each of mass M, and charges +q and -q respectively, are connectedby a massless rod of length, L. They are placed in a uniform electric field at an angle with

the E 00e j. Calculate the minimum time in which the system aligns itself parallel to the

E .

Solution : ( . )p E (If we assume angular displacement to be anti-clockwise, torque is clock-

wise)

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

------

--

)aa B

+q-q

Ar

---------->

Psin

Pcos

B p

)

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

+ -

-q

+q

qE

qE>

>

<

Ep

g


FHGIKJ

pE

I= - 2

As torque is proportional to ‘ ’ and oppositely directed, there will be an S.H.M. Here P =

q.L and moment of inertia, I M L M L ML ( / ) ( / ) /2 2 22 2 2

time period T = 2I

pE

The minimum time required to align itself is T/4

Potential due to an electric dipole

Vq

PB

q

PAP

LNM

OQP

1

4 0

When r >> a, PB = r-a cosPA = r + a cos

Vqa

r aP

1

4

2

02 2 2

.

cos

cose jAs r >>a, a2cos2 can be neglected in comparison to r2.

Vqa

r

p

rP

1

4

2

402

02

.

cos cos

.

Exercise 9: A electric dipole is placed in a non-uniform electric field.Is there a net force on it?

CapacitanceIf Q is the charge from given to a conductor and V is the potential to which it is raised by thisamount of charge, then it is found Q V of Q = CV, where C is a constant called capaci-

tance of the conductor.Capacitor

A pair of conducors separated by some insulating medium iscalled a capacitor. This medium is called dielectric of thecapacitor. If Q units of the charge is given to one of the con-ductors, and thereby a potential difference V is set up be-tween the conductors, the capacitance is then defined asC= Q/V

For parallel plate capacitor C = 0 r A

d

i.e. the capacitance does inded, depend only on geometrical factors namely, the plate areaand plate separation.

Exercise 9: Why capacitance is generally measured in F and pF ?

IIIustration 13:Find capacitance of a conducting sphere of radius R.Solution: Let charge q is given to sphere. The field outside the sphere at distance r is

E = kQ

r2 dV

drE

>

>

>---

------

----

)P

L E

+

-

p------------------------------------

----

----

----

----

----

----

----

--

------------A B-q +qa a

r

) )

+ Q -Q


dV Edr

v R

0

z z

V kQr

R

LNMOQP

1

VkQ

r since, C = Q/V = R/k

RR

1 44

00

/

Combination of capacitors

Series combinationIn series combination, each capacitors has equal chargefor any value of capacitance, provided that capacitorsare initially uncharged. Equivalent capacitance C isgiven by

1 1 1 1

1 2 3C C C C ...........

Parallel combinationIn parallel combination the potential differences of thecapacitor connectd in parallel are equa for any valueof capacitance. Equivalent capacitance is given by C= C

1+C

2+C

3+..........

Some times it may not be easy to find the equivalentcapacitance of a combination using the equations forseries parallel combinations. For any combination onecan proceed as follows:

Step-1:Connect an imaginary battery between the points across which the equivalent capacitance isto be calculated. Send a positive charge +Q from the positive terminal of the battery and -Qfrom the negative terminal of the battery.

Step-2: Write the charges appearing on each plates using charge conservation principle say, Q1, Q

2

............

Step-3:Assume the potential of the negative terminal of the battery be zero and that of positiveterminal to be V, and write the potential of each of the plates say v

1, v

2.......

Step-4:Write the capacitor equation Q= CV for each capacitor. Eliminate Q

1, Q

2 ......V

1, V

2.....etc to obtain the equivalent capacitance

C = Q/V

IIIustration 14:Four identical metal plates are located in air par-allel to each other and distance d from one another. Thearea of each plate is equal to A. The arrangement is sownin the figure. Find the capacitance of the system betweenpoints P and R.

Solution : If we get the charge on each face of all the plates,then we can easily get the equivalent capacitance. Toget charge on each face we will use Gauss’s theoremand principle of conservation of charge . Let the pointP and R have the potential V and 0 respectively and

VP = V(+)

c1 c

2 c3

q -q -q -qq q

V

P R1 2 3 4

PR

1 2 3 4

+q1

-q1

+q2

-q2 +q

1-q

1

VR=0

V

q

q

q

c1

c2

c3


the plates 1 and 4 are at the potential V1. Charge distri-

bution is shown in the figure.

q1 = C (V-V

1) from (4, 3) ........ (1)

q2 = CV from (3,2) ......... (2)

q1 = CV

1 from (2, 1) ........ (3)

From equation (1) and (2) , q1 =

CV

2.......... (4)

Charge supplied by the battery is

Q = q1 + q

2=

3

2CV using (2) and (4)

CV Q2

3............ (5)

If equivalent capacitance be C’ thenC’V = Q .............. (6)

Dividing (5) by (6) , C

C'

2

3 C C'

3

2=

3

2

oA

d

Aliter :By equivalent circuit method

Plates of parallel plate capacitor given different charges

Two identical plates of parallel plate capacitor aregiven the charges Q

1 and Q

2. Let the charges appear-

ing on the inner surface of Q1 be q, then the charges

appearing on other surfaces are as shown in the fig-ure. If we take a point P. inside the plate 1, thenelectric field at P should be zero. Suppose surfacearea of the each surface is A.

Using the equation E =

2 0

E E E E EP 1 2 3 4 ,E1 ,

E2 ,

E3 and

E4 are the electric field due to surfaces 1, 2, 3 and 4.

E

Q q

Ai

qi

qi

Q qiP

1

0 0 0

2

02 2 2 2

= Q q

A

Q qi1

0

2

02 2

FHG

IKJ

P

Q

1 2 3 42 3

3

32 21

4

C/2

C3C/2

CA

deq

3

20

+++++++++

++++++++++

++++++++

+--

-------

Q1

Q1-q

1 2 Q2

-q Q2+q

3 4

P

A B

+++++++++

++++++++++

++++++++

+--

-------

Q11 2 Q

23 4

P

A B

(Q1-Q

2)/2

(Q1-Q

2)/2 (Q

1+Q

2)/2

(Q1+Q

2)/2

-q


But E 0, which gives q =

Q Q1 2

2

Hence, change distribution on each surfaces are shown in figure.

Note : The charge on inner surface of plate A is t ehalf of the difference of the charge on plate A and

B (Q! and Q2 respectively)i.e. Q Q1 2

2

, and on inner surface of B is opposite of the chargg

appearing on inner surface o f A i.e. - Q Q1 2

2

FHG

IKJ; write the change on outer surfaces of A and

B is half of the sum of change on A and B i.e. Q Q1 2

2

Exercise 10: If one of the plates of a capacitor is halved, will there be any change in the capaci-tance of the capacitor ?

Dielectrics

When a dielectric is introduced between conductors of a capacitor, its capacitance increases.A dielectric is characterised by a constant ‘K’ callled dielectric constant .

Dielectric constantWhen a dielectric is placed in an external electric field, polarization occurs and it develops anelectric field in opposition to the external one. As a consequence total field inside it decreases. IfE be the total field inside the dielectric when it is placed in an external field E

0. then its dielectric

constant k is given as kE

E 0 (k > 1)

If a dielectric completely occupls the space between the conductors of a capacitor. its capacitanceincreases K times.Hence in presence of a dielectric with dielectric constant K, the capacitance of a parallel

plate capacitor = K A

d

0

Energy of a charged conductor or capacitorIf C is the capacity of a conductor or capacitor and V is the potential of the conductor (P.D.in case of a capacitor), then stored electrostatic energy is;

U =Q

CCV QV

22

2

1

2

1

2 . Also , as C = 0A d and V E.d/

U = 1

22CV =

1

20 2 A

dE.db g=

1

20

2FHG

IKJE A d.

u (energy density)= Energy per unit per volume = 1

20

2 E

If dielectric is introduced then, u = 1

20

2 E K

This energy is stored in a capacitor in the electric field between its plates.

Force on a dielectric in a capacitor >

<dx x

FForce on anexternal agent


Consider a differential displacement dx of the di-electric as shown in figure keeping the total forceon it always zero. ThenWhen battery is disconnected.W

Electrostatic +W

F= 0(where W denotes the work done in displacement dx)

WF = -W

Electrostatic

EF= U

LNMOQP F dx

Qd

c.

2

2

1 WQ

c

LNMM

OQPP

2

2

F dxQ

cdc.

2

22

FHGIKJF

Q

c

dc

dx

2

22

This is also true for the force between the plates of the capacitor.

If capacitor has battery connected to it , then as V = Q/C F Vdc

dxFHGIKJ

1

22

Exercise 11: A dielectric slab is inserted in one end of a charged parallel plate capacitor. (Theplates of the capacitor are horizontal and the charging battery having been disconnected)The electric field slab is released. Describe what happens. Neglect friction.

IIIustration 15:A capacitor is formed by two squaremetal plates of edge ‘l’ separated by a distanced. Dielectrics of dielectric constant K

1 and K

2

are filled in the gap as shown in t figure. Findthe capacitance.

Solution :Take a differential element of length dx at adistance x from either end.

dc1 =

K dx

d xC

K A

d1 0 0

LNM

OQPtanb g

dc2 =

K dx

x2 0

tanb gThese two capacitors are connected in series

LNM

OQP

1 1 1

0 1 2dc

d x

K

x

K dx

tan tan tan / d

L

N

MMM

O

Q

PPP1 1 1

0 1 2dc

d xd

K

xd

K dx

. .

tan LNM

OQP

d

z zdc

k K

d

dx

K K K x1 2 0

2

2 1 20

( )

LNMM

OQPPC

AK K

K K dk k

0 1 2

2 12 1b g b gln / [A = l2]

Student should note that the ‘k’ used here is different from k = 1

4 0 elsewhere

><

d

l

K1

K2

<

dK

1

K2

)

x dx

xtan

>

< >


SOLVED PROBLEMS

SUBJECTIVE

Problem 1:Four point charges 8C, -1C and +8C are fixed at the points - 27 2/ ,m - 3 2/ ,m

+ 3 2/ ,m and 27 2/ ,m respectively on the y axis. A particle of mass 6 x 10-4 kg and of

charge + 0.1 C moves along the -x direction. Its speed at x = + is v0 . Find the least value

of v0 for which the particle will cross the origin. Find also the kinetic energy of te particle

at the origin. Assume that space is gravity free. Given 1 / (4 0 )=9x109Nm2/C2.

Solution : E p AA( ') Electricity field at P due to charges

at A and A’ = 2

1

4 0 2 2 3 2

FHGIKJ

q x

y xi

A p

A pe je j/

Similarly

E

q x

y xip BB

B p

B p

( ') /

FHGIKJ

21

4 0 2 2 3 2 e jej

For Ep 0 x mp 2 5.

Hence , v0 should be just enough to enable the particle to reach P.

[Alternately V(x) = 2

4 0 2 2 1 2 2 2 1 2

Q

y x

Q

y x

B

B

A

A

F

HGG

I

KJJe j e j/ / ]

dV x

dx

( ) 0

5

2 x = 0 and x =

x = 5

20V x xbg ,

At x = 5

2, V(x) is positive V x( ) is maximum at x =

5

2

So, if the particle is able to cross x = 5

2 then it would be able to come to x = 0.]

L

NMMM

O

QPPP

1

2

2

402

02 2 2 2

mvq q

y x

q

y x

A

A p

B

B p

Solving we get, v0 = 3m/s

Kinetic energy at origin = loss of PE = (PE)P -(PE)

origin

LNM

OQP 1

2

2

42 5 100

2

0

4mvq q

y

q

yx JA

A

B

B. .

Problem 2: A uniform electric field, E exists between the plates of a capacitor . The plate length is

B

A

O

A’

B’

Y

X

P

x 5 2/ x 5 2/x = 0

v(x)


l and the separation of the plates is d.(a) An electron and a proton start from the negative plate and positive plate respectively andgo to the opposite plates. Which of them wins this race ?(b) An electron and a proton start from the midpoint of te separation of plates at one end of teplates. Which of the two willhave greater deviation when they start with the(i) same initial velocity(ii) same initial kinetic energy,and (iii) same initial momentum ?

Solution : (a) In the chosen reference frame there is no forcealong x axis. The accelerations of the electron andthe proton along y axis are as follows

ae =

eE

ma

eE

mep

p

,

Using S = ut + 1

22at

here S = d,u = 0

tS

ae

e

2and t

S

aP

P

2

or te=

2dm

eEe and t

dm

eEP

p

2

As me < m

p, t

e < t

p

Electron takes less time to cross over than the proton.

(b) When proton moves parallel to the plates, it is deflected to the negative plate.

Time taken by proton to cross over, t =

Vx

During this time, deflection along vertical direction

y = 1

22at or y =

1

2 0

2eE

m V

LNMOQP.

(i) Thus Yp=

1

2

12

2

eE

m VP x and YY

e =

1

2

2

2

eE

m Ve x

As mp > m

e, y

p < y

e of course, electron will be deflected in the opposite direction.

(ii) Also y =

1

22

2

2

2

eE

mVx

LNMOQP

= 1

4

2eE

k

where K =

1

22mvx , or initial kinetic energygy

y yp e

Also, as K = p

m

2

2, where p = momentum,

yeEI

p

m

eEI m

py

eEI m

py

eEI m

pp

pe

e 2

2

2

2

2

2

2

2

42

2 2 2( )

,

Thus for : mp > m

e , y

p>y

e

>d

e

+

-

P

Y

X

E

------------------------------ v

x

p

+

-

yp


Problem 3: Two spherical cavities f radii, a and b , are hollwed out from the interior of a neutralconducting sphere of radius R. At the centre of each cavity , a point charge is placed. Call thesecharges q

a and q

b

(a) Find the surface charges(b) What is the field outside the conductor ?(c) what is the field within each cavity ?(d) what is the force on q

a and q

b

(e) which of these answers would change if a third charge qc were brought near the conduc-

tor ?

Aaq

a

4 2 ,

bbq

b

4 2 ,

Ra bq q

R

4 2

(b) EK q q

rr R

a b

b gb g2,

(c) EKq

rr aa

a 2

,b g, EKq

rr bb

b 2

,b g(d)Force on q

a and q

b = 0 as both qa and qb are electrostatially shielded against each other.

(e) When a third charge is brought, only the field outside the conductor will change

Problem 4:What charges will flow after the shorting of the switch S in the circuit given in thefigure, through section 1 and 2 ?

Solution : When S is open, C1 and C

2 are in series and their equivalent capacitance is

C C

C C1 2

1 2

Charge on the plates 1 and 3 is

Q = +C C

C C1 2

1 2 .

when S is closed P.D. across C1 is zero.

Charge on the plate 1 is 0.If charge flown though 2 is q, then Q + q = 0

q = - Q = - C C

C C1 2

1 2 .

Charge on the plate 3 is C2 , which is also equal to the charge flown through 1.

Problem 5: When switch is swapped from 1 to 2,find the heat produced in the circuit.

Solution : Qa =

C C C

C CE

0

02.

Now, q

C

q

C

q q

C C

q

C Cq

C E

C Cb b a

b

0

0

0

0 0

2

02

.

When switch is shifted from 1 to 2, qa and q

b get interchanged. This can be seen from

symmetryPotential energy of the system is same before and after.

A charge q q qa b flows through the battery..

Work done by the battery = (q ).E = C . C0. E2/(2C+C

0).

>

>

qa

qb

a

b

R

< >

C1

C2

123

4

S

1 2

C0

Sq

0

C

qb

qa

1 2

E

C


Applying the first law of thermodynamics, QC C

C CE

LNM

OQP

..0

0

2

2

“-ve” sign means heat is liberated from te system

Problem 6:Three particles, each of mass 1 gm and carrying a charge, q, are suspended from acommon point by insulated massless strings, each 100 cm long. If the particles are in equilib-rium and are located at the corners of an equillateral triangle of side 3 cm, calculate thecharge, q on each particle (take g = 10ms-2)

Solution : F F F

q

aA AB AC

LNMM

OQPPLNMMOQPP

1

42

60

20

2

2

0

cos in the direction D to AA

= 2 1

4 0

2

2

q

a

LNMM

OQPP

3

2in the direction D to AA

For equilibrium, Tcos = mg ........ (1) & Tsin = FA

Divided (2) by (1), we get; tan = F

mgA

where, tan sin AD

AO

x

3 10

1

2

(AD = 2/3 of the median length , as D is the centroid)

or 21

4

3

20

2

2

LNMOQP

q

a=10-3 x ( 3 10 2x )

or q = 10 10 9x C

q 316 10 9. x C

Problem 7: Find the electric field caused by a disc of radius R with a uniform positive surfacecharge density at a point along the axis of the disc a distance x from its centre.

Solution : Consider a disc of surface charge density ' ' . Letus calculate the electric field due to a ring of chargesituated at a distance r, from the centre and having athickness, dr.Using the result of the previous question, we can say,

dE =kxdq

x r2 2 3 2e j /

Now, the area of the ring , dS = 2 r dr dq rdr. 2 ,

Thus, E dE kx

rdr

x rp

R

zz 2

2 2 3 2

0e j / = 1

4

2

0 2 2 3 2

0

xr

x rdr

R

./

ze j

E =

21

0 2 2 1 2

L

NMMM

O

QPPP

x

x Re j /

Also, as R , E =

2 0, which is the electric field in front of an infinite plane sheet of charge.

--------- -

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

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

------

-

FAC

FAB

FA

Tcos

)

(

T sin---------DA

C

B

T

O

a

mg

dr

r

x

dxP


Problem 8: A 4f capacitor is charged to 150 V and another 6f capacitor is charged to 200 V.V.

Then they are connected across each other. Find the potential difference across them. Calculatethe heat produced.

Solution: 4f charged to 150 V would have q1

=

C1V

1 = 600C

6f charged to 200 V would have q2 = C

2V

2 =

1200C

After connecting them across each other, theywill have a common potential difference VCharges will readjust as q

1’ and q

2’.

V = q

C

q

C

q q

C C

q q

C C1

1

2

2

1 2

1 2

1 2

1 2

, ' ' '

=

1800

4 6

C

fb g [Conservation of charge]

V = 180 volt.

Initial energy , U i =

1

2

1

2

1

24 150

1

26 200 01651 1

21 2

2 2 2C V C V f V f V J b gb g b gb g .

Final energy Uf =

1

2

1

24 6 180 01611 2

2 2C C V f f J . . . b gb g

Heat produced = U U Jf i 0 003.

Problem 9: Three charges q1, q

2 and q

3 are located at the vertices of an equilateral triangle of side a

Find the electric potential energy of the system.Solution : Taking datum at infinity. We can assume that the charge are brought one by one.

W = Vq

In absence of any charge, VA

=0,

W = V1 A V qb g1 0

FHG

IKJ W = V2 B V q

kq

aq

kq q

ab g2

12

1 20 ........ (2)

W = V3 c V qb g3

= kq

a

kq

aq

kq q

a

kq q

a1 2

31 3 1 30

FHG

IKJ ........ (3)

W = W1 +W

2 + W

3

Wkq q

a

kq q

a

kq q

a1 2 2 3 1 3

Problem 10 : In the above circuit, find the potentialdifference across AB.

Solution : Let us mark the capacitors as 1, 2, 3 and 4 for identifications. As is clear, 3 and 4 are inseries, and they are in parallel with 2. The 2, 3, 4 combine is in series with 1.

C34

= C C

C C3 4

3 4

.

= 4f , C2,34

= 8 +4 =12f

Ceq

= 8 12

8 124 8

x

. f , q =C

eq =4.8 x20 = 48C

q1

C1

q2 C

2

V

q1

q2 q

3

A

B C

a

1 23

4

P

Q

A

B

10

8 f 8 f

8 f8 f


The ‘q’ on 1 is 48 C , thus V1 = q/c = 6V [V

1=

48

86

c

FV ]

V VPQ 10 6 4

By symmetry of 3 and 4 , we say VAB

= 2V.

Problem11: What is VA - V

B in the arrangement shown? What is the condition such that V

A-

VB=0?

Solution : Let charge be as shown (Capacitors in series have the same charge)Take loop containg C

1,C

2 and E

q

C

q

CE

C

C C1 2

2

1 2

0

LNM

OQP

q = E.C1

From loop containing C3, C

4 and E

Similarly,

q

C

q

CE

C

C C

' '

3 4

4

3 4

0

LNM

OQP

q'= E.C3

Now VA-V

B =

q

C

q

CE.

C C C C2 4 1 2 3 4

LNM

OQP

' C C1 3

VA - V

B = 0 C

1C

4 = C

2C

3 =0

or C1 /C

2 = C

3 / C

4

OBJECTIVEProblem 1: Two point charges each of charge +q are

fixed at (+a, 0) and (-a, 0) . Another positivepoint charge q placed at the origin is free tomove along x-axis. The charge q a origin inequilibrium will have(a) maximum force and minimum potential energy(b) minimum force and maximum potential energy(c) maximum force & maximum potential energy(d) minimum force & minimum potential energy

Solution : The net force on q at origin is

F F F

q

ri

q

ri 1 2

0

2

20

2

2

1

4

1

40

. . ( )

The P.E. of the charge q in between the extreme charges at a distance x from the origin along+ve x axis is

U= 1

4

1

4

1

4

1 1

0

2

0

2

0

2

. .

q

a x

q

a xq

a x a x

LNMM

OQPPb g b g b gb g

dU

dx

q

a x a x

LNMM

OQPP

2

02 24

1 1

b g b g

For U to be minimum, dU

dx 0 ,

d U

dx

2

20 ,

a x a x a x a xb g b g2 2( )

x 0 , because other solution is relevant.

+ -

CC

C C

A

B

q q

q’ q’

E

--------------------------q q q

> x

y

(-a, 0) (+a, 0)


Thus the charged particle at the origin will have minimum force and minimum P.E. Dbg

Problem 2: Two concentric conducting spheres having radii a and b are charged to q1 & q2respectively. The potential difference between 1 & 2 will be

(a) q

a

q

b1

0

2

04 4 (b)

q

a b2

04

1 1

FHGIKJ

(c) q

a b1

04

1 1

FHGIKJ (d) none of these.

Solution :The potential on the surface of the sphere 1 is given by

vq

a

q

b1

0

1

0

21

4

1

4

The potential on the surface of the sphere 2 is given by,v = v

1-v

2

v2=

1

4

1

40

1

0

2

q

b

q

b v

q

a

q

b

1

4

1

40

1

0

1

FHGIKJv

q

a b1

04

1 1

(c)

Problem 3 The capacitance of the system of parallel plates shown in the figure is

(a) 2 0 1 2

1 2

A A

A A db g (b) 2 0 1 2

2 1

A A

A A db g

(c) 0 1A

d(d)

0 2A

d

Solution:Since the electric field between the parallelcharge plates is uniformned independent of thedistance, neglecting the fringe effect, the ef-fective area of the plate of area A

2 is A

1. Thus

the capacitance between the plates is

CA

d0 1

(C)

Problem 4:The charge flowing across the circuit onclosing the key k is equal to(a) CV (b) C/2V(c) 2CV (d) zero

Solution: When the key K is kept open the charge drawn from the source is Q = C’VWhere C’ is the equivalent capacitance given by C’ = C/2Therefore Q = (C/2)VWhen the key K is closed, the capacitor 2 gets short circuited and ths the charge in thatcapacitor comes back to the source.

Charge flowing isQ = (C/2)V (B)

Problem 5: The figure shows a spherical capacitor with inner sphereearthed. The capacitace of the system is

(a)4 0 ab

b a(b)

4 0 b

b a

2

q1

q2

12

ba>

>

>

............. ............

+

-

d

............. ............

A1

A2

0

E

CC

1 2

v+ -

k

ab

>

>

A1

A2

d


(c) 4 0 b ab g (d) none of these

Solution: Let V be potential on the outer sphere. Thus we can consider two capacitors between theouter sphere and inner sphere (a) and outer sphere and earth

Thus, C1 = 4 0

ab

b a; C

2 =

4 0 b

b a

2

(B)

Problem 6:The conducting spherical shells shown in the figure areconnected by a conductor. The capacitance of the system is

(a) 4 0ab

b a(b) 4 0 a

(c) 4 0 b (d) 4 0a

b a

2

Solution : As there will be no charge on the inner sphere, therefore the capacitance only will existdue to outer sphere. Hence the capacitance of the system is the capacitance due to outer

sphere of radius b, therefore C = 4 0b (C)

Problem 7 :A small charged particle of mass m and charge q is sus-pended by an insulated thread in front of a very large sheet ofcharge density . The angle made by the thread with thevertical in equilibrium is

(a) tan-1

g

mg2 0

FHG

IKJ (b) tan-1

g

mg0

FHGIKJ

(c) tan-1 2

0

g

mg

FHGIKJ (d) zero

Solution: In equilibrium, along x-axis Tsin = qE Tsin

= q

2 0......... (1)

Where T is the tension in the string.

Along y-axis in equilibrium, Tcos mg ....... (2)

From (1) and (2) we obtain, tan

FHG

IKJ

q

mg

q

mg2 20 0

= tan-1 Abg

Problem 8: A point charge q moves from point P topoint S. Along the path PQRS in a uniform

electric field E

pointing parallel to the posi-

tive direction of the x-axis. The coordinate ofthe points P, Q, R and S are (a, b, 0), (2a, 0, 0), (a, -b, 0) and (0, 0, 0) respectively. The workdone by the field in the above process is givenby the expression(a) qaE (b) -qaE

(c) q ( )a b E2 2FH IK (d) 3 2 2qE a b( )FH IKSolution : The work done is independent of the path followed and is equal to qE r

d i. , where r -=

>

>

a

b

+ + ++ + ++ + ++ + +

q

> >

>

>

><

>>

Tcos

+qE

mg

Tsin

(

y

x

...........................................(0, 0, 0)S)

>

>

>

>>

X

Y

........... .........

R(a, -b, 0)Q(2a, 0, 0)

p(a, b, 0)E


displacement vector PS ai bj , while E Ei

Work qE r qaE.

Bbg

ASSIGNMENTSSUBJECTIVE

LEVEL-I

1. Five point charges, each of magnitude +q are kept at thecorners of a regular hexagon of side a at C. Find theelectric field at the centre O of the hexagon.

2. The charges, q1 and q

2 are placed at the cor-

ners of a square as shown in the figure. Findq

2 such that the resultant force on q

1 is zero.

3. A plane surface of area S is inclined at an angle

with a uniform field E as shown in the fig.

Find the flux of E over S.

4. A small sphere of mass m carries a charge q.It hangs from a light inextensible thread oflength l making an angle with an infiniteline of charge as shown in the figure. Find thelinear charge density.

5. A circular wire of radius R carries a total charge Q distributed uniformly over its circumfer-ence. A small length of wire subtending angle at the centre is cut off. Find the electricfield at the centre due to the remaining portion.

6. Twenty-seven identical mercury droops are charged simultaneously to the same potential of10 volt. What will be the potential if all the charged drops are made to combine to form onelarge drop ? Assume all drops to be spherical.

7. Two charged particles having charge 1C and 1C and of mass 50 gm each are held at rest

while their separation is 2m. Find the speed of the particles when their separation is 1m.Neglect the effect of gravity.

8. A certain charge ‘Q’ is to be divided into two parts, q and Q-q. what is the relationship of‘Q’ to ‘q’. If the two parts, placed at a given distance ‘r’ apart, are to have maximumcoulombic repulsion? What is the work done in reducing the distance between them to halfits value?

9. A 1F and a 2F capacitor are connected in series across a 1200 V supply..

(a) Find the charge on each capacitor and voltage across each capacitor.(b) The charged capacitor are disconnected from the line and from each other, and are nowreconnected with terminals of like sign together. Find the final charge on each capacitor andthe voltage across each capaitor.

><

A

B C

D

E

O

aq1

q2

q1

q2

>>

>) E

S

+ +

+ +

+ +

+)

m, q


10. In the figure shoiwn, find the charge on each ca-pacitor (a) when the switch S is open (b)when the switch S is closed.

Level-II

1. Two small spherical bobs of same mass and radius having equal charges are suspended fromthe same point by strings or same length. The bobs are immersed in a liquid of relative

permitivity r and density 0 . Find the density of the bob for which the angle of divergence

of the strings is the same in the air and in the liquid ?

2. An infinite number of charges, each equal to q, are placed along the x-axis at x = 1, x = 2, x= 4, x = 8, ............. and so on. Find the potential and electric field at the point x =0 due tothis set of charges. When will be the potential and electric field if, in the above setup theconsecutive charges have opposite sign ?

3. The space between the plates of a parallel plate capacitor is filled with a dielectric as shown

in figures (1) & (2). The area of each plate is A and permitivity of the dielectric is r . Find

the capacitance in each case.

4. A 100 eV electron is projected directly toward a large metal plate that has a surface chargedensity of -2.0 x 10-6 C/m2. From what distance must the electron be projected, if it is tojust fall to strike that plate?

5. A 100 pF capacitor is charged to a potential difference of 24 V. It is connected to an un-charged capacitor of 20 pF. What will be the new potential difference across 100pF ?

6. In the figure shown, find(a) The charge(b) The potential difference and(c) The stored energy for each capacitor

7. Figure shows a parallel plate capacitor havingsquare plates of edge a and plate separation d.The gap between the plate is filled with a di-electric of dielectric constant k which variesfrom the left plate to the right plate as k = k

0

+x . Where k0 and are positive constants

and x is the distance from the left end. Calcu-late the capacitance.

8. A capacitance has squareplates, each of side a, making

S

V

C2

C1

V

>

> d

l/2l/2

>

>

>

>

d/2 d

F F

F

C1=10 C

2=5

C3=4

100v

K

>

>-----------------------)d


an angle as shown in the figure.Show that, for small , the ca-pacitance is C

= Ca

d

a

d

LNMOQP

02

12

9. In the arrangement shown in figure, three con-centric conducting shells are shown. Thecharge on the shell of radius b is q

0 . If the

innermost and outermost shells are connectedto the earth, findtheir charge densities and thepotential on the shell of radius b in terms of aand q

0. Given that a : b : c = 1 : 2 : 4

10. In the given circuit diagram, the switch SW

is shifted from position 1 to position 2. Find theamount of heat generated in the circuit.

LEVEL-III

1. Four identical metallic plates, each having areaof cross-section A, are separated by a distanced as shown in the figure. Plate 2 is given achange Q. Find the potential difference be-tween 2 & 3.

2. Five identical plates each of surface area A areplaced one above the other by same dielectricof thickness d and dielectric constant k. Findthe equivalent capacitance between a and b,the plates 1 and 4 and the plates 3 and 5 arejoined together.

3. An insulated conductor, initially free from charge, is charged by repeated contacts with aplate which, after each contact, is replenished to a charge Q from an electrophorus. If q isthe charge on the conductor after the first operation, prove that the maximum charge whichcan be given to the conductor inthis way is Qq/(Q-q).

4. Two perfectly insulated capacitors are connected in series. One is an air-capacitor withplate area 0.01 m2, the plates being 1 mm apart, while the other has a plate area of 0.001 m2,the plates separated by a solid dielectric of 0.1 mm thick with a dielectric constant 5. deter-mine the voltage across the combination, if the potential gradient in the air-capacitor is 200V/mm.

5. An electron files with an initial velocities v0 into a parallel plate capacitor in a direction

making an angle to the plates and leaves the capacitor at an angle to the plates . The

length of the capacitor plates is L. Find the intensity E of the field between the plates andthe kinetic energy of the electron as it leaves the capacitor. Take the mass and charge ofelectron as m and e respectively.

ab

c

1 2 3 4

1

2

345


6. A cone made of an insulating material has a total charge Q spread uniformly over its slopingsurface. Calculate the energy requied to bring a small test charge q form infinity to the apexA of the cone. The cone has a slope length L.

7. A non conducting sphere of radius R = 5cm has itscentre at the origin O of coordinate system as shownin figure. It has a spherical cavity of radius r = 1 cm,whose centre is at (0, 3cm). Solid material of sphere

has a uniform positive charge density 1/ c / m3 .

Calculate the potential at point P (4cm, 0).

8. The electric field vector is given by E a x i . Find

(a) the flux of E through a cube bounded by surfaces

x = l , x = 2l , y = 0, y = l, z = 0 and z = l.(b) The charge within the cube.

9. A charge Q is uniformly distributed over the volume of a sphere of radius R. Find theelectrostatic potential energy stored(a) in the surrounding space.(b) within the sphere

10. A spherical shell of radius r1 with a uniform charge Q has a point charge q at its centre . Find

the work done by an external agent in expanding the shell to a radius r2.

OBJECTIVELEVEL-I

1. A positive charged pendulum is oscillating ina uniform electric field as shown in figure. Itstime period, as compared to that whent wasuncharged;(a) Will increase (b) Will decrease(c) Will not change(d) Will first increase and then decrease

2. A and B are two concentric spheres. If A isgiven a charge Q while B is earthed as shownin figure(a) The charge density of A and B are same(b) The field inside and outside A is zero(c) The field between A and B is not zero(d) The field inside and outside B is zero

3. The maximum electric field intensity on the axis of a uniformly charged ring of change qand radius R will be

(a)1

4 0

q

R3 3 2 (b) 1

4 0

2

3 2

q

R

x

y

oP’

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

> > > > > >

+ ++++

++

++

+ ++

A

B

>> > >>

> > > > >


(c) 1

4 0

2

3 3 2

q

R(d)

1

4 0

3

2 2 2

q

R

4. The figure is a plot of the lines of force due to two charges q1 and q

2. The sign of the charges

are(a) both negative(b) Upper positive and lower negative(c) both positive(d) Upper negative and lower positive

5. There are two concentric metal shells of radii r1 and r

2 (> r

1) . If the outer shell has a charge

q and the inner shell is grounded, the charge on the inner shell is

(a) zero (b) r r q1 2/b g (c) r r q1 2 (d)

6. Electric charges q, q and -2q are placed at the corners of an equilateral triangle ABC of sideL. The magnitude of electric dipole moment of the system is

(a) qL (b) 2qL (c) 3e jqL (d) 4qL

7. In which of the following states is the potential energy of an electric dipole maximum

8. The effective capacitance between A and B willbe

(a) 0.5 F (b) 1.5F

(c) 2F (d) 2.5 F

9. In the electric field due to a point charge q, atest chargeis carried from A to the points B, C,D and E lying on the same circle around q.The work done is(a) the least along AB(b) the least along AD(c) zero along any one of the paths AB, AD,AC and AE(d) the least along AE.

10. The potential difference across the capacitor

of 2F is

(a) 10V (b) 60V(c) 28V (d) 56 V

LEVEL-II

1. Two connected charges of +q and -q respec-tively are at fixed distance AB a part in a non

................................................... ...................................................

<<

< <

-q

+q-q -q+q

+q+q

E E E E

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

1F-q 2F

2F

2F

1F

1F

A

B

C

D

E

+q

3F

F

F

F

3

6

70V

2

>>>>

+q

-qA

B


uniform electric field whose lines of force areshown in the figureThe resultant effect on the two charges is(a) a torque vector in the plane of the paper and no resultant force.(b) a resultant force in the plane of the paper and no torque.(c) a torque vector normal to the plane of the paper and no resultant force.(d) a torque vector normal to the plane of the paper and a resultant force in the plane of thepaper.

2. Two identical point charges are placed at a separation of l, P is a point on the line joining thecharges, at a distance x from any one charge. The field at is E. E is plotted against x forvalues of x from close to zero to slightly less than l. Which of the following best representsthe resulting curve?

3. The capacitance of two condensers in parallel is four times their capacitance when they areconnected in series. The ratio of the individual capacitances will be(a) 1 :2 (b) 1 : 1 (c) 2 : 1 (d) 4 : 1

4. Find the capacitance of a system of three parallel plates each of area A separated by dis-

tances d1 and d

2. The space between them is filled with dielectrics of relative permittivity 1

and 2 . The dielectric constant of free space is 0

(a)

1 2 0

1 2 2 1

A

d d (b)

1 2 0

1 1 2 2

A

d d

(c)

1 2

0 1 2 1 2

A

d d( )( ) (d) A

d d 0 1 2 1 1 2 2( )

5. A solid conducting sphere having a charge Q is surrounded by an uncharged concentricconducting spherical shell. Let the potential difference between the surface of the solidsphere and that of the outer surface of the shell be V. If the shell is now given a charge of -3Q, the new potential difference between the same two surface is:(a) V (b) 2V (c) 4V (d) -2V

6. A parallel plate capacitor is connected across a source of constant potential difference whena dielectric plate is introduced between the two plates then:(a) Some charge from the capacitor will flow back into the source.

E

O

........... ..................

>>x

(D)........... ..................

E

O >

>

x

(C)

E

O

................ .............

>

>

x

(A)E

O

................ .............

>

>

x

(B)


(b) Some extra charge from the source will flow into the capacitor.(c) the electric field intensity between the two plates increases.(d) energy of the capacitor does not change.

7. The figure shows a circuit of four capacitors .The effective capacitance between X and Y is

(a) 2F (b) 1F

(c) 3F (d) 1.5F

8. Two identical thin rings , each of radius R meters, are coaxially placed a distance R metersapart. If Q1 C and Q2C are respectively the charges uniformly spread on the two rings, thework done in moving a charge q from the centre of one ring to that fo the other is

(a) zero (b) q (Q1-Q

2)( 2 1 )/ 4 0 R

(c) q Q Q R2 41 2 0b g/ ( ) (d) q Q Q R1 2 02 1 24 b ge j/ ( )

9. Three capacitors are connected with the sourceof electromotive force E as shown in the fig-ure. Then the energy drawn from the sourceis

(a) 1

22CE (b)

3

22CE

(c) 1

42CE (d) 2CE2

10. The variation of potential with distance r froma fixed point is shown in figure. The electricfield at r = 3 cm and r = 5cm are, respectively,(a) 0, 2 V/cm (b) 2 V/cm, -2 V/cm(c) 0, -2 V/cm (d) 2 V/cm, 0

ANSWERSSUBJECTIVE

LEVEL-I

1. Eq

a

1

4 02

. in the direction CO 2. q1

2 23. ESsin

4.2 0 mg

q

sin tan5.

Q

R4 220

2

sinFHGIKJ 6. 90 V

7. 0.3 m/s 8. qQ

2

, Q

r

2

016 .

9. (a) 800C , 800V, 800C , 400V (b)1600/3V, 1600/3C , 3200

3C

1F

1F 1F

2F

X y

C C

C

B

>

543

2

1

1 2 3 4 5

v(volts)

r(cm)


10.C C

C CV1 2

1 2

FHG

IKJ (b) Zero on C

2, C

1V on C

1

LEVEL - II

1.

0

1r

r 2. 2 kg,

4

3

kg,

2

3

kg,

4

3

kg3.

2

10

r

r

A

d( )

4. 0.44mm 5. 20 V

6. (a) q1 =q

2 =3.3 x 10-4 C, q

3 = 4x 10-4C (b) V

1=33V , V

2 = 67V, V

3 =100V

(c) U1 = 5.4 x 10-3 J, U

2 = 10.9 x 10-3 J, U

3= 2 x 10-2J

7.

02

0

1

a

Ind

kFHG

IKJ

9.

inner

q

a 0

212,

outer

q

a 0

296,

q

a0

024

10.1

222CE

LEVEL -III

1.Qd

A3 02.

5

30K A

d

4. 240 V

5. E =mv

eL02 2cos

tan tan

b g ; K.E. = mv0

2 2

22

cos

cos

6.

Qq

L2 0

7. 35.16 V 8. a5 2 2 1/ e j , a5 202 1/ e j

9.Q

R

2

08(b)

Q

R

2

04010.

Q q Q

r r

FHG

IKJ

/ 2

4

1 1

0 1 2

b g

OBJECTIVELEVEL -I

1. A 2. C 3. C 4. A

5. B 6. C 7. A 8. C

9. C 10. B

LEVEL-II

1. D 2. D 3. B 4. A

5. A 6. B 7. A 8. B

9. B 10. A


REPORT COLUMN

1. TEST-1

DATE: :

SCORE :

MM : PARENT/GUARDIAN SIGNATURE

2. TEST-2

DATE: :

SCORE :

MM : PARENT/GUARDIAN SIGNATURE

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