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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 material covers 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 Joint Entrance Examination, it will also prove useful to students who are preparing for other Engineering examinations like AIEEE & BITSAT since the content of Physics, Chemistry and Mathematics remains 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

Note down the doubts and difficulties that you face in the study material and ensure that these are removed in the class.

Do full justice to the assignment problems. Even if you do not get the answer to a problem, keep trying on your own and only approach your friends or teachers after making lot of attempts. Do not look at the answer and try to work backwards. This would defeat the purpose of doing the problem. Remember the purpose of doing an assignment problem is not simply to get the answer (it is only evidence that you solved it correctly) but to develop your ability to think.

Before attempting the objective assignment make it sure that you have understood and solve the subjective ones of at least Level-I and II. It is advisable to solve the objective problems of each level at a single sitting.

In case you miss a class (which you should not unde any circumstances), go through the study material of that topic on your own, take the help of your friends in the class and pass on your problems to the faculty member in writing. Remember it is your sole responsibility to cover the topics that you have missed partly or completely.

It is possible that during your school - examinations, you might like to miss some classes at RADI-ANCE Generally each day of class at RADIRADI-ANCE must be supported by minimum 4 to 5 hr of study from your side. On top of this if you miss a class you must put at least 1 12 times of these hours to take care of the missed class. So never even think of missing any class at RADIANCE

We believe that you can produce the desired result provided you strictly follow our instructions. If you do not follow our instructions, you might still produce some success in IIT-JEE, but it would never really be your optimum performance.

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ELECTROSTATICS

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 electric field; Gauss’s law and its application in simple cases, such as, to find field due to infinitely long straight 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 the matter 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. To distinguish 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 CGS unit 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 or

proton). Charging of a body

Basically charging can be done by two methods;

1. Conduction 2. Induction

Ordinarily, matter contains equal number of protons and electrons. A body can be charged by the transfer of electrons or redistribution of electrons.

The process of charging from an already charged body can happen either by conduction or induction. Conduction from a charged body, involves transfer of like charges. A positively charged body can create more bodies, which are positively charged, but the sum of the total charge 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 charged bodies, without touching them or losing its own charge.

Properties of Electric Charge Quantization of charge

Charge exists in discrete packets rather than in continuous amount. i.e. charge on any body is the integral multiple of the charge of an electron

Q ne , where n = 0, 1, 2,...

Conservation of charge

Charge is conserved, i.e. total charge on an isolated system is constant. By isolated system we mean here a system through the boundary of which no charge is allowed to escape or enter.

This does not require that the amount of positive and negative charges are separately con-served only their algebraic sum is concon-served.

Charges on a conductor

Static charges reside on te surface of the conductor. Distribution of charges

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

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Exercise 1: Two identical metallic spheres of exactly equal masses are taken, one is given a positive charge 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 q2 at rest, separated by a distance r exert a force on each other whose magnitude is given by

F kq q r

 1 22 where k is a proportionality constant.

If between the two charges there is fre space then

k = 41 9 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 charges brought nearby?

IIIustration 1:Two particles A and B having charges 8 x 10-6 C and -2 x 10-6 C respectively are held

fixed with a separation of 20 cm. Where should a third charged particle be placed so that it does 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 be oppostie in direction. Hence, the particle should be placed on the line AB. As A and B have charges of opposite signs, C cannot be between A and B

Also A has larger magnitude of charge than B. Hence, C should be placed closer to B than A. The situation is shown in figure. Suppose BC = x and the charge on C is Q

 F x Q x i CA    1 4 8 0 10 0 2 0 6 2  . . 

e

j

b g

and  F x Q x i CB 1  4 2 0 10 0 6 2  . 

e

j

bg

   FCFCAFCB = 41 8 0 10 0 2 2 0 10 0 6 2 6 2  . . . x Q x x Q x    

L

N

M

M

e

b g

j

e

bg

j

O

Q

P

P

i But FC  0 Hence 1 4 8 0 10 0 2 2 0 10 0 6 2 6 2  . . . x Q x x Q x    

L

N

M

M

e

b g

j

e

bg

j

O

Q

P

P

=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, q2...qn, the the force on Q can be found out by calculating separately the forces F1, F2..., Fn, exerted by q1, q2....qn respectively on Q and then adding these forces vectorially. Their resultant F is the total force on Q due to all of charges.

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ELECTROSTATICS

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

Solution: Let the charge be Q

As ABCD is a square of side a

r a 2  a 2 2  F kq a i BA  2 2 ,  F kq a j BC   2 2 ,  F kq a i j BD  2 2 0 0 2 45 45 ( ) cos  sin 

e

j

 F kqQ a i j BQ  ( ) cos  sin  2 2 45 45 0 0

e

j

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

 F kq a kq a kqQ a i R   

F

H

G

22

I

K

J

2 2 0 2 0 2 45 2 45 ( ) cos ( ) cos  -  F kq a kq a kqQ a j R   

F

H

G

22

I

K

J

2 2 0 2 0 2 45 2 45 ( ) sin ( ) sin 

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

 FBXFBAFBDcos450FBQcos4500   

L

N

M

M

M

O

Q

P

P

P

 k q a q a Qq a 2 2 2 2 2 2 1 2 2 1 2 0

e j

.

e j

. Q q  4

e

1 2 2

j

Similarly, FBy = 0, if Q q  4

e

1 2 2

j

. Electric field

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

Electric field strength or Electric intensity

di

E at a point is t electric force experienced by a unit positive charge at that point. Mathematically,   E F q  0

where q0 is positive test charge.

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

     E k q r r r r    1 2 3

b g

1 2

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1E2...En

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

 E dq r r  1

z

40 2 A B X Y  r1

r

2   r1r2 ) A B C D q 450 q q q a a r Q x y  FBC  FBD  FBQ  FBA i

ej

j

ej

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where dq is the charge on one element of the charge distribution and r is the distance from the element to the point under consideration and r is the unit vector directed from the position of elemental 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 itself produces?

(ii) Two point charges q and -q are placed at a distance d apart. What are the points at which resultant 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. What is 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 of length ‘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 the electric field. The elementa charge, dq =dx

Then, dE = k dx x  2 or E = k x dx k x k a L a a a L a a L  12  1  1 1  

z

L

N

M

O

Q

P

   

L

N

M

O

Q

P

Thus, E =4 1 1 0 a L a  

L

N

M

O

Q

P

[ k =41 0  ]

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

Solution:Consider a different element of the ring of length ds. Charge on this element is dq = q

R ds 2

F

H

G IK

J

.

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. From symmetry 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  E

z

 E =

z

d cosE dE = 1 4 dq r = 1 4 qds 2 R . 1 R ; cos = x R 0 2 0 2 2    

F

H

G IK

J

F

H

G

I

K

J

 x2

e

x2

j

1 2/

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

Ex = d qx x ds  Ecos 1 4 0 2 R R2  

z

z

 2 3 2

e

j

/

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

---) x (R2 + x2) dE cos  y O x R .... .... .... .... .... .... .... ... dq 1/2 (  dE sin  P dE  a p ---dx x a P

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ELECTROSTATICS E = 1 40 R2 qx x  2 3 2

e

j

/

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

It has been found quite convenient to visualize the electric field pattern in terms of lines of force. The electric field pattern vector at a point is related to imaginary lines of force is two ways. The lin of force in an electric field is a curve such that the tangent at any point on it gives 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 is proportional 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’s surface and the electric field is zero, everywhere within the material of the conductor.

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

Gauss’s Law

The net “flow” of electric field through a closed surface depends on the net amount of electric charge containd within the surface. This ‘flow’ is described interms of the electric flux 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 elecelec-tric flux through a 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 various symmetry properties.

Mathematically

z

E ds .  q

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 symmetric that 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 equal charges located at the corners of an equilateral triangle ?

IIIustration 5: Figure shows a section of an infinite rod of charge having linear charge density which is constant for all points on 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 a Gaussian 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

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end by plane caps normal to the axis. 0 0 2 1   E ds qin E ds E ds E ds q C C C in . ; . . .

z

z

z

z

L

N

M

M

O

Q

P

P

Cylindrical plane surface, 0E

b g

2rI  E.ds.cos90o= I

E = I =

02 rI 02 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 charge means that the charge density at any point depends only onth distance of the point from the centre and not on the direction. Secondly, the object can not be a conductor, or else the excess carge will reside on its surface.

Now, apply Gauss’s law to a spherical Gaussian surface of radius r

r > R for point A 0 0E 4 r2 = q

b

g

, 

z

E ds . qen 

e j

 E=41 0 2  q

r where q is the total charge

For point B (r < R), 0

z

E ds . 0E

e j

4r2 q q’ = q r R 4 3 4 3 3 3   = q r R

F

H

GIK

J

3 E = 1 4 1 4 0 3 2 0 3   q r R r qr R

F

H

GIK

J

 Electric Potential

Potential at a point in an electric field, is the amount of work done by an external against electric forces in moving a unit positive charge with constant speed from infinitly to that point.

As work done by the external agent = - work done by electric force Hence the required potential V = -

   Ed q dr r q r r

z

z

  4 1 4 0 2 0  

The potential at a point due to a group of n point charges q1, q2, q3...qn

V = V1+V2+...+Vn (Scalar sum)  

41 0 1  q r i i i n R B r A

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ELECTROSTATICS

The electric potential due to a continuous charge distribution, V = 41

0

 dq

r

z

Potential due to a uniformly charged disc The elemental ring’s area = 2xdx

dV = 41 2 0 2 2   xdx r x

b g

 V = dV xdx r x R r r R

z z

  

L

N

M

 

O

Q

P

   2 0 2 2 2 2 0

Potential of a uniformly charged spherical shell If the charge on the shell = q

(i) for r > R E = kq r2     

z

V kq r dr kq r r 2 (ii) for r = R , V = kq R (iii)for r < R , E =kq r2 (R < r < ) , E = 0 (r < R)   

L

N

M

M

O

Q

P

P

z z

V Edr Edr R R r =  

z

kq r dr R 2 - 0 R r dr kq R

z

Potential of a uniformly charged spherical volume If the total charge = Q

(i) for r > R, Volume charge density 

Q R 4 3 3 E (r > R)= kQ r2 ; V = kQ r dr kQ r r 2 

z

 (ii) for r =R, V kQ r  E = 41 0 2  Q r (iii) r < R, E (r < R) = 3. r 0 = Q r R . . 4 3 3 3 0    E (r < R ) = Q r R . 4 30 = kQ. r R3 V = - Edr Edr R r 

z z

L

N

M

M

R

O

Q

P

P

= kQ r dr kQr R dr = kQ 2R 2 R 3 r 

z z

L

N

M

M

3

O

Q

P

P

2 2 r R R

For conducting solid sphere : (i) (r < R), E =0 , V= kQ R (ii) (r = R), E= kQ R2 , V= kQ R --- --- --- --- --dx x R r q R r p q R r --- --- --- --- --R r p Q

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(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 surfaces

The locus of points of equal potential is called an equipotential surface. The electric field is perpendicular to the equipotential at each point of the surface.

Exercise 7: A solid metallic sphere is placed in a uniform electric field. Which of these A, B, D and 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 the component of the field along that direction.

i.e.E dV

dr

r  

or we can say the electric field is along the direction in which the potential decreases at the maximum 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 

e j

A x i/ 3 .What is the potential in the region ?

Solution: The electric potential in the region

V(x, y, z) =      

z z

  E dr A x dx A x x y z x y z . , , , ,

b g

b g

3 2 2

IIIustration8: What is the electric potential on the axis of a uniformly charged ring of radius R containing 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,

V dq r r dq 

z

1 

z

4 1 4 0 0   Hence V = 41 0  q R2x2 1 2

e

j

/

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

great-Equipotential surface Electric field > > > > > > H A B C D > )  dr P P’ E 50V 40 V 30V 20V 10V A C B

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ELECTROSTATICS

est ? Solution:E dv

drn   

The potential difference betwen any two connective line dv = v1-v2 = 10 v constant and hence e will be maxium where the distance dr between the lines minimum.i.e. at B where the lines are closest.

IIIustration 10:Some equip-potential surface are shown in fig (1) and (2) what can you say about the 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

g

b

g

x -E .1/2=-10/0.10= E= 200 v/m Figure 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.

b g

 kq6

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

r Hence E= -dV dr = 6 2 r V m

F

H

GIK

J

/

Electrostatic potential energy

The electric potential energy of a system of point charges is the amount of work done in bringing the charges from infinity in order to form the system . Two point charges q1 and q2 are separatd at a distance r12, Electric potential energy of the system q1 and q2

U q q r  1 4 0 1 2 12 

For three particle system q1, q2 and q3

U q q r q q r q q r 

F

 

H

G

I

K

J

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 = UP/q; where UP, 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 of

an assembly of charges is given by

20V 30V q2 r23 1 2 3 4 a a a a - q +q +q - q q1 q1 q2 q3 r12 r 13 r12 ) 10 20 30 40 300 10V 20V30V y (cm) x (cm) 10 cm 20 cm 30 cm 60 V

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> < ---- +q -q   Equatorial line 1 40 q q r y j ij i 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 2 2 2 1

e j

e j

Electric Dipole

Two 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. The electric field at point P (x,0) will be then;

   E E E kq x a i kq x a i axial q q

b g

  

b g b g

     2 2  

, where i is the unit vector along axis.

= kq x a x a x a i kq x a x a i .  .       

b g b g

e

j

e

j

b gb g

e

j

2 2 2 2 2 2 2 2 2 2 = 2 2 2 2 kpx x a i 

e

j

 , as x >> a ,   E kp x 23 p = 2aq

Electric field on equatorial line At P,  E kq y a i j      2 2

e

j

e

cos  sin 

j

 E kq y a i j      2 2

e

j

e

cos  sin 

j

        E E E kq y a i p 2 2 2

e

j

cos     

L

N

M

M

O

Q

P

P

    E kq y a a y a i k aq y a i p 2 2 2 2 2 2 1 2 2 2 3 2

e

j e

.

j

/ 

e

b g

j

/  ... ... ... ... ... ... ... ... ... ... ... ... ... ... < < > > ( a E -E+ Ep P2(0, y) y x a y - +  > < > > < > -q +q Ep E+ E-(-a, 0) O y x (a, 0) (x,0) x

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ELECTROSTATICS 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 = 2k P3

r ( cos )

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

r ( sin )

3

E  E E  kp  r

R P2cos P2sin 3 4cos2 sin2

Resultant ,E  kp 

r

R 3 3cos 2 1 and tan

tan sin cos      E  E p p 2

Dipole in an external electric field

The net force on an electric dipole in a uniform external electric field is zero.. However the dipole in the presence of an external electric field experience a torque and has tendency to align itself along the external electric field.

Torque on dipole = force x force arm

= qE

b g bgb g

sin  q Esin pEsin

or   p  x E

The product of the charge q and separation l is the magnitude of a quantity called the electric dipole 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 external electric field.

W=

z z

Td pEsin d =- pE cos

1 2

2 0, dipole is perpendicular to the field

We assume, 1900(as the datum for measuring potential energy can be chosen anywhere)

U =pE cos or U= p E .

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

the E   0

e j

0 . 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) --- ---- ---- --) a a B +q -q A r  --- --- -> Psin Pcos B  p  ) -+ --q +q qE qE > > <  E p 

g

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 

F

H

G IK

pE

J

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( / )2 2M L( / )2 2 ML2/2

 time period T = 2

I pE

The minimum time required to align itself is T/4 Potential due to an electric dipole

V q PB q PA P 

L

N

M

O

Q

P

1 40

When r >> a, PB = r-a cos

PA = r + a cos V qa r a P   1 4 2 0 2 2 2    . cos cos

e

j

As r >>a, a2cos2

 can be neglected in comparison to r2. V qa r p r P  1 4 2 4 0 2 0 2     . cos cos .

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

Capacitance

If Q is the charge from given to a conductor and V is the potential to which it is raised by this amount 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 is called a capacitor. This medium is called dielectric of the capacitor. 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 as C= 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 area and 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 dr E > > > --- --- -) P L E + - p --- ---- ---- ---- ---- ---- ---- -- ---A B -q a a +q r ) ) + Q -Q

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ELECTROSTATICS  dV Edr v R 0

z z

    V kQ r R 

L

N

M

O

Q

P

 1 V kQ r  since, C = Q/V = R/k1 4R 4 R 0 0 /   Combination of capacitors Series combination

In series combination, each capacitors has equal charge for any value of capacitance, provided that capacitors are initially uncharged. Equivalent capacitance C is given by

1 1 1 1

1 2 3

C C C C ...

Parallel combination

In parallel combination the potential differences of the capacitor connectd in parallel are equa for any value of capacitance. Equivalent capacitance is given by C = C1+C2+C3+...

Some times it may not be easy to find the equivalent capacitance of a combination using the equations for series parallel combinations. For any combination one can proceed as follows:

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

Step-2: Write the charges appearing on each plates using charge conservation principle say, Q1, Q2 ...

Step-3:Assume the potential of the negative terminal of the battery be zero and that of positive terminal to be V, and write the potential of each of the plates say v1, v2...

Step-4:Write the capacitor equation Q= CV for each capacitor. Eliminate Q1, Q2 ...V1, V2...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. The area of each plate is equal to A. The arrangement is sown in the figure. Find the capacitance of the system between points P and R.

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

VP = V(+) c1 c2 c3 q -q q-q q -q V P R 1 2 3 4 P R 1 2 3 4 +q1 -q1 +q2 -q2 +q1 -q1 VR=0 V q q q c1 c2 c3

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the plates 1 and 4 are at the potential V1. Charge distri-bution is shown in the figure.

 q1 = C (V-V1) from (4, 3) ... (1)

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

q1 = CV1 from (2, 1) ... (3)

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

2 ... (4)

Charge supplied by the battery is Q = q1 + q2 = 3

2CV using (2) and (4)

CV 2Q

3 ... (5)

If equivalent capacitance be C’ then

C’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 are given the charges Q1 and Q2. 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, then electric field at P should be zero. Suppose surface area of the each surface is A.

Using the equation E = 2

0

    

EP E1E2E3E4,

E1, E2, E3 and E4 are the electric field due to surfaces 1, 2, 3 and 4.  E Q q A i q i q i Q qi P           1 0 0 0 2 0 2 2 2 2        = Q q A Q q i 1 0 2 0 2 2     

F

H

G

I

K

J

 P Q 1 2 2 3 3 4 3 3 2 2 1 4 C/2 C 3C/2 C A d eq  3 2 0  + + + + + + + + + ++ + ++ + ++ + + + + ++ ++ + + + -Q1 Q1-q 1 2 Q2 -q Q2+q 3 4 P A B ++ + ++ + ++ + ++ + + + + + + + + + + + + + + + + + -- -Q1 1 2 3 Q2 4 P A B (Q1-Q2)/2 (Q1-Q2)/2 (Q 1+Q2)/2 (Q1+Q2)/2 -q

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ELECTROSTATICS

But E  0, which gives q =Q1 Q2

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. Q1 Q2

2 

, and on inner surface of B is opposite of the chargg appearing on inner surface o f A i.e. - Q1 Q2

2 

F

H

G

I

K

J

; write the change on outer surfaces of A and B is half of the sum of change on A and B i.e. Q1 Q2

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 constant

When a dielectric is placed in an external electric field, polarization occurs and it develops an electric field in opposition to the external one. As a consequence total field inside it decreases. If E be the total field inside the dielectric when it is placed in an external field E0. then its dielectric constant k is given as k E

E

 0 (k > 1)

If a dielectric completely occupls the space between the conductors of a capacitor. its capacitance increases 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 capacitor

If 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 C CV QV 2 2 2 1 2 1 2

  . Also , as C = 0A d and V/ E.d

U = 1 2 2 CV = 1 2 0 2  A d

b g

E.d = 1 2 0 2 

F

H

G

E

I

K

J

A d.

u (energy density)= Energy per unit per volume = 1 2 0

2

 E

If dielectric is introduced then, u = 1

2 0

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

F Force on an external agent

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Consider a differential displacement dx of the di-electric as shown in figure keeping the total force on it always zero. Then

When battery is disconnected.

W Electrostatic +WF= 0(where W denotes the work done in displacement dx)

WF = -W Electrostatic EF= U   

L

N

M

O

Q

P

F dx Q d c . 2 2 1 W Q c 

L

N

M

M

O

Q

P

P

2 2  F dx Q c dc . 2 2 2 F Q

F

H

GIK

J

c dc dx 2 2 2

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12V2

F

H

GIK

dxdc

J

Exercise 11: A dielectric slab is inserted in one end of a charged parallel plate capacitor. (The plates 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 square metal plates of edge ‘l’ separated by a distance d. Dielectrics of dielectric constant K1 and K2 are filled in the gap as shown in t figure. Find the capacitance.

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

dc1 =

b

dK1 0xtandx

g

L

N

M

C K Ad0

O

Q

P

dc2 = K

b g

x2 0tandx

These two capacitors are connected in series

 

L

 

N

M

O

Q

P

1 1 1 0 1 2 dc d x K x K dx     tan tan  tand /    

L

N

M

M

M

O

Q

P

P

P

1 1 1 0 1 2 dc d x d K x d K dx     . .   tan  

L

N

M

d

O

Q

P

   

z

dc k K

z

d dx K K K x 1 2 0 2 2 1 2 0     ( )   

L

N

M

M

O

Q

P

P

C AK K K K d k k 0 1 2 2 1 2 1

b

g b

ln /

g

[A = l2]

Student should note that the ‘k’ used here is different from k = 41

0  elsewhere > < d l K1 K2 < d K1 K2 ) x dx xtan >  < >

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ELECTROSTATICS

SOLVED PROBLEMS SUBJECTIVE

Problem 1:Four point charges  8C,-1C and +8C 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

F

H

G

I

K

J

  q x y x i A p A p

e

j

/

e j

 Similarly  E q x y x i p BB B p B p ( ') /  

F

H

G

I

K

J

 2 1 40

e

2 2

j

3 2

ej

For Ep 0 xp  2 5. m

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

H

G

G

I

K

J

J

e

j

/

e

j

/ ] dV x dx ( ) 0  5 2 x = 0 and x = x = 5 2V

bg

x x 0, 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

N

M

M

M

O

Q

P

P

P

1 2 2 4 0 2 0 2 2 2 2 mv q 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

 

L

N

M

O

Q

P

  1 2 2 4 2 5 10 0 2 0 4 mv q q y q y x J A 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 = 0 x  5 2/ v(x)

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l and the separation of the plates is d.

(a) An electron and a proton start from the negative plate and positive plate respectively and go 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 t eplates. 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 force

along x axis. The accelerations of the electron and the proton along y axis are as follows

ae = meE a  meE e p p , Using S = ut + 1 2 2 at here S = d,u = 0 t  S a e e 2 and t S a P P  2 or te= 2dm eE e and t dm eE P p  2 As me < mp, te < tp

 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 = V

x

During this time, deflection along vertical direction y = 1 2 2 at or y =1 2 0 2 eE m V 

L

N

M

O

Q

P

. (i) Thus Yp= 12 1 2 2 eE mP Vx and YYe = 12 2 2 eE m Ve x 

As mp > me, yp < ye of course, electron will be deflected in the opposite direction.

(ii) Also y = 1 2 2 2 2 2 eE m Vx

L

N

M

O

Q

P

= 1 4 2 eE k  where K = 1 2 2

mvx, or initial kinetic energygy

yp ye Also, as K = p m 2 2 , where p = momentum, y eEI p m eEI m p y eEI m p y eEI m p p p e e      2 2 2 2 2 2 2 2 4 2 2 2 2 ( ) ,

Thus for : mp > me , yp>ye

> d e + -P Y X  E ---vx p + -yp

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ELECTROSTATICS

Problem 3: Two spherical cavities f radii, a and b , are hollwed out from the interior of a neutral conducting sphere of radius R. At the centre of each cavity , a point charge is placed. Call these charges qa and qb

(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 qa and qb

(e) which of these answers would change if a third charge qc were brought near the conduc-tor ?   A a q a   4 2 , b  b q b   4 2 , R  a b q q R   4 2 (b) E K q q r r R a b 

b

gb g

 2 , (c) E Kq r r a a a  2 ,

b g

, E Kq r r b b b  2 ,

b g

(d)Force on qa and qb = 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 the figure, through section 1 and 2 ?

Solution : When S is open, C1 and C2 are in series and their equivalent capacitance is CC C1 C2

1 2

Charge on the plates 1 and 3 is

Q = + C C

C C

1 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 C

1 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 C E   0 0 2 . Now, q C q C q q C C q C C q C E C C b b a b          0 0 0 0 0 2 0 2 .

When switch is shifted from 1 to 2, qa and qb get interchanged. This can be seen from symmetry

Potential energy of the system is same before and after. A charge qqaqb flows through the battery..

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

0). > > qa qb a b R < > C1 C2  1 2 3 4 S 1 2 C0 S q0 C qb qa 1 2 E C

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Applying the first law of thermodynamics, Q 

L

N

M

C CC C.0 .E

O

Q

P

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 a common 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 the charge, q on each particle (take g = 10ms-2)

Solution :    F F F q a A  AB AC

L

N

M

M

41

O

Q

P

P

2

L

N

M

M

O

Q

P

P

60 2 0 2 2 0

 cos in the direction D to AA

= 2 41 0 2 2  q a

L

N

M

M

O

Q

P

P

23in the direction D to AA

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

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

A

where, tan  sin AD AO x   3 10 1 2

(AD = 2/3 of the median length , as D is the centroid) or 2 1 4 3 2 0 2 2 

L

N

M

O

Q

P

q a =10 -3 x ( 3 10x 2) or q = 10 10 9 x  C  q316 10. x 9C

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

Solution : Consider a disc of surface charge density ' ' . Let

us calculate the electric field due to a ring of charge situated at a distance r, from the centre and having a thickness, dr.

Using the result of the previous question, we can say, dE =

kxdq x2r2 3 2

e

j

/

Now, the area of the ring , dS = 2r dr. dq 2 rdr,

Thus, E dE kx rdr x r p R   

z

z

 2 2 2 3 2 0

e

j

/ = 1 4 2 0 0 2 2 3 2  x r x r dr R . /

z

e

j

E =   2 0 1 2 2 1 2  

L

N

M

M

M

O

Q

P

P

P

x x R

e

j

/ Also, as R  , E =  

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

-- --- ---- ------ --- - FAC  FAB  FA T cos ) ( T sin  --- ---D A C B T O a mg   dr r x dx P

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ELECTROSTATICS

Problem 8: A 4fcapacitor is charged to 150 V and another 6f capacitor is charged to 200 V.V. Then they are connected across each other. Find the potential difference across them. Calculate the heat produced.

Solution: 4f charged to 150 V would have q1 = C1V1 = 600C

6f charged to 200 V would have q2 = C2V2 =

1200C

After connecting them across each other, they will have a common potential difference V Charges will readjust as q1’ and q2’.

V = q C q C q q C C q q C C 1 1 2 2 1 2 1 2 1 2 1 2 , ' ' '        = 1800 4 6   C f 

b g

[Conservation of charge] V = 180 volt. Initial energy , U i = 1 2 1 2 1 2 4 150 1 2 6 200 0 165 1 1 2 1 2 2 2 2 C V  C V 

b gb g b gb g

f V  f V  . J Final energy Uf = 1 2 1 2 4 6 180 0161 1 2 2 2 C C .V 

b

f f

gb g

.  . J Heat produced = UfUi  0 003. J

Problem 9: Three charges q1, q2 and q3 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

b

V q

g

10  W = V2 BV q 

F

H

G

kq 

I

K

J

 a q kq q a

b

g

2 1 0 2 1 2 ... (2) W = V3

b

cV q

g

3 = kq a kq a q kq q a kq q a 1 2 3 1 3 1 3 0  

F

H

G

I

K

J

  ... (3) W = W1 +W2 + W3 Wkq q   a kq q a kq q a 1 2 2 3 1 3

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

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

C34 = CC C3 C4 3 4 .  = 4f, C2,34 = 8 +4 =12f Ceq = 8 12 8 12 4 8 x   . f , q =Ceq =4.8 x20 = 48C q1 C1 q2 C 2 V q1 q2 q3 A B C a 1 2 3 4 P Q A B 10 8f 8f 8f 8f

(23)

The ‘q’ on 1 is 48 C, thus V1 = q/c = 6V [V1= 488 6 

c F  V] VPQ 10 6 4V

By symmetry of 3 and 4 , we say VAB = 2V.

Problem11: What is VA - VB in the arrangement shown? What is the condition such that VA -VB=0?

Solution : Let charge be as shown (Capacitors in series have the same charge) Take loop containg C1,C2 and E

q C q C E C C C 1 2 2 1 2 0     

L

N

M

O

Q

P

q = E. C1

From loop containing C3, C4 and E Similarly, q C q C E C C C ' ' 3 4 4 3 4 0     

L

N

M

O

Q

P

q' = E. C3 Now VA-VB = q C q C E. C C C C 2 4 1 2 3 4     

L

N

M

O

Q

P

' C1 C3 VA - VB = 0  C1C4 = C2C3 =0 or C1 /C2 = C3 / C4 OBJECTIVE

Problem 1: Two point charges each of charge +q are fixed at (+a, 0) and (-a, 0) . Another positive point charge q placed at the origin is free to move along x-axis. The charge q a origin in equilibrium 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 r i q r i  12     0 2 2 0 2 2 1 4 1 4 0  .   . ( )

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= 41 41 41 1 1 0 2 0 2 0 2  .  .  q a x q a x q a x a x       

L

N

M

M

O

Q

P

P

b g

b g

b g b g

dU dx q a x a x     

L

N

M

M

O

Q

P

P

2 0 2 2 4 1 1 

b g b g

For U to be minimum, dU dx  0, d U dx 2 2 0, 

b g

ax 2 (ax)2 ax 

b g

ax

x0, because other solution is relevant.

+ -C C C C A B q q q’ q’ E ---q q q > x y (-a, 0) (+a, 0)

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ELECTROSTATICS

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

bg

Problem 2: Two concentric conducting spheres having radii a and b are charged to q1 & q2 respectively. The potential difference between 1 & 2 will be

(a) q1a q b 0 2 0 4 4 (b) q a b 2 0 4 1 1  

F

H

G IK

J

(c) q1 a b 0 4 1 1  

F

H

G IK

J

(d) none of these.

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

v q a q b 1 0 1 0 2 1 4 1 4    

The potential on the surface of the sphere 2 is given by, v = v1-v2 v2= 1 4 1 4 0 1 0 2     q b q b v  q a q b 1 4 1 4 0 1 0 1   v q

F

H

G IK

J

a b 1 0 4 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 d

b

g

(b) 2 0 1 2 2 1  A A A A d

b

g

(c) 0A1 d (d) 0A2 d

Solution:Since the electric field between the parallel charge plates is uniformned independent of the distance, neglecting the fringe effect, the ef-fective area of the plate of area A2 is A1. Thus the capacitance between the plates is

C A d

 0 1  (C)

Problem 4:The charge flowing across the circuit on closing 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’V Where C’ is the equivalent capacitance given by C’ = C/2

Therefore Q = (C/2)V

When the key K is closed, the capacitor 2 gets short circuited and ths the charge in that capacitor comes back to the source.

 Charge flowing isQ = (C/2)V (B)

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

(a)4 ab0 ba (b) 4 b0 b a 2  q1 q2 1 2 b a > > > .... .... .... .... .... .... . + - d .... .... .... .... .... .... . A1 A20 E C C 1 2 v + -k a b > > A1 A2 d

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(c) 4 b a0

b g

 (d) none of these

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

Thus, C1 = 40 ab b a ; C2 = 4 b0 b a 2   (B)

Problem 6:The conducting spherical shells shown in the figure are connected by a conductor. The capacitance of the system is (a) 40 ab ba (b) 4 a0 (c) 4 b0 (d) 40 a b a 2 

Solution : As there will be no charge on the inner sphere, therefore the capacitance only will exist due 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 of charge density . The angle made by the thread with the vertical in equilibrium is (a) tan-1   g mg 2 0

F

H

G

I

K

J

(b) tan-1   g mg 0

F

H

G

I

K

J

(c) tan-1 2 0   g mg

F

H

G

I

K

J

(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  2qmg  

H

G

F

2qmg

I

K

J

0 0

= tan-1  A

bg

Problem 8: A point charge q moves from point P to point S. Along the path PQRS in a uniform electric field E  pointing parallel to the posi-tive direction of the x-axis. The coordinate of the points P, Q, R and S are (a, b, 0), (2a, 0, 0) , (a, -b, 0) and (0, 0, 0) respectively. The work done by the field in the above process is given by the expression

(a) qaE (b) -qaE

(c) q

F

H

(a2b2)E

IK

(d)

H

F

3qE (a2b2)

IK

Solution : The work done is independent of the path followed and is equal to

d i

qE r . , where r -= > > a b + + + + + + + + + + + +  q > > > > > < > > T cos +qE mg Tsin (  y x ... (0, 0, 0)S) > > > > > X Y ... ... ... ... ... ... .. R(a, -b, 0) Q(2a, 0, 0) p(a, b, 0) E

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ELECTROSTATICS

displacement vector PS aibj, while E Ei

WorkqE r. qaE  B

bg

ASSIGNMENTS SUBJECTIVE LEVEL-I

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

2. The charges, q1 and q2 are placed at the cor-ners of a square as shown in the figure. Find q2 such that the resultant force on q1 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 of length l making an angle  with an infinite

line of charge as shown in the figure. Find the linear 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 electric

field at the centre due to the remaining portion.

6. Twenty-seven identical mercury droops are charged simultaneously to the same potential of 10 volt. What will be the potential if all the charged drops are made to combine to form one large drop ? Assume all drops to be spherical.

7. Two charged particles having charge  1C and  1Cand 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 maximum coulombic repulsion? What is the work done in reducing the distance between them to half its value?

9. A 1F and a 2F 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 now reconnected with terminals of like sign together. Find the final charge on each capacitor and the voltage across each capaitor.

> < A B C D E O a q1 q2 q1 q2 > > > ) E S + + + + + + + )  m, q

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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 from the same point by strings or same length. The bobs are immersed in a liquid of relative permitivity rand 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 to this set of charges. When will be the potential and electric field if, in the above setup the consecutive 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 charge density of -2.0 x 10-6 C/m2. From what distance must the electron be projected, if it is to

just 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 having

square plates of edge a and plate separation d. The gap between the plate is filled with a di-electric of didi-electric constant k which varies from the left plate to the right plate as k = k0 +x. Where k0 and  are positive constants

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

8. A capacitance has square

plates, each of side a, making

S V C2 C1 V > > d l/2 l/2 > > > > d/2 d  F F F C1=10 C2=5 C3=4 100v K > >---) d 

(28)

ELECTROSTATICS

an angle as shown in the figure. Show that, for small , the

ca-pacitance is C =C a d a d 

L

N

M

O

Q

P

0 2  1 2

9. In the arrangement shown in figure, three con-centric conducting shells are shown. The charge on the shell of radius b is q0 . If the innermost and outermost shells are connected to the earth, findtheir charge densities and the potential on the shell of radius b in terms of a and q0. 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 the amount of heat generated in the circuit.

LEVEL-III

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

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

3. An insulated conductor, initially free from charge, is charged by repeated contacts with a plate which, after each contact, is replenished to a charge Q from an electrophorus. If q is the charge on the conductor after the first operation, prove that the maximum charge which can 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 with plate 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 200 V/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 and the kinetic energy of the electron as it leaves the capacitor. Take the mass and charge of electron as m and e respectively.

a b c 1 2 3 4 1 2 3 4 5

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

7. A non conducting sphere of radius R = 5cm has its

centre at the origin O of coordinate system as shown in 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 / m  3.

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

8. The electric field vector is given by Ea 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 the electrostatic 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.

OBJECTIVE LEVEL-I

1. A positive charged pendulum is oscillating in a uniform electric field as shown in figure. Its time period, as compared to that whent was uncharged;

(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 is

given a charge Q while B is earthed as shown in 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 q and radius R will be

(a)41 0  q R 3 3 2 (b) 1 40 2 3 2 q R x y o P’ --- ---> > > > > > + + + + + + + + + ++ + A B > > > > > > > > > >

References

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