unit - 11 electrostatics ug physics... · electrostatics 'rzqordghgiurpzzz vwxglhvwrgd\ frp. 2...

$: UNIT - 11 ELECTROSTATICS UG Physics... · ELECTROSTATICS 'RZQORDGHGIURPZZZ VWXGLHVWRGD\ FRP. 2 SUMMARY 1. Electric Charge : Just as masses of two particles are responsible for the$
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UNIT - 11ELECTROSTATICS

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SUMMARY1. Electric Charge : Just as masses of two particles are responsible for the gravitational force,

charges are responsible for the electric force. Electric charge is an intrinsic property of a particle.Charges are of two types : (1) Positive charges (2) Nagative charges.The force acting between two like charges is repulsive and two unlike charges it is attractivebetween .

2. Quantization of Electric Charge : The magnitude of all charges found in nature are an integralmultiple of a fundamental charge. ,neQ where e is the fundamental unit of charge.

3. Conservation of Electric Charge : Irrespective of any process taking place, the algebraic sumof electric charges in an electrically isolated system always remains constant.

4. Coulomb's Law : The electric force between two stationary point charges is directly propor-tional to the product of their charges and inversely proportional to the square of the distancebetween them.

221

02

21

rqq

41

rqqkF

If 0qq 21 then there is a repulsion between the two charges and for 0qq 22 , there is a attrac-tion between the charges.

5. Equation for Force using Columb’s Law, when two charges are placed in a medium havingdielectric constant k.

(1) The electric force F experienced by a test charge (q0) due to a source charge (q) whenboth are placed in a medium having dielectric constant k and separated by a dis-tance r, is given by :

r̂kr

qq4

1F 221

0

Fr

P

O (q)

(qo)

Here r̂ is the unit vector directed from q to q0.

(2) The equation of coulomb's force may be written as follows : r̂rk

qq4

1F2

21

0

(3) If the source charge and test charge are separated by a number of medium of thickness

1 2 3d , d , d ........ having dielectric constants ........k,k,k 321 respectively, then theelectric force on charge q0 due to a charge q is given by

0

2 2 20 1 1 2 2 3 3

qq1 ˆF r4 k d k d k d

0

20 i i

qq1 ˆF r4 k d

OR

In this equation ki is dielectric constant of medium which spreads through the distance dialong the line joining q and q0.



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For example, see the figure below :

Here, the space between the charges q and q0 is filled with medium (1, 2, 3). The thickness ofmedium 1 is d1 and its dielectric consant is k1 Similarly the thickness of medium 2 and 3 is d2 andd3 of medium 3 and their dielectric constants are k2 and k3 respectively.

6. Conditions for Equilibrium in Various Cases :Suppose three charges q1, q2 and q are situated on a straight line as shown below :

If q1 and q2 are like charges and q is of unlike charge then,

(1) Force on q1

2

12

21

1

0

11 r

qrr

q4

qF

(2) Force on q2

2

22

21

1

0

22 r

qrr

q4qF

(3) Force on q =

2

2

22

1

1

0 rq

rq

41F

Now, from above equations, it is clear that various equilibrium conditions can be as follows :

(a) Condition for 1F to be zero is, 221

21

221

22

1 rrr

qq

rrq

rq

(b) Condition for 2F to be zero is, 221

22

12

21

12

2 rrr

qq

rrq

rq

(c) Condition for F to be zero is, 22

22

1

1

rq

rq

22

21

2

1

rr

rq

If 21 q,q and q are of same type charges in nature, then,(1) Charge q will be in equilibrium, if

0rq

rq

4qF 2

2

22

1

1

0

21 2 1 12 2 2

1 2 2 2

q q q rr r q r

(2) Charges q1 and q2 will not be in equilibrium.



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7. Electric Field Intensity : The electric force acting on a unit positive charges at a given point inan electric field of a system of charges is called the electric field or the intensity of electric field

E at that point.

qFE

The SI unit of E is CN

or 1Vm .

If n21 r...,.........r,r are the position vectors of the charges n21 q.,.........q,q respectively, then the

resultant electric field at a point of position vector r is,

jn

1j3

j

j rrrr

qkE

8. Electric Dipole : A system of two equal and opposite charge, separated by a finite distance iscalled electric dipole.

Electric dipole moment a2qp

The direction of p is from the negative electric charge to the positive electric charge.9. Electric field of a dipole on the axis of the dipole at point z = z

^3

2kpE z p for z az

Electric field of a dipole on the equator of the dipole at point y = y

^3

kpE y p for y ay

10. The torque acting on the dipole place in an uniform the electric field at an angle ,

sinEp||,Ep

11. Electric Flux : Electric flux associated with surface of area A , placed in the uniform electricfield.

cosEAAE where, is the angle between AandE ,

Its SI unit is 2Nm

C or V.m.

12. Gauss's Law : The total electric flux associated with the closed surface,

0S

qE d a

where, q is the net charge enclosed by the surface.



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13. Electric field due to an infinitely long straight charged wire,

,r̂r1

2E

0

where, r is the perpendicular distance from the charged wire.

14. Electric field due to bending of charged rod,

15. Electric field due to uniformly charged thin spherical shell,

(1) Electric field inside the shell 0E (2) Electric field at a distance r from the centre outside the shell,

2

2

02 r

RrqkE

where, R = radius of spherical shell.

16. Electric field due to a uniformly charged density sphere of radius R,

(1) Electric field inside the region of the sphere, 30 0

Q r rE4 R 3



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(2) Electric field outside the sphere, 0

2

3

20 r3

Rrr

4QrE

where, Q is the total charge inside the sphere.17. The information about the work done to take an electric charge from one point to the other in a

given electric field, obtained from the quantities called electric potential and electric potentialenergy.

18. B

A

drE is the line-integral of electric field between point A and B and it shows the work done by

the electric field in taking a unit positive charge from A and B. Moreover, it does not depend on

the path and 0drE .

19. "The work required to be done against the electric field to bring a unit positive charge frominfinite distance to the given point in the electric field, is called the electric potential (V) at thatpoint".

Electric potential at point P is

P

p drEV

It unit is .voltcoulomb

Joule Symbolically C

JV

Its dimensional formula is 1321 ATLM

Absolute value of electric potential has no importance but only the change in it is important.20. "The work required to be done against the electric field to bring a given change (q) from infinite

distance to the given point in the electric field is called the electric potential energy of thatelectric charge at that point."

p

P

p qVdrEqU

The absolute value of electric potential energy has no importance, only the change in it is impor-tant.

21. Electric potential at point P, lying at a distance r from a point charge q is r

kqVp

22. The electric potential at a point at distance r from an electric dipole is

,rp

41rv 2

0 ( For r > > 2a)

Potential on its axis is ,rp

41V 2

0 Potential on its equator is 0V

23. Electric potential at a point r due to a system of point charge n21 q,.........q,q situated at position

at position n21 r,.........r,r is

n

1i i

i

rrkqV



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The electric potential at point r , due to a continuous charge distribution is

04

1rV

volume irr

'dr

The electric potential due to a spherical shell is

0

1 qV For r R4 r

and 0

1 qV For r R4 R

24. A surface on which electric potential is equal at all points is called an equipotential surface. Thedirection of electric field is normal to the equipotential surface.

25.dVEdl

gives the magnitude of electric field in the direction of dl

.

To find E from V, in general, we can use the equation

V V Vˆ ˆ ˆE i j kx y z

The direction of electric field is that in which the rate of decrease of electric potential with

distance dVdl

is maximum, and this direction is always normal to the equipotential surface.

26. The electrostatic potential energy of a system of point charges n21 q.,.........q,q situated at positions

n21 r,.......r,r is

ni j

i 1 iji j

kq qU

r

where ijij rrr

27. The electrostatic potential energy of an electric dipole in an external electric field E, is

U E p Ep cos

28. When a metallic conductor is placed in an external electric field,(i) A stationary charge distribution is induced on the surface of the conductor.(ii) The resultant electric field inside the conductor is zero.(ii) The net electric charge inside the conductor is zero.(iv) The electric field at every point on the outer surface of conductor is locally normal to thesurface.(v) The electric potential inside the region of conductor is the same every where.(vi) If there is a cavity in the conductor then, even when the conductor is placed in an external

electric field, the resultant electric field inside the conductor and also inside the cavity isalways zero.This fact is called the electrostatic shielding.



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When electric charge is placed on the metallic conductor :(i) The electric field inside the conductor is zero everywhere.(ii) The charge is distributed only on the outer surface of the conductor.

(iii) The electric field on the surface is locally normal, and is equal to n̂E0

.

(iv) If a charge is placed inside the cavity in the conductor, the electric field in the conductorremains zero.

29. "A device formed by two conductors seprated from each other is called a capacitor." Its capaci-

tance is VQC constant. The unit of C is volt

coulomb which is also called farad.

F10pF.F10F1 126

30. The effective capacitance in series connection is C then,

..........C1

C1

C1

C1

321

The effective capacitance in parallel connection is C then,

.........CCCC 321

31. The capacitance of the parallel plate capacitor is dAC 0 .

32. The energy stored in the capacitor is 2

VQ2

CVC2

QU22

and the energy density = energy

stored per unit volume ,E21 2

0 where E electric field.

33. When a dielectric is placed in an external electric field 0E , polarisation of dielectric occurs dueto electrical induction. The electric field produced by these induced charges is in the oppositedirection to the direction of external electric field. Hence the resultant electric field E, inside thedielectric is less than the external electric field 0E .

The dipole moment produced per uint volume is called the intensity of polarisation or in shortpolarisation

.n̂P b

Since EP , P E0 e ex x is called the electric susceptibility of the dielectric medium.

1 0 ex is called the permittivitty of the dielectric medium. 0

is called the relative

permittivity of that medium and it is also called the dielectric constant K.



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i.e. K

r

0

EK 1 , E .K

0ex Thus in the dielectric the electric field reduces to the thK part.

PED 0 is called the electric displacement. Gauss Law in the presence of dielectric is

written as qdsD , where q is only the net free charge.

34. when there is air (or vacuum) between the plates of a parallel plate capacitor, the capacitance is

dAC 0 . On placing a medium of dielectric constant K, the capacitance is .CK'C Thus the

capacitance becomes K times, due to the presence of the dielectric.

35. With the help of Van-De-Graf generator a potential differance of a few nillion volt can beestablished.



10

CO

NC

EPT

MA

P



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MCQFor the answer of the following questions choose the correct alternative from among thegiven ones.1. When a Piece of Polythene is rubbed with wool, a charge of –7–2 10 is developed on

polythene. The mass transferred to polythene is ..... kg.

(A) –1911.38 10 (B) –195.69 10 (C) –192.25 10 (D) –199.63 10

2. The protonic charge in 100 gm of water is .......... c

(A) 54.8 10 (B) 65.4 10 (C) 43.6 10 (D) 64.9 10

3. A copper sphere of mass 2 gm contains about 222 10 atoms. The charge on the nucleus ofeach atom is 29e. The fraction of electrons removed.

(A) –102 10 (B) –121.19 10 (C) –111.25 10 (D) –112.16 10

4. The rate of alpha partical falls on neutral spheare is 1012 per second. The time in whichsphere gets charged by 2µc is ......... sec.

(A) 2.25 (B) 3.15 (C) 6.25 (D) 1.665. A charge Q is divided into two parts and then they are placed at a fixed distance. The force

between the two charges is always maximum when the charges are .........

(A) Q Q,3 3

(B) Q Q,2 2

(C) Q 3Q,4 4

(D) Q 4Q,5 5

6. Two point charges repel each other with a force of 100 N. One of the charges is increasedby 10% and other is reduced by 10%. The new force of repulsion at the same distance wouldbe ........ N.(A) 121 (B) 100 (C) 99 (D) 89

7. Given that q1 + q2 = q if the between q1 and q2 is maximum, 1qq ...............

(A) 1 (B) 0.75 (C) 0.25 (D) 0.5

8. Two small conducting sphere of equal radius have charges +1µc and – 2µc respectively andplaced at a distance d from each other experience force F1. If they are brought in contact andseparated to the same distance, they experience force F2. The ratio of F1 to F2 is ..........(A) –8 : 1 (B) 1 : 2 (C) 1 : 8 (D) –2 : 1

9. Three charges, each of value Q, are placed at the vertex of an equilateral triangle. A fourthcharge q is placed at the centre of the triangle. If the charges remains stationery then,q = ...............

(A) Q2 (B)

Q–3 (C)

Q–2 (D)

Q3



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10. Tw o small charged spheres repal each other w i th a force –32 10 N . The charge on one sphereis twice that of the other. When these two spheares displaced 10 cm further apart the forceis –45 10 N , then the charges on both the spheres are ........

(A) –19 –191.6 10 C, 3.2 10 C (B) –19 –193.4 10 C,11.56 10 C

(C) –19 –1933.33 10 C, 66.66 10 C (D) –19 –192.1 10 C, 4.41 10 C

11. Three charges –q1, + q2 and –q3 are placedas shown in figure. The x component ofthe force on –q1 is proportional to .........

(A) 322 2

qq – sinθb a

(B) 322 2

qq – cosθb a

(C) 322 2

qq + sinθb a

(D) 322 2

qq + cosθb a

12. Two equal negative charges –q are fixed at points (o, a) and (o, –a). A positive charge Q isreleased from rest at the point (2a, o) on the X - axis. The charge Q will ..........(A) move to the origin and remain at rest there(B) execute simple harmonic motion about the origin(C) move to infinity(D) execute oscillations but not simple harmonic motion

13. Four charges, each equal to –Q, are placed at the corners of a square and a charge +q isplaced at its centre. If the system is in equilibrium, the value of q is .........

(A) Q 1 2 24

(B) Q– 1 2 24

(C) Q– 1 2 22

(D) Q 1 2 22

14. For the system shown in figure, if the

resultant force on q is zero, then

Q = ...............

(A) –2 2Q (B) 2 2Q

(C) 2 3Q (D) –3 2Q

1–q1 +q2

X

Y

ba

–q3

Q qFA

FAa

q(0, a) Q

a



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15. Two point positive charges q each are placed at (–a, o) and (a, o). A third positive charge qo isplaced at (o, y). For which value of y the force at qo is maximum .........

(A) a (B) 2a (C) a2 (D)

a3

16. Two identical charged spheres suspended from a common point by two massless strings oflength l are initially a distance d (d << l ) apart because of their mutual repulsion. The chargebegins to leak from both the spheres at a constant rate. As a result the spheres approach eachother with a velocity υ . Then function of distance x between thembecomes ...........

(A) αv x (B) –12αv x (C) –1αv x (D)

12αv x

17. Three identical spheres each having a charge q and radius R, are kept in such a way thateach touches the other two spheares. The magnitude of the electric force on any spheredue to other two is ...........

(A) 2

0

1 52 4π 4 RR q

(B)

2

0

1 28π 3 R

q

(C) 2

0

1 34π 4 R

q

(D) 2

0

1 3–8π 2 R

q

18. Two equal negative charges –q are fixed at points (o, a) and (o, –a) on the Y axis. Apositive charge q is released from rest at the point x (x < < a) on the X-axis, then thefrequency of motion is .........

(A) 2

30

qπ ma (B)

2

30

2q4π ma (C)

2

30

4q2π ma (D)

2

30

q2π ma

19. Two identical balls having like charges and placed at a certain distance apart repel eachother with a certain force. They are brought in contact and then moved apart to a distanceequal to half their initial separation. The force of repulsion between them increases 4.5times in comparison with the initial value. The ratio of the initial charges of the balls is........(A) 4 : 1 (B) 6 : 1 (C) 3 : 1 (D) 2 : 1

20. A point charge q is situated at a distance r from one end of a thin conducting rod of lengthL having a charge Q (uniformly distributed along its length). The magnitude of electricforce between the two, is ...............

(A) 2kqQ

r(r + L) (B) kqQ

r(r + L) (C) kqQ

r(r – L) (D) kQ

r(r + L)

21. Two point charges of +16µc and –9µc are placed 8 cm apart in air. ............... distance ofa point from –9µc charge at which the resultant electric field is zero.

(A) 24 cm (B) 9 cm (C) 16 cm (D) 35 cm



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22. Point charges 4µc and 2µc are placed at the vertices P and Q of a right angle triangle PQR

respectively. Q is the right angle, –2PR = 2 10 m and –2QR = 10 m . The magnitude anddirection of the resultant electric field at c is .........

(A) 9 –1 04.28×10 NC , 45 (B) 8 –1 02.38×10 NC , 40.9

(C) 4 –1 01.73×10 NC , 34.7 (D) 10 –1 04.9 ×10 NC , 34.7

23. An inclined plane making an angle of 30o with the horizontal is placed in an uniformelectric field E = 100 Vm–1. A particle of mass 1 kg and charge 0.01 c is allowed to slidedown from rest from a height of 1m. If the coefficient of friction is 0.2 the time taken bythe particle to reach the bottom is .......... sec.(A) 2.337 (B) 4.337 (C) 5 (D) 1.337

24. A small sphere whose mass is 0.1 gm carries a charge of –103 10 C and is tieup to one endof a silk fibre 5 cm long. The other end of the fibre is attached to a large vertical conductingplate which has a surface charge of –6 –225 10 Cm , on each side. When system is freelyhanging the angle fibre makes with vertical is ...............(A) 41.80 (B) 450 (C) 40.80 (D) 45.80

25. A Semicircular rod is charged uniformly with a total charge Q coulomb. The electric fieldintensity at the centre of curvature is .......

(A) 2

2KQπr (B) 2

3KQπr (C) 2

KQπr (D) 2

4KQπr

26. The electron is projected from a distance d and withinitial velocity 0υ parallel to a uniformly charged flatconducting plate as shown in figure. It strikes the plateafter travelling a distance l along the direction. The sur-face charge density of conducting plate is equal to

(A) 0 02d mυe

l (B) 2

00d mυe

l

(C) 0 0d mυe

l (D)

200

2

2d mυe

l

27. Two point masses m each carrying charge –q and +q are attached to the ends of a masslessrigid non-conducting rod of length l. The arrangement is placed in a uniform electric fieldE such that the rod makes a small angle 50 with the field direction. The minimum timeneeded by the rod to align itself along the field is ........

(A) 2mt = π3qE

l(B)

π mt =2 2qE

l(C)

mt =qE

l(D)

mt = 2πE

l

v0p

X

Y

----l-----



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28. Two uniformaly charged spherical conductors A and B having radius 1mm and 2mm areseparated by a distance of 5 cm. If the spheres are connected by a conducting wire then inequilibrium condition, the ratio of the magnitude of the electric fields at the surfaces ofspheres A and B is .........(A) 4 : 1 (B) 1 : 2 (C) 2 : 1 (D) 1 : 4

29. Let 4

QP(r) = rπR be the charge density distribution for a solid sphere of radius R and total

charge Q. For a point ‘P’ inside the sphere at distance r1 from the centre of the sphere themagnitude of electric field is

(A) 20 1

Q4π r (B)

21

40

Qr4π R (C)

21

40

Qr3π R (D) 0

30. Two point charges 1q 2µc and 2q 1µc are placed atdistance b = 1cm and a = 2 cm from the origin on the yand x axes as shown in figure. The electric field vectorat point P (a, b) will subtend an angle with the X - axisgiven by,(A) tan θ 4 (B) tan θ 1

(C) tan θ 3 (D) tan θ 2

31. A simple pendulum consists of a small sphere of mass m suspended by a thread of lengthl. The sphere carries a positive charge q. The pendulum is placed in a uniform electricfield of strength E directed Vertically upwards. If the electrostatic force acting on thesphere is less than gravitational force the period of pendulum is

(A)

12

qEm

T = 2πg –

l

(B) 12

T = 2π lg

(C)

12

T = 2π qEm

lg +

(D) 12mlT = 2

qE

32. Consider a system of three charges q/3, q/3 and –2a/3placed at points A, B and C respectively as shown in thefigure. It the radius of the circle is R and 0CAB = 60then the electric field at centre 0 is ........

(A) 20

q8π R (B)

2

20

q54π R

(C) 20

q6π R (D) 0

O

B

C

A

600

X

Y

q1

b

a q2

P(a, b)



16

33. In Millikan’s oil drop experiment an oil drop carrying a charge Q is held stationary by a p.d.2400 v between the plates. To keep a drop of half the radius stationary the potential differ-ence had to be made 600 v. What is the charge on the second drop ?

(A) 3Q2 (B)

Q4 (C) Q (D)

Q2

34. Equal charges q are placed at the vertices A and B of an equilateral triangle ABC of side a.The magnitude of electric field at the point c is .........

(A) 2

Kqa (B) 2

3Kqa

(C) 2

2Kqa

(D) 20

q2 t a

35. An electric dipole is placed along the x-axis at the origin o. A point P is at a distance of 20

cm from this origin such that OP makes an angle π3 with the x-axis. If the electric field at

P makes an angle with the x-axis, the value of would be ...........

(A) –1π 3tan3 2

(B)

π3 (C) 2π

3(D) –1 3tan

2

36. A particle having a charge of –191.6 10 C enters between the plates of a parallel platecapaciter. The initial velocity of the particle is parallel to the plates. A potential differenceof 300v is applied to the capacitor plates. If the length of the capacitor plates is 10cm andthey are separated by 2cm, Calculate the greatest initial velocity for which the particlewill not be able to come out of the plates. The mass of the particle is –2412 10 kg .

(A) 4 m10s (B) 2 m10

s (C) –1 m10s (D) 3 m10

s

37. A charged particle of mass 1 kg and charge 2µc is thrown from a horizontal ground at an

angle o = 45 with speed 20m/s. In space a horizontal electric field 7E = 2 10 V/m exist.The range on horizontal ground of the projectile thrown is ............... .(A) 100 m (B) 50 m (C) 200 m (D) 0 m

38. If electron in ground state of H-atom is assumed in rest then dipole moment of electronproton system of H-atom is ............... .

Orbit radius of H atom in ground state is 0

0.56 A .

(A) –290.253×10 cm (B) –290.848×10 cm (C) –290.305 ×10 cm (D) –281.205×10 cm

39. At what angle θ a point P must be located from dipole axis so that the electric fieldintensity at the point is perpendicular to the dipole axis ?(A) 530 to 540 (B) 500 to 510 (C) 450 to 460 (D) 520 to 530



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40. A n electric dipole is placed at an angle of 60o with an electric field of intensity 105 NC–1.It experiences a torque equal to 8 3Nm . If the dipole length is 2cm then the charge on thedipole is ........... c.

(A) 3–8 10 (B) –48.54 10 (C) –38 10 (D) –60.85 10

41. An electric dipole coincides on z axis and its mid point is on origin of the cartesianco-ordinate system. The electric field at an axial point at a distance z from origin is E(z)

and electric field at an equatorial point at a distance y from origin is (y)E

(z)

(y)

EE

(y = z > > a) = ...............

(A) 1 (B) 2 (C) 4 (D) 342. An oil drop of 12 excess electrons is held stationary under a constant electric field of

4 –12.55×10 Vm . If the density of the oil is 1.26 gm/cm3 then the radius of the drop is.......... m.

(A) –79.81 10 (B) –79.29 10 (C) –89.38 10 (D) –89.34 10

43. A Charge q is placed at the centre of the open end of cylindrical vessel. The flux of theelectric field through the surface of the vessel is ...........

(A) 0

q (B)

0

q2 (C)

0

2q (D) Zero

44. The inward and outward electric flux for a closed surface in units of Nm2/C are respec-tively 38 10 and 34 10 . Then the total charge inside the surface is ........... c.

(A) 3

0

–4 10

(B) 3–4 10 (C) 34 10 (D) 30–4 10

45. A sphere of radius R has a uniform distribution of electric charge in its volume. At adistance x from its centre, (for x < R), the electric field is directly proportional to ..........(A) x (B) x–1 (C) x–2 (D) x2

46. The electric flux for gaussian surface A that enclose the chrged particles in free space is.............(given q1 = –14nc, q2 = 78.85 nc, q3 = –56nc)

(A) 104 Nm2/C (B) 103 Nm2/C (C) 3 26.2 ×10 Nm /C (D) 4 26.3×10 Nm /C



18

47. A hollow cylinder has a charge q coulomb within it. If is the electric flux in units of volt-meter associated with the curved surface B, the flux linked with the plane surface A in unitsof volt-meter will be .............

(A) 0

1 q –2

(B) 0

q2

(C) 3

(D) 0

q –

48. An infinitly long thin straight wire has uniform linear charge density of 1 c/m3 . Then, the

magnitude of the electric intiensity at a point 18 cm away is .......... NC–1.

(A) 110.66 10 (B) 111.32 10 (C) 110.33 10 (D) 113 10

49. Two points are at distances a and b (a < b) from a long string of charge per unit length .The potential difference between the points in proportional to .............

(A) blna

(B)

2

2

blna (C)

02π

blna (D)

02π

blna

50. A long string with a charge of per unit length passes through an imaginary cube of edgel. The maximum possible flux of the electric field through the cube will be ...........

(A) 0

l3 (B)

0

l (C)

0

l2 (D)

2

0

6 l

51. Two Points P and Q are maintained at the Potentials of 10 v and –4 v, respectively. Thework done in moving 100 electrons from P to Q is ..............

(A) –162.24 10 J (B) –17–9.60 10 J (C) –16–2.24 10 J (D) –179.60 10 J

52. The electric Potential V at any Point o (x, y, z all in metres) in space is given by V = 4x2

volt. The electric field at the point (1m, 0.2m) in volt/metre is ......(A) 8, along negative x - axis (B) 8, along positives x - axis(C) 16, along negative x - axis (D) 16, along positives x - axis

53. Charges of –910+ 10 C3 are placed at each of the four corners of a square of side 8cm. The

potential at the intersection of the diagonals is ....

(A) 150 2 Volt (B) 900 2 Volt (C) 1500 2 Volt (D) 900 2 2 Volt

C

AB



19

54. Three charges 2q, –q, –q are located at the vertices of an equilateral triangle. At the centreof the triangle. (A) The Field is Zero but Potential is non - zero(B) The Field is non - Zero but Potential is zero(C) Both field and Potential are Zero(D) Both field and Potential are non- Zero

55. In the electric field of a point charge q, acertain charge is carried from point A to B, C,D and E. Then the work done ....(A) Is least along the Path AB(B) Is least along the Path AD(C) Is Zero along all the Path AB, AC, and(D) Is least along AE

56. Three concentric spherical shells have radii a, b and c (a < b < c) and have surface chargedensities σ , –σ and σ respectively. If VVA, VVB and VVC denote the Potentials of the threeshells, then for c = a + b, we have

(A) VC = VB = VA (B) c B AV = V V (C) c B AV V V (D) c A BV V V

57. The electric Potential at a point P (x, y, z) is given by V = –x2y – xz3 + 4 The electric field

E at that point is ..........

(A) 3 2 2i (2 + ) + j + k3 xy z x xz (B) 2 2 2i 2 + j ( + ) + k (3 – ) xy x y xy y

(C) 3 2i + j + k z xyz z (D) 3 2 2i (2 – ) + j + k3 xy z xy z x

58. Three particles, each having a charge of 10µc are placed at the corners of an equilateraltriangle of side 10 cm. The electrostatic potential energy of the system is

9 2 2

o

1Given = 9 × 10 N.m /c4π

.

(A) 100 J (B) 27 J (C) Zero (D) Infinite59. Four equal charges Q are placed at the four corners of a square of each side is ‘a’. Work

done in removing a charge - Q from its centre to infinity is ...

(A) 0 (B) 2

0

2Qπ a (C)

2

0

2Q4π a (D)

2

0

Q2π a

BC D

E

A

+q



20

60. Two charged spheres of radii R1 and R2 having equal surface charge density. The ratio oftheir potential is ...

(A) 2

1

RR (B)

2

2

1

RR

(C) 2

1

2

RR

(D) 1

2

RR

61. Two equal charges q are placed at a distance of 2a and a third charge -2q is placed at themidpoint. The potential energy of the system is ....

(A) 2

0

9q8π a (B)

2

0

q8π a (C)

2

0

7q–8π a (D)

2

0

6q8π a

62. Two point charges 100µc and 5µc are placed at points A and B respectively with AB = 40cm. The work done by external force in displacing the charge 5µc from B to C where BC

= 30 cm, angle πABC = 2 and

9 2 2

0

1 9 10 Nm /c4π

.

(A) 9 J (B) 9 J25 (C)

81 J20 (D)

9– J4

63. The electric potential V is given as a function of distance x (metre) by V = (5x2 + 10x – 9)volt. Value of electric field at x = 1 is .....

(A) v–20m (B)

v6m (C)

v11m (D)

v–23m

64. A sphere of radius 1cm has potential of 8000 v, then energy density near its surfacewill be ...

(A) 53

J64 ×10m (B) 3

J2.83m (C) 3

3

J8×10m (D) 3

J32m

65. If a charged spherical conductor of radius 10cm has potential v at a point distant 5 cmfrom its centre, then the potential at a point distant 15cm from the centre will be .....

(A) 1 V3 (B)

3 V2 (C) 3V (D)

2 V3

66. Electric charges of +10µc, 5µc, –3µc and 8µc are placed at the corners of a square of side

2m the potential at the centre of the square is ......

(A) 1.8 V (B) 51.8 ×10 V (C) 61.8 ×10 V (D) 41.8 ×10 V

67. Two positive point charges of 12µc and 8µc are 10 cm apart each other. The work done inbringing them 4cm closer is .....(A) 5.8 J (B) 13eV (C) 5.8eV (D) 13 J



21

68. The displacement of a charge Q in the electric field 1 2 3E = e i + e j + e k

is r = ai + b j .

The work done is .....

(A) 2 21 2Q e + e a + b (B) 2 2

1 2Q e + e a + b

(C) 1 2Q ae + be (D) 2 21 2Q ae + be

69. If an electron moves from rest from a point at which potential is 50 volt to another pointat which potential is 70 volt, then its kinetic energy in the final state will be .....

(A) 1 N (B) –183.2 ×10 J (C) –103.2 ×10 J (D) 1 dyne

70. Three charges Q, + q and + q are placed at the verticlesof a right-angled triangle as shown. The net electrostaticenergy of the configuration is zero if Q is equal to .....

(A) –2q (B) –q

1 + 2

(C) + q (D) –2q

2 + 271. Two electric charges 12µc and –6µc are placed 20cm apart in air. There will be a point P

on the line joining these charges and outside the region between them, at which the elec-tric potential is zero. The distance of P from –6µc charge is ........

(A) 0.20 m (B) 0.10 m (C) 0.25 m (D) 0.15 m72. In the rectangle, shown below, the two corners have

charges 1q = –5µc and 2q = +2.0µc . The work done inmoving a charge 3µc from B to A is

10 2 2

0

1take = 10 Nm /c4π

.

(A) 5.5 J (B) 2.8 J (C) 3.5 J (D) 4.5 J73. 4 Points charges each +q is placed on the circumference of a circle of diameter 2d in such

a way that they form a square. The potential at the centre is ......

(A) 0 (B) 4dq (C)

q4d (D)

4qd

74. Three identical charges each of 2µc are placed at the vertices of a triangle ABC as shownin the figure. If AB + AC = 12 cm and AB . AC = 32cm2, the potential energy of the chargeat A is .....(A) 1.53 J (B) 5.31 J (C) 1.35 J (D) 3.15 J

+q+q

Q

a

A

q2

q1

B------ 15 cm -------



22

75. A ball of mass 1 gm and charge 10–8 c moves from a point A, where the potential is 600 voltto the point B where the potential is zero. Velocity of the ball of the point B is 20cm/s. Thevelocity of the ball at the point A will be .....

(A) m16.8s (B)

cm22.8s (C)

cm228s (D)

m168s

76. Three charges Q, +q and +q are placed at the vertices of anequilateral triangle of side l as shown in the figure. It thenet electrostatic energy of the system is zero, then Q is equal(A) –q (B) +q

(C) Zero (D) q–2

77. Electric potential at any point is V = –5x + 3y + 15z , then the magnitude of the electricfield is ..... N/C.

(A) 3 2 (B) 4 2 (C) 7 (D) 5 2

78. A small conducting sphere of radius r is lying concentrically inside a bigger hollow con-ducting sphere of radius R. The bigger and smaller sphere are charged with Q and q (Q >q) and are insulated from each other. The potential difference between the sphers will be......

(A) 0

1 q Q–4π r R

(B) 0

1 q q–4π r R

(C) 0

1 Q q+4π R r

(D) 0

1 q Q–4π R r

79. If 3 charges are placed at the vertices of equilateral triangle of charge ‘q’ each. What is thenet potential energy, if the side of equilateral triangle is l cm.

(A) 2

0

1 3q4π l (B)

2

0

1 2q4π l (C)

2

0

1 q4π l (D)

2

0

1 4q4π l

80. If identical charges (–q) are placed at each corner of a cube of side b, then electric poten-tial energy of charge (+q) which is placed at centre of the cube will be .....

(A) 2

0

4q–3π b (B)

2

0

8 2q4π b (C)

2

0

–8 2qπ b (D)

2

0

–4 2q4π b

81. A simple pendulum of period T has a metal bob which is negatively charged. If it isallowed to ascillate above a positively charged metal plate, its period will ........(A) Remains equal to T (B) Less than T(C) Infinite (D) Greater than T

+q +q

Q

l l

l



23

82. A charged particale of mass m and charge q is released from rest in a uniform electric fieldE. Neglecting the effect of gravity, the kinetic energy of the charged particale after ‘t’second is ......

(A) 2

2

Eq m2t

(B) 2 2 2E q t2m

(C) 2 22E t

qm (D) Eqm

t

83. A thin spherical conducting shell of radius R has a charge q. Another charge Q is placed at

the centre of the shell. The electrostatic potential at a point p a distance R2 from the centre

of the shell is .....

(A) 0

(q + Q) 24π R (B)

0 0

2Q 2q–4π R 4π R (C)

0 0

2Q q4π R 4π R

(D)

0

2Q4π R

84. Two point charges –q and +q are located at points (o, o, –a) and (o, o, a) respectively. Thepotential at a point (o, o, z) where z > a is ......

(A) 2 20

2qa4π z a (B)

0

q4π a (C) 2

0

qa4π z (D) 2 2

0

2qa4π z – a

85. Point charges 1q = 2µc and 2q = –1µc are kept at points x = 0 and x = 6 respectively..Electrical potential will be zero at points .....(A) x = –2, x = 2 (B) x = 1, x = 5 (C) x = 4, x = 12 (D) x = 2, x = 9

86. Two thin wire rings each having a radius R are placed at a distance d apart with their axescoinciding. The charges on the two rings are +q and –q. The potential difference betweenthe centres of the two rings is ....

(A) 0 (B) 2 20

Q 1 1–2π R R + d

(C) 2 2

0

Q 1 1–4π R R + d

(D) 2

0

QR4π d

87. Tw o charges q1 and q2 are placed 30cm apart, as shownin the figure. A third charge q3 is moved along the arc ofa circle of radius 40 cm from C to D. The change in the

potential energy of the system is 3

0

q K4π , where k is .....

(A) 8q2 (B) 8q1 (C) 6q2 (D) 6q1

q3

q2B

DAq1

40 c

m

30 cm

C



24

88. Figure shows a triangular array of three pointcharges. The electric potencial v of thesesource charges at the midpoint P of the base

of the triangle is 9 2 2

0

1 9 ×10 Nm /c4π

(A) 55 KV (B) 63 KV(C) 49 KV (D) 45 KV.

89. Charges +q and –q are placed at point A and Brespectively which are a distance 2L apart, C is themidpoint between A and B. The work done inmoving a charge +Q along the semicircle CRDis .......

(A) 0

qQ2π L (B)

0

–qQ6π L (C)

0

qQ6π L (D)

0

qQ4π L

90. N identical drops of mercury are charged simultaneously to 10 volt. when combined toform one large drop, the potential is found to be 40 volt, the value of N is ......(A)4 (B)6 (C) 8 (D) 10

91. Tw o paral lel plate ai r capaci tors have thei r plate areas 100 and 500 cm2 respectively. Ifthey have the same charge and potential and the distance between the plates of the firstcapacitor is 9.5 mm, what is the distance between the plates of the second capacitor ?(A) 0.25 cm (B) 0.50 cm (C) 0.75 cm (D) 1 cm

92. The effective capacitances of two capacitors are 3µF and 16µF , when they are connectedin series and parallel respectively. The capacitance of each capacitor is

(A) 2µF,14µF (B) 4µF,12µF (C) 6µF, 8µF (D) 10µF, 6µF

93. An electrical technician requires a capacitance of 2µF in a circuit across a potential dif-ference of 1KV. A large number of 1µF capacitors are available to him, each of which canwithstand a potential difference of not than 400 V. suggest a possible arrangement thatrequires a minimum number of capacitors.(A) 2 rows with 2 capacitors (B) 4 rows with 2 capacitors(C) 3 rows with 4 capacitors (D) 6 rows with 3 capacitors

94. Two spherical conductors of radii r1 and r2 are at potentials V1 and V2 respectively, thenwhat will be the common potential when the conductors are brought in contant ?

(A) 1 1 2 2

1 2

r v + r vr + r

(B) 1 1 2 2

1 2

r v + r vr – r

(C) 1 1 2 2

1 2

r v – r vr + r

(D) None of these

q = 3 10 C3 -6

l l

|– 0.2 m –| |– 0.2 m –| q = -2 10 C2 -6q = 1 10 C1 -6

0.3m

A C B D

R



25

95. A 5µF capacitor is charged by a 220 V supply. It is then disconnected from the supply and is

connected to another uncharged 2.5µF capacitor. How much electrostatic energy of thefirst capacitor is lost in the form of heat and electromagnetic radiation ?

(A) 0.02 J (B) 0.121 J (C) 0.04 J (D) 0.081 J

96. Find the equivalent capacitance of thesystem across the terminals A and B. Allthe capacitors have equal capacitances.

(A) 2C (B) 4C(C) 3C (D) 5C

97. Capacitance of a parallel plate capacitor becomes 43 times its original value if a dielectric

slab of thickness t = d/2 is inserted between the plates (d is the separation between theplates). The dielectric constant of the slab is(A) 8 (B) 4 (C) 6 (D) 2

98. The plates of a parallel capacitor are charged up to 100 V. If 2 mm thick plate is insertedbetween the plates, then to maintain the same potential difference, the distance betweenthe capacitor plates is increased by 1.6mm the dielectric constant of the plate is(A) 5 (B) 4 (C) 1.25 (D) 2.5

99. A parallel plate air capacitor has a capacitance 18µF . If the distance between the plates is

tripled and a dielectric medium is introduced, the capacitance becomes 72µF . The dielec-tric constant of the medium is(A) 4 (B) 12 (C) 9 (D) 2

100. Taking earth to be a metallic spheres, its capacity will approximately be

(A) 66.4 ×10 F (B) 700 F (C) 711 F (D) 700 pF

101. A parallel plate capacitor has the space between its plates filled by two slabs of thickness

d2 each and dialectric constant K1 and K2. If d is the plate separation of the capacitor, then

capacity of the capacitor is ...............

(A) 0 1 2

1 2

2d K + KA K K

(B) 0 1 2

1 2

2A K Kd K + K

(C) 0 1 2

1 2

2A K + Kd K K

(D) 01 2

2A K + Kd

A

B

C

C C

C CC



26

102. For the circuit shown in figure the charge on4µF capacitor is

(A) 20µc (B) 40µc

(C) 30µc (D) 54µc

103. The capacitors of capacitance 4µF, 6µF and 12µF are connected first in series and then inparallel. What is the ratio of equivalent capacitance in the two cases ?(A) 2 : 3 (B) 11 : 1 (C) 1 : 11 (D) 1 : 3

104. Large number of capactors of rating 10µF/200V V are available. The minimum number ofcapacitors required to design a 10µF/700V capacitor is

(A) 16 (B) 8 (C) 4 (D)7105. A variable condenser is permanently connected to a 100 V battery. If capacitor is changed

from 2µF to 10µF . then energy changes is equal to

(A) –22 ×10 J (B) –22.5 ×10 J (C) –26.5×10 J (D) –24 ×10 J

106. Two positive point charges of 12µc and 8µc are placed 10 cm apart in air. The work doneto bring them 4 cm closer is(A) Zero (B) 4.8 J (C) 3.5 J (D) 5.8 J

107. 1000 similar electrified rain drops merge together into one drop so that their total chargeremains unchanged. How is the electric energy affected ?(A) 100 times (B) 200 times (C) 400 times (D) 102 times

108. There are 10 condensers each of capacity 5µF . The ratio between maximum and mini-mum capacities obtained from these condensers will be(A) 40 : 1 (B) 25 : 5 (C) 60 : 3 (D) 100 : 1

109. A parallel plate capacitor is made by stocking n equally spaced plates connected alter-nately. If the capacitance between any two plates is x, then the total capacitance is,

(A) nx (B) nx2 (C) nx (D) (n – 1)x

110. For the circuit shown figure, which ofthe following statements is true ?

(A) With S1closed V1 = 15V, V2 = 20V

(B) With S3 closedV1 = V2 = 20V

(C) With S1 and S3 closed V1 = V2 = 0

(D) With S1and S3closed V1 = 30V, V2 = 20V

1 F

5 F

4 F

3 F

10V

+ –

S1 S3 S2v1=30v v2=20v

C1=2pf C2=3pf



27

111. Two identical metal plates are given positive charges Q1 and Q2 (< Q1) respectively. If theyare now brought close to gether to form a parallel plate capacitor with capacitance C, thepotential difference between them is

(A) 1 2(Q Q )2c (B) 1 2(Q Q )

c (C) 1 2(Q – Q )

2c(D) 1 2(Q – Q )

c

112. In the circuit arrangement shown in figure, the valueof C1 = C2 = C3 = 30 pF and C4 = 120 pF. If thecombination of capacitors is charged with 140VDC supply, the potencial differences across the fourcapacitors will be respectively

(A) 80, 40, 40 and 20V (B) 20, 40, 40 and 80V

(C) 35, 35, 35 and 35V (D) 80, 20, 20 and 20V

113. In the arrangement of capacitors shownin figure, each capacitor is of 9µF , then

the equivalent capacitance between thepoints A and B is

(A) 18µF (B) 9µF

(C) 15µF (D) 4.5µF

114. The electric potencial V at any point x, y, z (all in metre) in space is given by V = 4x2 volt.The electric field at the point (1m, 0, 2m) in Vm–1 is

(A) + 8i (B) – 8i (C) – 16i (D) +16i

115. Two air capacitors A = 1 µF , B = 4 µF are connected in series with 35 V source. When a

medium of dielectric constant K = 3 is introduced between the plates of A, change on thecapacitor changes by

(A) 16 µc (B) 32 µc (C) 28 µc (D) 60 µc

116. A parallel plate capacitor with air between the plates has a capacitance of 9 pF. Theseparation between its plates is d. The space between the plates is now filled with twodielectrics. One of the dielectric constant K1 = 3 and thickness d/3 while the other one hasdielectric constant K2 = 6 and thickness 2d/3. Capacitance of the capacitor is now

(A) 1.8 pF (B) 20.25 pF (C) 40.5 pF (D) 45 pF

C1

140V

+ –

C4

C3

C2

C1

C2 C3

C4

A

B



28

117. A thin spherical shell of radius R has change Q spread uniformly over its surface. Which ofthe following graphs, figure most closely represents the electric field E (r) produced by theshell in the range 0 r < , where r is the distance from the centre of the shell.

(A) (B)

(C) (D)

118. A parallel plate condenser with dielectric of constant K between the plates has a capacity Cand is charged to potential V volt. The dielectric slab is slowly removed from between theplates and reinserted. The network done by the system in this process is

(A) Zero (B) 21 (K – 1)cv2 (C) (K – 1)cv2 (D) 2 (K – 1)cv

K

119. Charges are placed on the vertices of a square as shown. Let E be the electric field and V

the potential at the centre. If the charges on A and B are interchanged with those on D andC respectively then

(A) EChange V remains unchanged

(B) E remains unchanged , V changes

(C) Both E

and V change

(D) E

and V remain unchanged

120. The potencial at a point x (measured in µm ) due to some charges situated on the x-axis is

given by 2

20V( ) = Volt– 4

xx . The electric field at = 5µmx is given by

(A) 53 Vm–1 and in positive x - direction

(B) 109 VVm–1 and in positive x - direction

(C) 109 VVm–1 and in positive x - direction

(D) 53 VVm–1 and in positive x - direction

E

r R

E

r R

E

r R

E

r R

A B

C D

q q

-q -q



29

121. A battery is used to charge a parallel plate capacitor till the potential difference between theplates becomes equal to the electromotive force of the battery. The ratio of the energystored in the capacitor and work done by the battery will be

(A) 12 (B)

21 (C) 1 (D)

14

122. Two spherical conductors A and B of radii 1mm and 2mm are separated by a distance of5mm and are uniformly charged. If the spheres are connected by a conducting wire thenin equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces ofsphere of A and B is(A) 1 : 2 (B) 2 : 1 (C) 4 : 1 (D) 1 : 4

123. The following arrangement consists of fiveidentical metal plates marked 1, 2, 3, 4 and 5parallel to each other. Area of each plate is Aand separation between the successive platesis d. The capacitance between P and Q is

(A) 0A5d

(B) 0A73 d

(C) 0A5

3 d

(D) 0A43 d

124. A parrallel plate capacitor of capacitance 5µF and plate separation 6 cm is connected to a1 V battery and charged. A dielectric of dielectric constant 4 and thickness 4 cm is intro-duced between the plates of the capacitor. The additional charge that flows into the ca-pacitor from the battery is(A) 2µc (B) 5µc (C) 3µc (D) 10µc

125. For circuit the equivalent capacitance betweenpoints P and Q is(A) 6 C (B) 4 C

(C) 3 C2 (D)

6 C11

126. Four identical capacitors are connected in series with a 10 V battery as shown in thefigure. Potentials at A and B are(A) 10 V, 0 V (B) 5 V, –5V (C) 7.5 V, –2.5 V (D) 7.5 V, 2.5 V

127. 64 identical drops of mercury are charged simultaneously to the same potential of 10 volt.Assuming the drops to be spherical, if all the charged drops are made to combine to formone large drop, then its potential will be(A) 100 V (B) 320 V (C) 640 V (D) 160 V

128. Two metal plate form a parallel plate capacitor. The distance between the plates is d. Ametal sheet of thickness d/2 and of the same area is introduced between the plates. Whatis the ratio of the capacitance in the two cases ?(A) 4 : 1 (B) 3 : 1 (C) 2 : 1 (D) 5 : 1

12345

p

q

P Q



30

129. The circular plates A and B of a parrallel plate air capacitor have a diameter of 0.1 m and are–32 10 m apart. The plates C and D of a similar capacitor have a diameter of 0.12 m and

are –33 10 m apart. Plate A is earthed. Plates B and D are connected together. Plate C isconnected to the positive pole of a 120 V battery whose negative is earthed, The energystored in the system is

(A) 0.1224µJ (B) 0.2224µJ (C) 0.4224µJ (D) 0.3224µJ

130. Two parallel conducting plates of area A = 2.5m2 eachare placed 6 mm apart and are both earthed. A thirdplate, identical with the first two, is placed at a dis-tance of 2 mm from one of the earthed plates and isgiven a charged of 1 C. The potencial of the centralplate is

(A) 76 ×10 V (B) 73×10 V (C) 72 ×10 V (D) 74 ×10 V

131. What is the energy stored in the capacitor between terminals Aand B of the network shown in the figure ? (Capacitance of eachcapacitor C = 1 F )

(A)Zero (B) 50 µJ

(C) 12.5 µJ (D) 25 µJ

132. Two identical capacitors 1 and 2 are connected in series to a battery as shown in figure.Capacitor 2 contains a dielectric slab of constant K. Q1 and Q2 are the charges stored in 1and 2. Now, the dielectric slab is removed and the corresponding charges are Q’1 and Q’2.Then

(A) 1

1

1

Q K + 1=Q K

(B) 1

11

1

Q K=Q 2

(C) 1

2

2

Q K +1=Q 2K

(D) 1

2

2

Q K +1=Q K

133. A parallel plate capacitor has plate of area A and separation d. It is charged to a potentialdifference Vo. The charging battery is disconnected and the plates are pulled apart to threetimes the initial separation. The work required to separate the plates is

(A) 2

0 0A Vd

(B) 2

0 0A V2d (C)

20 0A V

3d (D)

20 0A V

4d

134. Two identical capacitors have the same capacitance C. one of them is charged to a potentialV1 and the other to V2. The negative ends of the capacitors are connected together. Whenthe positive ends are also connected, the decrease in energy of the combined system is

(A) 2 21 2

1 c v – v4 (B) 2 2

1 21 c v + v4 (C) 2

1 21 c v – v4 (D) 2

1 21 c v + v4

10v+ –

C

C

C

C

C

C

1 2

k

E+ –

1

2

3

2mm

4mm



31

135. A parallel plate air capacitor has a capacitance C. When it is half filled with a dielectric ofdielectric constant 5, the percentage increase in the capacitance will be(A) 200 % (B) 33.3 % (C) 400 % (D) 66.6 %

136. A network of six identical capacitors, each of value C is made as shown in figure. Equiva-lent capacitance between points A and B is

(A) C4 (B)

3C4

(C) 4C3 (D) 3 C

137. The capacities of three capacitors are in the ratio 1 : 2 : 3. Their equivalent capacity when

connected in parallel is 60 µF11 more then that when they are connected in series. The

individual capacitors are of capacities in F

(A) 4, 6, 7 (B) 1, 2, 3 (C) 1, 3, 6 (D) 2, 3, 4138. In the given arrangement of capacitors equivalent

capacitance between points M and N is

(A) 5 c4 (B)

4 c5

(C) 4 c3 (D)

3 c4

139. An electric circuit requires a total capacitance of 2µF across a potencial of 1000 V. Largenumber of 1µF capacitances are available each of which would breakdown if the poten-tial is more then 350 V. How many capacitances are required to make the circuit ?(A) 24 (B) 12 (C) 20 (D) 18

140. Read the assertion and reason carefully to mark the correct option out of the options givenbelow :(a) If both assertion and reason are true and the reason is the correct explanation of the

assertion.(b) If both assertion and reason are true but reason is not the correct explanation of the

assertion.(c) If assertion is true but reason is false.(d) If the assertion and reason both are false.(e) If assertion is false but reason is true.

A

B

C

C

C

C

C

C

CC

C

CC

CM N



32

1. Assertion : The coulomb force is the dominating force in the universe.Reason : The coulomb force is weaker than the gravitational force.

2. Assertion : If three capacitors of capacitance C1 < C2 < C3 are connected in parallel thentheir equivalent capacitance Cp > C3

Reason :p 1 2 3

1 1 1 1C C C C

3. Assertion : A metalic shield in form of a hollow shell may be built to block an electricfield.

Reason : In a hollow shherical shield, the electric field inside it is zero at every point.4. Assertion : Electrons move away from a low potential to high potential region.

Reason : Because electrons have negative charge5. Assertion : If the distance between parallel plates of a capacitor is halved and dielectric

constant is made three times, then the capacitance becomes 6 times.Reason : Capacity of the capacitor does not depend upon the nature of the material.

6. Assertion : A parallel plate capacitor is connected across battery through a key. A di-electric slab of constant K is introduced between the plates. The energy whichis stored becomes K times.

Reason : The surface density of charge on the plate remains constant or unchanged.7. Assertion : Electric lines of force cross each other.

Reason : Electric field at a point superimpose to give one resultant electric field.8. Assertion : If a proton and an electron are placed in the same uniform electric field.

They experience different acceleration.Reason : Electric force on a test charge is independent of its mass.

9. Assertion : Dielectric breakdown occrus under the influence of an intense light beam.Reason : Electromagnetic radiations exert pressure.

10. Assertion : When charges are shared between any two bodies, no charge is really lost,but some loss of energy does occur.

Reason : Some energy disappeares in the form of heat, sparking etc.



33

1(A) 17(C) 33(D) 49(D) 65(D) 81(B) 97(D) 113(C) 129(A)

2(B) 18(A) 34(C) 50(A) 66(B) 82(B) 98(A) 114(B) 130(A)

3(D) 19(D) 35(A) 51(A) 67(D) 83(C) 99(B) 115(C) 131(C)

4(C) 20(B) 36(A) 52(A) 68(C) 84(D) 100(B) 116(C) 132(C)

5(B) 21(A) 37(C) 53(C) 69(B) 85(C) 101(B) 117(B) 133(A)

6(C) 22(B) 38(B) 54(B) 70(D) 86(B) 102(B) 118(A) 134(C)

7(D) 23(D) 39(D) 55(C) 71(A) 87(A) 103(C) 119(A) 135(D)

8(A) 24(C) 40(C) 56(D) 72(B) 88(D) 104(A) 120(B) 136(C)

9(B) 25(A) 41(B) 57(A) 73(D) 89(B) 105(D) 121(A) 137(B)

10(C) 26(D) 42(A) 58(B) 74(C) 90(C) 106(D) 122(B) 138(A)

11(C) 27(B) 43(D) 59(B) 75(B) 91(A) 107(A) 123(C) 139(D)

12(D) 28(C) 44(D) 60(D) 76(D) 92(C) 108(D) 124(B)

13(B) 29(B) 45(A) 61(C) 77(C) 93(D) 109(D) 125(D)

14(A) 30(D) 46(B) 62(D) 78(B) 94(A) 110(D) 126(C)

15(C) 31(A) 47(A) 63(A) 79(A) 95(C) 111(D) 127(D)

16(B) 32(C) 48(C) 64(B) 80(A) 96(A) 112(A) 128(C)

KEY NOTES

1(D) 2(C) 3(A) 4(A) 5(B)

6(C) 7(E) 8(B) 9(B) 10(B) 140 :



35

(9) Distance OA = 23

a sin 60c h

= 23

32

. .a

= a 3

The force on the Charge at A due to those at B and C have magnitude F = 2

2

Qka

The Resultant of theses force isF F1 2 30 cos

F1 = 2

2

322

kQa

F KQq2

2

23

The force on the charge Q at A due to Charge q at o is, F2 = 2

3KQqa

Now for equilibrium of the Charges, F1 2 F Now Calculate q.(10) Let q and 2q be the Charges and r the distance between them.

Then 2 10 232

2 kqr

and 5 10 20 1

42

2

kqr( . )

So, 205

= rr 0 1 2

2

.b g

2 = r

r 0 1

2.

r m 0 1. .

Now, Substitating the Value of r So obtained

2 10 3 = 9 109 2

2

2(0.1)

q

q21410

9

q 10

3

7

Now find out q and 2q.

(11) Force on -q1 due to q2 is F12 = kq q

b1 22 along axis

Force on -q1 due to q3 is F kq qa13

1 32 at negative direction of yaxis.

x component of force on -q, is Fx = F F12 13 sin

= Fq qb

qa1

22

32L

NMOQPsin

i.e = sin32x 2 2

qqFb a

C

Oq

F2

Q

QB

AF

F1

F

a



36

(12) Force exerted by charges -q at A and B onCharge Q are F1 amd F2 which are equal and

have a magnitude 2

KQqFr

The resultant of these equal forces equallyinclined with the X-axis is along the -X di-rection to wards the origin.so, ' cos2 2 2 2F F F 2F = 2 12F ( cos )

since Fr

F12

1, is also Proportional to 12r ,

Hence charge Q will move towards the origin and because of its inertia it willovershoot the origin o. Thus charge Q will oscillate about o but its motion is notsimple harmonic.

(13) aOA OC r2

F F KQa1 2

2

2 F KQa3

2

22 4 2

2KQqFa

The resultant of F F1 2& is F F KQa

2 31

2

2

F and F3 act along AP So, F F F13 find.

Now for equilibrium F F14 .

From this find out q.(14) Net attraction force on q due to Q = repulsion force due to q

2 1F FA from find out q.

(15) 0 0

1 22 2 2 2

kq q kq qF Fa y a y

F F1 2

By symmetry, the x components of force willcancel each other while along y axis will add up The resultant force on +q is

F F 2 1 cos .02 2 2 2

2kqq ya y a y

Now, Force on charge q0 will be maximum, when dFdy = 0

32

32

0 2 2 52 22

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

kq qaa

1 3

52

02 2

32

2

2 2a y

ya yc h c h

Now find out y.



37

(16) Tcos = mg and Tsin = F

tan Fmg

=

2

22 x q

x q x2 3

q x2 32

12dq 3 dxx

dt 2 dt

12V x

Constantdqdt

(17) FRAB

14

94

2

2 (along

BA )

FRAC

14

94

2

2 (along

CA )

F F FAB AC

F F F F1 2 2 22 60cosNow find out F1 .

(18) Force on q is F kqa x

kqa x

2

2

2

2( ) ( )

ma = F kqa x a x

LNM

OQP

22 2

1 1( ) ( )

2

3q xama

Now compare with a = –2x and find out frequenly.(19) Suppose the balls having charges Q1 and Q2 respectively..

Initially,....... F = 2

1 2kQ Qr

It is given that F F1 4 5 .

K Q Qr

1 22

2

b g = 4.5 1 22

Q Qr

Q Q1 22

b g = 4.5 Q Q1 2 Now solve it, and find QQ

1

2.



38

(20) Consider a small element of the rod of length dx, at a distance x from the pointcharge q.The force between q and charge element will be,

dF =kqdQ

x2 But dQ = QL

dx ,

dF kqQL

dxx2

L + L

r r

2r r

kqQ 1F dF dxL x

Now solve it.

(21)k q

xkqx

.( . )

12

220 08

then find x

(22) pE NC( )

p 12

kqPQ

QE NC( )

Q 12

kqQR

in PQR, QRPR

cos 12

= 60

0 cos 60 and2 2p Q p QE E 2E E

Q sin tan

cos p QE

find out E and .(23) From the fig, Net force F acting

along the inclined plance ...(i)F = mgsin - qEcos - fma = mg sin - q E cos–So, ma = mg sin – qE cos

- (mg cos + q E sin)from this, find out a.

Now, 2o

1d v t at2

use this equation and calculate t.

(24) From the fig sino

ET qE

cos T mg cos q T

mg

Now calculate .



39

(25) Let be the charge Per unit length so, charge on Portion PQ ...

= Intensity at o is dE = k rd

r

k ddEr

Total intensity E = 0

dE

and

total cahrge on the rod, Q = r

Now find out E.

(26) Here E = and / ot l v

Along y axis vo = 0, a = FM

eE M /

So, 2o

1d v t at2

Now Put the values(27) qE l sinθ

qElθ The moment of inertia of the rod is

I m l m l ml F

HGIKJ F

HGIKJ

2 2 2

2 2 2

Now, = Iα α=I

Calculate and compale with 2 then qEml

So, 2πT =

and rod will become Paralled to E in a time Tt =4

Now calculate.

(28) After Connection V V1 2 , 1 2 1 2

1 2 2 1

KQ KQ Q r=r r Q r

The ratio of electric fields

12 2

1 1 1 22

22 1 222

KQE r Q r= = ×KQE r Q

r

Now Calculate



40

(29) Consider a Spherical Shell of thickness dx and radius x.The area of this Spherical shell = 4 2x dx

QxR

x dx QR

x dx

42

434 4L

NMOQP

The Charge enclosed in a Sphere of

radius r1 is 1rr 4

3 414 4 4

0 0

4Q 4Q Q= d = = rR R 4 R

xx x

= 1

44 1

4

12

.

QR

r

r

L

N

MMM

O

Q

PPPNow find out E.

(30) 11 2

KqE =a

and 22 2

KqEb

Now, 2

2 22

1 1

E qtanθ =E q b

a Now, Put the values.

(31) Net acceleration g g gEm

1

So, '

lTg

Now Put the Value of 'g .

(32) Net electric field due to both charges q3

will get cancelled.

So. electric field of O is

2

q

ER

= 2

2q

6 R(33) In balance condition

3V 4QE mg Q r gd 3

Q rV

3

Now find out Q2

(34) Use equation, E E E EB A 2 2 2 60cos

(35)

3,tan tan

12 3

tan 1 32

So,

33

21tan



41

(36) 21d at2

2

1 qE x2 m v

2

2 x0

qEV2md

Now find out vo.

(37) Time of flight is tvgF

oy2

Now range 2Fox F x

1R V t a t2

Where xqEam

Now, Calculate range.

(38) P e ro Calculate P..

(39) As tan tan ,

2 Shown in fig 90 tan( ) tan90

2

cot tan

21

2tantan

tan2 2

tan ( )1 2 52 53to(40) PE sin = q(2q)E sin

( ) sinq

2a E

Now find out q.

(41) Theory related quesition.(42) As the drop is stationary, weight of drop = froce due to electric field.

43

3 r g = neE So, 3 neEr4 g

now find out r..

(43) Theory related question.

(44) = 3 3

0 00

Q enclosed Q enclosed f ( 8 10 4 10 )

Now find Q.

(45) Applysing gauss law

E x.4 2 3q v 4 x3

= E x

(46) Applying gauss law 0

q

(47) total = A B Cq

Assume b = and A = C=1

So, 2

= 12 0

q



42

(48) . qE da

q.0

E 2 rl

qE2 rl

0

Er

Now find out E.

(49) E dvdr

0

dvr

2 r

dvdr

vb b

va a

1dv dr2 r

Now calculate.(50) maximum length of string = 3l

maximan enclosed charge = 3 l l

(51) (A) W = 100e (-4-10) = -1400ev= -1400(-1.6 10-19) J = 2.24 10-16 J

(52) (A) The electric Potential V ( , , )x y z x 4 2 Volt

Now ˆˆ ˆdv dv dvE i j kdx dy dz

Now dvdx = 8x ,

dvdy

0 and dvdz

0

Hence E

= ˆ8xi , So at Point (1m,0,2m)

E

= ˆ8i Volt/metre or 8 along negative x-axis

(53) (c) Potential at the centre o, q1V 4 4 2

Where Q = 103 10 9 C and a = 8cm = 8 10 2 m

So = 4 9 10 103

109 9

8 102

2

= 1500 2 Volt

(54) Obviously, from charge configuration, at thecentre electric field is non - zero. Potential

at the centre due to 2q charge 2qqVr

and

Potential due to –q Charge qV q r

(r = distance of centre Point)

Total Potential 2q q qV V V V = 0



43

(55) ABCDE is an equipotential surface, on equipotential surface no work is done inshifing a chrge from one place to another.

(56) V a b cA

( )

V ab

b cB FHG

IKJ

2

VC = ac

bc C

2 2 e j

on Putting C = a +b VA = VC VB

(57) E dvdx

xy zx

2 3

Ey dvdy

x

2

Ez dvdz

xz

3 2

ˆ ˆˆ ˆ ˆ ˆ E 3 2 2x y zE i E j C k xy z i x j 3xz k

(58) (B) For Pair of Charge

U = 14

10 10 10 1010

100

6 6

L

NMM

OQPP +

10 10 10 1010

100

6 6 LNMM

OQPP

+ 10 10 10 10

10100

6 6 LNMM

OQPP

= 3 9 10 100 10 10010

9 2

= 27J

(59) (B) Potential at Centre o of the square Vo = a2

Q44

Work done in shifting (-Q) Charge from Centre to infinity

00W Q V V QV

2 2Q Q

4 a a

(60) (D) Let Q1, and Q2 are the Charges on Sphere of radii R1 and R2 respectively

Surface Charge density ch eArea

arg

According to given Problem, 1 2

= QR

QR

1

2

2

24 4

QQ

1

2 =

RR

12

22 .....(1)



44

In Case of a charged sphere, vs = 1

4 QR

V1 = 1

4

QR

1

1 , V2 =

14

QR

2

2

VV

QR

RQ

1

2

1

1

2

2

= QQ

RzR

1

2 1

= RR

RR

RR

1

2

2

2

1

1

2

FHG

IKJ .....(using (i) )

(61) Usystem = 1

4

( )( ) ( )( ) ( )( )

q 2q 1 2q q 1 q qq 4 a 4 2a

= o

27q8 a

(62) Work done indisplacing charge of 5lic from B to C is W V Vc b 5 109 ( ) where

V vB

9 10 100 10

0 494

109 6

6

.

V vC

9 10 100 10

0 595

109 6

6

.

So W J FHG

IKJ

5 10 95

10 94

10 94

6 6 6

(63) 2dv dE 5x 10x 9 10x 10dx dx

( )E x

v10 20m

(64) Energy density ue = 12

12 8 86 102 12

2

E vr

FHGIKJ

. = 2 83 3. /J m

(65) Potenti al i nsi de the Sphere w i l l be same as that on i ts Surf ace i .e. v = V surface

= q volt

10

Vout = q

15 volt

vout

v23

Vout V23

(66) Length of each side of square is 2m

so distance of it’s centre from each cornet is 22

1 m

Potential at the centre

V

L

NMOQP

9 10 10 101

5 101

3 101

8 101

96 6 6 6

5V 1.8 10 V



45

(67) W Uf U Q Qr r

LNM

OQP1

91 2

2 1

9 10 1 1

= W

LNM

OQP

9 10 12 10 8 10 1

4 101

10 109 6 6

2 2

= 12 96 10 13. J J

(68) By using W = . ΔQ E r

ˆˆ ˆ ˆ ˆ . 1 2 3 1 2W Q e i e j e k ai bj Q e a e b

(69) K.E. = ( ) . ( )6 180 A Bq V V 1 6 70 50 10 J

(70) Net electrostatic energy 2KQq Kq KQqU 0

a a a 2

Kq QQ qa 2

Now find out Q.(71) Point P will lie near the Charge which is smaller in magnitude i.e. -6lic Hence

Potential at P.

( ) ( )( . )

6 66 1 12Vx 4 0 2 x

x m 0 2.

(72) Calculate as MCQ 67.

(73) Calculate as MCQ 66.

(74) (C) AB+AC = 12 cm .......(i)= AB . AC = 32 cm2

= AB AC AB AC AB AC ( ) .2 4= AB AC 4From equation (i) and (ii)AB = 8 cm; AC = 4cmPotential energy at Point A

A 1 21 1V q q

AB AC

V JA 1 35.

(75) Use the equation 12 m V V1

22

2d i = QV



46

(76) Potential energy of the system

U kQql

kql

kqQl

2

0

( )KQ Q q Q 0l

qQ2

(77) ( ) x

dvE 5dx also find Ey an Ez

Enet = 2 2 2x y zE E E = ( ) ( )5 3 15 72 2 2

d i(78) Van - d - glaph generater principal

(79) U Q Qr

Q Qr

Q Qr

LNM

OQP

14

1 2

1

2 3

2

1 3

3

Unet = 2

0

134

.q

πε l

(80) Length of the diagonal of a cube having each side b is 3b. So distance of centre

of cube from each vertex is 32

b .

Hence Potential energy of the given system of charge is

U = 8 0

1 ( )( )4 3 2

q qbπε =

24q3 b

(81) When a negatively charged Pendylum oscillates overa Positively charged Plate

then effective Value of 9 increases so, according to 2Tg

decreases

(82) When charge q is released in uniform electric field E then its acceleration

a = qEm (is constant) so it’s motion will be uniformly accelerated motion and it’ss

Velocity after time is given by

tqEV atm

= K = 12 mv2

2 2 2 21 qq q E tt2 m 2m

(83) Electric Potential at P

R2

KQ Kq Q qVR 4 R 4 R



47

(84) Potential at P due to (+q) charge

. ( )1

0

qVz a

Potential at P due to (-q) Charge

= .( )2

1 qV4 z a

Total Potential at P due (AB) electric dipole V V V 1 2

Now calculate(85) (C) Potential will be zero at two Points .....

At internal Point(M) = 1

42 10

61 10

06 6

L

NMM

OQPP

( )l lc h

l 2So distance of M from origin; x 6 2 4 At exteriot Point (N)

( ) '( )

6 61 2 1 l4 6 l l'

So distance of N from origin x 6 6 12(86) Potential at the centre of rings are

. ( )

1O 2 2

k q k qVR R d

, ( )

2O 2 2

k q kqVR R d

1 2O O 2 2 2 2

1 1 q 1 1V V 2kqR 2 RR d R d

(87) (A) Change in Potential energy (u) = uf - ud

Now calculate

(88) Theory base Question

(89) Potential at C, VC = 0

Potential difference VD -VC = 23

kqL

( ) D CW Q V V put the values

(90) V N V N 2

32

340 10

N N N2

3 24 64 8 (91) C C1 2

Ad

Ad

1

1

2

2

d AA

d22

11 Now find d1.



48

(92) Here s 1 p 1C 3l F and C 16 l F Now in Parallel connection C C Cp 1 2

So, C C1 2 16 11

in series connection C C CC C5

1 2

1 2

So, =

C CC C

1 2

1 2 .....(2)

or C C1 2 Now, C C1 22

b g = C C C C1 22

24 b g , orC C1 2 8 ........(3)Adding the question (1) and (3) we haveC C1 2b g+ C C1 2b g16+8

1 & C 2a F C F (93) Suppose that the techinician makes a combination use of N Capacitors and

connects them in m rows, each row having m Capacitors Then N=mnCapacitance of each capacitor = 1 l F1 required Capacitance of the combinationC = 2l F1

Voltage rating of each < apalitor = 400Vrequired Voltage rating of the combination = 103VWhen capacitars are connected in series P.d. a cross their Plates get added. forn capacitorsVoltage ...to 400 nv 400 103n

n 10400

2 53

. or n 3

The total Capucitae 1 1

111

11

31C ' 1C

3

The total Capacitanae of m rous C ' OR '

Cmc mC

(94) Common Potential VCV C V

C Cr v r v

r r

1 1 2 2

1 2

1 1 2 2

1 2

4 44 4

V r v r vr r

1 1 2 2

1 2

(95) U C V J1 1 121

20 121 .

Total charge on the two Capacitors q CV C V C 1 1 2 2

411 10C C V l F l F l F F

1 2 1 1 165 2 5 7 5 7 5 10. . .

VC

V 9 440

3

U CV J221

20 081 . Now, energy lost = U U1 2 find it.



49

(96) Apply wheatstone bridge law.

(97) Cair Ad

with dielectlic slab C A

d t t k1

( / ) Now C = 43 C

F

HIK

A

d t tk

Ad

43

= K = 4 4( / 2)

4 4( / 2 )

t q

t d d d = 2

(98) As P.d. remains the same, Capacity must remain the same,

x = t11

k x = 1.5 mm t= 2mm Now find out k.

(99) C Ado 18 C k A

d

372 Now, find out K.

(100) C = 4R Rk

. Put the Values of R&K.

(101) 1 0 1 01

ε 2 ε

2

k A k AC d d

C k d2

222

Now,

1 1 1

3 1 2C C C

(102) Apply lawa of series & Parallel conection of Capacitor

(103)1 1 1 1

5 1 2 3C C C C

C C C Cp 1 2 3 Now fiud out s

p

CC

(104) Number of Capacitors to be connected in series

V voltage rat g requiredvoltage rat g required

intint

.700200

3 5

i e. .4

C r Feq 104

2 5 1.

Number of rows required = Capacityrequired

Capacityof each row 102 5

4.

Total number of Capacitor 4 4 = 16

(105) U C V U C V U U U1 12

2 22

1 212

12

&



50

(106) Work done = 1 22 1

2 1

q q 1 1U U4 r r

Now Calculate.

(107) Volume of big drop = 1000 volume of small drop 43

1000 43

3 2 R l

Now find out initial energy

U qC1

2

10002

then final energy U qC2

2

1

10002

b g then

UU

2

1

R = 10r so, C C1 10(108) minimum capacity is Cs (Series Connection) & maximum Capaciy is CP (Parallel

Connection) Now p

s

CC Calculate

(109) Theory base question(110) With S S1 3& Closed , the Capacitors C C1 2& are in series arrangenent. So

111

2

VV

CC

V V V V V2

12

12

11 2

32

30 20 50 & Now simplify..

(111) E QA

QA

1 2 1 1

2 2 2 2 Now Simplify..

(112) C F23 30 30 60 Now,,1 1 1 1

1 23 4C C C C

C pf120

7 Total charge Q =CV

V QC1

1

, V V V QC

V QC2 3 23

234

4

&

(113) The arrangement can be redrawn asshown in figule Now find out.

(114) V = 4x2 E = –

dvdr = –8x Now Put the value of x.

(115)1 1 1

5 1 2C C C and q C V 5

(116) C Ad

PF

9 1 01

3

k ACd and 2 0

23

2

k AC

Now 1 1 1

3 4 2C C C



51

(117) Knowledge base question.(118) Knowledge base question.(119) Knowledge base question.

(120) Use equation E dVdx

(121) W QV CV V CV ( ) 2 and 12

2CV Now find Uw

(122) When Spheres are connected by a conducting wire their Potentials become equal.

1 1 1 1 1

2 2 2 2 2

C r q C V C1 1= = = = =C r 2 q C V C 2

Now, 1

21 1

2222

KqE r

KqEr

Now find out rativ..

(123)

(124) Q = CV = 5c

= C A

d t tk

1 FH

IK

= A d

t tk

d

/

1F

HGGIKJJ

Now Put the value.

(125) Knowledge base question.

(126)qc

qc

qc

qc

10 or

CV10

42 5.

Now AV – 0 = 7.5 V VA 7 5.Again V VN B ..... O VB 2 5. V VB 25

(127) V = 64

4q

R =

644 4

qr ( )

C Ad

(128) C A

d tk

1

1 1

FH

IK

Use this equation



52

(129) C Ad11

1

54

and C2

65

Now C C CC C

1 2

1 2

(130) C = 32

Ad and V

QC

(131) Use wheatston’s bridge equation eqC C and Q CVCharge on Capacitor between the terminals

A and B is Q CV2 2 = Enegy stored in capacitor =

2Q2

2C(132) Theory base question.

(133) Work done = Final energy - initial energy = QC

QC

2

1

2

2 2

(134) U C V Vi 12 1

22

2d i and 1 2 1 2

1 2

q + q V +VV = =c + c 2

U C Vf 12

2 2( ) Now find U Ui f

(135) C Ad1 10

C A

d2 2

C C5

53

Percentage increase in Capacitance = C C

C5 100

%

(136) Use Series and Parallel Connection law.(137) Theory base question.(138) Theory base question.

(139)1000n = = 2.8 = 3350

S1C = µF3 m= µF

3

m µF = 2µF3

m = 6total no = mn = 3 6 = 18



unit - 11 electrostatics ug physics... · electrostatics 'rzqordghgiurpzzz vwxglhvwrgd\ frp. 2...

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