ELECTROSTATICS JEE ADVANCED-VOL - VI ELECTROSTATICS · 2020. 4. 13. · Narayana Junior Colleges...

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ELECTROSTATICS JEE ADVANCED-VOL - VI

Narayana Junior Colleges

ELECTROSTATICSLEVEL-IV

SINGLE ANSWER QUESTIONS

CHARGE AND ITS PROPERTIES1. The linear charge density of a thin metallic

rod varies with the distance ‘x’ from one end

as 20 0 .x x l The total charge on

the rod is

A) 3

03l

B) 3

04l

C) 3

023

lD)

302l

2. The linear charge density of a uniformsemicircular wire varies with ' ' shown inthe figure as 0 cos . If R is the radiusof the loop, the charge on the wire is

++++++ +++ + +

++++

R+

+++++ +++ + +

++++

R

A) 0R B) 03 R C) 023

RD) 02 R

3. For the hemisphere of radius ‘R’ shown inthe figure the surface charge density' ' varies with ' ' (shown in figure) as

0 cos . The total charge on thesphere is

++++++ +++ + +

++++

++++++ +++ + +

++++

A) 202 R B)

20

2R

C)2

03

R D) 2

0 R

COULOMB’S LAW & PRINCIPLE OFSUPERPOSITION

4. A charge ‘+Q’ is uniformly distributed alongthe circular arc of radius ‘R’ as shown in thefigure. The magnitude of the forceexperienced by the point charge +q placed

at the centre of cuvature is 0

14

k

+ + + + + + + + + ++

R R

+ + + + + + + + + ++

R R

A) 2

sin2

kQq

R

B)

2

2 sin2

kQq

R

C)2

2 cos2

kQq

R

D)

2

2 tan2

kQq

R

5. A thin circular wire ring of radius ‘r’ has a

charge ‘Q’. If a point charge ‘q’ is placed atthe centre of the ring then the tension in thewire will increase by

A) 2 208

qQr B) 2 2

04qQ

r

C) 204

qQr

D) 2 2

02qQ

r

6. Two metallic spheres each of mass M andradius R charged separately to 1Q and 2Q ,are at a distance ‘d’ between their centres(d>2R). The force between them will be

A) 1 2

20

14

Q Qd B)

1 22

0

14

Q Qd

C)1 2

20

14

Q Qd

D) zero

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ELECTROSTATICSJEE ADVANCED-VOL - VI

2 Narayana Junior Colleges

7. A positive point charge +Q is fixed in space.A negative point charge -q of mass m re-volves around a fixed charge in elliptical or-bit. The fixed charge +Q is at one focus ofthe ellipse.The only force acting on nega-tive charge is the electrostatic force due topositive charge.Then which of the followingstatement is true.

A) Linear momentum of negative point charge isconserved.

B) Angular momentum of negative point chargeabout fixed positive charge is conserved.

C) Total kinetic energy of negative point chargeis conserved.

D) Electrostatic potential energy of system of bothpoint charges is conserved.

8. Find the force experienced by thesemicircular rod charged with a charge q,placed as shown in figure. Radius of the wireis R and the line of charge with linear chargedensity is passing through its centre andperpendicular to the plane of wire.

(A) R2q

02

(B) Rq

02

(C) R4q

02

(D) R4q

0

ELECTRIC FIELD AND LINES OF FORCE9. Two identical point charges are placed at a

separation of l. P is a point on the line joiningthe charges, at a distance x from any one

charge. The field at P is E and E is plottedagainst x for values of x from close to zeroto slightly less than l. Which of the followingbest represents the resulting curve?

(A) (B)

(C) (D)

10. Two infinitely large charged planes havinguniform surface charge density + and –are placed along x-y plane and yz planerespectively as shown in the figure. Then thenature of electric lines of force in x-z planeis given by :

z–

+ X

(A)

Z

X

(B)

Z

X

(C)

Z

X

(D)

Z

X

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ELECTRIC POTENTIAL & POTENTIALENERGY

11. A conducting bubble of radius “a” andthickness t t a has a potential ‘V’.Now the bubble collapses into a droplet (ofradius R). The electric potential on thedroplet is

A)1

2

3aVt

B) 1

323aVt

C)1

3

3aVt

D)V

12. In the following diagram a uniform electricfield of magnitude 325 NC-1 is directed along

ve y-direction. The co-ordinates of pointA are (-20, -30) cm and those of point B are(40, 50)cm. The potential difference alongthe path shown is

E

x

y

BC

A E

x

y

BC

A

A) 260V B) 260V C) 130V D) 130V13. An infinite by long plate has surface charge

density . As shown in the fig. a point chargeq is moved from A to B. Net work done byelectric field is :

(A) 0

q2 (X1 – X2) (B)

0

q2 (X2 – X1)

(C) 0

q (X2 – X1) (D)

0

q ( 2 r r )

14. A conducting disc of radius R rotates aboutits axis with an angular velocity . Then thepotential difference between the centre ofthe disc and its edge is (no magnetic field ispresent) :

(A) zero (B) 2 2em R2e

(C) 3em R3e

(D) 2eem R2

15. A neutral conducting spherical shell is keptnear a charge q as shown. The potential atpoint P due to the induced charges is

q r

r'C

P

(A) kqr (B)

kqr '

(C) kq kq–r r ' (D)

kqCP

16. The following diagram represents a semicircular wire of linear charge density

0 sin where 0 is a positive constant.The electric potential at ‘o’ is

0

14

take k

A) 0 sink B) 0 cos2

k

C) zero D) 0 cosk

17. The electric potential at the centre ofhemisphere of radius ‘R’ having uniformsurface charge density is

A) 0

R B)

0

2 R C)

02R D)

03R

18. For an infinite line of charge having linearcharge density ' ' lying along x-axis, thework done in moving a charge ‘q’ from C toA along CA is

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CB

aA

a

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

CB

aA

a

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

A) 0

ln 22

q B)

0ln 2

4q

C) 0

ln 24

q D)

0ln 0.5

2q

19. The diagram shows three infinitely longuniform line charges placed on the X, Y andZ axis. The work done in moving a unitpositive charge from (1, 1, 1) to (0, 1, 1) isequal to

(A) (l ln 2) / 2 0 (B) (l ln 2) / 0(C) (3l ln 2) / 2 0 (D) None

20. Two positively charged particles X and Y areinitially far away from each other and at rest.X begins to move towards Y with some initialvelocity. The total momentum and energyof the system are p and E.(A) If Y is fixed, both p and E are conserved.(B) If Y is fixed, E is conserved, but not p.(C) If both are free to move, p is conserved butnot E.(D) If both are free, E is conserved, but not p.

21. A metal ball of radius R is placedconcentrically inside a hollow metal sphereof inner radius 2R and outer radius 3R. Theball is given a charge +2Q and the hollowsphere a total charge – Q. The electrostaticpotential energy of this system is :

(A) R24Q7

0

2

(B) R16Q5

0

2

(C) R8Q5

0

2

(D) None

RELATION BETWEEN E & V22. Figure shows an electric line of force which

curves along a circular arc. The magnitudeof electric field intensity is same at all pointson this curve and is equal to E. If thepotential at A is V, then the potential at B is

AE

BR

(A) V – ER (B) V – E2R sin 2

(C) V + ER (D) V + 2ER sin 2

23. Uniform electric field of magnitude 100 V/min space is directed along the line y = 3 + x.Find the potential difference between pointA (3, 1) & B (1, 3)

(A) 100 V (B) 200 2 V(C) 200 V (D) 0

24. Electrical potential 'V' in space as a function

of co-ordinates is given by, V = 1x

+ 1y

+ 1z

.

Then the electric field intensity at (1, 1, 1) isgiven by :(A) – ˆ ˆ ˆ(i j k) (B) ˆ ˆ ˆi j k

(C) zero (D) 1 ˆ ˆ ˆ( i j k)3

25. A graph of the x component of the electricfield as a function of x in a region of space isshown. The Y and Z components of the

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electric field are zero in this region. If theelectric potential is 10 V at the origin, thenpotential at x = 2.0 m is :

1 2 3 4 5x(m)

20

E (N/C)x

–20

(A) 10 V (B) 40 V (C) – 10 V (D) 30 V26. If the electric potential of the inner metal

sphere is 10 volt & that of the outer shell is5 volt, then the potential at the centre willbe :

(A) 10 volt (B) 5 volt(C) 15 volt (D) 0

27. Three concentric metallic spherical shells A,B and C of radii a, b and c (a < b < c) havesurface charge densities – , + , and – respectively. The potential of shell AAis :(A) 0 [a + b – c] (B) 0 [a – b + c]

(C) 0 [b – a – c] (D) none28. There are four concentric shells A, B, C and

D of radii a, 2a, 3a and 4a respectively.Shells B and D are given charges +q and –qrespectively. Shell C is now earthed. Thepotential difference VA – VC is :

(A) Kq

a2(B)

Kqa3

(C) Kq

a4(D)

Kqa6

29. At distance 'r' from a point charge, the ratio

2uv

(where 'u' is energy density and 'v' is

potential) is best represented by :

u/v2

r

u/v2

r

(A) (B)

u/v u/v2

r

u/v2

r

(A) (B) (C) (D)

30. Six charges of magnitude + q and –q arefixed at the corners of a regular hexagon ofedge length a as shown in the figure. Theelectrostatic interaction energy of thecharged particles is :

+q

+q

+q –q

–q

–q

(A)

2

0

q 3 15–a 8 4 (B)

2

0

q 3 9–a 2 4

(C)

2

0

q 3 15–a 4 2 (D)

2

0

q 3 15–a 2 8

MOTION OF CHARGED PARTICLE INUNIFORM & NON UNIFORM

ELECTRIC FIELD31. A particle of mass ‘m’ carries an electric

positive charge Q and is subjected to thecombined action of gravity g and a uniformhorizontal electric field of strength E. It isprojected with speed u in the vertical planeparallel to the electric field and at an angle to the horizontal. The maximum distance theparticle travels horizontally when it is at thesame level as its starting point is

A) 2u

g B) 2

2

u EQ mgmg

C) 2

2 22

u EQ mg EQmg

D) 2

2

EQumg

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32. The charge and mass of two particles are+Q, M and -q, m respectively. The particlesseparated by a distance L, are released fromrest in a uniform electric field E. The elec-tric field is parallel to line joining both thecharges and is directed from negative topositive charge. For the separation betweenparticles to remain constant, the value of L

is 0

14

k

.

A)

M m KQqE qM Qm

B)

M m KQqE qm QM

C) mMKQq

E qM Qm D) mMKQq

E QM qm

33. A bullet of mass m and charge q is firedtowards a solid uniformly charged sphere ofradius R and total charge + q. If it strikes thesurface of sphere with speed u, find theminimum speed u so that it can penetratethrough the sphere.(Neglect all resistive forces or friction actingon bullet except electrostatic forces)

(A) mR2q

0 (B) mR4q

0

(C) mR8q

0 (D) mR4q3

0

34. In space a horizontal EF (E = (mg)/q) existsas shown in figure and a mass m attached atthe end of a light rod. If mass m is releasedfrom the position shown in figure find theangular velocity of the rod when it passesthrough the bottom most position

(A) lg

(B) lg2

(C) lg3

(D) lg5

35. A unit positive point charge of mass m isprojected with a velocity V inside the tunnelas shown. The tunnel has been made insidea uniformly charged non conducting sphere.The minimum velocity with which the pointcharge should be projected so that it canreach the opposite end of the tunnel, is equalto

(A) [rR2/4m0]1/2 (B) [rR2/24m0]1/2

(C) [rR2/6m0]1/2

(D) zero because the initial and the final pointsare at the same potential.

36. A particle of charge – q & mass m moves ina circle of radius r around an infinitely longline charge having linear charge density + .Then time period will be.

r

+

–q

(A) T =

m2 r2k q (B) T2 =

234 mr

2k q

(C) T =

1 2k q

2 r m (D) T = 1 m

2 r 2k q

where k = 0

14

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37. A point mass of charge ‘-q’ and mass ‘m’ isreleased from rest from a distance ‘R’ alongthe axis of a uniform circular disc of charge‘Q’. Assuming gravity free situation thevelocity attained by point mass when itreaches centre of disc is (Radius of disc-R)

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

+

R

,q m

Q

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

+

R

,q m

Q

A) 1/2

0

2 2 1Qq

mR

B) 1/2

0

2 1Qq

mR

C) 1/2

0

2 1

2

Qq

mR

D) 1/2

0

3 1

2

Qq

mR

38. The adjoining figure represents a semi-circular loop of charge ‘Q’ and of radius ‘R’.The minimum velocity with which a pointmass ‘m’ of charge ‘-q’ should be projectedso that it becomes free from the influenceof the loop is

+ + + + + ++

++

++++++

R

+ + + + + ++

++

++++++

+ + + + + ++

++

++++++

R

A) 0

2QqmR B)

0

QqmR

C) 02

QqmR D)

04Qq

mR

39. A charge ‘q’ of mass ‘m’ is projected from along distance with speed ‘v’ towards anotherstationary particle of same mass andcharge. The distance of closest approach ofthe particles is

A) 2

202

qmv

B) 2

20

2qmv

C) 2

20

32

qmv

D)2

20

qmv

40. Two balls of charges 1q and 2q in magnitudehave velocities of equal magnitude (v) andsame direction. After a uniform electric fieldis applied for a certain interval of time thedirection of first ball changes by 060 andvelocity magnitude is reduced by half, thedirection of velocity of second ball changesby 090 then magnitude of charge to massratio of second ball if that of first ball is ' ' is

A) 43

B)23

C) 3

D)

41. In the diagram shown the charge +Q is fixed,another charge +2q is projected from adistance ‘R’ from a fixed charge. Minimumseparation between the charges, if thevelocity at this moment is half of velocity ofprojection in magnitude (Assume gravity tobe absent )

Q, 2m q030

RQ

, 2m q030

R

A) 3R B) 32

RC) 2

RD) 4R

ELECTRIC DIPOLE42. The dipole moment of a system of charge

+q distributed uniformly on an arc of radiusR subtending an angle /2 at its centrewhere another charge -q is placed is :

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(A)

qR22(B)

qR2

(C)

qR(D)

qR2

43. As shown in the figure an electric dipole liesat a distance ' 'x from the centre of a chargedring of radius ' 'R with charge ' 'Q uniformlydistributed over it . The net force acting onthe dipole is

2ax

q q

2ax

q q

A) 2 2

5/22 20

24aQq R x

R x

B) 2 2

5/22 20

22aQq R x

R x

C) 2 2

5/22 20

24aQq R x

R x

D) 2 2

5/22 20

22aQq R x

R x

44. Two point dipoles p k & p k2 are located at

(0, 0, 0) & (1m, 0, 2m) respectively. Theresultant electric field due to the two dipolesat the point (1 m, 0, 0) is :

(A) 0

9p k32 (B) 0

–7p k32

(C) 0

7p k32 (D) none of these

45. A dipole of dipole moment p is kept at thecentre of a ring of radius R and charge Q.The dipole moment has direction along theaxis of the ring. The resultant force on the

ring due to the dipole is :

(A) zero (B) 3kpQR

(C) 32kpQR

(D) 3kpQR

only if the charge is uniformly

distributed on the ring.46. A point electric dipole is placed at the origin

' 'O and is directed along the ve x axis .At a point ' 'P faraway from the dipole , theelectric field is parallel to the y axis . Theline ' 'OP makes an angle ' ' with thex axis . ThenA ) tan 3 B) tan 2

C) 45 D) 1tan2

47. A large sheet carries uniform surface chargedensity ' ' A rod of length '2 'l has a linearcharge density ' ' on one half and onthe other half . The rod is hinged at mid point' 'O and makes an angle ' ' with normal tothe sheet . The torque experienced by therod is

(A) 2

0cos

2l (B)

22

0cosl

(C) 2

0sin

2l (D) zero

GAUSS LAW

48. Two infinite line charges, each havinguniform charge density , pass through themid points of two pairs of opposite faces ofa cube of edge ‘L’ as shown in figure. The

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modulus of the total electric flux due to boththe line charges through the face

E

B

F

D

HG

A

a

A) CDHG is 02

L

B) AEHD is 03

L

C) ABCD is 02

L

D) CDHG is 03

L

49. The figure shows an imaginary cube oflength L/2 and a uniformly charged rod oflength L touching the centre of the right faceof the cube normally. At time t = 0, the rodstarts moving to the left slowly at a constantspeed ‘v’. The electric flux (F) through thecube is plotted against time (t). The correctgraph showing the variation of flux with timeis

A) B)

C) D)

50. A ring of radius R is placed in the plane withits centre at origin and its axis along the x-axis and having uniformly distributedpositive charge. A ring of radius r(<<R) andcoaxial with the larger ring is moving alongthe axis with constant velocity then thevariation of electrical flux ( ) passingthrough the smaller ring with Position willbe best represented by :

rv

Rx

y

(A) (B) (C)

(B) (C) (D)

51. A point charge ‘q’ is placed at a distance ' 'lfrom the centre of a disc of radius ‘R’ alongits axis. The electric flux through the disc is

A) 2 201q l

R l

B) 2 201

2q l

R l

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C) 2 20

12 2q l

R l

D) 2 20

12

q l

R l

52. Two point charges ' ' ' 'q and q areeseparated by '2 'l . The electric flux ofelectric field strength vector through a circleperpendicular to the line joining the chargesas shown in the figure is

060060q q

2ll

l

060060q q

2ll

l

A) 02

q B)

0

q C)

0

2q D) Zero

53. A point charge ' 'q is placed at the centreof a cylinder of length ‘L’ and radius ‘R’ theelectric flux through the curved surface ofthe cylinder is

A) 2 20

q L

L R

B) 2 20 2

q L

L R

C) 2 20 4

q L

L R

D) 02

q

54. Two charges 1q and 2q are placed at AAand B respectively. A line of force emanatesfrom 1q at angle ' ' with line AB and itterminates at an angle ' ' as shown then

(A) 1 1

2sin sin

2qq

(B) 1 1

22sin sinq

q

(C) 1 2

12sin sin

2qq

(D) 1 1

22sin sin

2qq

55. The potential inside a charged ball depends

only on the distance ‘r’ of the point from itscentre according to the following relation

2V Ar B volts . The charge densityinside the ball will beA) 06 A B) 06 A

C) 03 Ar

D) 06 A

r

56. Two very large thin conducting plates havingsame cross-sectional area are placed asshown in figure. They are carrying charges'Q' and '3Q' respectively. The variation ofelectric field as a function of x (for x = 0 to x= 3d) will be best represented by.

Y

Q 3Q

(d, 0) (2d, 0) (3d, 0) X

E

d 2d 3d X

(A)

(d, 0) (2d, 0) (3d, 0)

2d 3d

E

d 2d 3d X

(B)

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d 2d 3d X

E

d

2d 3d X(C)

2d 3d d 2d 3d X

E

d

2d 3d X2d 3d(D)

57. Inside a ball charged uniformly with volumedensity ' ' there is a spherical cavity. Thecentre of the cavity is displaced with respectto the centre of the ball by a distance a. Thenthe field strength inside the cavity is

A) 0 B) 0

a2 C)

0

a3 D)

0

a

58. Two spherical, nonconducting, and very thinshells of uniformly distributed positivecharge Q and radius d are located a distance10d from each other. A positive point chargeq is placed inside one of the shells at adistance d/2 from the center, on the lineconnecting the centers of the two shells, asshown in the figure. What is the net force onthe charge q?

(A) 20d361

qQ to the left

(B) 20d361

qQ to the right

(C) 20d361

qQ362 to the left

(D) 20d361

qQ360 to the right

59. A point charge ‘q’ is located at the centre ofthe spherical layers of uniform isotropicdielectric with relative permitivity ‘K’. Theinside radius of the layer is equal to ‘a’ andthe outside radius is equal to ‘b’. Theelectrostatic energy inside the dielectriclayer is

A)2

20

1 1 14 2

qa bK

B) 2

20

1 1 14

qb aK

C) 2

20

1 1 14 2

qb aK

D) 2

20

1 2 14

qb aK

PROPERTIES OF CONDUCTORS &

CHARGE DISTRIBUTION60. The figure shows a charge q placed inside a

cavity in an uncharged conductor. Now if anexternal electric field is switched on :

cq

(A) Only induced charge on outer surface willredistribute.(B) Only induced charge on inner surface willredistribute.(C) both induced charge on outer and innersurface will redistribute.(D) force on charge q placed inside the cavitywill change.

61. The following diagram shows three metalballs. Ball ‘A’ is charged to ‘Q’ coulombs andB,C are uncharged. The charges on balls Band C when the switches 1S and 2S areclosed are respectively

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Q2aB

A C

5a

Q5a

5a

a

1S

2S

Q2aB

A C

5a

Q5a

5a

a

1S

2S

A) 8 3,23 23Q Q

B) 3 8,23 23Q Q

C)8 3,23 23

Q Q D)

3 8,23 23

Q Q

62. In the following diagram the conductingshells are concentric. The amount of chargethat flows through the switch(S) after closingit is

R

ABC

2R

4R

S

Q2Q

3Q

R

ABC

2R

4R

S

Q2Q

3Q

A) Q B) 2Q

C) 3Q

D) 4Q

63. In the following diagram the plates areconducting and are held parallel,then amountof charge that flows through the switch afterclosing it is

Q

2 d

2Q

4 d

3Q 4Q

d

Q

2 d

2Q

4 d

3Q 4Q

d

A) Q B) 52Q

C) 72Q

D) 2Q

64. Two concentric thin conducting sphericalshells having radii R and 2R are as shownin the figure. A charge +Q is given to theshell ‘A’ and -4Q is given to the shell ‘B’. IfA and B are connected by a thin conductingwire then the amount of heat produced afterthe connection is

R

BA2R

Q

4Q

R

BA2R

Q

4Q

A)2

0

34

QR B)

2

016Q

R

C)2

0

58

QR D)

2

08Q

R

MULTI ANSWER QUESTIONS

65. Five balls numbered 1 to 5 are suspendedusing separate threads. Pairs (1,2), (2,4) and(4,1) show electrostatic attraction while pairs(2,3) and (4,5) show repulsion. Therefore ball1 must beA) positively charged B) negatively chargedC) neutral D) made of metal

66. A point charge ‘Q’ is placed at origin ‘O’.

Let ,A B cE E and E

represent electric fieldsat A,B and C respectively. If coordinates ofA,B and C are respectively

1,2,3 , 1,1, 1m m and 2, 2, 2 m then

A) A BE E

B) ||A CE E

C) 4B CE E

D) 8B CE E

67. A negative point charge placed at the pointA is

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(A) in stable equilibrium along x-axis(B) in unstable equilibrium along y-axis(C) in stable equilibrium along y-axis(D) in unstable equilibrium along x-axis

68. Four charges of 1 mC, 2 mC, 3 mC, and –6mC are placed one at each corner of thesquare of side 1m. The square lies in the x-y plane with its centre at the origin.(A) The electric potential is zero at the origin.(B) The electric potential is zero everywherealong the x-axis only on the sides of the squarewhich are parallel to x and y axis.(C) The electric potential is zero everywherealong the z-axis for any orientation of the squarein the x-y plane.(D) The electric potential is not zero along the z-axis except at the origin.

69. Two fixed charges 4Q (positive) and Q(negative) are located at A and B, the distanceAB being 3 m.

(A) The point P where the resultant field due toboth is zero is on AB outside AB.(B) The point P where the resultant field due toboth is zero is on AB inside AB.(C) If a positive charge is placed at P anddisplaced slightly along AB it will executeoscillations.(D) If a negative charge is placed at P anddisplaced slightly along AB it will executeoscillations.

70. Two identical charges +Q are kept fixedsome distance apart. A small particle P withcharge q is placed midway between them. IfP is given a small displacement D, it willundergo simple harmonic motion if(A) q is positive and D is along the line joiningthe charges.(B) q is positive and D is perpendicular to theline joining the charges.(C) q is negative and D is perpendicular to theline joining the charges.(D) q is negative and D is along the line joiningthe charges.

71. Select the correct statement : (Only forceon a particle is due to electric field)

(A) A charged particle always moves along theelectric line of force.(B) A charged particle may move along the lineof force(C) A charge particle never moves along the lineof force(D) A charged particle moves along the line offorce only if released from rest.

72. Two point charges Q and – Q/4 are separatedby a distance x. Then

(A) potential is zero at a point on the axis whichis x/3 on the right side of the charge – Q/4(B) potential is zero at a point on the axis whichis x/5 on the left side of the charge – Q/4(C) electric field is zero at a point on the axis which isat a distance x on the right side of the charge – Q/4(D) there exist two points on the axis whereelectric field is zero.

73. Three point charges Q, 4Q and 16Q areplaced on a straight line 9 cm long. Chargesare placed in such a way that the system hasminimum potential energy. Then(A) 4Q and 16Q must be at the ends and Q at adistance of 3 cm from 16Q.(B) 4Q and 16Q must be at the ends and Q at adistance of 6 cm from 16Q.(C) Electric field at the position of Q is zero.

(D) Electric field at the position of Q is 04

Q .

74. Two infinite sheets of uniform chargedensity + and – are parallel to eachother as shown in the figure. Electric fieldat the

++ ––––––––

+++++++

+ –

(A) points to the left or to the right of the sheetsis zero.(B) midpoint between the sheets is zero.(C) midpoint of the sheets is / o and is directedtowards right.(D) midpoint of the sheet is 2 / o and is directedtowards right.

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75. If we use permittivity e, resistance R,gravitational constant G and voltage V asfundamental physical quantities, then(A) [angular displacement] = e0R0G0V0

(B) [Velocity] = e–1R–1G0V0

(C) [dipole moment] = e1R0G0V1

(D) [force] = e1R0G0V2

76. A thin-walled, spherical conducting shell Sof radius R is given charge Q. The sameamount of charge is also placed at its centreC. Which of the following statements arecorrect?(A) On the outer surface of S, the charge density

is 2R2Q

.

(B) The electric field is zero at all points inside S.(C) At a point just outside S, the electric field isdouble the field at a point just inside S.(D) At any point inside S, the electric field is inverselyproportional to the square of its distance from C.

77. A hollow closed conductor of irregular shapeis given some charge. Which of the followingstatements are correct?(A) The entire charge will appear on its outersurface.(B) All points on the conductor will have the samepotential.(C) All points on its surface will have the samecharge density.(D) All points near its surface and outside it willhave the same electric intensity.

78. Three point charges are placed at thecorners of an equilateral triangle of side Las shown in the figure.

(A) The potential at the centroid of the triangle iszero.(B) The electric field at the centroid of the triangleis zero.(C) The dipole moment of the system is qL2

(D) The dipole moment of the system is qL3 .

79. An electric dipole is placed at the centre ofa sphere. Mark the correct answer(A) the flux of the electric field through the sphereis zero(B) the electric field is zero at every point of thesphere.(C) the electric potential is zero everywhere onthe sphere.(D) the electric potential is zero on a circle onthe surface.

80. An electric field converges at the originwhose magnitude is given by the expressionE = 100rN/C, where r is the distancemeasured from the origin.(A) total charge contained in any sphericalvolume with its centre at the origin is negative.(B) total charge contained in any sphericalvolume, irrespective of the location of its centre,is negative.(C) total charge contained in a spherical volumeof radius 3 cm with its centre at the originhas magnitude3 ×10–13C.(D) total charge contained in a spherical volumeof radius 3 cm with its centre at the originhas magnitude 3 × 10–9 C.

81. A conducting sphere A of radius a, withcharge Q, is placed concentrically inside aconducting shell B of radius b. B is earthed.C is the common centre of the A and B.

(A) The field at a distance r from C, where

a r b is 20 r

Q4

1 .

(B) The potential at a distance r from C, where

a r b , is rQ

41

0.

(C) The potential difference between A and B is

b1

a1Q

41

0(D) The potential at a distance r from C, where

a r b is,

b1

r1Q

41

0.

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82. Two thin conducting shells of radii R and 3Rare shown in the figure. The outer shellcarries a charge + Q and the inner shell isneutral. The inner shell is earthed with thehelp of a switch S.

(A) With the switch S open, the potential of the innersphere is equal to that of the outer.(B) When the switch S is closed, the potential ofthe inner sphere becomes zero.(C) With the switch S closed, the charge attainedby the inner sphere is – q/3.(D) By closing the switch the capacitance of thesystem increases.

83. X and Y are large, parallel conducting platesclose to each other. Each face has an areaA. X is given a charge Q. Y is without anycharge. Points A, B and C are as shown infigure.

(A) The field at B is A2Q

0

(B) The field at B is AQ

0

(C) The fields at A, B and C are of the samemagnitude.(D) The field at A and C are of the samemagnitude, but in opposite directions.

84. A particle of mass m and charge q is thrownin a region where uniform gravitational fieldand electric field are present. The path ofparticle(A) may be a straight line(B) may be a circle(C) may be a parabola(D) may be a hyperbola

85. Two large, parallel conducting plates areplaced close to each other. The inner surfaceof the two plates have surface chargedensities and . The outer surfacesare without charge. The electric field has amagnitude of

A) 0

in the region between the plates

B) 0

in the region outside the plates

C) 0

2 in the region between the plates

D) Zero in the region outside the plates86. The electric potential in the region of space

is given by: 2 ,V x A Bx Cx whereV is in volts, x is in meters and A, B, C areeconstants. Then,A) E

varies linearly with xB) the unit of E

is N/C

C) E

is in the negative x-directionD) the electric field E

in this region is constant

87. A dipole is placed in x y plane parallel tothe line 2y x . There exists a uniformelectric field along z axis . Net force actingon the dipole will be zero. But it canexperience some torque. We can show thatthe direction of this torque will be parallelto the lineA) 2 1y x B) 2y x

C) 12

y x D) 1 22

y x

88. 10 C of charge is given to a conductingspherical shell and a 3C point charge isplaced inside the shell. For this arrangement,markout the correct statement(s).A) The charge on the inner surface of the shellwill be 3C and it can be distributed uniformlyor non-uniformly.B) The charge on the inner surface of the shellwill be 3C and its distribution would beuniform.C) The net charge on outer surface of the shellwill be 7C and its distribution can be uniformor non-uniform.D) The net charge on outer surface of the shellwill be 7C and its distribution would be uniform.

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89. For Gauss’s law, mark out the correctstatement(s).A) If we displace the enclosed charges (within aGaussian surface) without crossing the boundary,then E

and both remain same.

B) If we displace the enclosed charges withoutcrossing the boundary, then E

changes but

remains the same.

C) If charge crosses the boundary, then both E

and would change.D) If charge crosses the boundary, then changes but E

remains the same.

90. Two infinite, parallel, non- conducting sheetscarry equal positive charge density . Oneis placed on the other at distance x=a. Takepotential V = 0 at x = 0. Then:

A) 0 ,For x a potential 0xV

B) For ,x a potential 0

xV x a

C) For ,x a potential 0

xV x a

D) For ,x a potential 0

xV x

91. Consider two large, identical, parallelconducting plates having surface X and Y,facing each other. The charge per unit areaon X is 1 and charge per unit surface area

on Y is 2 . Then,

A) 1 2 only if a charge is given to eitherX or Y

B) 1 2 0 if equal charges are given toboth X and Y

C) 1 2 if X is given a charge more thanthat given to Y

D) 1 2 in all cases

92. A particle having specific charge ‘s’ startsfrom rest in a region where the electric fieldhas constant direction but magnitudevarying with time t as 0E E t . In time t, itis observed that the particle acquires avelocity v after covering a distance x, then

A) 30

12

x E s t B) 30

16

x E s t

C) 30v E s t D) 2

012

v E s t

93. Two charges 1Q and 2Q are placed at thepoints A and B having separation r lyinginside and outside the uncharged conductingshell. The force on 1Q is F and that on 2Q isf. Then

A) 0F B) 1 2

20

14

Q QFr

C) f 1 2

20

14

Q Qr

D) f = 0

94. In a parallel plate capacitor, the potentialdifference between the plates is V. A particleof mass m and charge -Q leaves the negativeplate and reaches the positive plate atdistance d in time t with a momentum p. Then

A) p mQV B) 2p mQV

C) 2mdt

QV D)

22mdtQV

COMPREHENSION TYPE QUESTIONSPASSAGE : I

The sketch below shows cross-sectionsof equipotential surfaces between twocharged conductors that are shown insolid black. Some points on the equipo-tential surfaces near the conductors aremarked as A,B,C,........ . The arrange-ment lies in air. (Take

0 = 8.85 × 10–12

C2/N m2]

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E

B CD

Solid conducting sphere

0.3mLarge conducting plate

–30V –20V –10V 10V 20V 30V 40V

A

95. Surface charge density of the plate isequal to(A) 8.85 × 10–10 C/m2 (B) –8.85 × 10–10 C/m2

(C) 17.7 × 10–10 C/m2 (D) –17.7 × 10–10 C/m2

96. A positive charge is placed at B. When itis released :

(A) no force will be exerted on it.(B) it will move towards A.(C) it will move towards C.(D) it will move towards E.

97. How much work is required to slowly movea – 1mC charge from E to D ?

(A) 2 × 10–5 J (B) –2 × 10–5 J(C) 4 × 10–5 J (D) –4 × 10–5 J

PASSAGE-II:A particle of mass 2' 'm carrying a charge

2Q is fixed on the surface of the earth .Another particle of mass 1' 'm and charge

1Q is positioned right above the first one at

an altitude h R R is radius of earth.

The charges 1Q and 2Q are of same sign .Then

98. The velocity of 1m at a point P very close to

2Q at a distance 1h from the surface of theearth (if the initial velocity of 1' 'm was zero,air drag and earths magnetic field beingignored) is

A) 1 2

0 1 12

2Q Qgh

h m

B) 1 2

20 1 1

24

Q Qghh m

C) 2gh D) 1 2

0 1 12

4Q Qgh

h m

99. The magnitude of charge 2Q at which the

velocity of 1' 'm at an altitude 2h is zero isgiven by

A) 0 1 2

21

m ghhQQ

B)

0 1 22

1

4 m ghhQQ

C) 0 1 2

21

2 m ghhQQ

D)

0 1 22

14m ghhQ

Q

100. At what altitude 3h will object 1m be inequilibrium and what will be the nature ofobjects in motion if it is disturbed slightlyfrom equilibrium.

A) 1 23

0 1, ,

4Q Qh periodic not oscillatory

m g

B) 1 23

0 1,

2

Q Qh periodic and oscillatorym g

C) 1 23

0 1,

4

Q Qh periodic and oscillatorym g

D) 1 23 4 0 1

Q Q

hm g

non- periodic and non

oscilllatoryPASSAGE - III

There is a fixed semicircular ring of radius‘R’ lying in yz plane , with centre at originand it is uniformly charged with charge QA pipe is fixed along x axis from the origin. The inner surface of pipe is smooth and ismade of insulating material . A small ball ofcharge q and mass ' 'm is projected in thepipe with negligible velocity , ball can movein the pipe. Whole arrangement is in gravityfree space.

y

x

z

y

x

z

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101. The maximum acceleration of the ball in thepipe is

(A) 204Qq

mR (B) 2012 3

QqmR

(C) 206 3

QqmR (D) 2

012 2Qq

mR102. The kinetic energy of the particle when the

acceleration is maximum is

(A) 0

1 214 3

QqR

(B)

0

1 14 2

QqR

(C) 0

1 114 3

QqR

(D)

0

1 114 3

QqR

103. Normal reaction exerted by pipe on ball

when the ball is moving in pipe is along(A) Z a x is always(B) Z a x is always(C) Initially along y axis and then alongZ axis(D) y axis always

PASSAGE-IV:

Suppose electric potential varies along thex axis as shown in the above figure. Thepotential does not vary in y or z direction of theintervals shown (ignore the behaviour at the endpoints of the intervals), the field xE has a

maximum absolute value 11" "FB Vm in the

region 2" "FB . Its value in the region cd is1

3" "FB Vm then

104. The value that fills 1" "FB is

A) 252 B) 25 C)

152 D) -50

105. The region that fills 2" "FB isA) ab B) de C) bc D) dc

106. The value that fills 3" "FB isA) 1 B) 2 C) 3 D) zero

107. The plot xE vs x is

A)

B)

C)

D)

108. In the V vs x curve , the potentialpossesses a zero value at

A) 125

x m B) 513

x m

C) 135

x m D)

145

x m

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PASSAGE : VThe law governing electrostatics iscoulomb’s law. In principle coulomb’s lawcan be used to compute electric field dueto any charge configuration. In practice,however, it is a formidable task to com-pute electric fields due to uniform chargedistributions. For such cases Gauss pro-posed a theorem which states that

0

qE.ds

Where ds is an element of area directedalong the outward normal for the surfaceat every point. The integral is calledelectric flux. Any convenient surface thatwe choose to evalvate the surface integralis called the Gaussian surface.

109. Gauss theorem in electrostatics states

that electric flux s 0

qE.ds . Where S is

the Gaussian surface . The electric field Eis due to all the chargesA) both inside and outside the surface SB) outside the surface SC) inside the surface SD) on the surface S

110. The SI unit of electric flux isA) 1Vm B) 2Vm C) 1 2NC m D) 1 2NC m

111. Consider a uniform electric field E =3 1ˆ3 x10 i NC . What is the flux through a

square of side 10 cm whose plane isparallel to the y-z plane?

A) 1 230 NC m B) 1 220 NC m

C) 1 210 NC m D) zeroPASSAGE -VI

Figure shows concentric sphericalconductors A,B and C with radii

, 2 4R R and R respectively. A and C areeconnected by a conducting wire and B ishaving uniform charge ,Q then

112. The charge on conductor A is Aq and that

on C is Cq then

A) / 3A C

q q Q B) 2,

3 3A C

Q Qq q

C) ,3 3A C

Q Qq q D) ,3 3A C

Q Qq q

113. The electric potential of ' 'A is

A) 04

QR

B)

012Q

R

C) 016

QR D)

08Q

R

114. The electric potential of ‘B’ is

A) 04

QR B)

0

512

QR

C) 0

36

QR D)

0

548

QR

PASSAGE - VII:

CB

Aa

2a

3a

S

CB

Aa

2a

3a

S

Three conducting concentric shells A,B andC of radius a , 2a and 3a are as shown inthe figure . The charge on shell B is ‘Q’.The switch ‘S’ is closed. Then after closingthe switch

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115. The charge on innermost shell is

(A) 4Q

(B) 4Q

(C) 2Q

(D) 2Q

116. The charge on outermost shell is

(A) 4Q

(B) 4Q

(C) 2Q

(D) 2Q

117. If ,A B Cand are the surface chargedensities on A,B and C respectively then

: :A B C is(A) 9 :9 :1 (B) 9 : 9 :1(C) 9 :9 :1 (D) 1:1:1

ASSERTION AND REASON QUESTIONSA) Both A and R are True and R is the correctexplanation of AB) Both A and R are True but R is not the correctexplanation of AC) A is True, R is FalseD) A is False, R is True

118. Assertion(A) : An applied field will polarizethe polar dielectric material.Reason(R) : In polar dielectrics, each moleculehas a permanent dipole moment but these arerandomly oriented in the absence of an externallyapplied electric field.

119. Assertion(A) : Total work done by non-uniformelectric field on a charged particle starting fromrest to any time is non-negative.(Assume no otherforces act on the charged particle.)Reason(R) : The angle between electrostaticforce and velocity of the charged particle releasedfrom rest in non-uniform electric field is alwaysacute. (Assume no other forces act on thecharged particle.)

120. Assertion(A) : A metallic shield in the form of ahollow shell may be built to block an electric field.Reason(R) : In a hollow spherical conductingsphere , the electric field inside it is zero at everypoint.

121. Assertion(A) : E

in the outside vicinity of aconductor depends only on the local chargedensity and it is independent of the othercharges present anywhere on the conductor.Reason(R) : E

in outside vicinity of a conductor

is given by 0

122. Asser t ion(A ) : If Gaussian surface does notenclose any charge, then E

at any point on the

Gaussian surface must be zero.Reason(R) : No net charge is enclosed byGaussian surface, so net flux passing through thesurface is zero.STATEMENT TYPE QUESTIONS

A) If both statements are TRUE andSTATEMENT 2 is the correct explanation ofSTATEMENT 1.B) If both statements are TRUE butSTATEMENT 2 is not the correct explanationof STATEMENT1.C) If STATEMENT 1 is TRUE andSTATEMENT 2 is FALSED) If STATEMENT 1 is FALSE butSTATEMENT 2 is TRUE

123. Statement - 1: The potential of an unchargedconducting sphere of radius R, for a point chargeq located at a distance r from its centre

04

qr R isr

Statement - 2: Electric field intensity inside theconductor is zero therefore potential at eachpoint on the conductor is zero.

124. Statement - 1: Any charge will move fromelectric potential 1V to 2V by its own only

when 1V 2V .Statement - 2: Electron moves from

1 2V V towards 2 4V V125. Statement - 1: Electric field intensity at surface

of uniformly charged spherical shell is E. If theshell is punctured at a point then intensity at

punctured point becomes 2E

.

Statement - 2: Electric field intensity due tospherical charge distribution can be found outby using Gauss Law.

126. Statement - 1: If a point charge q is placed infront of an infinite grounded conducting planesurface, the point charge will experience a force.Statement - 2: This force is due to the inducedcharge on the conducting surface, which is at zeropotential.

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127. Statement - 1: The surface of a chargedconductor is always an equipotential.Statement - 2: Electric field lines are alwaysperpendicular to the equipotential surface.

MATRIX-MATCHING QUESTIONSThis section contains 4 questions. Each questioncontains statements given in two columns whichhave to be matched. Statements (A, B, C, D) inColumn I have to be matched with statements(p, q, r, s) in Column II. The answers to thesequestions have to be appropriately bubble asillustrated in the following example. If the correctmatches are A–p, A–s, B–q, B–r, C–p, C–q andD–s, then the correctly bubbled 4 × 4 matrixshould be as given :

128. Two charges q and Kq are placed atorigin O and point A:. P is a point on a lineparallel to y-axis where potential has a zerovalue.

Column-I Column-IIA) If 1 2,OP r AP r p) 1Kr

and ,OA d then 2r is given by

B) If ,OA d then q) 1 sinKr

1PP is equal to

1PAP

C) If 6OA m and r) 1 24; 1, 1x x

2,K then locus of point P is a circle of radius and centre given by 1x and 2xrespectively.

D) For some value of s)none of theseOA (> 0) and K(>0), locus of point P will bea circle of radius andcentre given by

1x and 2x respectively..129. The variation of electric field (E) with

distance (x) from its centre is shown. Thesphere has a positive charge.

Column - I Column - IIA) The sphere is p)non-conducting material

made of aB) The dielectric q)different

constant of thesphere and itssurrrounding are

C) Charge density r)non-uniformthroughout thevolume of thesphere is

D) Electric potential s)none of the aboveat the centredue to sphere is

130. In options (A) to (C) spherical conductor ishollow and neutral and in (D), it is hollowand charged. Match the columns:Column - I Column - II

A) p)Potential inside the

conductor is varying.

B) q) E inside the

conductor is zero.

C) r) E

inside the

conductor is constant.

D) s) E

inside theconductor is varying.

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INTEGER TYPE QUESTIONS

131. A uniform surface charge density 8 existsover the entire x-y plane except for a circularhole of radius a centred at the origin. The

electric field at a point 0,0, 3P a on the

z-axis is found to be 0

2 x . Find the value

of x .132. Two fixed, equal, positive charges, each of

magnitude 55 10 C are located at pointsA and B separated by a distance of 6m. Anequal and opposite charge moves towardsthem along the line COD , the perpendicularbisector of the line AB. The moving chargewhen it reaches the point C at a distance of4 m from O , has a kinetic energy of 4 joules.The distance of the farthest point D , inmetre, where the negative charge will reachbefore returning towards C is 6 x . Find x .

133. A thin fixed ring of radius 1 meter has apositive charge 51 10 coulomb uniformlydistributed over it. A particle of mass 0.9gand having a negative charge of 61 10 coulomb is placed on the axis at a distanceof 1 cm from the centre of the ring. Themotion of the negatively charged particle isapproxi mately simple harmonic. Hence the

time period of oscillations is k

. Find k .

134. Two concentric spherical shells have radii0.15 m and 0.3 m. The inner sphere is given

a charge 26 10 .C An electron escapesfrom the inner sphere with negligible speed.Assuming that the region between thespheres is vacuum, the speed with which it

will strike the outer sphere is 6 110 ,x msFind the value of x?

11 11.76 10e for electron Ckgm

135. A particle having a charge of 8.85q Cis placed on the axis of a circular ring ofradius R = 30 cm at a point P at a distanceof a=40 cm from the centre of the ring. Theelectric flux passing through the ring is

510 / .x N C Find the value of ?x136. It is required to hold four equal point

charges each having a charge

8 1 2 27

Q C in equilibrium at the

corners of a square. Find the point charge,in coulomb, that will do this if placed at thecentre of the square.

137. A ball of mass 2 kg, charge 1 C is droppedfrom top of a high tower. In space electricfield exists in horizontal direction away fromthe tower which varies as

6 1E= 5 - 2x 10 .Vm Find maximumhorizontal distance that the ball can go fromthe tower.

138. A solid sphere of radius R has charge Qdistributed in its volume with a charge

density ,akr where k and a areconstants and r is the distance from its

centre. If the electric field at 1

2 8Rr is

times that at r=R , find the value of a.

139. A charge 5

100Q nC is distributed over

two concentric hollow spheres of radii r=3cm and R=6 cm such that the densities areequal. Find the potential, in volt, at thecommon centre.

140. A point charge + Q is placed at the centroidof an equilateral triangle. When a secondcharge + Q is placed at a vertex of thetriangle, the magnitude of the electrostatic

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force on the central charge is 8 N. Themagnitude of the net force on the centralcharge when a third charge + Q is placed atanother vertex of the triangle is :

141. Two very small particles 1 and 2 each ofmass 0.5kg and of charges 1 1q mc and

2 1q c respectively are connected by amass less spring of spring constant

1200Nm and placed on a horizontal roughsurface. The particle 1 is fixed and 2 is freeto move. Initially the spring is in its naturallength (2m) when 2' 'q is released. Ifcoefficient of static friction between 2 andhorizontal surface is 0.5 the separationbetween the charges when they are inequilibrium will be (Neglect gravitationalinteractions between 1 and 2) g=10ms-2.

LEVEL - IV - KEY

1) A 2) D 3) D 4) B5) A 6) C 7) B 8) B9) D 10) C 11) A 12) C13) A 14) B 15) C 16) C17) A 18) A 19) B 20) B21) A 22) A 23) D 24) B25) D 26) A 27) C 28) D29) B 30) D 31) C 32) A33) B 34) B 35) A 36) A37) A 38) C 39) D 40) A41) B 42) A 43) D 44) B45) B 46) B 47) C 48) C49) C 50) C 51) B 52) A53) C 54) D 55) A 56) C57) A 58) C 59) A 60) A61) C 62) C 63) D 64) B65) C,D 66) C 67) C,D 68) A,C69) A,D 70) A,C 71) B 72) A,B,C73) B, C 74) A, C

75) A,C,D 76) A,C,D77) A,B 78) A, D79) A, D 80) A, B, C81) A, C, D 82) A, B, C, D83) A,C,D 84) D85) A,D 86) A,B,C87) C,D 88) A,D89) B,C 90) A,B,D91) A,C 92) B,D93) A,C 94) B,D

PASSAGE-I95) A 96) B 97) DPASSAGE-II98) A 99) B 100) CPASSAGE-III101) C 102) A 103) APASSAGE-1V104) B 105) A 106) D107) D 108) CPASSAGE-V109) A 110) D 111) APASSAGE-VI112) D 113) C 114) DPASSAGE-VII115) B 116) A 117) C

118) A 119) C 120) A121) D 122) D 123) C124) D 125) B 126) A127) A 128) A-p, B-q, C-s, D-r129) A-p, B-q, C-s, D-r130) A-p,s;B-q,r; C-p,s; D-p,s131) 3 132) 2 133) 5134) 8 135) 1 136) 2137) 5 138) 2 139) 9140) 8 141) 2

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LEVEL - IV HINTS1.

x

dx

x

dx

dQ dx

20x dx

2.

20

0

lx dx

303.lQ

d d

dQ R d

0 cos Rd

20

2

cosQ R d

= 20 2sinR

02 R3.

rdA

rdA

2dA rRd

22 sinR d

202 sin cosdQ R d

20 sin 2R d

/22

00

sin 2Q R d

2

/200cos2

2R

20Q R

4.

++ + + ++ + + + ++

/2

/2

dF cosdF

++ + + ++ + + + ++

/2

/2

dF cosdF 5

The force on the pt. charge q due to the

differential element is

2 2

k R d qkqdqdFR R

2

.cos cos

k q R ddF

R

/2/2

coskqF dR

2 sin2

kq

R

but QR

2

2 sin2

kQqF

R

5.

TT2

d2

d2cos dT

2cos dT

2sin dT

TT2

d2

d2cos dT

2cos dT

2sin dT

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20

.2 sin2 4

d q dQTr

20

22 4

d q rdTr

204

q rTr

20(2 )4

qQrTr r

2 208

qQTr

6.d

1Q

2Q

d

1Q

2Q

Due to induction the charge distribution will beas shown

++++++

+ ++++

++

+++++++

+ ++++

++

+

The distance between the centres of charge willbe >dThe force between them will be less than

1 22

0

14

Q Qd

7 Only angular momentum x & total energy areconserved but not kinetic energy

8.

d

sin F

cosdF

dF

d

02

ER

dF = E dq1 E Rd

From symmetry cos 0 dF

sin F dF

1

0

ER d

1

0 0

sin2

R dR

on solving 20

qFR

9. Diagram is the solution10. The electric field intensity due to each uniformly

charged infinite plane is uniform. The electricfield intensity at points A, B, C and D due toplane 1, plane 2 and both planes are given byE1, E2 and E as shown in figure 1. Hence, theelectric lines of forces are as given in figure 2.

E1

E2

EB

C D

A

E1E

E2

E2

E1E

2

z

1 xE2

EE1

+

–

(figure 1)(figure 2)

Z

X

11.0

14

qVa

04q aV equating volumes of bubble and droplet

2 3443

a t R

1

2 33R a t

0

14

x qVR

01

0 2 3

414

3

aV

a t

13

3aVt

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12. . .C B

B AA C

V V E dr E dr

. 0B

CE dr

.C

B AA

V V E dr

0.5

o

0.3325 dl cos180

325 0.8

=260V13. (A) Wnet =

q E.d

where E =

0

i2

net 1 20

W q (X X )2

14. (B) eE = mew2r

Edr = R2

e

0

m r dre V = 2 2

em R2e

15. (C) Vp = in

ka Vr = VC =

kqr

Vin =

kq kqr r

16. 0 sin

dQ dl

0 sin R d

0

0

sin14

R ddV

R

/2

00 /2

1 sin4

V d

/20/2

0

cos4

= 0

17. The area of shaded element is2dA rRd

22 sinR d

0

14

dqdVR

2

0

1 2 sin4

R dR

0sin

2R d

/2

0 0

sin2

RV d

02

R

18.

0

,2

Ey

For a small displacement dr dy j dxi

.dw E dr

0

.2

j dy j dxiy

02

dyy

20

ln2

a

a

qW y

0

ln 0.52

q

0

ln 22

q

19. Conceptual20. (A) vA – vB = work done by electric field on

+ 1 coul. charge from A to B = E Rq vB = vA – E Rq = v – E Rq

21.Usystem = Uself + Uint

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22.

V V VE i j kx y z

23. No change in potential for the wire placed onX axis. So W = 0 and for wires on Y & Z axesis

2

0 1

n2

rW r lr

W = W1 + W2

0

4 22 1

W l n

0

2

W l n

24. (B) E = = 2 2 2

1 1 1ˆ ˆ ˆi j kx y z

= ˆ ˆ ˆi j k

25. (D) B A xV V E dx

= – [Area under Ex – x curve]

VB – 10 = 1.2.( 20)2

= 20

VB = 30 V.

26. 1 2 10

q qK vb a

1 2 5

q qK vb

1 2 ?

cq qV Kb a

Clearly Vc = 10V27.

A BC

C B Ac

q q qV Kc b a

2 2 2

0

1 4 4 44A

c b aVc b a

0

( )b c a

28.

A BC

D

+q4q -q

4(3 ) 2 4

Aq q qK Va a a

6A

KqVa

29. (B) u = 20

1 E2 =

2 20

4K Q1

2 2 r

V = KQr

2u

V =

2

20 4

2 2

2

1 QK2 r

K Qr

= 0

212 2r

because 2 2u 1

V rso the correct option is B.

30. (D) Interaction energy of system of charges is

= 1 61 [U ... U ]2 = 1

1 [6U ]2 = 3U1

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=

2 2 2 13 kqa 2aa 3 =

2

0

q 3 15a 2 8

31. 21 EQx ucos t2 m

21y usin t gt2

2usintg

2 2 2

2

u 1 E 4u sinx sin2 .g 2 m g

2

4

dx 0d

then mgtan2EQ

22 2

max 2

ux EQ mg EQmg

32. In order to maintain constant separation, the par-ticles must have the same acceleration.

Assuming the system of both charges to accel-eration towards left. Applying Newton’s secondlaw to the left particle we get

2

KQqQE MaL

............... (1)

Under given condition the acceleration of bothcharges should be same and should also be equalto acceleration of centre of mass of both thecharges.

net Q q EFatotal mass m M

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

Hence from equation (1) and (2) we get

M m KQqL

E qM Qm

33. (TE)surface = (PE)centreVelocity at centre is nearly zero then it cancross the sphere

2

0 0

1 . 1 3 14 2 2 4

q q qmvR R

34. . E gW W K EUse work energy theorem

21sin mg (1 cos )2

Eq l l I

35. Use conservation of energy

at surface at contrect path

PE KE PE

36. (A) We have centripetal force equation

2kqr =

2mvr

so v = 2kq

m Now,

T = 2 rv =

m

2k q

where, k = 0

14

37.R PR P

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The charge on the shaded part

2 2QdQ rdrR

22Q r drR

Potental due to dQ at P

2 2 2

0

2

4

r drQdVR R r

2 2 202

Q r drR R r

2 2 102

R

o

Q r drVR R r

2 22 002

RQ R rR

20

2 12

Q RR

0

2 1

2

Q

R

0

2 1

2

QqU of q

R

Similarly the potential due to dQ at centre of discis

2

0

24r drQdV

rR

202

Q drR

02QV

R

U of q at centre of disc is

02QqU

R

Applying mechanical energy conservation

2

0 0

12 12 2 2

Qq QqO mvR R

2

0

1 1 2 12 2

QqmvR

2

0

1 2 22 2

QqmvR

2

0

2 2 112 2

Qqmv

R

1/ 2

0

2 2 1Qqv

mR

38.2

0

1 1 02 4

QqmvR

0 0

24 2

Qq QqvmR mR

39. As the first particle approaches the second dueto repulsion it recedes away from it along thesame l ine of apprach. The separation betweenthem is minimum when their velocities are equal.Applying conservation of linear momentum

f fmv mv mv

/ 2fv v

2

2 2

0

1 12 2 4f

qmv m m vr

2 22

0

1 1 22 2 4 4

v qmv mr

22

0

14 4

qmvr

2

20

qrmv

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40. 1q

E

v v

E

2q

y

x

1q

E

v v

E

2q

y

x

Let x yE E i E j

By impulse - momentum theoremImpulse=change in momentum

Let the final velocities of balls be 1 2&v v

1 2vv

1 1 1cos60 sin60

2 2x yv vq E i E j t m i j m vi

2

2 2 2 2cos90 sin 90

x yq E i E j t

m v i v j m vi

11

11

3 ,4

34

x

y

m vq E t

m vq E t

2 2 2 2 2,x yq E t m v q E t m v

1 1

2 2

34x

q Ex t mq E t m

2

2

43

qm

41.

Separation between them is minimum whenvelocity is perpendicular to line joining the twochargesApplying conservation of angular momentum

0 sin 30mV R mVd .

012 3

oVmV R m d 32

Rd

42. dP = R dq R dx R Rd

sin yP dP

/22

0

sin

R d 2 R

22

qRR

2

qR

2

xqRP

x yP P i P j43. Field due to ring is given by

3/22 204

QxER x

Force on dipole is given by

dEF pdx

3 / 22 20

2 .4

d QxF a qdx R x

5/2 3/22 2 2 2

0

2 3 24 2aq Q x R x x R x

5/2 3/22 2 2 2 2

0

2 34aq Q x R x R x

5/22 2 2 2 2

0

2 34aqQ R x x R x

2 2

5 / 22 20

22aqQ R xF

R x

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44. (B) The given point is at axis of p2

dipole

and at equatorial line of p dipole so total

field at given point is.

E =

3 3kp 2k(p / 2)(1) (2) =

1kp –18

=

0

732

p

45. (B) Electric field at each point on the surface of

ring due to dipole is E = 3k pR

in direction

opposite to the dipole moment. (figure below)

Hence, net force on ring is F = QE = 3kpQR

Alternate solutionElectric field due to ring at point P on its axisat a distance x from the centre O of the ring is

E =

2 2 3 / 2

at x 0

Qx dEk ;dx(x R )

= 3kQR

Force on dipole = dEdx = 3

k QpR

46.

x

x

1 1tan tan

2

given / 2

1 1tan tan2 2

1 tan cot2

tan 2

47.

By symmetry2

sin2 o

l

48. The charge inside the cube due to each line is L .Flux due to horizontal wire through the facesAEDH, BCGF is zero as the electric fieldproduced will be parallel to these faces.

But the total flux due to this line is 0

L as it is

equally distributed among the remaining fourfaces. Similarly flux due to vertical coire throughthe faces DHGC,AEFB is zero.

Flux through CDHG = 0 04 4

L LO

Flux through AEHD = 0 04 4

L LO

Flux through ABCD = 0 0 04 4 2

L L L

49. Flux first increases linearly ( v is constant),remains constant (when length L/2 is inside thecube) and then decreases linearly (as rodmoves out).

50. (C) f =

E dssince r << R so we can consider electric fieldis constant throughtout the surface of smallerring,

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hence

f = 2 2 3 / 2

xE(R x )

So, the best represented graph is C.51.

E

qq

dA

l

E

qq

dA

l

2dA r dx

2 20

14

qEr l

cosd E dA

2 2 2 20

1 24

q lr dxr l r l

3/22 202ql r dr

r l

3/22 20 02

Rql r dr

r l

2 20 0

12

r R

r

ql

r l

2 20

1 12ql

l R l

2 201

2q l

R l

52. 202 1 cos60

q

02q

53. q

d A

E

dx

x

q

d A

E

dx

x

2 20

14

qER x

2 2 2 20

1 24

q Rd R dxR x R x

2

3/22 202qR dx

R x

/22

3/22 20 /22

L

L

qR dx

R x

2 20 4

q L

L R

54.

1 2

0 0

1 cos 1 cos2 2q q

2 21 22sin 2sin

2 2q q

2 21

2

sin sin2 2

qq

1 1

2

2sin sin2

qq

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55.

rr

2V Ar B dVEdr

2E Ar

0. qE ds

32

0

42 4 .3

rAr r 06 A

56. Two very large thin conducting plates having samecross-sectional area are placed as shown infigure. They are carrying charges 'Q' and '3Q'respectively. The variation of electric field as afunction of x (for x = 0 to x = 3d) will be bestrepresented by.

Y

Q 3Q

(d, 0) (2d, 0) (3d, 0) X

E

d 2d 3d X

E

d

2d 3d X

(A)

(C)

2d 3d

E

d 2d 3d X

E

d

2d 3d X2d 3d

(B)

(D)

57. Electric field at any point inside the cavity =Electric field due to entire sphere at that point-the field at the same point due to a solid spherewhich can tightly fit into the cavity

O1ra

p2r

0

1 2E E E

According to Guass's Law

1 10

E r3

1 2

0E (r r )

3

2 20

E r3

= 1 2

0a ( a r r )

3

58.conceptual

59.a x

K

b

dx

a x

K

b

dx

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consider a small elemental shell of thickness ‘dx’Volume 24dv x dx

Eletric field at 20

14

qxKx

Energy density 20

12

E

0 2 2 4

0

1 12 4

qK x

Energy in elemental shell is

22

0 2 2 40

1 1 42 4

qdU x dxK x

2

2 20

14 2

b

a

q dxUK x

2

20

1 1 14 2

qUa bK

.60. The figure shows a charge q placed inside a cavity

in an uncharged conductor. Now if an externalelectic field is switched on :(A) Only induced charge on outer surface willredistribute.(B) Only induced charge on inner surface willredistribute.(C) both induced charge on outer and innersurface will redistribute.(D) force on charge q placed inside the cavitywill charge.

61.

Q2aB

A C

5a

Q5a

5a

a

1S

2S

Q2aB

A C

5a

Q5a

5a

a

1S

2S

Let charges on B and C are respectively 1q and

2q

0BV

1 2

0 0 0

1 1 1 04 5 4 2 4 5

q qQa a a

1 2 0 12 5 5q q Qa a a

0cV

1 2

0 0 0

1 1 1 04 5 4 5 4

q qQa a a

1 2 0 25 5q q Qa a a

1 2 1 22 5 5q q q qa a a a

1 2 122 5 5

q q q q

1 23 4

10 5q q

1 23 8q q

1 13 02 40 5q q Qa a a

1 132 40 5q q Q

12340 5

q Q

1823Qq

12

38qq

323Q

62.

2 1 2q q Q

1 2 3 4qQ q q Q

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1 1 2

32

4 4 2 4 2

4 4 4 4

o o o

o o

q q qR R R

qqR R

AV

1 1 2 2 314 4 c

o

q q q q q VR

A CV V

3 31 2 21 2 2 4 4 4

q qq q qq

1 2 02 4q q

2 12q q

1 12 2q q Q

123Qq

the charge that flows through the switch is Q/3.63.

1 2 5q q Q q q Q

1 2 6q q Q

22 4 0

o o o

q Q Q qq d d dA A A

2 4 4 4 0q q Q Q q 0q

0AE

1 21 20

2 2o o

q q q qA A

1 2 3q q Q

q that flows through switch is 2Q.

64.2 2 2

0 0 0

16 1 48 16 4 2i

Q Q QUR R R

2 2 2

0

2 16 816

Q Q QR

2

0

58

QR

2

0

916f

QUR

Heat produced 2 2

0 0

5 98 16

Q QR R

20/16Q R

MULTI ANSWER QUESTIONS

65. CONCEPTUAL

66. .CONCEPTUAL

67. CONCEPTUAL

68. CONCEPTUAL

69. CONCEPTUAL

70. CONCEPTUAL

71. CONCEPTUAL

72. CONCEPTUAL

73. CONCEPTUAL

74. CONCEPTUAL.

75. CONCEPTUAL

76. CONCEPTUAL

77. CONCEPTUAL

78. CONCEPTUAL

79. CONCEPTUAL

80. CONCEPTUAL

81. CONCEPTUAL

82. CONCEPTUAL

83. CONCEPTUAL84.

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Vectorially 30

14

qE rr

^

2 2 20

1 2 24 1 2 3

AQE i j k

^

30

1 2 24 14

AQE i j k

3

2 2 20

14 1 2 3

BQE i j k

3

0

14 3

BQE i j k

3

2 2 20

1 2 2 34 2 2 2

CQE i j k

3

30

2 2 314 2 3

C

Q i j kE

. 0A BE E

A option is correct14C BE E

C option is correct

85.

0 0

02 2PE

0 0

02 2QE

0 0 02 2QE

86. / 2E V x B Cx i.e., E varies linearly with x and is along negativex-direction. It is also clear that B has gotdimensions or units of E, i.e., 1NC .

87. Torque will be perpendicular to the line y = 2xand it should be in x - y plane, because electricfield in z-direction. The lines in option (c) and(d) both are perpendicular to y = 2x.

88. Due to induction, charge on various faces are asshown.

Charge on the inner surface of shell = +3 CNet charge on outer surface of shell =

3 10 7C C C Distribution of charge on inner surface would beuniform if charge is placed at the centre,otherwise non-uniform.On outer surface, chargewould be always uniformly distributed asdisplacement of inside charges does not affectthe distribution of the outer charge.

89. crossing through Gaussian surface does not

depend on the location of charge, while E

depends on it. If q crosses the boundary, then

enclosedq and hence the flux and E

also change.

90. 0 ;x a

00

0x

x xV E dx V

(as xE = 0)

;x a 0

x

x x aV E dx V

xV

0 0 0

x

aa

dx V x a x a

0; xx V

000

x

x xE d x V E

00 0

xV x V x

91. Conceptual

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92. 0E E t

since 00

qE qE ta E stm m

0 0

dv E st dv E stdtdt

20 0

0

12

t

v E s tdt E st

20

12

dx E stdt

20

12

dx E st dt

2 30 0

0 0

1 12 6

x t

dx E s t dt x E st 93. Conceptual

94. 212

mv QV

2

22p QV p mQVm

2

{ 0 }

Q Vv mt v a t

Q vam d

22mdtQV

COMPREHENSION TYPE QUESTIONSPASSAGE-I:

95. E = 3.01040

= 100 V/m

(near the plate the electric field has to beuniform \ it is almost due to the plate).For conducting plate

E = 0

0 E

Therefore s = 8.85 × 10–12 × 100=8.85 × 10–10 C/m2

96. Direction of E.F. at B is towards A that willexert force in this direction only, causing thepositive charge to move. [E

is perpendicular

to equipotential surface and its direction isfrom high potential to low potential.]

97. W = q. V= –1 × 10–6[20 – (–20)]= – 4 × 10–5 J.

PASSAGE-II:

98.

h<< R where R is radius of earth.Applying mechanical energy conservation

21 2 1 21 1 1 1

0 0 1

1 1 14 4 2

QQ QQm gh m gh mvh h

2 1 21 1 1

0 1

1 1 12 4

Q Qm v m g h hh h

2 1 2 11

0 1 1

224

Q Q h hv g h hm hh

2 1 2 11

0 1 1

224

Q Q h hv g h hm hh

1

2 1 1 2

0 1 1

122 1

4

hhh Q Q hv ghh m hh

since 1h h then 1hh can be neglected

2 1 2

0 1 12

2Q Qv gh

m h

1 2

0 1 12

2Q Qv gh

m h

99.1 2 1 2

1 1 20 0 2

1 14 4

q q q qm gh m ghh h

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( KE is zero at both positions)

1 1 20 2

1 1 124

q ql m g h hh h

21 2 1 2

0 2

14

h hq q m g h hh h

0 1 22

1

4 m ghhqq

100.1 2

20 3

14

q q mgh

1 23

04q qh

mg

PASSAGE - III101,102,103

x xdF q dE

2 2

1 cos4 o

dQqR x

3/22 2

14x

o

QqxFR x

3/22 20

14

Qqxam R x

for a to be maximum

3/22 20d x

dx R x

5/22 2

3/22 2

1 31 2 02

x R x xR x

2

3/2 5/22 2 2 2

1 3x

R x R x

2 2 23R x x

/ 2x R

max 3/222

1 / 24

2o

QqRam RR

2

1 (2 2)4 2 3 3)o

QqmR

26 3 o

QqmR

Applying mechanical energy conservation

2

2 2

1 1 14 4 2o o

Qq Qq mvR R x

KE 21 1 21

2 4 3o

QqmvR

PASSAGE-IV:

104,105,106,107,108

x MaxE value 1' 'FB in the region 2FB

xE value in cd is 13 .FB Vm

dVEdx

25 251abE

5 15 101 2 3bcE

0cdE

15 5 103 2deE

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125x MaxE Vm 2FB is ab

Taking the values of , , , ,ab bc cd deE E E E The plot

of xE Vs x will be as shown in ‘D’ option.For a b part 25 65V x

when 65 130

25 5V x m

PASSAGE : V109. (a) Conceptual

110. (d) = E .S

= F Sq

1 2NC m

111. (a) E S cos

is angle between area vector and field vector

2 2S 10cm 10cm 10 m

3 1E 3 10 NC

1 2E S Cos 30 NC m

PASSAGE -VI112. A CV V

0 0

1 14 2 4 4 4

A B C A B Cq q q q q qR R R R

4 2A B C A B Cq q q q q q 3 A Bq q Q

3AQq

3CQq

113.0

14 2 4

CA BA

qq qVR R R

0

14 3 2 12

Q Q QR R R

0

1 1 4 64 12

Q Q QR

0 0

1 1 34 12 16

QQR R

.

114.0

14 2 2 4

CA BB

qq qVR R R

0

14 6 2 12

Q Q QR R R

0

14 6 2 12

Q Q QR R R

0

1 2 6 14 12

QR

0

548

QR

PASSAGE - VII:115.

When switch is closed the inner most and outermost shells will acquire same potential.Let the charge on outer shell be ‘q’ and that onthe inner shell be -q, the total charge on innerand outer shells is zero.

A CV V

0 0

1 14 2 3 4 3

q Q q Qa a a a

3 2 3q Q Qq

23 3 2 6q Q Q Q

4Qq

.

4AQq

116. 4CQq

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117. 24 4AQ

a

24 4BQ

a

244 9CQ

a

1 1 1: : : :4 4 36A B C 9: 9:1

ASSERTION AND REASON QUESTIONS

118. E

in outside vicinity of conductor’s surfacedepends on all the charge present in the space,

but expression 0

E .

119. E at any point on Gaussian surface may be dueto outside charges also.

120. 00

arg14

induced ch eqVr r

00

1 04

qVr

Hence, potential is constant121. Electron being a negative charge, will be move

from lower to higher potential122. Conceptual

STATEMENT TYPE QUESTIONS

123. Apply the concept of electric image124. Since field is zero for a charged conductor on

the surface and inside it also. So the surface of acharged conductor is always equipotential. Also,

for equipotential surface. Hence . 0E dl

.E dl

Hence both are true and Statement- 2 is the correct explanation to Statement - 1.

125. (A)

0 1 2

1 04P

KqqVr r

or 2 1r Kr

(B) Now,

1 1 1 1 2sin sinPP POP r PAP r

1 2 1

1 1 1

sinsin

POP r Kr KPAP r r

1sin sinPOP K or 1

1

sinPP Kr

(C) 1 1, 6OP x P A x 2 2 2

1r x y

22 22 6r x y

2 1 2 12r r from r Kr

2 2 2 26 4x y x y

On solving, 2 2 22 0 4x y

Locus of point P is a circle of radius 1 4x m

and centre 2 2,0x (D) The values of OA and K should be known

126. E is proportional to distance upto ,x R so thesphere has uniformly distributed charge over itsvolume and because of this reason, the sphere ismade of non-conducting material.The electric field is same just inside and outsideof the sphere, therefore dielectric constants ofthe materials of sphere and surrounding aresame.Potential at the centre is maximum.

127. For option (B), ther e is no charge in theconductor, thus E

inside the conductor is zeroand the potential is same as potential of theconductor and is constant.For other cases, E

inside the conductor is non-zero and varying. Potential inside the conductoris also varying.MATRIX-MATCHING QUESTIONS

128. Conceptual129 Conceptual130. Conceptual

INTEGER TYPE QUESTIONS

131.0

'2net discE E

2 20 0

' ' 12 2net

zEz a

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2 20

'

2net

zEz a

0 0 0

8 312 2 3

2 2a

Ea

3x

132. From figure 2 2AC AO OC

2 23 4 5AC m Similarly, BC = 5 m

P.E. at C is 92 9 10cq q

UAC

K.E at C is 4 JTotal energy at C is . .cE P E K E .

9 4 5CE J

Potential energy at D is ,DU given by

29 52 9 10 5 10DU

r

where AD = BD = r (say)

45DU J

r

K.E at D = 0So, total energy at D is

45 450DE Jr r

By Law of Conversation of Energy,

C C D DU K U K

45 5 9r mR

2 2NowOD AD AO

2 29 3 72OD m

6 2OD 2x

133. Let us first the force on a-q charge placed at adistance x from centre of ring along its axis.Figure shows the respective situation. In this caseforce on particle P is

PF qE

3/ 22 204Pq QxF

x R

For small x, x<<R, we can neglect x, com

pared to R. So

30

14

qQxFR

Now, acceleration of particle is

30

14

F qQa xm mR

30

04

qQx xmR

which is a standard SHM equation. So

304

QqmR

302 42 mRT

Qq

33

9 5 60.9 10 1

29 10 10 10

T

55

T s k

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134. Gain in KE = 2 1e V V

2

0

1 1 1 12 4

mv e qa b

2

0

2 14

e abv qm b a

6 1 68 10 10v ms x x=8

135. 2 20

12E

q aR a

6

12

8.85 10 401502 8.85 10

E

610 412 5E

65 110 1 10

2 5E NC

5 51 10 10E x x=1

136. Let us consider the equilibrium of charge Qplaced at 2. Then for equilibrium

2 0F

2 2 0x yF i F j

2 0xF

24 25 2321 cos45 cos45 cos90 0F F F F

2 2 20 0

0

1 0 04 24 2 4

2

Q Q Qpa aa

2 02 2QQ q

2 2 1 4 0Q q

2 2 14Qq

8 1 2 27

Since Q

1 8 1 2 2 1 2 24 7

q

1 8 7 24 7

q C

137. xqEam

5 2xdv q xdt m

At time tv 65 2 10E x

5 2dv qv xdx m

2

252xv q x x

m

2 22 5xqv x x

m

22 5dx q x xdt m

For x to be maximum 0dxdt

25 0 5x x x m

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138. Total charge

2 3

0

443

r Ra a

r

kQ dV Kr r dr Ra

32

' 2

0

443 2

Rr aa

r

k RQ dV Kr r dra

According to question

'

2 20 0

1 1 14 8 4

2

Q QRR

putting the value of Q and 'Q geta = 2

139. Let 1 2q and q be the respective chargesdistributed over two concentric spheres ofradii r and R such that

1 2q q Q As surface densities are given to be equal,therfore

1 2

1 22 24 4

q qr R

21

22

q rq R

2 2 21 1 2

2 22 2

1 1q r q q r Rq qR R

using (1), we get2 2

22

Q r Rq R

This gives 2

2 2 2Rq Q

r R

Therefore 1 2q Q q

2 2

1 2 2 2 2R rq Q Q Q

r R r R

the potential 1V at common centre due to

charge 1q is given by

11

0

14

qVr

2

1 2 20

14

r QVrr R

1 2 20

14

QrVr R

The potential 2V at common centre due tocharge 2q is

2 2 20

14

QRVr R

Net potential at common centre,

1 2V V V

2 20

14

QRV r Rr R

2 2

0

14

Q R rV

R r

9 95 99 10 10 100100 45

V

9V V .

140. (D) R = 2 28 8 2.8.8 cos 120 = 8 N

8N

60°

R 60°

120°

8N

141. Based on friction and Coulombs Law9 99 10 10 2.254

F N

which is less mg acting on 2 final separation =2m only

ELECTROSTATICS JEE ADVANCED-VOL - VI ELECTROSTATICS · 2020. 4. 13. · Narayana Junior Colleges...

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