Electrostatics Solved QBpdf

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ELECTROSTATICS

EASY QUESTIONS E 1.

Sol.:

In the circuit shown, a potential difference of 60V is applied across AB. The potential difference between the points M and N is (a) 10 V (b) 15 V (c) 20 V (d) 30 V

Charge on capacitor C between MN is  60 

C 2

 30 C

E 2.

Sol.:



VMN



(d)

30 C   30 V C

2C

A 60 V

C

C

B

2C

A

2C

60 V B

M

N M C

C N

2C

A positively charged thin metal ring of radius R is fixed in x – y plane with its centre at the origin O. A negatively charged particle P is released from rest at the point (0, 0, Z0). Then the motion of P is (a) periodic for all values of Z0 (b) SHM for all values of Z0 satisfying 0 < Z0 < R (c) approximately SHM, provided Z >> R (d) can’t be said (a)

E 3.

A capacitor of capacitance ‘C’ is connected with a battery of emf ε as shown. After full charging a dielectric of same size of capacitor and dielectric constant k is inserted then choose correct statements. (capacitor is always connected to battery) (a) electric field between plates of capacitor remain same (b) charge on capacitor is Cε (c) energy on capacitor decreased (d) electric field between plates of capacitor increased.

Sol.: E

P.D. same and electric field same  (a)

C



Question Bank

ELECTROSTATICS

4.

Shown in the figure flux of electric field surface is (a) 3 q / 0 (c) q / 0

Sol.: E 5.

Sol.:

E 6.

is a distribution of charges. The due to these charges through the (b) 2 q / 0 (d) zero

Sol.:

E

–q

Two spherical conductors A and B of radii 1 mm and 2 mm are separated by a distance of 5 cm and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surface of spheres A and B is (a) 1 : 4 (b) 4 : 1 (c) 1 : 2 (d) 2 : 1 E A rB 2   E B rA 1  (d)

A point charge + q is fixed at point B. Another point charge + q at A of mass m vertically above B at height h is dropped from rest. Choose the correct statement (a) It will collide with B (b) It will execute S.H.M

+q

A h

+q B

q2 < mgh2 4 0

(d) go down up to a point and then come up. For charge + q at A to come down, Fe < mg q2   mg 40 h 2 

E 7.

–q

(d)

(c) It will go down only if

Sol.:

+q

(c)

The electric field intensity at a point at a distance 2 m from a charge q is E. The amount of work done in bringing a charge of 2 coulomb from infinity to this point will be E E (a) 2E joules (b) 4E joules (c) joules (d) joules 2 4 q Potential at this point, V   rE  2 E 40 r Work done = qV = 4E joules.  (b)

Question Bank

ELECTROSTATICS

8.

Sol.:

A simple pendulum of length l has a bob of mass m, with a charge q on it. A vertical sheet of charge, with surface charge density  passes through the point of suspension. At equilibrium, the string makes an angle  with the vertical, then q q q q (a) tan   (b) tan   (c) cot   (d) cot   2 0 mg  0 mg 2 0 mg  0 mg tan  

Fe qE  mg mg

tan  

q / 2 0 q  mg 2 0 mg



(a)

O+ + + + + + + + + C+ +

 T

B

A

E

Sheet of charge

E 9.

Sol.:

A charged particle of mass m and charge q is released from rest in an electric field of constant magnitude E. The kinetic energy of the particle after time t will be 2 E 2t 2 E q2 m E 2 q2 t 2 Eqm (a) (b) (c) (d) 2 mq 2m 2t 2t (c)

E 10.

If a positively charged pendulum is oscillating in a uniform field as shown, then its time period as compared to that when it was uncharged will (a) increase (b) decrease (c) not change (d) none of these Sol.: (a) E 11.

A and B are two concentric metallic hollow spheres. If A is given a charge q while B is earthed as shown in figure, then (a) charge density of A and B are same (b) field inside and outside A is zero (c) field between A and B is not zero (d) field inside and outside B is zero Sol.: (c)

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

q

+++ + A + + + ++ ++

B

ELECTROSTATICS

E 12.

Sol.: E 13.

Sol.: E 14.

Sol.:

A table tennis ball which has been covered with a conducting paint is suspended by a silk thread so that it hangs between two metal plates. One plate is earthed. When the other plate is connected to a high voltage generator, the ball (a) is attracted to the high voltage plate and stays there (b) hangs without moving (c) swings backward and forward hitting each plate in turn (d) is repelled to the earthed plate and stays there (c)

A spring block system undergoes vertical oscillations above a large horizontal metal sheet with uniform positive charge. The time period of the oscillation is T. If the block is given a charge Q, its time period of oscillation (a) remains same (b) increases (c) decreases (d) increases if Q is positive and decreases if Q is negative (a)

There is an electric field E in X-direction. If work done in moving a charge 0.2 C through a distance of 2 m along a line making an angle of 60° with X-axis is 4.0 joule. The value of E is (a) 3 N / C (b) 4 N / C As W  F ds cos   qE ds cos 4  0.2 E  2 cos 60, E  20 N/C. 

E 15.

Sol.: E 16.

Question Bank

(c) 5 N / C

(d) 20 N / C.

(d)

In electrolysis, the amount of mass deposited or liberated at an electrode is directly proportional to (a) amount of charge (b) square of current (c) concentration of electrolyte (d) square of electric charge (a)

The ratio of the forces between two small spheres with same charges when they are in air to when they are in a medium of dielectric constant K is (a) 1 : K (b) K : 1 (c) 1 : K2 (d) K2 : 1

Question Bank

ELECTROSTATICS

Sol.:

In air F1 

1 q2 4 0 r 2

In medium (k) F2  

F 1 q2 k  1  2 4 0 k r F2 1

(b)

E 17.

A charge Q is divided into two parts of magnitude q and Q –q. If the coulomb repulsion Q between them when they are separated at some distance is to be maximum, the ratio of q should be  (a) 2 (b) 1/2 (c) 4 (d) 1/4 1 q Q  q  Sol.: F  4 0 r2 dF  Q  2q  0 For maximum repulsion force dq Q 2  2 q 1  (a) E 18.

There are two charges +1 C and +5 C. The ratio of the forces acting on them will be  (a) 1 : 5 (b) 1 : 1 (c) 5 : 1 (d) 1 : 25 F 1 Sol.: F12  F21 then 12  F21 1  (b) E 19.

The electric potential V is given as a function of distance x (metre) by V = (5x2 +10x –9)volt. Magnitude of electric field at x = 1 is (a) 20 V/m (b) 6 V/m (c) 11 V/m (d) –23 V/m dV Sol.: E   10 x  10 x 1  20 V/m dx  (a) E 20.

If specific resistance of a wire is , its volume is 3m3 and its resistance is 3 omhs, then its length will be  (a)

1 

(b)

3 

(c)

1 3 

(d) 

1 3

Question Bank

ELECTROSTATICS

Sol.:

Volume = Al = 3  A  Now R   

E 21.

Figure shows the electric lines of force emerging from a charged body. If the electric field at A and B are EA and EB respectively and if the displacement between A and B is r then  (a) E A  E B (b) E A  E B

g'

 E 22.

Sol.:

E 23.

9 l 3  l2   l  A  

(b)

(c) E A 

Sol.:

EB r

(d) E A 

r

A

B

EB r2

g R n  1  d  g 1   , d  n n  n (b)

The magnitude of electric field intensity E is such that, an electron of mass m and charge e placed in it would experience an electrical force equal to its weight is given by  mg e2 e (a) mge (b) (c) (d) g e mg m2 eE = mg mg E e  (b)

Figure shows three points A, B and C in a region of  uniform electric field E . The line AB is perpendicular and BC is parallel to the field lines. Then which of the following holds good. Where VA, VB and VC represent the electric potential at points A, B and C respectively  (a) V A  VB  VC (b) V A  VB  VC (c) V A  VB  VC

Sol.:

3 l

(b)

(d) V A  VB  VC

A

B

C

Question Bank

ELECTROSTATICS

E 24.

ABC is an equilateral triangle. Charges +q are placed at each corner. The electric field intensity at centroid O will be 1 q 1 3q (a) (b)  2  2 2 0 r 2 0 r

A

+q r

O 3q 1 (d) zero  2 +q +q C B 2 0 r Electric field due each charge have same magnitude and makes an angle of 1200 with each other.  (d)

(c)

Sol.:

E 25.

A charged particle of mass m and charge q is released from rest in an electric field of constant magnitude E. The kinetic energy of the particle after a time t is

E 2q 2t 2 m v  u  at , u = 0 qE v t m 1 2 q 2 E 2t 2 mv  2 2m  (c) (a)

Sol.:

E 26.

(c)

E 2 q 2t 2 2m

Three charges Q, +q and +q are placed at the vertices of a right angle triangle (isosceles triangle) as shown. The net electrostic energy of the configuration is zero, if Q is equal to (a)

q 1 2

(c) –2q Sol.: k  E

2E 2 q 2 t 2 m

(b)

(b)

Q

 2q +q

2 2

+q a

(d) +q

Qq kq 2 kQq 2q   0, Q a a 2a 2 2



(b)

4E 2 q 2 t 2 m

(d)



Question Bank

ELECTROSTATICS

27.

A spherical conductor A of radius r is placed concentrically inside a conducting shell B of radius R(R > r). A charge Q is given to A, and then A is joined to B by a metal wire. The charge flowing from A to B will be  R   Rr

 r   Rr

(a) Q  Sol.:

E 28.

(b) Q

(c) Q

(d) zero

When the two are joined by a metal wire, they become a single conductor. As charge can reside only on the outer surface of a conductor, the entire charge Q must flow to the outer sphere.  (c)

An electron moves in a circular orbit at a distance from a proton with kinetic energy E. To escape to infinity, the energy which must be supplied to the electron is (a) E

(b) 2 E

(c) 0.5 E

(d)

2E

Sol.: Total energy = – kinetic energy = -E So energy E should be supplied  (a) E 29.

Sol.:

A capacitor is connected to a battery. The force of attraction between the plates when the separation between them is halved (a) remains the same

(b) becomes eight times

(c) becomes four times

(d) becomes two times

Distance between plates halved capacitance becomes doubled also charge becomes doubled, Force of attraction  Q2 Force becomes four time. 

E 30.

(c)

  The electric E is given by E  aiˆ  bˆj (where a and b is constant and iˆ, ˆj are unit vector along x and y axis respectively), the flux passing through a square area of side l and parallel to y-z plane is (a) bl2

Sol.:

(b) al2

 Area vector A  l 2 iˆ     E . A  al 2 

(b)

(c)

a

2



 b2 l 2

(d)

a

2



 b2 l 2

Question Bank

ELECTROSTATICS

E 31.

Sol.:

Four charge particle of charges – q, – q, – q and 3q are placed at the vertices of a square of side a as shown. The magnitude of dipole moment of the arrangement is

–q

(a) 2 2q a

(b) q a

–q

(c) 3 qa

(d) 2  2 qa





–q

 p  qa

pnet = 2 2 qa 

3q

–q

(a)

q,q,q

 p1  qa 2  p  qa –q

E 32.

A cylinder of radius R and length L is placed in a uniform electric field E parallel to the axis of cylinder. The total flux through the curved surface of the cylinder is given by (a) 2 R2E

Sol.: E 33.

Sol.:

(c) E2RL

(d) zero

An uniform electric field in positive x-direction exists in a region. Let A be the origin, B be the point on the x-axis at x = +1 cm and C be the point on the y-axis at y = +1 cm. The potential at the points A, B and C are VA, VB and VC respectively, then (b) V A  VB

(c) V A  VC

(d) V A  VC

In the direction of electric field potential decreases and perpendicular to electric field it remain same, hence V A  VC  VB 

E 34.

(b) 2 R2 /E

(d)

(a) V A  VB Sol.:

–q

(b)

The electric field at the origin is along the positive X-axis. A small circle is drawn with the centre at the origin cutting the axes at points A, B, C and D having coordinates (a, 0) (0, a), (–a, 0), 0, –a) respectively. Out of the points on the periphery of the circle, the potential is minimum at (a) A (b) B (c) C (d) D Clearly potential decreases along the direction of electric field  Vmin is at A (a, 0)  (a)

Question Bank

ELECTROSTATICS

E 

35.



Charge Q is given a displacement r  aiˆ  bˆj in an electric field E  E1iˆ  E 2 ˆj . The work done is (b) Q ( E1 a) 2  ( E 2 b) 2

(a) Q(E1a + E2b)

(d) Q E12  E 22 )

(c) Q(E1 + E2) a 2  b 2 



dw = Q E . d r

Sol.:

a

a2  b2

= Q ( E1iˆ  E 2 ˆj ) . ( dxiˆ  dyiˆ) b

= Q E1  dx  QE 2  dy = Q (E1a + E2b) 0

 E 36.

Sol.:

0

(a)

In the circuit shown, the equivalent capacitance between the points A and B is 10 15 (a) F (b) F 3 4 12 25 (c) F (d) F 5 6 3 6 Since  The circuit is treated as wheatstone bridge. 2 4 3 6 2  4  Ceq =  3 6 2 4 = 2+ 

3 F 5 F 6 F A

2 F

4 F

B

8 20 10  = F 6 6 3

(a)

E 37.

A non-conducting ring of radius R has charge Q distributed uniformly over it. If it rotates with an angular velocity , the equivalent current will be (a) zero

Sol.:

E

(b) Q

(c) Q

 2

(d) Q

 2R

With each rotation, charge Q crosses any fixed point P near the ring. Number of rotations per second = /2. 

Charge crossing P per second current =



(c)

Q 2

Question Bank

ELECTROSTATICS

38.

A battery consists of a variable number n of identical cells having internal resistance r each. They are connected in series. The terminal of the battery are joined by a conducting wire and the current i is measured. Which of the graphs as shown in figure gives correct relationship between i and n? Y

Y

i

i

(a)

(b) O

n

O

X

i

i

(c)

(d) O

Sol.:

X

Y

Y

E 39.

n

n

O

X

n

X

A half ring of radius R has a charge of  per unit length. The potential at the center of the half ring is    (a) k (b) k (c) k (d) k R R R kdQ k Rd dV =  R R v



 dv  k  d 0

0

= k  (d) E 40.

Sol.:

  1 If electric field is given by E   2  iˆ V/m, the magnitude of potential difference between x  points x = 10 cm and x = 20 cm is (a) 1 V (b) 2 V (c) 5 V (d) 10 V 0.2 1 dV = –  E . dx  –  2 dx  – 5 V 0.1 x

 E 41.

(c)

A cube of side b has a charge q at each of its vertices. The electric potential at the centre of the cube is (a)

4q 3 0 b

(b)

3q  0 b

(c)

2q  0 b

(d) zero

Question Bank

ELECTROSTATICS

Sol.:

V 

 E 42.

Sol.:

1 q 4q . 8  4 0  b 3  3  b    2    (a)

Four equal charges Q are placed at the four corners of a square of side a. The work done in removing a charge Q from the centre of the square to infinity is Q2 2 Q2 2 Q2 (a) zero (b) (c) (d) 4 0 a  0 a 2 0 a W   Q V   QV f  Vi   Q0  Vi  ,

 E 43.

Sol.:

E 44.

 1 Q 2 2 Q2   W = Q. .   4   a   4  a 

(c)

In a regular polygon of n sides, each corner is at a distance r from the center. Identical charges of magnitude Q are placed at (n –1) corners. The field at the center is Q Q n Q n 1 Q (a) k 2 (b) (n  1) k 2 (c) k 2 (d) k 2 n 1 r n r r r Q From symmetry electric field at the centre is k 2 r  (a)

Seven point charges each of charge q is placed at the seven corners of a cube of side a (one corner is empty). Find the magnitude of electric field at centre of cube. (a) zero

Sol.: E 

(b)

1 q 4 0 a 2

(c)

1 q 3 0 a 2

(d)

1 7q 4 0 a 2

1 q 1 q  2 4 0  a 3  3 0 a 2    2   

 (c) E 45.

A charged sphere of diameter 4 cm has a charge density of 10-4 coulombs/cm2. The work done in joules when a charge of 40 nano-coulombs is moved from infinity to a point which is at a distance of 2 cm from the surface of the sphere, is (a) 14.4  (b) 28.8  (c) 144  (d) 288 

Question Bank

ELECTROSTATICS

10 4 C × 16 cm2 = 16 × 10-4 C 2 cm Q qQ  Work done W = qV = V 4 0 r 4 0 r

Sol.: Q  A 

 (a) E 46.

The magnitude of electric intensity at a distance x from a charge q is E. An identical charge is placed at a distance 2 x from it. Then the magnitude of the force it experience is qE qE (a) qE (b) 2 qE (c) (d) 2 4 q Sol.: Given E = . 4 0 x 2 Hence the magnitude of the electric intensity at a distance 2x from charge q is q q 1 E E'     2 2 4 4 4 0 (2 x ) 4 0 x Therefore, the force experienced by a similar charge q at a distance 2x is qE F  qE '  4  (d) E 47.

A conductor of resistance 3  is stretched uniformly till its length is doubled. The wire is now bent in the form of an equilateral triangle. The effective resistance between the ends of any side of the triangle in ohms is (a)

Sol.: R =

9 2

(b)

8 3

(c) 2

(d)1

l l 2    2     l , RAB = 8/3  A Al  V 

4

4

 (b) A

E 48.

4

B

Eight dipoles of charges of magnitude e are placed inside a cube. The total flux coming out of the cube equals to 8e 16e e (a) (b) (c) (d) zero  0 0 Sol.: Net charge inside the cube is zero  (d)

Question Bank

ELECTROSTATICS

E 49.

Four point charges q1, q2, q3 and q4 are placed at the corners of the square q1 of side a, as shown in figure. The potential at the centre of the square is (Given : q1 = 1 × 10–8 C, q2 = –2 × 10–8 C, q3  3  10 8 C , a

q2

a P

a

8

q 4  2  10 C , a = 1 m) (a) 507 V (c) 550 V 1  q1 q 2 q 3 q 4  Sol.: V      = 507 V 4 0  r r r r   (a)

q3

a

q4

(b) 607 V (d) 650 V

E 50.

If there are n capacitors in parallel connected to V volt source, then total energy stored is equal to 1 1 (a) CV (b) nCV 2 (c) CV 2 (d) CV 2 2 2n 1 1 Sol.: Energy Stored = (C net )V 2  nCV 2 2 2  (b) E 51.

There is an air-filled 1 pF parallel plate capacitor. When the plate separation is doubled and the space is filled with wax, the capacitance increases to 2pF. The dielectric constant of wax is (a) 2

Sol.: E 52.

Sol.:

E

(b) 4

(c) 6

(d) 8

(b)

For the circuit shown in the adjoining figure, the charge on 4 F capacitor is (a) 30 C (b) 40 C (c) 24 C (d) 54 C

q1 q1   10 4 6 q1  24 μC  (c)

1F 4F 5F 10V 3F

q1 4F

q1 q2

3F

6F 10V

Question Bank

ELECTROSTATICS

53.

Two conducting spheres of radii r1 and r2 are at the same potential. The ratio of their charges is  r2   r2  r r (a)  12  (b)  22  (c) 1 (d) 2 r2 r1  r2   r1  1 q Sol.: Potential (V) = 4 0 r q1 r1  q 2 r2  (c) E 54.

Six point charges are arranged at the vertices of regular hexagon of side length a (shown in figure). The magnitude of electric field at the centre of regular hexagon is q (a) (b) zero 4 0 a 2 q (c) (d) none of these 2 0 a 2 2q Sol.: Resultant field = 2 E 0  4 0 a 2  (c) E 55.

In the given figure, find the equivalent capacitance between A and B. 2C 5C (a) (b) 3 4 (c)

Sol.:

3C 2

(d)

4C 5

–q

+q O

+q +q

+q

+q

C C

C C

A

B C

C

C

C

Reduced circuit is

C

C

C eq 

3 C 2

C

C



(c) C

C

Question Bank

ELECTROSTATICS

E 56.

Sol.:

Two large plate separated by a distance d in vertical plane and connected to battery as shown. An electron of charge e and mass m is at rest between the plates. Find the value of potential difference of battery. (a)

mge d

(b)

2mge d

(c)

mgd e

(d)

2mgd e

mg  eE  mg  e



electron d

V mgd ,V  d e

(c)

E 57.

Two plates are 2 cm apart. A potential difference of 10V is applied between them, the electric field between the plates is (a) 20 N/C

Sol.:

d = 2cm = 2  10–2 m, 

E 58.

Sol.:

(c) 5N/C

V = 10 volt , E 

(d) 250N/C

V 10   500 N/C d 2  10 2

(b)





Suppose the electrostatic potential at some points in space are given by V  x 2  2 x . The electrostatic field strength at x = 1 is (a) zero (b) -2 (c) 2 (d) 4 dv E  x    , E  x   2  2 x  , at x = 1 E(x=1) = 0 dx 

E 59.

(b) 500 N/C

(a)

d

A slab of copper of thickness b is inserted in between the plates of parallel plate capacitor as shown in figure. d The separation between the plates is d if b  , then 2 the ratio of capacities of capacitors after and before inserting the slab will be

(a)

2 :1

(b) 2 : 1

(c) 1 : 1

b (d) 1 :

2

Question Bank

ELECTROSTATICS

Sol.:

E 60.

Sol.:

C1 

 A 0 A , C2  0 , d  t  d



(b)

Sol.: E 62.

Sol.: E 63.

C A

C C

(b)

An electric dipole is placed at an angle of 300 to a non-uniform electric field. The dipole will experience (a) a translational force only in the direction of the field (b) a translational force only in a direction normal to the direction of the field (c) a torque as well as a translational force (d) a torque only (c) Coulomb’s law is applicable to (a) point charges (b) spherical charges (c) like charges

(d) all of these

(a) A point charge q and a charge –q are placed at x = -a and x = +a respectively. Which of the following represents a part of E-x graph? (a)

E

–a O +a

x

(b) –a O

E x

(c) O +a

x

(d) all of these

(d)

E 64.

B

One single capacitor is in short circuit and the remaining two capacitors are in parallel.

E

Sol.:

C 2 d given  2  2 C1 1

The equivalent capacity of the combination shown in figure is (a) C (b) 2C (c) 3/2C (d) C/2

 E 61.

t b 

The charge on any one of the 2F capacitors and 1F capacitor will be given respectively (in C) as  (a) 1 , 2 (b) 2 , 1 (c) 1, 1 (d) 2,2

2F

2F 1F 2V

Question Bank

ELECTROSTATICS

Sol.:

Potential difference across the both the line is same i.e. 2V. Hence charge flowing line 2 2 Q     2  2 C 2 So charge on each capacitor in line (2) is 2 C  (d)

2F

2F 1F

Line (2) Line (1)

2V

E 65.

The electric potential V (in volt) varies with x (in metre) according to the relation V  5  4 x 2 . The force experienced by a negative charge of 2  10–6 C located at x = 0.5 m is (a) 2  10–6 N

Sol.:

Electric field E  

(b) 4  10–6 N

(c) 6  10–6 N

(d) 8  10–6 N

dV d  5  4 x 2  8 x dx dx





Force on charge  q   qE  8qx At x = 0.5 m, force = 8  2  10–6  0.5 = 8  10–6 N 

(d)

E 66.

A charge q is placed at the centre of the line joining two equal charges Q. The system of the three charges will be in equilibrium if q is equal to (a) – (Q/4)

Sol.:

(b) – (Q/2)

(c) (Q/2)

(d) (Q/4)

Net force on any charge = 0. Force on any charge Q at end

FK 

Q 2 KqQ  2  0. 4x2 x

Hence, q 

Q 4

(a)

E 67.

The electric potential at a point situated at a distance r on the axis of a short electric dipole of moment p will be 1/4 (0) times (a) p/r3

Sol.:

(b) p/r2

Potential at axial point V  

(b)

1 P 40 r 2

(c) p/r

(d) none of the above

Question Bank

ELECTROSTATICS

E 68.

The charge per unit length for a very long straight wire is . The electric field at points near the wire (but outside it) and far from the ends varies with distance r as (a) r

Sol.:

E

(b) 1/r  . 20 r



Hence E 

(c) 1/r2

(d) 1/r3

1 r

(b)

E 69.

A body has a charge of one coulomb. The number of excess (or lesser) electrons on it from its normal state will be (b) 1.6  10–19

(a)  Sol.:

n

Q = ne, 

(c) 1.6  1019

(d) 6.25  1018

(c) 2q

(d) zero

1  6.25  1018 1.6  10 19

(d)

E 70.

The net charge on a condenser is (a) Infinity

Sol.:

(b) q/2

Net charge  Q   Q   0 

(d)

E 71.

Sol.:

A conducting hollow sphere of radius 0.1 m is given a charge of 10 C. The electric potential on the surface of sphere will be (a) zero

(b) 3  105 V

Vsurface =

1 Q 9  10 9 10  10 6   9  10 5 V 40 r 10 1



(c) 9  105 V

(d) 9  109 V

(c)

E 72.

Sol.:

The effective capacitance between the points x and y in the will be (a) 1 F

(b) 1.5 F

(c) 2 F

(d) 4 F

 1 1  Ce  1     1.5 F 1 1



(b)

x 1F 1F 1F y

Question Bank

ELECTROSTATICS

E 73.

The equivalent capacitance between the points X and Y in the figure will be (a) 2C/3

(b) C/3

(c) 3C/2

(d) 3C

X C + – Y C/2

Sol.:

Capacitors in series, Ce  

C1C 2 C  C1  C 2 3

(b)

E 74.

Sol.:

+ + S + +  + +

A charged ball hangs from a silk thread of length l. It makes an angle  with a large charged conducting sheet P as shown in the figure. The surface charge density  of the sheet is proportional to (a) cos 

(b) cot 

(c) sin 

(d) tan 

q , T cos   Mg 20 tan  

T sin  



Tcos

,

 q.

Tsin

(d)

 2 0

Mg

E 75.

Sol.:

A charge q is placed at a distance a/2 above the centre of a horizontal square surface of edge a as shown in figure. The electric flux through the square surface is (a) Q/20

(b) Q/0

(c) Q/60

(d) Q/80

Complete the cube, adding other five faces. q q ,  face   cube  0 6 0 

(c)

q a/2 a a

Question Bank

ELECTROSTATICS

E 76.

In the figure shown (circuit), the capacitance between the points A and B will be (a) 1F

(b) 2F

(c) 3F

(d) 4F

2F A 1F

2F 1F

B 2F

Sol.:

Ceff = 2 F Simplifying the circuit) 

(b)

E 77.

Two equal positive charges +q each are fixed a certain distance apart. A third equal positive charge +q is placed exactly mid-way between them. Then the third charge will (a) move at an angle of 450 to the line joining the two charges (b) move at an angle of 900 to the line joining the two charges (c) move along the line joining the two charges (d) stay at rest

Sol.:

Mid point will be neutral point, hence third charge will experience no force and will stay at rest. 

(d)

E 79.

Two equal negative charges –q are fixed at points (0, a) and (0, –a) on the y-axis. A positive charge Q is released from rest at a point (2a, 0) on the x-axis. The charge Q will (a) execute simple harmonic motion about the origin (b) move to the origin and remain at rest there (c) move to infinity (d) execute oscillatory but not simple harmonic motion

Sol.:

Resultant force on Q will be always towards origin. It will undergo oscillatory motion, but not SHM (amplitude being comparable to other dimension and not small). 

(d)

E 80.

Two point charges +4q and +q are placed 30 cm apart. At what point on the line joining them is the electric field zero? (a) 15 cm from charge 4q

(b) 20 cm from charge 4q

(c) 7.5 cm from charge q

(d) 5 cm from charge q

Question Bank

ELECTROSTATICS

Sol.:

K 4q  K q   2 r12 r2 r1  2 and r1  r2  30 r2



Hence, r1 = 20 cm 

(b)

E 81.

The effective capacitance of two capacitors of capacitances C1 and C2 (with C2>C1) 25 connected in parallel is times the effective capacitance when they are connected in 6 series. The ratio C2/C1 is (a)

Sol.:

3 2

(b)

Cparallel = C1 + C2,

4 3

Cseries =

(c)

5 3

(d)

25 6

C1C2 C1  C2

25 Cparallel and C2 > C1 6 C 3 Solving 2  C1 2

Given Cseries 



(a)

E 82.

Sol.:

E 83.

A capacitor of capacitance 4 F is charged to 80 V and another capacitor of capacitance 6 F is charged to 30 V. When they are connected together, the energy lost by the 4 F capacitor is (a) 7.8 mJ (b) 4.6 mJ (c) 3.2 mJ (d) 2.5 mJ C V  C2V2 Vcommon  1 1  50V C1  C 2 1 1 For 4 F capacitor Ei  C1V 2 , E f  C1VC2 2 2 1 Energy loss  Ei  E f  C V 2  VC2   7.8  10–3 J 2  (a) Two concentric spheres are of radii r1 and r2. The outer sphere is given a charge q. The charge q' on the inner sphere will be (inner sphere is grounded) (a) q

(b)  q

(c)  q

r1 r2

q' r1

(d) zero

r2

Question Bank

ELECTROSTATICS

Sol.:

q' q  0 r1 r2  (d)

E 84.

The equivalent capacitance between points M and N is 10 (a) C0 (b) 2C0 11 (c) C0 (d) none of these

C0

M

C0

N

C0

C0

C0 C0

Sol.:

1 1 3   C e 2C0 5C0 10 Ce  C0 11  (a)

2C0

(5/3)C0 N

M

E 85.

Q As shown in the figure. If value of Q' is then what is 2 the value of dielectric constant k is (a) 3 (b) 1/2

Q

Q

(d) 

(c) 2

Metal plate

Sol.:

1 Q 1   We know Q '  Q1   ,  Q1    k = 2 k 2 k  



E 86.

(c)

In the given circuit diagram, initially battery was connected. Find the work done by battery if capacitor is completely filled with a dielectric of dielectric constant k = 3. (a)

1 CV 2 2

(b) CV 2

3 CV 2 2 Q  ( kCV  CV )  ( k  1)CV  2CV ,

(c) 2CV 2 Sol.:

Dielectric



(c)

(d)

W = QV  2CV 2

C

V

Question Bank

ELECTROSTATICS

E 87.

A parallel plate capacitor of capacitance C is connected to a battery of emf V. If a dielectric slab is completely inserted between the plates of the capacitor and battery remains connected, then electric field between plates (a) decreases

(b) increases

(c) remains constant

(d) may be increase or may be decrease

Sol.:

(c)

E 88.

A uniform electric field E  E0 (iˆ  ˆj ) exists in the region. The potential difference VQ  VP  between point P(0, 0) and Q (a, 0) is (a)  E0 a

Sol.:

(b) E0 2 a

(c)  E0 a

(d)  E0 2 a

VQ  VP   E0 (iˆ  ˆj )  (a iˆ)   E0 a . 

(a)

E 89.

A long string with a charge of  per unit length passes through an imaginary cube of edge a. The maximum flux of the electric field through the cube will be (a) a /  0

Sol.:

(b)

2 a 0

(c)

6a 2 0

(d)

3 a 0

The maximum length of the string which can fit into the cube is 3a , equal to its body diagonal. The maximum charge inside the cube is 3a , and hence the maximum flux through the cube is 

3a 0

(d)

E 90.

The electric potential V at any point x, y, z (all in metres) in space is given by V = 4x2 volts. The electric field (in V/m) at the point (1 m, 0, 2 m) (a)  8 iˆ

Sol.:

(b) 8 iˆ

V  8 x x E y  E z  0 as V is independent of y and z.  For x = 1, E  8iˆ Ex  



(a)

(c)  16 iˆ

(d) 8 5 iˆ

Question Bank

ELECTROSTATICS

E 91.

In a parallel-plate capacitor of capacitance C, a metal sheet is inserted between the plates, parallel to them. The thickness of the sheet is half of the separation between the plate. The capacitance now becomes (a) 4C

(b) 2C

(c) C/2

(d) C/4

Sol.: d A

A

d/4

d/2

C

C

d/4

d/4

=

d/4

Before the metal sheet is inserted, C =

0 A d

After the sheet is inserted, the system is equivalent to two capacitors in series, each of capacitance C  

0 A  4C . ( d / 4)

The equivalent capacity is now 2C.  E 92.

(b)

Which one of the following statement is incorrect? (a) A moving charged particle produced electric and magnetic field both. (b) Equipotential surface is always perpendicular to electric field. (c) Kirchhoff’s junction law follows conservation of charge. (d) electric field inside the conductor is always zero.

Sol.:

(d)

E 93.

Two capacitors of capacitance 3 F and 6 F are charged to a potential of 12 V each. They are now connected to each other, with the positive plate of one to the negative plate of the other. Then (a) the potential difference across 3 F is zero (b) the potential difference across 3 F is 4 V (c) the charge on 3 F is zero (d) the charge on 3 F is 10 C

Sol.:

(d)

Question Bank

ELECTROSTATICS

E 94.

Four charges +2q, –2q, –3q and +3q are kept in the corners of a square of side a. The total field at the centre O is, 2 2q (a) zero (b) 4 0 a 2 2q (c) 4 0 a 2

Sol.:

O +3q

10 2 q (d) 4 0 a 2

     E  E  2 q  E  2 q  E 3 q  E 3 q

+2q

E  2 E3q  E 2q  E 2

5q

 a  40    2 10 2 q E 4 0 a 2 

–2q

+2q

2

+3q

a

–3q

 –2q E  2 q E 3q O  E 2 q E3 q a

–3q

A 

 B

(d)

E 95.

Sol.:

Figure shows some of the electric field lines corresponding to an electric field. The figure suggests that (E = electric field, V = potential) (a) VA = VB

(b) EA = EB

(c) VA  VB

(d) VA  VB

Potential decreases in the direction of electric field. VA  VB  (d)

E 96.

Two spheres of radii r and R carry charges q and Q respectively. When they are connected by a wire, there will be no loss of energy of the system if (a) qr = QR

Sol.:

E 97.

(b) qR = Qr

(c) qr2 = QR2

(d) qR2 = Qr2

There will be no loss of energy if the potential of the spheres is the same i.e. if q Q q Q or  V   r R 4 0 r 4 0 R  (b) If three moles of monatomic gas is mixed with 1 moles diatomic gas, the resultant value of  for the mixture is (a) 1.66 (b) 1.50 (c) 1.40 (d) 1.57

Question Bank

ELECTROSTATICS



Sol.: 

n1C p1  n2C p2 nC v1  n2C v2



3(3  2)  1(5  2)  1.57 3(3)  1(5)

(d)

E 98.

Sol.:

4 times its original value if a dielectric slab of 3 thickness t = d/2 is inserted between the plates (d = separation between the plates). The dielectric constant of the slab is (a) 2 (b) 4 (c) 6 (d) 8  A 0 A 4C 4 d 2d C 0 ,  ,  , 2d   3d , k=2 d d d  3 d d 3 k   d  2  2k  2 2k

Capacitance of a capacitor becomes

 E 99.

Sol.:

In the figure shown, conducting shells A and B have charges Q and 2Q distributed uniformly over A and B .Value of VA – VB is Q Q (a) (b) 4 0 R 8 0 R 3Q 3Q (c) (d) 4 0 R 8 0 R

VA 

1  Q 2Q  1 2Q     4 0  R 2 R  4 0 R

VB 

1 3Q , 4 0 2 R

 E 100.

Sol.:

(a)

V A  VB 

B 2Q Q A R 2R

Q 8 0 R

(b)

Three point charges q, –2q and –2q are placed at the vertices of an equilateral triangle of side a. The work done by some external force to slowly increase their separation to 2a will be q2 1 2q 2 1 3q 2 (a) (b) (c) (d) zero . . 4 0 a 4 0 a 4 0 3R Ui 

Uf 

q  2q   q  2q    2q  2q  4 0 a

q  2q   q  2q    2q  2q   0 4 0 2a 

Wext = U = 0 

=0

(d)

Question Bank

ELECTROSTATICS

E 101.

Sol.:

Two identical charges are placed at the two corners of an equilateral triangle. The potential energy of the system is U. The work done in bringing an identical charge from infinity to the third vertex is (a) U (b) 2U (c) 3U (d) 4U W  U f  U i  3U  U  2U

 E 102.

Sol.:

A unit positive charge has to be brought from infinity to a midpoint between two charges 20C and 10C separated by a distance of 50 m. How much work will be required? (a) 10.8 × 104 J (b) 10.8 × 103 J (c) 1.08 × 106 J (d) 0.54 × 105 J Work done in taking unit positive charge from infinity to that point = potential at that point  20  10 6 10  10 6   J  10.8  10 3 J  9  10 9   25 25   

E 103.

Sol.:

E 104.

E 105.

(b)

The electric potential at a point (x, y) is given by: V = –Kxy. The electric field intensity at a distance r from the origin varies as (a) r2 (b) r (c) 2r (d) 2r2   V ˆ V ˆ  E   i j   K yiˆ  xˆj  E  E x2  E y2  ( Ky ) 2  ( Kx ) 2 = Kr y   x i.e., Er  (b)





The inward and outward electric flux for a closed surface in units of N-m2/coulomb are respectively 8 × 103 and 4 × 103. Then the total charge inside the surface is (a) 4 × 103 coulomb (b) – 4 × 103 coulomb (c) 

Sol.:

(b)

4  10 3 coulomb 0

(d) – (4 × 103 )0 coulomb

  4  10 3  8  10 3 = – 4  103 N-m2/coulomb According to Gauss’s law, q = – (4 × 103) 0 coulomb  (d) In the circuit shown in the figure, C = 6F. The charge stored in the capacitor of capacity C is (a) zero (b) 90 C (c) 40 C (d) 60 C

C

2C

10V

Question Bank

ELECTROSTATICS

Sol.:

Both the capacitors are in series. Therefore, charge stored on them will be same. C ( 2C ) 2 2 Net capacity   C   6 F = 4 F C  2C 3 3 Potential difference = 10 V 

E 106.

Two small spheres each having the charge +Q are suspended by insulating threads of length L from a hook. This arrangement is taken in space where there is no gravitational effect, then the angle between the two thread and the tension in each will be

(c) 180,

Sol.:

1 Q2 4 0 2 L2

1 Q2 (b) 90, 4 0 L2 (d) 180,

The position of the balls in no gravity space will be as shown 1 Q2  = 180° and Force =  40 ( 2 L) 2 

E 107.

q = CV = 40 C

(c)

1 Q2 (a) 180, 4 0 (2 L) 2

Sol.:



1 Q2 4 0 L2

L

180°

L

(a)

A solid conducting sphere of radius 5 cm is charged so that the potential on its surface is 10V. The potential at the centre of the sphere is (a) 5 V (b) 10 V (c) 15 V (d) zero Charge resides on the outer surface of a conducting sphere Therefore, potential at the centre will be same as that on the surface.  (b)

E 108.

Sol.:

The distance between the plates of an isolated parallel plate condenser is 4mm and potential difference is 60 volts. If the distance between the plates is increased to 12mm, then (a) The potential difference of the condenser will become 180 volts. (b) The P.D. will become 20 volts. (c) The P.D. will remain unchanged. (d) The charge on condenser will reduce to one third V d V V 60  12 For capacitor 1  1  V2  1 2   180V V2 d 2 d1 4 

(a)

Question Bank

ELECTROSTATICS

E 109.

Sol.:

E 110.

Sol.:

4F Four condensers each of capacity 4F are connected as 4F shown in figure. V P  VQ  15 volts. The energy stored in 4F Q P the system is 4F (a) 2400 ergs (b) 1800 ergs (c) 3600 ergs (d) 5400 ergs 8 Total capacitance of given system C eq  μF 5 1 1 8 U  C eqV 2    10 6  225  180  10 6 J  180  10 6  10 7 erg  1800 erg 2 2 5  (b)

Three charges 2q, –q, –q are located at the vertices of an equilateral triangle. At the centre of 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 Obviously, from charge configuration, at the centre electric field is non-zero. Potential at the centre due to 2q charge V2 q

2q r

2q  . r

r

and potential due to – q charge



q (r = distance of centre point) r Total potential V  V2 q  V q  Vq  0



(b)

Vq  

–q

r

E–q E2q E–q

–q

E 111.

The charge on 4F capacitor in the given circuit is (a) 12 C (b) 24 C (c) 36 C (d) 32 C

4F

1F

3F 5F 10V

Sol.:

Equivalent capacity between A and B 6 4   2.4F 10 Hence charge across 4F (Since in series combination charge remains constant) or 6F  2.4  10  24 C  (b)

6F 1F A

4F 5F 3F

10V

A

B

6F

4F 3F

 10V

B

Question Bank

ELECTROSTATICS

E 112.

Sol.:

E 113.

The electric potential V at any point P (x, y, z) in space is given by V = 4x2 (where V is in volt and x in meter). The electric field at the point (1m, 2m) in volt/metre is (a) 8 along negative x-axis (b) 8 along positive x-axis (c) 16 along negative x-axis (d) 16 along positive z-axis The electric potential V ( x, y , z )  4 x 2 ,   V ˆ E i  8 xiˆ  E (1, 2)  8iˆ x  (a) A particle A has charge +q and a particle B has charge +4q with each of them having the same mass m. When allowed to fall from rest through the same electric potential difference, v the ratio of their speed A will become vB (a) 2 : 1

Sol.:

Using v  

E 114.

v 2QV v Q  A  m vB

(c) 1 : 4 QA  QB

(d) 4 : 1

q 1  4q 2

(b)

Two infinitely long parallel conducting plates having surface charge densities + and – respectively, are separated by a small distance. The medium between the plates is vacuum. If 0 is the dielectric permittivity of vacuum, then the electric field in the region between the plates is  (a) zero (b) volts/meter 2 0 (c)

Sol.:

(b) 1 : 2

 volts/meter 0

Electric field between the plates is  ()   2 0 2 0  volt/meter  0 

(c)

(d)

2 volts/meter 0

–



EE

ELECTROSTATICS

E 115.

Sol.:

Question Bank

Two identical capacitors, have the same capacitance C. One of them is charged to potential V1 and the other to V2. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is 1 1 (a) C V12  V22  (b) C V12  V22  4 4 1 1 (c) C V1  V2 2 (d) C (V1  V2 ) 2 4 4 1 1 Initial energy of the system U i  CV1 2  CV 2 2 2 2 When the capacitors are joined, common potential CV1  CV 2 V1  V2 V  2C 2 Final energy of the system 2

1 1  V  V2  1 2 U f  (2C )V 2  2C  1   C (V1  V2 ) 2 2  2  4 1 Decreased in energy  U i  U f  C (V1  V2 ) 2 4  (c)

E 116.

Sol.:

If an insulated non-conducting sphere of radius R has charge density . The electric field at a distance r from the centre of sphere (r < R) will be R r r 3R (a) (b) (c) (d) 3 0 0 3 0 0 For non-conducting sphere Ein  

E 117.

Sol.:

k .Qr r  3 0 R3

(c)

Two spheres A and B of radius 4cm and 6cm are given charges of 80C and 40C respectively. If they are connected by a fine wire, the amount of charge flowing from one to the other is (a) 20 C from A to B (b) 16 C from A to B (c) 32 C from B to A (d) 32 C from A to B

 r  Total charge Q  80  40  120C. By using the formula Q1   Q  1 .  r1  r2   r   4  New charge on sphere A is Q A  Q  A   120  48C.  4  6   rA  rB  Initially it was 80C i.e., 32C charge flows from A to B.  (d)

Question Bank

ELECTROSTATICS

MODERATE QUESTIONS M 118.

A point charge q is placed at a height a from vertex of square of side a as shown. The electric flux through the square is q (a) 0

(c) Sol.:

q a

(b) zero

a a

q 6 0

(d)

q 24 0

a a

(d)

M 119.

(a)

3 CV 2 2

(c) CV 2 Sol.:

C

In the given circuit, find the heat generated if switch S is closed. (b)

1 CV 2 2

(d)

1 CV 2 3

C C S

V

Ceq = C So, work done by battery = CV2 Heat generated = 

M 120.

1 CV 2 2

(b)

E

If net electric field E due to dipole at point P makes an angle 30° with the line OP as shown, then find the value of angle .  1  (a) tan 1  (b) tan 1 2   2

30°

 

 2  (c) tan 1    3 Sol.:

tan 30° = 

(c)

1 tan  , 2

 3  (d) tan 1   2  

tan  =

2 3

P  q

O

q

Question Bank

ELECTROSTATICS

M 121.

Let V and E be the potential and the field respectively at a point. Which of the following assertion is correct? (a) If V = 0, E must be zero

(b) If V  0, E cannot be zero

(c) If E  0, V cannot be zero

(d) none of these

Sol.:

(d)

M 122.

Sol.:

A charge –q is placed at the axis of a charged ring of radius r at a distance of 2 2 r as shown in figure. If ring is fixed and carrying a charge Q, the kinetic energy of charge –q when it is released and reaches the centre of ring will be, (a)

qQ 4 0 r

(b)

qQ 12 0 r

(c)

qQ 6 0 r

(d)

qQ 2 0 r

K.E. of –q at O = qVO  VP   Q Q = q   4 0 r 40 r 2  (2 2 r ) 2  2 qQ qQ =  3 40 r 60 r



   

+ + Q+ + + r + + + + + + + +++

+ + Q+ + + r + + + O + + + + +++

–q 2 2r

–q P

2 2r

(c)

M 123.

Sol.:

Figure shown in five capacitors connected across a 12 V power supply. What is the potential drop across the 2F capacitor? (a) 2 V

(b) 4 V

(c) 8 V

(d) 10 V

1 F, 2 F and 3F are in parallel and their equivalent capacitance is 6F.

6F

6F 1F

2F

12V

6F

6F

Hence, each 6 F capacitor has potential difference of 4V. 12V



(b)

3F

6F

Question Bank

ELECTROSTATICS

M 124.

A parallel plate capacitor is maintained at a certain potential difference. When a dielectric slab of thickness 3 mm is introduced between the plates, the plate separation had to be increased by 2 mm in order to maintain the same potential difference between the plates. The dielectric constant of the slab is (a) 2

Sol.:

M 125.

(b) 3

A point charge q is placed inside a conducting spherical shell of inner radius 2R and outer radius 3R at a distance of R from the centre of the shell. The electric potential at the centre

(a)

q 2R

VC =

Sol.:

1 times (potential at infinity is zero) 4 0

(b)

4q 3R

(c)

5q 6R

(d)

2q 3R

1 q q q  1  5q    =     4 0  R 2 R 3R  4 0  6 R 

 M 126.

(d) 5

 1   Q d 't 1    Q Q Qd  K  , V   V ' 1   0 A C 0 A C 1  Now, V '  V  d  d   t 1    K 1  d 'd  2 mm , t  3 mm  2  31    K =3  K  (b)

of shell will be

Sol.:

(c) 4

(c)

Two conducting plates X and Y, each having large surface area A (on one side) are placed parallel to each other. The plate X is given a charge Q whereas the other is neutral. The electric field at a point in between the plates is given by (a)

Q 2A

(b)

Q towards left 2 A 0

(c)

Q towards right 2 A 0

(d)

Q towards right 2 0

Charge on inner face, q = 

Electric field =



(c)

Q0 Q  2 2

Q 2 A 0

Q

Question Bank

ELECTROSTATICS

M 127.

Two concentric conducting shells A and B are of radii R and 2R. A charge q is placed at the centre of shells. Shell B is earthed and a charge q is given to shell A. If charges on outer surface of A and B are QA and QB respectively, then (a) QA = q, QB = 0

(b) QA = 2q, QB = q

(c) QA = 2q, QB = 0

(d) QA = q, QB = q

Sol.: M 128.

(c) A solid hemispherical uniform charged body having charge Q is kept symmetrically along the y-axis as shown in figure. The electric potential at a distance d from the origin along the x-axis at point P will be (a)

1 Q 4 0 d

(c) more than Sol.:

y

P (d,0,0)

(0,0,0)

1 Q 4 0 d 2 Q (d) more than 4 0 d

(b) less than 1 Q 2 Q and less than 4 0 d 4 0 d

Assuming sphere is complete then charge on it = 2Q So potential at point P due to this spherical charge = Hence potential due to hemisphere = 

1 2Q 4 0 d

1 Q 4 0 d

(a)

M 129.

Two large conducting parallel plates having equal charge Q are placed very close to each other and distance between the plates is d. Find the work done by external agent to increase the separation between plates by d (area of plate is A)

Q

Q

d

(a) Sol.:

Q2d 2 A 0

(b) 

Q2d 2 A 0

(c) 

 Q  Q Force on first plate due to second plate  2 A  0   2 Q d Wext = – W electric field   2 A 0  (b)

3Q 2 d 2 A 0

(d) 

Q 2d A 0

Q

Q

d

x

Question Bank

ELECTROSTATICS

M 130.

Sol.:

A capacitor of capacitance C having initial charge 2Q0, is connected to a battery of potential Q difference V  0 as shown, then work done by the C battery is Q02 3Q02 (a) (b) 2C 2C 2 3Q0 2Q02 (c) (d) C 3C

Sol.:

3Q 0 Q 0 C

Two point charges +8q and -2q are located at x = 0 and x = L respectively. The location of a point on the x-axis at which the net electric field due to these two point charges is zero, is

 

M 132.

V

(c)

(a) 2 L Sol.:

S

Charge through the battery  Q0   2Q0   3Q0 . Work done by battery  

M 131.

2Q0 + – + – + – + –

(b)

2q 4 0  x  L 

2



L 4

(c) 8 L

(d) 4 L

8q  0 or x = 2 L 4 0 x 2

(a)

Four conducting plates are placed parallel to each other. Separation between them is d and area of each plate is A. Plate number 1 and 3 are connected to each other and plate number 2 and 4 are connected to a battery of emf . Charge flowing through the battery is  A 2 0 A 3 0 A (a) 0  (b)  (c)  d 2 d d There are three capacitors C1 2 , C2 3, C3 4 2 0 A equivalent capacitance is . 3 d 2 0 A So charge flown through battery is  3 d  (d)

1 2 3



4

(d)

1 2

2 0 A  3 d

2 3

3 4



Question Bank

ELECTROSTATICS

M 133.

Sol.: M 134.

Sol.:

In the given circuit the value of charge across capacitor AB as a function of time is t t       2 RC  2 RC    (a) CE 1  e (b) 2CE 1  e       2 t 2 t       (c) CE / 21  e RC  (d) 2CE 1  e RC     

A

B C

C

R

D

C E

Charge flow through battery as a function of time , Q  2CE (1  e  (b)



t 2CR

)

Two identical conducting spheres having unequal charges q1 and q2 separated by distance r. If they are made to touch each other and then separated again to the same distance. The electrostatic force between the spheres in this case will be (neglect induction of charges) (a) less than before (b) same as before (c) more than before (d) zero Since the two spheres are identical, the final charges on each of the sphere after they are  q  q2  made to touch will be  1   2  Change in electrostatic force between them 2

 q  q2  K 1  Kq1 q 2 2  K K  2 2  F2  F1    2 q1  q 2   4q1 q 2  2 q1  q 2   0 2 2 r r 4r 4r  F2  F1







(c)

M 135.

A small ball of mass m and charge +q tied with an insulating string of length l, is rotating on a vertical circular path under gravity and in a uniform horizontal electric field E as shown. The tension in the string will be minimum for

 qE   (a)   tan 1   mg  (c)  

 qE     tan 1  2  mg 

(b)   

 qE   (d)     tan 1   mg 

E 

m,q

Question Bank

ELECTROSTATICS

Sol.:

The geffective will be directed along the resultant force of mg and qE. The angle made by the direction of geff with  qE   . vertical  tan 1   mg 

qE mg

The tension in the string will be minimum when the bob is just opposite to the direction of geff.

geff

 qE   Hence the required angle     tan 1   mg  

(d)

M 136.

A dipole of dipole moment p is kept at the centre of a ring of radius R and charge q. The dipole moment has direction along the axis of the ring. The resultant force on the ring due to the dipole is (a) zero (c)

Sol.:

pq 2 0 R 3

(b)

pq 4 0 R 3

(d)

pq only if the charge is uniformly distributed on the ring. 4 0 R 3

The electric field due to the dipole on the circumference of the ring E 

p and it is 4 0 R 3

directed normal to the plane of charged ring. Hence force on the charged ring F  qE  

pq 4 0 R 3

(b)

M  137. An electric field given by E  4iˆ  3( y 2  2) ˆj pierces Gaussian cube of side 1m placed at origin such that its three sides represents x, y and z axes. The net charge enclosed within the cube is (a) 4 0 (b) 3 0 (c) 50 (d) zero Sol.:

Net flux in x-direction = 0 Net flux in y-direction  A[3(1) 2  2]  A[3(0)  2] q   3A 0 

q  3 0 A  3 0 , (as A  1m 2 )



(b)

y

x z

Question Bank

ELECTROSTATICS

M 138. A parallel plate capacitor of capacitance C is charged with a battery of potential V. The battery is then disconnected and electromagnetic waves are incident on negative plate of the capacitor. As a result the negative plate starts emitting electrons towards the positive plate. The current which flows between the plates remains constant till time t1 and then starts decreasing. The potential difference between the plates at time t1 is (Assume plates of capacitor are close to each other) (a) V (b) V/2 (c) 2V (d) zero Sol.: The current starts decreasing when the potential difference between the plates becomes zero.  (d) M 139. n identical charge particle are placed on the vertices of a regular polygon of n sides of side length a. One of the charge particle is released from polygon. When this particle reaches a far of distance, another particle adjacent to the first particle is released. The difference of kinetic energies of both the particles at infinity is k. Magnitude of charge is (a) Sol.:

Sol.:

M 141.

k 4 0 a

(c)

k a

(d)

ka

(a)

A dielectric slab of thickness d is inserted in a parallel plate capacitor whose negative plate is at x = 0 and positive plate is at x = 3d. The slab is equidistant from the plates. The capacitor is given some charge. As x goes from 0 to 3d. (a) the magnitude of the electric field remains the same (b) the direction of the electric field changes continuously (c) the electric potential increases continuously (d) the electric potential increases at first, then decreases and again increases (c) A circular ring carries a uniformly distributed positive charge and lies in X–Y plane with centre at origin of co-ordinate system. If at a point (0, 0, z), the electric field is E, then which of the following graphs is correct? Y T

(a)

Y T

E

O z Y T E

Z

O z

Z

(c) Sol.:

(b)

q2 k 40 a 

M 140.

4 0 ak

(c)

(b)

E

O z

Z

Y E

(d)

O z

Z

Question Bank

ELECTROSTATICS

M 142.

Four metallic plates, each with surface area of one side A, are placed at a distance d from each other. The plates are connected as shown in figure. Then the capacitance of the system between P and Q is

1 2

P(+) 3

4

3 0 A 2 0 A (b) d d 2 A C  2C 2 C PQ   C 0 C  2C 3 3d  (c) (a)

Sol.:

M 143.

Sol.:

(c)

2 0 A 3d

(d) P(+)

2 1

Q(–)

3 0 A 2d 34

Q(–)

2 3

A semi-circular arc of radius a is charged uniformly and the charge per unit length is . The electric field at its centre is     (a) (b) (c) (d) 4 0 a 4 0 a 2 0 a 2 0 a (d)

M 144. Two identical parallel plate capacitors are placed in series and connected to a constant voltage source of V0 volt. If one of the capacitors is completely immersed in a liquid with dielectric constant K, the potential difference between the plates of the other capacitor will change to K 1 K K 1 2K (a) V0 (b) V0 (c) V0 (d) V0 K K 1 2K K 1 Sol.: (b) M 145.

Sol.:

The equivalent capacitance of the network (with all capacitors having the same capacitance C) is (a)  (b) zero  3 1   3 1    (c) C  (d) C   2  2     1 1 2 3C  2C     C  C  C C C (C   C )  1  3   2C2  2CC   C 2  0  C   C  2    (c)

A B

Upto 

Question Bank

ELECTROSTATICS

M 146. A hollow conducting sphere is placed in an electric field produced by a point charge placed at P as shown in figure. Let VA, VB, VC be the potentials at points A, B and C respectively. Then  (a) VC  V B (b) V B  VC (c) V A  V B Sol.: M 147.

Sol.:

A

C

P

B

(d) V A  VC

(a)

A current I is passing through a wire having two sections P and Q of uniform diameters d and d/2 respectively. If the mean drift velocity of electrons at P and Q is denoted by vp and vQ respectively, then  1 1 (a) v p  v Q (b) v p  v Q (c) v p  v Q (d) v p  2v Q 2 4 1 i 1 Drift velocity v d   vd  or v d  2 neA A d 2

 dQ   d / 2 2 v 1   Then P    v  vQ  P 4 v Q  d P   d   (c) M 148.

Three large thin sheets with surface charge density , –2 and – are placed as shown in figure. Electric field intensity at the point P is 4 ˆ 4 ˆ (a)  k (b) k 0 0 2 ˆ 2 ˆ (c)  k (d) k 0 0

kˆ



z  3a P

–2

z a

–

z 0

  2   ˆ 2 ˆ EP     k  k   2  2  2    0 0 0 0 (c)

 

Sol.:

 M 149.

Two charges, each of magnitude q = 2 C are placed at the vertices Q and R of the triangle as shown in the figure. The sum of the sides PQ and PR is 12 cm and their product is 32 cm2. The potential at point P would be (a) 6.00  105 V

(b) 6.25  105 V

P

Q

(c) 6.50  105 V

R

(d) 6.75  105 V

Question Bank

ELECTROSTATICS

Sol.:

x  y  12 , xy  32 kq kq xy = 6.25  105 Vp    kq x y xy  (b)

M 150.

A particle carrying a charge q is moving with a speed v towards a fixed particle carrying a charge Q at large distance. It approaches Q up to a certain distance r and then returns as shown in the figure. If charge q were moving with a speed 2v at large distance , the distance of the closest approach would be (a) r

Sol.:

M 151.

(b) 2r

(c)

Q r

r 2

(d)

r 4

1 2 KqQ mv  2 r 1 KqQ r 2  r1  m2v   4 2 r1  (d)

Three infinitely long charge sheets are placed as shown in figure. The electric field at point P is

 ˆ (a)  k 0

2 ˆ (b)  k 0

 



 

 

P

 –2

  Ep   kˆ 0

 

(a)

Three concentric conducting spherical shells have radii r, 2r and 3r and charges q1, q2 and q3 respectively appeared on the shells when innermost and outermost shells are earthed as shown in the figure. Then (a) q1  q3   q 2 (c)

z

–

4 ˆ 4 ˆ (c) k (d) – k 0 0   2 ˆ  Sol.: E p   kˆ  k   kˆ , 2 0 2 0 2 0 M 152.

v

q

q3 3 q1

(b) q1   (d)

q2 4

q3 1  q2 3

q3 q2 q1 r 2r 3r

Z=3a Z=0 x Z=–a

Question Bank

ELECTROSTATICS

Sol.:

k q1  q 2  kq3   voltage of outersurface. Which is connected to the earth 3r 3r k q1  q 2  kq3    0  q1  q 2  q3  0 3r 3r  (a)

M

 153. The electric field in a region is given by E  4iˆ  ˆj N/kg. Work done by this field is zero when a particle is moved along the line (a) y + 4x = 2 (b) 4y + x = 6 (c) x + y = 5 (d) x – y = 5 Sol.: Work done will be zero when displacement is perpendicular to the field. The field makes an angle





1 1  tan 1   with positive x-axis 4

while the line y + 4x = 2 makes an angle  2  tan 1  4  with positive x-axis 

1   2  90 0



(a)

 i.e. the line y + 4x = 2 is perpendicular to E .

M 154. A charge +q is fixed at each of the points x = x0, x = 3x0, x = 5x0 …..  on the x-axis and a charge –q is fixed at each of the points x = 2x0, x = 4x0, x = 6x0 …… . Here, x0 is a positive constant. Take the electric potential at a point due to charge Q at a distance r from it to be Q/40r. Then the potential at the origin due to the above system of charges is q q ln 2  (a) zero (b) (c)  (d) 8 0 x0 ln 2 4 0 x 0  kq  kq kq kq Sol.: V        ............... ,  x 0  2 x 0 3x 0 4 x 0

V

q ln 2 4 0 x 0

 (d) M 155. Figure shows three spherical and equipotential surfaces 1, 2 and 3 round a point charge q. The potential difference V1 –V2 = V2 –V3. If t1 and t2 be the distance between them. Then (a) t1 = t2

(b) t1 > t2

(c) t1 < t2

(d) t1  t2

3 2 1 q

t1 t2

Question Bank

ELECTROSTATICS

1 1  r  r  Sol.: V1  V2  Kq     Kq  2 1   r1 r2   r1r2  so, r2  r1   r1r2

[ potential difference is constant]

 t2 > t1  (c) M 156. If the capacitance of each capacitor is C , then effective capacitance of the shown network across any two junctions is (a) 2C

(b) C

(c) C/2

(d) 5C

Sol.: The given network is a balanced wheatstone bridge with one capacitor in parallel with this bridge.  M 157.

(a)

V2=20V

V1=20V

For the given diagram which of the following statement is true? (V1 and V2 are the potential difference across capacitor C1 and C2 respectively)

S1

C1=2pF

S3

C2=3pF

S2

(a) with S1 closed, V1 = 15V, V2 = 20V (b) with S3 closed, V1 = V2 = 25V (c) with S1 and S2 closed, V1 = V2 = 0 (d) with S1 and S3 closed, V1 = 20 V, V2 = 20V Sol.: M 158.

(d)

2V

A part of the circuit is shown in the figure. All the capacitors have capacitance of 2F. Then (a) charge on capacitor C1 is zero. (b) charge on capacitor C2 is zero. (c) charge on capacitor C3 is zero. (d) charge on capacitors cannot be determined.

C2 3V

2V C1

2V

C3

Question Bank

ELECTROSTATICS

Sol.:

2 X  2   2 X  2   2 X  2  3  0

2V

6 X  18

C2

X  3V Potential difference across C3 is zero.

3V

(X–2)

2V X

C1

2V

C3

Hence charge on C3 is zero.  M 159.

(c)

A parallel plate capacitor has two layers of dielectrics as shown in figure. This capacitor is connected across a battery, then the ratio of potential difference across the dielectric layers is (a) 4/3

Sol.:

K1=2 K2=6

(b) 1/2

d

(c) 1/3 (d) 3/2 Electric field in dielectrics are E1 and E2 V1 Ed 3  1  V2 E 2 2d 2

K1=2 K2=6



Sol.:

E2

E1

E  K   1  2  3   E2 K1 

M 160.

2d

d

2d

(d)

Q

An air capacitor consists of two parallel plates A and B as shown in the figure. Plate A is given a charge Q and plate B is given a charge 3Q. P is the median plane of the capacitor. If C0 is the capacitance of the capacitor, then (VP, VA and VB are potential at P, A and B respectively) (a) VP  V A 

Q 4C 0

(b) VP  V A 

Q 2C 0

(c) VP  V A 

Q C0

(d) VP  VB 

Q 4C 0

Q C0 Q VB  V P  2C 0 Q VP  V A  2C 0

A



(b)

3Q

Q

VB  V A 

–Q

2Q

Q 2Q A

P

B

3Q

P

B

Question Bank

ELECTROSTATICS

M 161.

 An electric field is given by E  yiˆ  xˆj N/C . The work done by electric field in moving a   1C charge from position vector rA  2iˆ  2 ˆj m to rB  4iˆ  ˆj m is







(a) + 4 J Sol.:

(b) – 4 J

Welectric field = 







(c) + 8 J



(d) zero

 4 ,1  4, 1   4,1 ˆ ˆ ˆ qE. dxi  dyj  dzk   y dx  xdy   d  xy   xy 2, 2   0





2,2 

 2, 2 

(d)

M 162.

A ball of mass 2 kg having charge 1 C is dropped from the top of a high tower. In space electric field exist in horizontal direction away from tower which varies as E  5  2 x   10 6 V/m (where x-is horizontal distance from tower), the maximum horizontal distance ball can go from the tower is (a) 5m

Sol.:

(b) 2.5m





dv 5  2x 5  2x , vx x  2 dx 2 X max

0



2v x dv x 

0

 5  2 x dx

 5X

max

2  X max 0

0

The maximum electric field intensity on the axis of a uniformly charged ring of charge q and radius R will be (a)

Sol.:

(d) 15m

Fx  1  10 6 5  2 x   10 6  5  2 x ax 

M 163.

(c) 10m

1 q 4 0 3 3R 2

(b)

R dE  0, dx

Emaximum 



(c)

1 4 0

(c)

Kqx

E at axis of ring 

For Emax,

1 2q 4 0 3R 2

2

 x2



3 2

R 2

x qR/ 2

 2 R2   R   2  

3 2



1 2qR 4 0 3 3R 2

1 2q 4 0 3 3R 2

(d)

1 2q 4 0 2 2 R 2

Question Bank

ELECTROSTATICS

M 164.

Two capacitors of capacitances 3 F and 6 F are charged to a potential of 12 V each. They are now connected to each other, with the positive plate of each joined to the negative plate of the other. The potential difference across each will be (a) zero (b) 3 V (c) 4 V (d) 6 V 36 Sol.: Charge stored in capacitor 3F = 36 C + – charge stored on capacitor 6 F = 72 C + – 36  Q 72  Q  3 6

i.e., – 72 + 2Q = 72 – Q 3Q = 2 × 72 Q = 48 48  36  Vcommon = =4V 3  (c)

– – –

+ + +

+ + + + +

– – – – –

72

M 165. Two identical metal plates are given positive charges Q1 and Q2 (< Q1) respectively. If they are brought close together to form a parallel plate capacitor with capacitance C, the potential difference between them is Q  Q2 Q  Q2 Q  Q2 Q  Q2 (a) 1 (b) 1 (c) 1 (d) 1 2C C C 2C Sol.: Within the plates electric fields due to charges Q1 and Q2 are Q Q E1  1 and E 2  2 2 0 A 2 0 A E  E1  E 2 

V = Ed =

1 (Q1  Q2 ) 2 0 A

Q  Q2 d (Q1  Q2 )  1 2 0 A 2C

 (d) M 166. Two identical thin rings, each of radius R are coaxially placed at a distance R apart. If Q1 and Q2 are the charges uniformly spread on the two rings, the work done by the electric field in moving a charge q from the centre of first ring to the centre of the second ring is q (a) zero (b) (Q1  Q2 )( 2  1) 4 0 2 R (c)

q 2 (Q1+Q2) 4 0 R

(d)

( 2  1)q (Q1  Q2 ) 2 4 0 R

Question Bank

ELECTROSTATICS

Sol.: V1 =

1 4 0

 Q1 Q    2 ,  R 2 R  

 Work done = q (V1 – V2) = q

=

4 0 2 R



V2 = q 4 0

1 4 0

 Q2 Q    1   R 2 R  

 Q1 Q  Q Q   2    2  1   2 R   R 2 R    R

(Q1 – Q2) ( 2  1)

(b)

M 167. Two equal point charges are fixed at x = -a and x = + a on the x-axis. Another point charge Q is placed at the origin. The change in the electrical potential energy of system, when it is displaced by a small distance x along the x-axis, is approximately proportional to (a) x (b) x2 (c) x3 (d) 1/x qQ qQ 2qQ Sol.: Potential energy of the system when charge Q is at O is U 0    a a a When charge Q is shifted to position O’, the potential energy will be qQ qQ qQ (2a ) 2qQ  x 2 U   2   1  2 ( a  x) ( a  x) ( a  x 2 ) a  a

2qQ  x 2 =  1  a  a 2

  

 U = U – U0 =

  

1

( x  a )

2qQ  x 2 1  a  a 2

 2qQ 2qQ 2    3 (x ) a a 

Hence U U  x 2 . (b) M 168. A charge q is placed at the midpoint of the line joining two equal charges Q. The system of three charges will be in equilibrium when q has the value (a) Q/4 (b) Q/2 (c) –Q/4 (d) –Q/2 Sol.:

Net force on a charge placed at A is zero Q2 4Qq  0 2 4 0 d 4 0 d 2 Q q 4  (c)

A

Q

q B d

Q

C

Question Bank

ELECTROSTATICS

M 169. A fully charged capacitor has a capacitance C. It is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity s and mass m. If the temperature of the block is raised by T, the initial potential difference V across the capacitance is 2mCT 2msT mCT msT (a) (b) (c) (d) s s C C Sol.: M 170.

1 Energy Stored = Loss of Heat , CV 2  msT , 2  (d)

M 171.

1 C 3

M 172.

(d)

1 C 6

(b)

Two concentric spherical shell of radius R and 2R having initial charges Q and 2Q respectively as shown. On closing the switch S charge flow from outer sphere to earth is (a) Q (b) –Q (c) –3Q

Sol.:

2msT C

Three large conducting parallel plates are separated by a small distance. A charge 1.0 C is given to middle plate. Find the charge on outer surface of the upper plate. 1 (a) 1 C (b) C 2 (c)

Sol.:

V

2Q

Q 2R R

S

(d) 3Q

On closing the switch potential of outer shell is zero KQ KQ1   0 , Q1 = –Q charge flow = 3Q 2R 2R  (d)

Three unchanged capacitor of capacitance 1F, 2F and 3F are connected as shown in figure. If potentials at point P, Q and R are 1V, 2V, 3V respectively. Then potential at O is 18 11 (a) V (b) V 11 18 (c) 5 V (d) none of these

P 1 F 3 F R

O 2 F Q

Question Bank

ELECTROSTATICS

Sol.:

1v  1  2v  2   3v  3  0 7 v V 6



P 1V 1 F O

(d)

3 F

2 F

R 3V

Q 2V

M 173.

A positively charged ball hangs from a silk thread. We put a positive test charge q0 at a F point and measure , then it can be predicted that the electric field strength E becomes q0 (a) greater than (c) less than

Sol.:

F q0

(b) Equal to

F q0

F q0

(d) cannot be estimated

(a)

M 174.

Two point charges +4e and e are kept at distance x apart. At what distance a charge q must be placed from charge +e, so that q is in equilibrium (a)

Sol.:

K 4e q

x  y 

2

 M 175.

Sol.:

x 2



Kqe , y2

(b)

2x 3

y

x 3

(c)

x 3

(d)

x 6

(c)

Consider two concentric metal spheres. The outer sphere is given a charge Q > 0, then (a) the inner sphere will be polarized due to field of the charge Q. (b) the electrons will flow from inner sphere to the earth if S is shorted. (c) the shorting of S will produce a charge of –Qb/a on the inner sphere (d) none of the above

A B b a

S

When S is open, spheres A and B have same +ve potential. When S is closed, potential of B becomes zero. So, the potential of B decreases. This decreases of potential takes place due to flow of electrons from earth to inner sphere B. 

(c)

Question Bank

ELECTROSTATICS

M 176.

In the given diagram, the potential difference between A and B is V1, when switch is open. The potential difference between A and B when switch is closed is V2. Then Q0 Q , V2  0 2C C Q (c) V1  0 , V2  zero 2C

M 177.

S

– – – –

Q0 + + B + + 2C

(b) V1  zero , V2  zero

(a) V1 

Sol.:

+ + + +

A

Q0 – – – – C

Q0 Q , V2  0 2C 2C Q Charge flow is zero after closing the switch, so, V1  V2  0 2C  (d)

(d) V1 

Two concentric hollow spherical shell of radius R and 2R having charges Q and 2Q as shown. Find the total electric potential energy of arrangement.

5 KQ 2 (a) 2 R

KQ 2 (b) R

3KQ 2 (c) R

2 KQ 2 (d) R

2Q 2R R Q

2

Sol.:

KQ 2 K 2Q  3 KQ 2   2R 22 R  2 R KQ 2Q  KQ 2 Interaction energy   2R R 2 5 KQ Total  2 R Self energy 



(a)

M 178.

Sol.:

The dielectric strength of air is 3.0  106 NC–1. The largest charge that a 0.30 cm radius metal sphere can hold without sparking is (a) 9 nC (b) 8.2 nC (c) 6 nC  | E | on the surface, should be less than 3  106 N/C K .Q  3  10 6 , 2 r Q  3  10 9 , Qmax  3nC

Hence,



(d)

9  10 9 

Q  3  10 6 6 9  10

(d) 3 nC

Question Bank

ELECTROSTATICS

M 179.

Sol.:

If each capacitor has capacitance C then find CAB (a) C

(b) C/2

(c) 3C/2

(d) none of these

Effective circuit 3C C eff  2 

B

A

C

C

A

B C

A

C

C

B

=

C

C

(c)

C

M 180.

Sol.:

Two insulating plates are both uniformly charged in such a way that the potential difference between them is V2 –V1 = 20 V. (i,e, plate 2 is at a higher potential). The plates are separated by d = 0.1m and can be treated as infinitely large. An electron is released from rest on the inner surface of plate 1. What is its speed when it hits plate 2? ( e = 1.6  10–19 C, m0 = 9.11  10–31 kg) (a) 2.65  106 m/s

(b) 7.02  1012 m/s

(c) 1.87  106 m/s

(d) 32  10–19 m/s

1 m0V 2  eV  , 2

V  2.65  10 6 m/s



Y 0.1m X 2

1

(a)

M 181.

Sol.:

A condenser of capacitance C1 is charged to V0 volt. The energy stored in it is U0. It is connected in parallel to another uncharged condenser of capacitance C2. The energy loss in the process is 2

C 1C 2U 0 (a) 2C1  C 2 

 C  C2   U 0 (b)  1 C  C  1 2 

Cfinal = C1 + C2,

 CV Vcommon   1 0  C1  C 2

Efinal

 CV 1  C1  C 2   1 0 2  C1  C 2

E initial 



  

2

(c)

C1U 0 C1  C 2

  

C12V02 CU   1 0 2C1  C 2  C1  C 2

CU 1 C1V02  U 0 , Eloss = Einitial -Efinal = 2 0 2 C1  C 2

(d)

(d)

C 2U 0 C1  C 2

Question Bank

ELECTROSTATICS

M 182.

Sol.:

In the adjoining figure. A section of a complicated circuit is shown in which E = 10 volt, C1 = 2F, C2 = 3F and (VB – VA) = 10 volt. The potential on C1 will be (a) 0 volt

(b) 4 volt

(c) 12 volt

(d) 16 volt

V1 

C1

–

10V

10V

C 2Vnet 3  20  =12 Volt C1  C 2 5

C1

+ –

A V1

(Vnet = V1 + V2 = 20 V) 

+

A

B C2

C2

10V

B V2

(c)

M 183.

If n drops, each of capacitance C and charged to a potential V, coalesce to form a big drop, the ratio of the energy stored in the big drop to that in each small drop will be (b) n4/3: 1

(a) n : 1 Sol.:

Charge on big drop = n (CV) Radius of big drop R  n1 / 3 r Energy of small drop E s  Energy of big drop Eb 

(c) n5/3: 1

(d) n2: 1

(charge conservation) (volume conservation)

1 q2 2 4 0 r

1 n2q 2 2 40 n1 / 3

Eb n 5 / 3  Es 1 

(c)

M 184.

Two concentric spheres of radii r1 and r2 carry charges q1 and q2 respectively. If the surface charge density () is the same for both spheres, the electric potential at the common centre will be (a)

Sol.:

 r1 .  0 r2

Vcommon 



(b)

q1 q2  4 0 r1 4 0 r2

q1 q  22 2 4r1 4r2

 r2 .  0 r1

(c)

 r1  r2  0

(d)

 r1  r2  0

Question Bank

ELECTROSTATICS

Vcommon = 

q r   1  q1 r1  2 22   r1  r2   2  0  4r1 4r2   0

(d)

M 185.

A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. The potential difference between the surface of solid sphere and the outer surface of the hollow shell is V. If the shell is now given a charge of –3Q , the new potential difference between the same two surfaces is (a) V

Sol.:

(b) 2 V

(c) 4 V

(d) –2 V

In the given situation potential difference is independent of charge on outer shell. 

(a)

M 186.

Two parallel plate capacitors of capacitances C and 2C are connected in parallel and charged to a voltage V with the help of a battery. The battery is then disconnected and the space between the plates of the first capacitor is filled with a dielectric of dielectric constant k. The voltage across the combination now will be (a) V

Sol.:

(b)

V 3k



Sol.:

3V k 1

(d)

3V k 2

Total charge  CV  2CV  3CV Final capacitance  KC  2C  C  K  2  Common voltage across combination VC 

M 187.

(c)

3CV 3V  C K  2 K  2

(d)

Consider the situation shown in the figure. The capacitor A has a charge q on it whereas B is un charged. The charge appearing on the capacitor B a long time after the switch is closed is q 2

(a) zero

(b)

(c) q

(d) 2q

q + + + + + + +

A

– – – – – – –

B

When switched is closed, inner plates of two capacitor get connected with each other. But outer plates are not connected. It means circuit is not completed. Therefore, no current will flow or no charge will appear on B.  (a)

Question Bank

ELECTROSTATICS

M 188.

Sol.:

Q

A thin, metallic spherical shell contains a charge Q on it. A point charge q is placed at the centre of the shell another charge q1 is placed out side as shown. All three charges are positive. The electrostatic force on the charge at the centre is (a) towards left (d)

(b) towards right

(c) upward

q1

q

(d) zero

M 189.

Sol.:

Four positive charges of same magnitude Q are placed at the four corners of a rigid square frame as shown in the figure. The plane of the frame is perpendicular to z-axis. If a negative charge –q is placed at a distance z away from the centre above frame (z E in dielectric  |slope in air| > | slope in dielectric| O



Sol.:

x

P

Q C

C

C

C

C

C

(d)

Seven capacitors, a switch S and a source of e.m.f. are connected as shown in the figure. Initially, S is open and all capacitors are uncharged. After S is closed and steady state is attained, the potential difference in volt across the plates of the capacitor A is (a) 12

d 2d 3d 4d 5d

(b)

For circuit the equivalent capacitance between points P and Q is (a) 6C (b) 4C (c) 3C/2 (d) 3C/4 1 1 1 3C    C eff  C eff 3C C 4 

M 199.

x

V

(c)

M 198.

d 2d 3d 4d 5d

(b) 15

(c) 17

59V 21F 7F 28F

A

3F

4F

2F

1F

(d) 19

Question Bank

ELECTROSTATICS

Sol.:

M 200.

The equivalent circuit is If q be the charge on each capacitor, then q q q q q      59 21 7 28 3 7 or q = 84 C  Potential difference across A q 84    12 volt. 7 7  (a)

59V 21F 7F 28F

A

S

3F

4F

2F

1F

59V 21F 7F 28F

S 7F

A

3F

Two spherical conductors A and B of radii a and b (b > a) are placed concentrically in air. The two are connected by a copper wire as shown in figure. Then the equivalent capacitance of the system is

B

b a

A

4 0 ab (b) 4 0 (a  b) (c) 4 0 b (d) 4 0 a ba All the charge given to inner sphere will pass on to the outer one. So, capacitor of that outer one = 40b.  (c) (a)

Sol.:

M 201.

Two charges q1 and q2 are placed 30 cm apart as shown. A third charge q3 is moved along the arc of a circle of radius 40 cm from C to D. The change in the q3 potential energy of the system is k , where k is 4 0 (a) 8q2 (c) 6q2

Sol.:

(b) 8q1 (d) 6q1

Change in potential energy ( U )  U f  U i  

 q1q3 q 2 q3   q1q3 q 2 q3   0.4  0.1    0.4  0.5      q 1 U  [8q 2 q3 ]  3 (8q 2 ) 4 0 4 0 U 

1 4 0



k  8q2



(a)

C q3 40cm 30cm q2 q1 B

A

D

q3

40cm 50cm q1

q2 30cm 40cm

D 10cm

Question Bank

ELECTROSTATICS

M 202.

Sol.:

M 203.

The plates of parallel plate capacitor are charged upto 100V. A 2mm thick dielectric plate is then inserted between the plates of capacitor. Then to maintain the same potential difference, the distance between the plates is increased by 1.6 mm. The dielectric constant of the plate is (a) 5 (b) 1.25 (c) 4 (d) 2.5 In air the potential difference between the plates  …(i) Vair  .d 0 In the presence of partially filled medium potential difference between the plates  t …(ii) Vm  ( d  t  ) 0 K Potential difference between the plates with dielectric medium and increased distance is  t  Vm   (d  d )  t   …(iii) 0  K t According to question Vair  Vm  which gives K  t  d 2 Hence K  5 2  1. 6  (a)

Four metallic plates, each with a surface area of one side A, are placed at a distance d from each other. The plates are connected as shown in the figure. Then the capacitance of the system between a and b is 3 0 A d 2 A (c) 0 3d

a

b

2 0 A d 3 A (d) 0 2d

(a)

(b)

Sol.:

1 2 3 4 C eq 



3 0 A 2 d (d)

a

a

 Equivalent circuit diagram is

b

2 3 21

b 4 3

Question Bank

ELECTROSTATICS

DIFFICULT QUESTIONS D 204.

Sol.:

Half part of ring is uniformly positively charged and other half is uniformly negatively charged. Ring is in equilibrium in uniform electric field as shown and free to rotate about an axis passing through its centre and perpendicular to plane. The equilibrium is (a) stable

(b) unstable

(c) neutral

(d) can be stable or unstable

E0

– – –

–

–

–

–– ++

+ + + + +

  Assuming ring as dipole, then dipole moment P and E are in same direction, so potential energy U = –PE  (a)

D 205.

– ++

In the arrangement shown all the plates have equal area A and spacing d between them. The equivalent capacitance between point P and Q will be, (a) (c)

0 A d

(b)

0 A 2d

(d)

0 A 3d

P

Q

2 0 A d

Sol.: 1

P

2

2 1

P

3 4

4 5

Q

P

3 4 5

Q

6

1 1 1 1 4     Ceq 2C C 2C 2C C  A C eq   0 2 2d  (c)

2 3

6 5

C

0 A d

Question Bank

ELECTROSTATICS

D 206.

A solid sphere of radius R, and dielectric constant k has spherical cavity of radius R/4. A point charge q1 is placed in the cavity. Another charge q2 is placed outside the sphere at a distance of r from q1. Then Coulombic force of interaction between them is found to be ‘F1’. When the same charges are separated by same distance in vacuum then the force of interaction between them is found to be F2, then (a) F1 

Sol.:

F2 k

(b) F2 

F1 k

(c) F1  F2 

1 k

(d) F1  F2

Coulombic force between them remains same.  (d)

D 207.

The magnitude of electric field intensity at point B (2, 0, 0) due to a dipole of dipole  moment, P  iˆ  3 ˆj kept at origin is (assume that the point B is at large distance from the 1 dipole and k  ) 4 0 (a)

Sol.:

13k 8

(b)

13k 4

(c)

7k 8

(d)

The dipole moment makes an angle 600 with x-axis and lies in x-y plane as shown.

 P  = 600

The electric field a point B due to dipole is kP E  3 1  3 cos 2  where  = 600 r   D 208.

E

7k 4

B(2,0,0)

7k 8

(c)

Two identical parallel plate capacitors A and B are connected in series through a battery of potential difference V (see figure). Area of each plate is a and initially plates of capacitors are separated by a distance d. Now, separation between plates of capacitor B starts increasing at constant rate v, find the rate by which work is done on the battery when separation between plates of capacitor B is 2d.

A

a 0 vV 2 (a) d2

2a 0vV 2 (d) d2

a 0vV 2 (b) 2d 2

a 0vV 2 (c) 9d 2

B

V

Question Bank

ELECTROSTATICS

Sol.:

Let at any instant separation between plates of capacitor B is x, then a 0 a 0 . a 0 x  a 0 (dx )  Ceq  d a 0 a 0 dx (d  x) dx  d x a 0 Q  CeqV  V dx dQ a 0 dx a 0  V  Vv 2 dt ( d  x) dt ( d  x) 2 a 0 vV 2  dQ  Rate of work done on the battery   V  (when x = 2d)  9d 2  dt   (c)

D 209. For spherical charge distribution given as   r   0 1   ,  3    0 , 

when r  3 when r  3

(where r is the distance from the centre of spherical charge distribution) The electric field intensity is maximum for the value of (a) r  1.5 (b) r  2 (c) r  4.5 (d) at infinity 1 Sol.: By Guass theorem,  EdS  0  dq

40 0

 x 2  0 1  3 x dx dE for max. electric field, 0 dr 1 r 6   0  r 2 3 6 3  (b) 

E  4r 2 

r



E

0 0

r r2   3  12   

D 210. Four concentric conducting shells A, B, C and D are arranged as shown in figure. A charge + q is given to the outer most shell D. Shell B and C are connected by conducting wire while Shell A and C is earthed. The equivalent capacitance of system is (a) 640 a

(b) 160a

(c) 40 a

(d) none of these

D

C

4a

+q B 3a

A 2a

a

Question Bank

ELECTROSTATICS

Sol.:

The equivalent capacitance 40 (3a  4a) C  40 (4a ) (4a  3a )  640 a

 (a) D 211. A bob of mass m = 100 gm having charge q = 50C connected with a non-conducting string of length l = 1m is whirled in vertical plane with minimum speed such that it can complete a circular path in space where electric field of strength E = 2 × 104 N/C is switched on as shown in figure. The velocity of bob at the point where tension in string become zero is   (a) v  ( 5iˆ  3 ˆj ) (b) v  10iˆ

 5 (c) v   (iˆ  3 ˆj ) 2 Sol.:

y 1m

x E

m +q



u

 5 (d) v   ( 3iˆ  ˆj ) 2

The centripetal force,



30º

mv 2 T  mg cos 60º qE cos 60º  l 2 mv mg  (as T  0 and qE  mg ) l v  gl

qE 30º 60º 60º

V

mg

 10 m/sec the velocity vector  v  v cos 60iˆ  v sin 60 ˆj

 D 212.

An electric field line emerges from a positive point charge +q at angle  to the straight line connecting it to a negative point charge  2q as shown in figure. At what angle  will the field line enter the charge  2q ?

+  q



– –2q

 1   1   1  (b) 2 sin 1  sin  (c) sin 1  sin  (d) sin 1  sin  2 2 2  2  2 2 2 The electric field lines connect the two charges of equal magnitude, the number of lines emerging from the charge  q within the angle 2 is equal to the number of lines entering the charge  2q at an angle 2 . Consequently, | q | (1  cos  ) | 2q | (1  cos ), hence

(a)  Sol.:

 5 v   [iˆ  3 ˆj ] 2 (c)

Question Bank

ELECTROSTATICS

1    |q| sin =  sin 2 2 2 | 2q | 2 1    2 sin 1 sin 2 2  (b) sin



D 213. A cavity of radius r is made inside a solid sphere. The volume charge density of the remaining sphere is . An electron (charge e, mass m) is released inside the cavity from point P as shown in figure. The centre of sphere and centre of cavity are separated by a distance a. The time after which the electron again touches the sphere is 6 2 r 0 m ea ea F  eE  3 0 1F 2 r 2 t 2m

(a) Sol.:

t

(b)

2 r 0 m ea

(c)

6r 0 m ea

P 45° a

(d)

r 0 m ea

P 45° a

6 2r 0 m ea

 (a) D 214.

A parallel plate capacitor having square plate of area a and separation between plates d is completely filled with a dielectric of dielectric constant k. Capacitor is connected with a battery of potential difference V. Now dielectric is pulled with constant speed u. Find the initial value of current.  0 au  au (k  1) (b) 0 (k  1) 2d d k a ( a  x ) 0 a x 0 C  d d Q  CV

(a) Sol.:

I



  k a 0 a 0  dQ dC  V   u u dt dt d d    au = 0 (k  1) d (b)

(c) zero

u

V

(d)

2 0 a u (k  1) d u

V

Question Bank

ELECTROSTATICS

D 215.

Sol.:

A charge q is placed at some distance along the axis of a uniformly charged disc of surface charge density . The flux due to the charge q through the disc is . The electric force on charge q exerted by the disc is    (a)  (b) (c) (d) 2 4 3  Electric field due to disc at position of q is E  , F  qE   q  (a)

D 216.

The diagram shows an arrangement of identical metal plates placed parallel to each other and the variation of potential between the plates by dotted line. Using the details given in diagram, the equivalent capacitance connected across the battery is equal to (Separation between the plates = L, area of each plate = A)

3 0 A 3 A (b) 0 L 4L Circuit can be redrawn as (a)

Sol.:

Ceq =  D 217.

(d)

X

0 A L

0V6 5

2 3

3 4

6C 6 0 A  5 5L

6 7

10V 0V –10V

10V 2 1

–10V

7 6

(c)

E l

1 2 3 4 5

6 0 A 5L

A cylindrical conductor AB of length l and area of cross-section a is connected to a battery having emf E and negligible internal resistance. The specific conductivity of cylindrical conductor varies as l   0 , where 0 is constant and x is distance x from end A. What is the electric field just near the end B of cylinder? (a)

Sol.:

(c)

Y

(b)

3E 2l

Resistance of cylinder R  

l

0

I

E R

1 dx 2 l   a 3a 0

(c)

2E 3l

B

A

E

(d)

E 2l

Question Bank

ELECTROSTATICS

Electric field 

 D 218.

J I  x E( l ) 3E      a Ra 0 l 2l 2 l a 0 l 3a 0

(b)

A uniform electric field E is present horizontally along the paper throughout region but uniform magnetic field B0 is present horizontally (perpendicular to plane of paper in inward direction) right to the line AB as shown. A charge particle having charge q and mass m is projected vertically upward and crosses the line AB after time t0. Find the speed of projection if particle moves after t0 with constant velocity. (given qE = mg)

A E  B0

q, m B

(a) gt0 (b) 2gt0 gt (c) 0 2 (d) particle can’t move with constant velocity after crossing AB Sol.:

When crosses AB qvB0 cos   mg qvB0 sin = qE qE tan   1 mg   4

qvB

A

v 

qE

mg B

qE t0 m u  gt 0  v sin 

along horizontal v cos  

qE   u  g   t 0 = 2gt0 m 



(b)

D 219. A conducting disc of radius R is rotating with an angular velocity . Allowing the fact that electrons are the current carries in conductor, the potential difference between the centre of the disc and edge is (mass and charge of electron is m and e and neglect gravity) m2 R m2 R 2 m2 R m2 R 2 (a) V  (b) V  (c) V  (d) V  2e 2e e 4e Sol.: Centripetal force on electron at distance r from centre of disc. m2 r 2  E  m r  eE e

Question Bank

ELECTROSTATICS

dV m 2 r  dr e 2 m R m 2 R 2 V  rdr   e 0 2e (b)

   D 220.

Sol.:

A capacitor is to be designed to operate, with constant capacitance, in an environment of fluctuating temperature. As shown in the figure, the capacitor is a parallel plate capacitor with ‘spacer’ to change the distance for compensation of temperature effect. If 1 be the co-efficient of linear expansion of plates and 2 that of spacer, the condition for no change in capacitance with change of temperature is (The capacitance of the capacitor is equal to C and spacer have insulated ends) (a) 1   2 (b) 1  2 2 (c) 21   2  A C 0 L  log C  log 0  log A  log L dC dA dL   C A L dC  21 dT   2 dT  0 C  21   2 

D 221.

(d) 21  3 2

(c)

An electric dipole is placed along the x-axis at the origin O. A point P is at a distance of  20cm from the origin such that OP makes an angle with the x-axis. If the electric field at 3 P makes an angle  with the positive direction of x-axis, the value of  would be (a)

Sol.:

spacer

 3

(b)

 3    tan 1   3 2  

 1 1    where tan   tan   tan 3 2 2 3 3  3    tan 1 so,    tan 1 2 3 2  (b)

(c)

 3  (d) tan 1   2  

2 3

Y



–Q

E P  /3 +Q

O

P



X

Question Bank

ELECTROSTATICS

D 222.

In the given figure two semicircular wire is connected which are in x-y and x-z plane respectively. And +2q0 charge is distributed over it, then what will be the magnitude of electric field intensity at the origin. (a) (c)

Sol.: E y   D 223.

q0

(b)

 0 R 2 q0 2

2  0 R

2

z ++ +++ + + + x + + ++ + + + y

4q02  0 R 2

q0

(d)

2 0 R 2

q0 q , Ez  2 0 2 , 2 2  0 R 2  0 R

Enet  E y2  Ez2 

2

q0 2 2  0 R 2

(c) 

There is uniform electric field E  E0iˆ as shown in figure. Two charged particles q1 and q2 and masses m1 and m2 are projected from point O with velocities   v1 and v2 at t = 0. At t = 2t0 their velocities become       v1 ' and v2 ' . Then find | (m1v1'  m2v2' )  (m1v1  m2v2 ) | (gravity is absent) (a) (q1+q2) E0 t0 (c)

Sol.: F 

1 (q1+q2) E0 t0 2

 E

x

O z

(b) 2(q1+q2) E0 t0 (d) zero

P , F = (q1 + q2) E0, t



y

t = 2t0,

(q1+q2) E0 =

P , 2t0

P = 2 (q1 + q2) E0 t0

(b)

D 224.

If the area of each plate is A and the successive separation are d, 2d and 3d. Then find the equivalent capacitance across A and B in the given figure.

(a)

0 A 4d

(b)

3 0 A 4d

(c)

0 A d

d

A 2d A

3d

(d)

4 0 A d

Question Bank

ELECTROSTATICS

Sol.:

D 225. Sol.:

D 226

Sol.: D 227.

0 A  A , C2  0 d 2d  A and C3  0 3d C2 is short-circuited. C1 and C3 are in series. CC Hence, C AB  1 3 C1  C 3  A Substituting the values, we get C AB  0 4d  (a) C1 

2d A

3d

One brass plate is inserted between two charges. The force between two charges will (a) remain the same (b) increase (c) decrease (d) fluctuate (b)

There is an air-filled 1 pF parallel plate capacitor. When the plate separation is doubled and the space is filled with wax, the capacitance increases to 2pF. The dielectric constant of wax is (a) 2 (b) 4 (c) 6 (d) 8 (b) Two charges q1 and q2 are placed 30 cm apart as shown. A third charge q3 is moved along the arc of a circle of radius 40 cm from C to D. The change in the q3 potential energy of the system is k , where k is 4 0 (a) 8q2 (c) 6q2

Sol.:

d

A

(b) 8q1 (d) 6q1

Change in potential energy ( U )  U f  U i  

 q1q3 q 2 q3   q1q3 q 2 q3   0.4  0.1    0.4  0.5      q 1 U  [8q 2 q3 ]  3 (8q 2 ) 4 0 4 0 U 

1 4 0



k  8q2



(a)

C q3 40cm 30cm q2 q1 B

A

D

q3

40cm 50cm q1

q2 30cm 40cm

D 10cm

Question Bank

ELECTROSTATICS

D 228.

Sol.:

Four plates of equal area A, are separated by equal distances d and are arranged as shown in the figure. The equivalent capacity between x and y is  2 0 A 4 0 A (a) (b) d d 3 A  A (c) 0 (d) 0 d d The equivalent circuit is as shown in figure, 2 A 0 C eff  2C  d 

x

y

y

x

(a)

D 229.

–Q

Charges Q, 2Q and –Q are given to three concentric conducting spherical shells A, B and C respectively, as shown in the figure. The ratio of charges on the inner and the outer surfaces of the shell C will be

C

2Q B Q

A

3 3 3 3 (b)  (c) (d)  4 4 2 2 The electric field inside a conductor is zero. So if we take a Gausian spherical surface just covering the inner surface of the shell C, the flux on it will be zero. Hence from Gauss’s law, the charge on the inner surface of shell C should be –3Q, and from conservation of charge, charge on the outer surface of C will be +2Q.

(a)  Sol.:



(d)

D 230. A point charge +Q is projected with velocity V from a distance R from an infinitely long fixed line charge of linear charge density . What is the tangential velocity of the point charge when its radial distance from the line charge becomes eR ?

(a) Sol.:

V e

V1 R1  V2 R2  (a)

(b) eV

(c) e 2V

+ + + + + + +

V R

Q

(d) V/e2

Question Bank

ELECTROSTATICS

D 231. In the diagram shown, the charge +Q is fixed. Another charge +2q, is projected from a distance R from the fixed charge. Minimum separation between the two charges if the velocity becomes half of the projected velocity, at this moment is (Assume gravity to be absent) (a)

3R

3 R  Mvd  2 d  3R  (a)

Sol.: Mv

D 232.

3 v R d 2 2

(b) –W

+Q

1 R 2

(c) W 

Q2 4 0 R

M, +2q

R

(d) 4R

y

x Q z

(d)  W 

Q2 4 0 R

(b)

n identical charge particle are placed on the vertices of a regular polygon of n sides of side length a. One of the charge particle is released from polygon. When this particle reaches a far of distance, another particle adjacent to the first particle is released. The difference of kinetic energies of both the particles at infinity is k. Magnitude of charge is (a)

Sol.:

v

(c)

60°

Work done to rotate the ring is equal to work done to return the charge at its initial position. 

D 233.

3 R 2

The semicircular ring shown in the figure has radius R and carries a charge Q distributed uniformly over it. A point charge Q is taken from the point (0, 0, 2R) to the point (0, 2R, 0). The work done in doing this is equal to W. After fixing the charge at its new position, the ring is rotated in anticlockwise sense about x-axis through an angle /2. The work done in rotating the ring is (a) W

Sol.:

(b)

V

4 0 ak

q2 k 40 a 

(a)

(b)

k 4 0 a

(c)

k a

(d)

ka

Question Bank

ELECTROSTATICS

D 235.

A point charge is placed at a distance r from center of a conducting neutral sphere of radius R (r  R). The potential at any point P inside the sphere at a distance r1 from point charge due to induced charge of the sphere is given by [k = (a) kq/r1 (c) kq/r – kq/r1

Sol.:

P

r1

R

q

r

C

1 ] 4 0

(b) kq/r (d) –kq/R

Potential of centre of sphere =

Kq Kq  Vi  r r

where Vi = potential due to induced charge at centre = 0 [ qi = 0 and all induced charges are equidistance from centre]

D 236.



potential at point P =



q q Vi  K     r r1 



(c)

Kq Kq   Vi (For conductor all points are equipotential) r r1

In the circuit shown in figure if switch S is closed, the charge that flows through battery is equal to (a) 5 C (b) 7.5 C (c) 10 C (d) 15 C

3F

S

15C, 3F _ + _ + _ +

10 V

Sol.:



Ceq = 3/2 F

Charge flow q = Ceq (10 – 

15 3 ) = × 5 = 7.5 C 3 2

(b)

D 237. Four concentric conducting shells A, B, C and D are arranged as shown in figure. A charge + q is given to the outer most shell D. Shell B and C are connected by conducting wire while Shell A and C is earthed. The equivalent capacitance of system is (a) 640 a

(b) 160a

(c) 40 a

(d) none of these

D

C

4a

+q B 3a

A 2a

a

Question Bank

ELECTROSTATICS

Sol.:

The equivalent capacitance 40 (3a  4a) C  40 (4a ) (4a  3a )  640 a 

D 238.

(a)

In the circuit shown, the capacitors C1 and C 2 have capacitance C each. The switch S is closed at time t  0. Taking Q0  CE and   RC, the charge on C 2 after time t will be (a) Q0 (1  e  t / 2 )

Sol.:

C1 C2

(b) Q0 (1  e  t /  )

S

E

Q (c) 0 (1  e t / 2 ) (d) Q0 (1  e 2t /  ) 2 Rearranging the circuit, we observed that C1 is joined directly to the cell and acquires its full charge when S is closed. It plays no part in the charging of C2 through R. So, q2  Q0 (1  e t /  )



R

C2 R E

(b)

C1 D 239.

Two large identical plates are placed in front of each other at x = d and x = 2d as shown in figure. If charges on plates are Q and -5Q, the potential versus distance graph for region x = 0 to x = 3d is (d is very small and potential at x = 0 is v0) V

V

d 2d 3d

x

x

3d

V v0

(c) d 2d 3d

x

Slope of potential from x = 0 to x = d is 

Slope of potential from x = 2d to x = 3d is

v0

(d)

d 2d 3d

4Q 2Q  2 0 A 0 A

Slope of potential from x = d to x = 2d is 

(d)

-5Q 2d

d

V

(b)

(a)



0

v0

v0

Sol.;

Q

3Q 0 A

2Q 0 A

x

d 2d 3d

x