Electrostatics Questions — JEE Physics

Q1

An electric dipole of mass

m

, charge

q

, and length

l

is placed in a uniform electric field

\vec{E} = E_0\hat{i}

. When the dipole is rotated slightly from its equilibrium position and released, the time period of its oscillations will be :

A

\dfrac{1}{2\pi} \sqrt{\dfrac{ml}{2qE_0}}

B

2\pi \sqrt{\dfrac{ml}{2qE_0}}

C

2\pi \sqrt{\dfrac{ml}{qE_0}}

D

\dfrac{1}{2 \pi} \sqrt{\dfrac{2 \mathrm{~m} l}{\mathrm{q} \mathrm{E}_0}}

Correct Answer

Option B

Solution

\begin{aligned} & \mathrm{I} \omega 2 \theta=\mathrm{q} \ell \mathrm{E}_0 \theta \\ & 2 \mathrm{~m}\left(\frac{\ell}{2}\right)^2 \omega^2=\mathrm{q} \ell \mathrm{E}_0 \\ & \omega^2=\frac{2 \mathrm{qE}_0}{\mathrm{~m} \ell} \\ & \mathrm{~T}=2 \pi \sqrt{\frac{\mathrm{~m} \ell}{2 \mathrm{qE}_0}} \end{aligned}

Q2

If a charge

q

is placed at the center of the line joining two equal charges

Q

such that the system is in equilibrium then the value of

q

is

A

Q/2

B

- Q/2

C

Q/4

D

- Q/4

Correct Answer

Option D

Solution

For equilibrium of charge

Q

K{{Q \times Q} \over {{{\left( {2x} \right)}^2}}} + K{{Qq} \over {{x^2}}} = 0 \Rightarrow q = - {Q \over 4}

Q3

If the electric potential at any point (x, y, z) m in space is given by V = 3x2 volt. The electric field at the point (1, 0, 3) m will be :

A 3 Vm

-

1, directed along positive x-axis.

B 3 Vm

-

1, directed along negative x-axis.

C 6 Vm

-

1, directed along positive x-axis.

D 6 Vm

-

1, directed along negative x-axis.

Correct Answer

Option D

Solution

\overrightarrow E = -{{dV} \over {dx}}\widehat i

\overrightarrow E = - 6x\widehat i

So,

\overrightarrow E

at (1, 0, 3) is

\overrightarrow E = - 6\widehat i

V/m

Q4

Two point charges A and B of magnitude +8

\times

10

-

6 C and

-

8

\times

10

-

6 C respectively are placed at a distance d apart. The electric field at the middle point O between the charges is 6.4

\times

104 NC

-

1. The distance 'd' between the point charges A and B is :

A 2.0 m

B 3.0 m

C 1.0 m

D 4.0 m

Correct Answer

Option B

Solution

Electric field at P will be

E = {{kq} \over {{{(d/2)}^2}}} \times 2 = {{8kq} \over {{d^2}}}

So,

{{8 \times 9 \times {{10}^9} \times 8 \times {{10}^{ - 6}}} \over {{d^2}}} = 6.4 \times {10^4}

So,

d = 3

m

Q5

On moving a charge of

20

coulomb by

2

cm,

2

J

of work is done, then the potential differences between the points is

A

0.1

V

B

8

V

C

2V

D

0.5

V.

Correct Answer

Option A

Solution

We know that

{{{W_{AB}}} \over q} = {V_B} - {V_A}

$\therefore$

{V_B} - {V_A} = {{2J} \over {20C}} = 0.11J/C = 0.1V

Q6

If the electric flux entering and leaving an enclosed surface respectively is

{\phi _1}

and

{\phi _2},

the electric charge inside the surface will be

A

\left( {{\phi _2} - {\phi _1}} \right){\varepsilon _0}

B

\left( {{\phi _2} + {\phi _1}} \right)/{\varepsilon _0}

C

\left( {{\phi _1} - {\phi _2}} \right)/{\varepsilon _0}

D

\left( {{\phi _1} + {\phi _2}} \right){\varepsilon _0}

Correct Answer

Option A

Solution

The flux entering an enclosed surface is taken as negative and the flux leaving the surface is taken as positive, by convention.

Therefore the net flux leaving the enclosed surface

= {\phi _2} - {\phi _1}

$\therefore$ the change enclosed in the surface by Gauss's law is

q = { \varepsilon _0}\,\left( {{\phi _2} - {\phi _1}} \right)

Q7

A thin spherical conducting shell of radius

R

has a charge

q.

Another charge

Q

is placed at the center of the shell. The electrostatic potential at a point

P

a distance

{R \over 2}

from the center of the shell is

A

{{2Q} \over {4\pi {\varepsilon _0}R}}

B

{{2Q} \over {4\pi {\varepsilon _0}R}} - {{2q} \over {4\pi {\varepsilon _0}R}}

C

{{2Q} \over {4\pi {\varepsilon _0}R}} + {q \over {4\pi {\varepsilon _0}R}}

D

{{\left( {q + Q} \right)2} \over {4\pi {\varepsilon _0}R}}

Correct Answer

Option C

Solution

Electric potential due to charge

Q

placed at the center of spherical shell at point

P

is

{V_1} = {1 \over {4\pi {\varepsilon _0}}}{Q \over {R/2}} = {1 \over {4\pi {\varepsilon _0}}}{{2Q} \over R}

Electric potential due to charge

q

on the surface of the spherical shell at any point inside the shell is

{V_2} = {1 \over {4\pi {\varepsilon _0}}}{q \over R}

Q8

Two spherical conductors

B

and

C

having equal radii and carrying equal charges on them repel each other with a force

F

when kept apart at some distance. A third spherical conductor having same radius as that

B

but uncharged is brought in contact with

B,

then brought in correct with

C

and finally removed away from both. The new force of repulsion between

B

and

C

is

A

F/8

B

3

F/4

C

F/4

D

3

F/8

Correct Answer

Option D

Solution

F \propto {{{Q_A}{Q_C}} \over {{x^2}}}

x

is distance between the spheres. After first operation charge on

B

is halved i.e

{Q \over 2}.

and charge on third sphere becomes

{Q \over 2}.

Now it is touched to

C

, charge then equally distributes them selves to make potential same, hence charge on

C

becomes

\left( {Q + {Q \over 2}} \right){1 \over 2} = {{3Q} \over 4}.

$\therefore$

{F_{new}} \propto {{{Q_C}{Q_B}} \over {{x^2}}}

= {{\left( {{{3Q} \over 4}} \right)\left( {{Q \over 2}} \right)} \over {{x^2}}}

= {3 \over 8}{{{Q^2}} \over {{x^2}}}

or,

{F_{new}} = {3 \over 8}F

Q9

A charge particle

'q'

is shot towards another charged particle

'Q'

which is fixed, with a speed

'v'

. It approaches

'Q'

upto a closest distance

r

and then returns. If

q

were given a speed of

'2v'

the closest distances of approaches would be

A

r/2

B

2r

C

r

D

r/4

Correct Answer

Option D

Solution

{1 \over 2}m{v^2} = {{kQq} \over r}

\Rightarrow {1 \over 2}m{\left( {2v} \right)^2} = {{kqQ} \over {r'}}

\Rightarrow r' = {r \over 4}

Q10

Four charges equal to -

Q

are placed at the four corners of a square and a charge

q

is at its center. If the system is in equilibrium the value of

q

is

A

- {Q \over 2}\left( {1 + 2\sqrt 2 } \right)

B

{Q \over 4}\left( {1 + 2\sqrt 2 } \right)

C

- {Q \over 4}\left( {1 + 2\sqrt 2 } \right)

D

{Q \over 2}\left( {1 + 2\sqrt 2 } \right)

Correct Answer

Option B

Solution

Net field at A should be zero

\sqrt 2 \,{E_1} + {E_2} = {E_3}

$\therefore$

{{kQ \times \sqrt 2 } \over {{a^2}}} + {{kQ} \over {\left( {\sqrt 2 a} \right)}} = {{kq} \over {{{\left( {{a \over {\sqrt 2 }}} \right)}^2}}}

\Rightarrow {{Q\sqrt 2 } \over 1} + {Q \over 2} = 2q

\Rightarrow q = {Q \over 4}\left( {2\sqrt 2 + 1} \right).