Electrostatics — Page 2 | JEE Physics

Q11

A charged oil drop is suspended in a uniform field of

3 \times {10^4}

v/m

so that it neither falls nor rises. The charge on the drop will be (Take the mass of the charge

= 9.9 \times {10^{ - 15}}\,\,kg

and

g = 10\,m/{s^2}

)

A

1.6 \times {10^{ - 18}}\,C

B

3.2 \times {10^{ - 18}}\,C

C

3.3 \times {10^{ - 18}}\,C

D

4.8 \times {10^{ - 18}}\,C

Correct Answer

Option C

Solution

At equilibrium, electric force on drop balances weight of drop.

qE = mg \Rightarrow q

= {{mg} \over E} = {{9.9 \times {{10}^{ - 15}} \times 10} \over {3 \times {{10}^4}}}

= 3.3 \times {10^{ - 18}}C

Q12

Two thin wire rings each having a radius

R

are placed at a distance

d

apart with their axes coinciding. The charges on the two rings are

+q

and

-q.

The potential difference between the centres of the two rings is

A

{q \over {2\pi \,{ \in _0}}}\left[ {{1 \over R} - {1 \over {\sqrt {{R^2} + {d^2}} }}} \right]

B

{{qR} \over {4\pi \,{ \in _0}\,{d^2}}}

C

{q \over {4\pi \,{ \in _0}}}\left[ {{1 \over R} - {1 \over {\sqrt {{R^2} + {d^2}} }}} \right]

D zero

Correct Answer

Option A

Solution

{V_A} = {V_{self}} + {V_{due}}

to

(2)

\Rightarrow {V_A} = {1 \over {4\pi {\varepsilon _0}}}\left[ {{q \over R} - {q \over {\sqrt {{R^2} + {d^2}} }}} \right]

{V_B} = {V_{self}} + {V_{due}}

to

(1)

\Rightarrow {V_B} = {1 \over {4\pi {\varepsilon _0}}}\left[ {{{ - q} \over R} + {q \over {\sqrt {{R^2} + {d^2}} }}} \right]

\Delta V = {V_A} - {V_B}

= {1 \over {4\pi {\varepsilon _0}}}\left[ {{q \over R} + {q \over R} - {q \over {\sqrt {{R^2} + {d^2}} }} - {q \over {\sqrt {{R^2} + {d^2}} }}} \right]

= {1 \over {2\pi {\varepsilon _0}}}\left[ {{q \over R} - {q \over {\sqrt {{R^2} + {d^2}} }}} \right]

Q13

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

A

{L \over 4}

B

2

L

C

4

L

D

8

L

Correct Answer

Option B

Solution

{{ - K2q} \over {{{\left( {x - L} \right)}^2}}} + {{K8q} \over {{x^2}}} = 0 \Rightarrow {1 \over {{{\left( {x - L} \right)}^2}}} = {4 \over {{x^2}}}

or,

{1 \over {x - L}} = {2 \over x} \Rightarrow x = 2x - 2L

or,

x=2L

Q14

An electric dipole is placed at an angle of

{30^ \circ }

to a non-uniform electric field. The dipole will experience

A a translation force only in the direction of the field

B a translation 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

Correct Answer

Option C

Solution

The electric field will be different at the location of the two charges.

Therefore the two forces will be unequal.

This will result in a force as well as torque.

Q15

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 surfaces of spheres

A

and

B

is

A

4:1

B

1:2

C

2:1

D

1:4

Correct Answer

Option C

Solution

After connection,

{V_1} = {V_2}

\Rightarrow K{{{Q_1}} \over {{r_1}}} = K{{{Q_2}} \over {{r^2}}}

\Rightarrow {{Q{}_1} \over {{r_1}}} = {{{Q_2}} \over {{r_2}}}

The ratio of electric fields

{{{E_1}} \over {{E_2}}} = {{K{{{Q_1}} \over {r_1^2}}} \over {K{{{Q_2}} \over {r_2^2}}}} = {{{Q_1}} \over {r_1^2}} \times {{r_2^2} \over {{Q_2}}}

\Rightarrow {{{E_1}} \over {{E_2}}} = {{{r_1} \times r_2^2} \over {r_1^2 \times {r_2}}} \Rightarrow {{{E_1}} \over {{E_2}}} = {{{r_2}} \over {{r_1}}} = {2 \over 1}

Since the distance between the spheres is large as compared to their diameters, the induced effects may be ignored.

Q16

An electric charge

{10^{ - 3}}\,\,\mu \,C

is placed at the origin

(0,0)

of

X-Y

co-ordinate system. Two points

A

and

B

are situated at

\left( {\sqrt 2 ,\sqrt 2 } \right)

and

\left( {2,0} \right)

respectively. The potential difference between the points

A

and

B

will be

A

4.5

volts

B

9

volts

C zero

D

2

volts

Correct Answer

Option C

Solution

The distance of point

A\left( {\sqrt 2 ,\sqrt 2 } \right)

from the origin,

OA = \left| {\overrightarrow {{r_1}} } \right|

= \sqrt {{{\left( {\sqrt 2 } \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}}

= \sqrt 4 = 2

The distance of point

B(2,0)

from the origin,

OB = \left| {\overrightarrow {{r_2}} } \right| = \sqrt {{{\left( 2 \right)}^2} + {{\left( 0 \right)}^2}} = 2

units. Now, potential at

A,

{V_A} = {1 \over {4\pi { \in _0}}}.{Q \over {\left( {OA} \right)}}

Potential at

B,

{V_B} = {1 \over {4\pi { \in _0}}}.{Q \over {\left( {Ob} \right)}}

$\therefore$ Potential difference between the points

A

and

B

is zero.

Q17

The potential at a point

x

(measured in

\mu \,m

) due to some charges situated on the

x

-axis is given by

V\left( x \right) = 20/\left( {{x^2} - 4} \right)

volt The electric field

E

at

x = 4\,\mu \,m

is given by

A

(10/9)

volt /

\mu

m

and in the

+ ve

x

direction

B

\left( {5/3} \right)

volt/

\mu

m

and in the

-ve

x

direction

C

\left( {5/3} \right)

volt/

\mu

m

and in the

+ve

x

direction

D

\left( {10/9} \right)

volt/

\mu \,m

and in the

-ve

x

direction

Correct Answer

Option A

Solution

Here,

V\left( x \right) = {{20} \over {{x^2} - 4}}volt

We know that

E = - {{dV} \over {dx}} = {d \over {dx}}\left( {{{20} \over {{x^2} - 4}}} \right)

or,

E = + {{40x} \over {{{\left( {{x^2} - 4} \right)}^2}}}

At

x = 4\mu m,

E = + {{40 \times 4} \over {{{\left( {{4^2} - 4} \right)}^2}}}

= + {{160} \over {144}} = + {{10} \over 9}volt/\mu m.

Positive sign indicates that

\overrightarrow E

is in

+ve

x

-direction.

Q18

A charge

Q

is placed at each of the opposite corners of a square. A charge

q

is placed at each of the other two corners. If the net electrical force on

Q

is zero, then

Q/q

equals:

A

-1

B

1

C

- {1 \over {\sqrt 2 }}

D

- 2\sqrt 2

Correct Answer

Option D

Solution

Let

F

be the force between

Q

and

Q.

The force between

q

and

Q

should be attractive for net force on

Q

to be zero. Let

F'

be the force between

Q

and

q.

For equilibrium

\sqrt 2 F' = - F

\sqrt 2 \times k{{Qq} \over {{\ell ^2}}} = - k{{{Q^2}} \over {{{\left( {\sqrt 2 \ell } \right)}^2}}}

\Rightarrow {Q \over q} = - 2\sqrt 2

Q19

Let

P\left( r \right) = {Q \over {\pi {R^4}}}r

be the change density distribution for a solid sphere of radius

R

and total charge

Q

. For a point

'p'

inside the sphere at distance

{r_1}

from the center of the sphere, the magnitude of electric field is :

A

{Q \over {4\pi \,{ \in _0}\,r_1^2}}

B

{{Qr_1^2} \over {4\pi \,{ \in _0}\,{R^4}}}

C

{{Qr_1^2} \over {3\pi \,{ \in _0}\,{R^4}}}

D

0

Correct Answer

Option B

Solution

Let us consider a spherical shell of thickness

dx

and radius

x.

The volume of this spherical shell

= 4\pi {r^2}dr.

The charge enclosed within shell

= {{{Q_r}} \over {\pi {R^4}}}\left[ {4\pi {r^2}dr} \right]

The charge enclosed in a sphere of radius

{r_1}

is

{{4Q} \over {{R^4}}}\int\limits_0^{{r_1}} {{r^3}} dr

= {{4Q} \over {{R^4}}}\left[ {{{{r^4}} \over 4}} \right]_0^{{r_1}}

= {Q \over {{R^4}}}r_1^4

$\therefore$ The electric field at point

p

inside the sphere at a distance

{r_1}

from the center of the sphere is

E = {1 \over {4\pi { \in _0}}}{{\left[ {{Q \over {{R^4}}}r_1^4} \right]} \over {r_1^2}}

= {1 \over {4\pi { \in _0}}}{Q \over {{R^4}}}r_1^2

Q20

Two points

P

and

Q

are maintained at the potentials of

10

V

and

-4

V

, respectively. The work done in moving

100

electrons from

P

to

Q

is :

A

9.60 \times {10^{ - 17}}J

B

- 2.24 \times {10^{ - 16}}J

C

2.24 \times {10^{ - 16}}J

D

- 9.60 \times {10^{ - 17}}J

Correct Answer

Option C

Solution

{{{W_{PQ}}} \over q} = \left( {{V_Q} - {V_P}} \right)

\Rightarrow {W_{PQ}} = q\left( {{V_Q} - {V_P}} \right)

= \left( { - 100 \times 1.6 \times {{10}^{ - 19}}} \right)\left( { - 4 - 10} \right)

= + 2.24 \times {10^{ - 16}}J