Electrostatics — Page 5 | JEE Physics

Q41

A solid conducting sphere, having a charge Q, is surrounded by an uncharged conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of –4 Q, the new potential difference between the same two surfaces is :

A V

B 2V

C –2V

D 4V

Correct Answer

Option A

Solution

Initially when uncharged shell encloses charge Q, charge distribution due to induction will be as shown, The potential on surface of inner shell is

{V_A} = {{kQ} \over a} + {{k( - Q)} \over b} + {{kQ} \over b}

..... (i) where, k = proportionality constant. Potential on surface of outer shell is

{V_B} = {{kQ} \over b} + {{k( - Q)} \over b} + {{kQ} \over b}

..... (ii) Then, potential difference is

\Delta {V_{AB}} = {V_A} - {V_B} = kQ\left( {{1 \over a} - {1 \over b}} \right)

Given,

\Delta {V_{AB}} = V

So,

kQ\left( {{1 \over a} - {1 \over b}} \right) = V

....... (iii) Finally after giving charge $-$ 4Q to outer shell, potential difference will be

\Delta {V_{AB}} = {V_A} - {V_B}

= \left( {{{kQ} \over a} + {{k( - 4Q)} \over b}} \right) - \left( {{{kQ} \over b} + {{k( - 4Q)} \over b}} \right)

= kQ\left( {{1 \over a} - {1 \over b}} \right) = V

[from Eq. (iii)] Hence, we obtain that potential difference does not depend on the charge of outer sphere, hence potential difference remains same.

Q42

The bob of a simple pendulum has mass 2g and a charge of 5.0 μC. It is at rest in a uniform horizontal electric field of intensity 2000 V/m. At equilibrium, the angle that the pendulum makes with the vertical is : (take g = 10 m/s2)

A tan–1(5.0)

B tan–1(2.0)

C tan–1(0.5)

D tan–1(0.2)

Correct Answer

Option C

Solution

Tcos $\theta$ = mg Tsin $\theta$ = qE tan $\theta$ =

{{qE} \over {mg}}

tan $\theta$ =

{{5 \times {{10}^{ - 16}} \times 2000} \over {2 \times {{10}^{ - 3}} \times 10}} = {1 \over 2}

$\Rightarrow$

{\tan ^{ - 1}}\left( {{1 \over 2}} \right) = {\tan ^{ - 1}}\left( {0.5} \right)

Q43

Choose the incorrect statement : (1) The electric lines of force entering into a Gaussian surface provide negative flux. (2) A charge 'q' is placed at the centre of a cube. The flux through all the faces will be the same. (3) In a uniform electric field net flux through a closed Gaussian surface containing no net charge, is zero. (4) When electric field is parallel to a Gaussian surface, it provides a finite non-zero flux. Choose the most appropriate answer from the options given below

A (3) and (4) only

B (2) and (4) only

C (4) only

D (1) and (3) only

Correct Answer

Option C

Solution

Since,

\phi = \overrightarrow E \,.\,\overrightarrow A = EA\cos \theta

$\theta$ = 90

^\circ

$\therefore$ $\phi$ = 0

Q44

An electric field of 1000 V/m is applied to an electric dipole at angle of 45o. The value of electric dipole moment is 10–29 C.m. What is the potential energy of the electric dipole?

A - 7

\times

10–27 J

B

-

9

\times

10–20 J

C

-

10

\times

10–29 J

D

-

20

\times

10–18 J

Correct Answer

Option A

Solution

U = $-$

\overrightarrow P .\overrightarrow E

= $-$ PE cos $\theta$ = $-$ (10 $-$ 29) (103) cos 45o = $-$ 0.707 $\times$ 10 $-$ 26 J = $-$ 7 $\times$ 10 $-$ 27 J.

Q45

Consider the force F on a charge 'q' due to a uniformly charged spherical shell of radius R carrying charge Q distributed uniformly over it. Which one of the following statements is true for F, if 'q' is placed at distance r from the centre of the shell?

A

{1 \over {4\pi {\varepsilon _0}}}{{qQ} \over {{R^2}}} > F > 0

for r < R

B

F = {1 \over {4\pi {\varepsilon _0}}}{{qQ} \over {{r^2}}}

for r > R

C

F = {1 \over {4\pi {\varepsilon _0}}}{{qQ} \over {{r^2}}}

for all r

D

F = {1 \over {4\pi {\varepsilon _0}}}{{qQ} \over {{R^2}}}

for r < R

Correct Answer

Option B

Solution

Inside the shell for r < R E = 0 hence F = 0 Outside the shell E =

{1 \over {4\pi {\varepsilon _0}}}{Q \over {{r^2}}}

hence F =

{1 \over {4\pi {\varepsilon _0}}}{{Qq} \over {{r^2}}}

for r > R

Q46

Two identical electric point dipoles have dipole moments

{\overrightarrow p _1} = p\widehat i

and

{\overrightarrow p _2} = - p\widehat i

and are held on the x axis at distance '

a

' from each other. When released, they move along the x-axis with the direction of their dipole moments remaining unchanged. If the mass of each dipole is 'm', their speed when they are infinitely far apart is :

A

{p \over a}\sqrt {{3 \over {2\pi { \in _0}ma}}}

B

{p \over a}\sqrt {{1 \over {\pi { \in _0}ma}}}

C

{p \over a}\sqrt {{1 \over {2\pi { \in _0}ma}}}

D

{p \over a}\sqrt {{2 \over {\pi { \in _0}ma}}}

Correct Answer

Option C

Solution

Using energy conservation : KEi + PEi = KEf + PEf 0 +

{{2KP} \over {{a^3}}} \times P

=

{1 \over 2}m{v^2} \times 2 + 0

$\Rightarrow$ v =

\sqrt {{{2{P^2}} \over {4\pi {\varepsilon _0}{a^3}m}}}

=

{P \over a}\sqrt {{1 \over {2\pi {\varepsilon _0}am}}}

Q47

Ten charges are placed on the circumference of a circle of radius R with constant angular separation between successive charges. Alternate charges 1, 3, 5, 7, 9 have charge (+q) each, while 2, 4, 6, 8, 10 have charge (–q) each. The potential V and the electric field E at the centre of the circle are respectively. (Take V = 0 at infinity)

A V = 0; E = 0

B

V = {{10q} \over {4\pi {\varepsilon _0}R}}

;

E = {{10q} \over {4\pi {\varepsilon _0}{R^2}}}

C

V = {{10q} \over {4\pi {\varepsilon _0}R}}

; E = 0

D V = 0;

E = {{10q} \over {4\pi {\varepsilon _0}{R^2}}}

Correct Answer

Option A

Solution

Net charge = 5q - 5q = 0 Potential of centre = V =

{{K\sum q } \over r}

VC =

{{K\left( 0 \right)} \over r}

= 0 Let E be electric field produced by each charge at the centre, then resultant electric field will be EC = 0, Since equal electric field vectors are acting at equal angle so their resultant is equal to zero.(

From symmetric property of vector)

Q48

A particle of charge q and mass m is subjected to an electric field E = E0 (1 –

a

x2) in the x-direction, where

a

and E0 are constants. Initially the particle was at rest at x = 0. Other than the initial position the kinetic energy of the particle becomes zero when the distance of the particle from the origin is :

A

a

B

\sqrt {{2 \over a}}

C

\sqrt {{3 \over a}}

D

\sqrt {{1 \over a}}

Correct Answer

Option C

Solution

W = \Delta KE

As inital and final kinetic energy both are zero so

\Delta KE

= 0 $\therefore$ W = 0 $\Rightarrow$

\int\limits_0^x {Fdx} = 0

$\Rightarrow$

q\int\limits_0^x {{E_0}\left( {1 - a{x^2}} \right)dx} = 0

$\Rightarrow$

q{E_0}\left[ {\int\limits_0^x {dx - a} \int\limits_0^x {{x^2}dx} } \right] = 0

$\Rightarrow$

q{E_0}\left[ {x - {{a{x^3}} \over 3}} \right] = 0

$\Rightarrow$

x\left( {1 - {{a{x^2}} \over 3}} \right) = 0

$\Rightarrow$

x = 0,

{1 - {{a{x^2}} \over 3}}

= 0 $\Rightarrow$

{{{a{x^2}} \over 3}}

= 1 $\Rightarrow$

{x = \sqrt {{3 \over a}} }

Q49

Concentric metallic hollow spheres of radii R and 4R hold charges Q1 and Q2 respectively. Given that surface charge densities of the concentric spheres are equal, the potential difference V(R) – V(4R) is :

A

{{3{Q_2}} \over {4\pi {\varepsilon _0}R}}

B

{{3{Q_1}} \over {4\pi {\varepsilon _0}R}}

C

{{3{Q_1}} \over {16\pi {\varepsilon _0}R}}

D

{{{Q_2}} \over {4\pi {\varepsilon _0}R}}

Correct Answer

Option C

Solution

\sigma = {{{Q_1}} \over {4\pi {R^2}}} = {{{Q_2}} \over {4\pi 16{R^2}}}

$\Rightarrow$

16{Q_1} = {Q_2}

$\therefore$

{V_R} - {V_{4R}} = {{K{Q_1}} \over R} + {{K{Q_2}} \over {4R}} - {{K{Q_1}} \over {4R}} - {{K{Q_2}} \over {4R}}

= {{3K{Q_1}} \over {4R}} = {{3{Q_1}} \over {16\pi {\varepsilon _0}R}}

[As K =

{1 \over {4\pi {\varepsilon _0}}}

]

Q50

Two isolated conducting spheres S1 and S2 of radius

{2 \over 3}R

and

{1 \over 3}R

have 12

\mu

C and –3

\mu

C charges, respectively, and are at a large distance from each other. They are now connected by a conducting wire. A long time after this is done the charges on S1 and S2 are respectively :

A 4.5

\mu

C on both

B +4.5

\mu

C and –4.5

\mu

C

C 6

\mu

C and 3

\mu

C

D 3

\mu

C and 6

\mu

C

Correct Answer

Option C

Solution

q1 + q2 = 12 - 3 = 9 $\mu$ C .....(1) and V1 = V2 $\Rightarrow$

{{K{q_1}} \over {{{2R} \over 3}}} = {{K{q_2}} \over {{R \over 3}}}

$\Rightarrow$ q1 = 2q2 .....(

2) Now puttion the value of (2) in (1), we get 2q2 + q2 = 9 $\Rightarrow$ 3q2 = 9 $\Rightarrow$ q2 = 3 $\mu$ C $\Rightarrow$ q1 = 6 $\mu$ C