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 :

Electrostatics PYQ — JEE Main 2019 · 8 Apr · Shift 1 · Q56 with Solution

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Correct answer: A

Initial configuration

Let the radius of the solid conducting sphere be $a$ , and the inner and outer radii of the hollow conducting shell be $b$ and $c$ respectively.

Charge on solid sphere = $Q$
Hollow shell is initially uncharged.

Because of electrostatic induction:

Charge on inner surface of shell = $-Q$
Charge on outer surface of shell = $+Q$

Initial potential difference

We need the potential difference between:

surface of solid sphere $(r=a)$
outer surface of shell $(r=c)$

Now compute potentials.

Potential at the surface of the solid sphere

Potential at $r=a$ is due to:

solid sphere charge $Q$ : $\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{a}$
inner surface charge $-Q$ : $\dfrac{1}{4\pi\varepsilon_0}\dfrac{-Q}{b}$
outer surface charge $+Q$ : $\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{c}$

So,

V_a = \frac{1}{4\pi\varepsilon_0}\left(\frac{Q}{a}-\frac{Q}{b}+\frac{Q}{c}\right)

Potential at the outer surface of shell

At $r=c$ :

due to $Q$ at center: $\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{c}$
due to inner surface charge $-Q$ : $\dfrac{1}{4\pi\varepsilon_0}\dfrac{-Q}{c}$
due to outer surface charge $+Q$ : $\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{c}$

Thus,

V_c = \frac{1}{4\pi\varepsilon_0}\frac{Q}{c}

Therefore the potential difference is

V_a - V_c = \frac{1}{4\pi\varepsilon_0}\left(\frac{Q}{a}-\frac{Q}{b}\right)

This is given as $V$ .

So,

V = \frac{1}{4\pi\varepsilon_0}\left(\frac{Q}{a}-\frac{Q}{b}\right)

After giving the shell charge $-4Q$

Now total charge on shell becomes $-4Q$ .

The solid sphere still has charge $Q$ , so to keep electric field inside the conductor zero:

charge on inner surface of shell must still be $-Q$

Hence remaining charge on outer surface of shell is:

-4Q - (-Q) = -3Q

So now:

solid sphere: $Q$
inner surface of shell: $-Q$
outer surface of shell: $-3Q$

New potentials

Potential at the surface of the solid sphere

V_a' = \frac{1}{4\pi\varepsilon_0}\left(\frac{Q}{a}-\frac{Q}{b}-\frac{3Q}{c}\right)

Potential at the outer surface of shell

At $r=c$ :

due to $Q$ : $\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{c}$
due to inner surface charge $-Q$ : $\dfrac{1}{4\pi\varepsilon_0}\dfrac{-Q}{c}$
due to outer surface charge $-3Q$ : $\dfrac{1}{4\pi\varepsilon_0}\dfrac{-3Q}{c}$

So,

V_c' = \frac{1}{4\pi\varepsilon_0}\left(\frac{Q}{c}-\frac{Q}{c}-\frac{3Q}{c}\right) = \frac{1}{4\pi\varepsilon_0}\left(-\frac{3Q}{c}\right)

New potential difference

V_a' - V_c' = \frac{1}{4\pi\varepsilon_0}\left(\frac{Q}{a}-\frac{Q}{b}-\frac{3Q}{c}+\frac{3Q}{c}\right)

V_a' - V_c' = \frac{1}{4\pi\varepsilon_0}\left(\frac{Q}{a}-\frac{Q}{b}\right)

But this is exactly the old value $V$ .

Hence the new potential difference remains unchanged:

\boxed{V}

Option check

A: $V$ ✅
B: $2V$ ❌
C: $-2V$ ❌
D: $4V$ ❌

Therefore, the correct answer is A.

More from Electrostatics

View written solutionFree

Correct answer: A

Initial configuration

Let the radius of the solid conducting sphere be $a$ , and the inner and outer radii of the hollow conducting shell be $b$ and $c$ respectively.

Charge on solid sphere = $Q$
Hollow shell is initially uncharged.

Because of electrostatic induction:

Charge on inner surface of shell = $-Q$
Charge on outer surface of shell = $+Q$

Initial potential difference

We need the potential difference between:

surface of solid sphere $(r=a)$
outer surface of shell $(r=c)$

Now compute potentials.

Potential at the surface of the solid sphere

Potential at $r=a$ is due to:

solid sphere charge $Q$ : $\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{a}$
inner surface charge $-Q$ : $\dfrac{1}{4\pi\varepsilon_0}\dfrac{-Q}{b}$
outer surface charge $+Q$ : $\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{c}$

So,

V_a = \frac{1}{4\pi\varepsilon_0}\left(\frac{Q}{a}-\frac{Q}{b}+\frac{Q}{c}\right)

Potential at the outer surface of shell

At $r=c$ :

due to $Q$ at center: $\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{c}$
due to inner surface charge $-Q$ : $\dfrac{1}{4\pi\varepsilon_0}\dfrac{-Q}{c}$
due to outer surface charge $+Q$ : $\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{c}$

Thus,

V_c = \frac{1}{4\pi\varepsilon_0}\frac{Q}{c}

Therefore the potential difference is

V_a - V_c = \frac{1}{4\pi\varepsilon_0}\left(\frac{Q}{a}-\frac{Q}{b}\right)

This is given as $V$ .

So,

V = \frac{1}{4\pi\varepsilon_0}\left(\frac{Q}{a}-\frac{Q}{b}\right)

After giving the shell charge $-4Q$

Now total charge on shell becomes $-4Q$ .

The solid sphere still has charge $Q$ , so to keep electric field inside the conductor zero:

charge on inner surface of shell must still be $-Q$

Hence remaining charge on outer surface of shell is:

-4Q - (-Q) = -3Q

So now:

solid sphere: $Q$
inner surface of shell: $-Q$
outer surface of shell: $-3Q$

New potentials

Potential at the surface of the solid sphere

V_a' = \frac{1}{4\pi\varepsilon_0}\left(\frac{Q}{a}-\frac{Q}{b}-\frac{3Q}{c}\right)

Potential at the outer surface of shell

At $r=c$ :

due to $Q$ : $\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{c}$
due to inner surface charge $-Q$ : $\dfrac{1}{4\pi\varepsilon_0}\dfrac{-Q}{c}$
due to outer surface charge $-3Q$ : $\dfrac{1}{4\pi\varepsilon_0}\dfrac{-3Q}{c}$

So,

V_c' = \frac{1}{4\pi\varepsilon_0}\left(\frac{Q}{c}-\frac{Q}{c}-\frac{3Q}{c}\right) = \frac{1}{4\pi\varepsilon_0}\left(-\frac{3Q}{c}\right)

New potential difference

V_a' - V_c' = \frac{1}{4\pi\varepsilon_0}\left(\frac{Q}{a}-\frac{Q}{b}-\frac{3Q}{c}+\frac{3Q}{c}\right)

V_a' - V_c' = \frac{1}{4\pi\varepsilon_0}\left(\frac{Q}{a}-\frac{Q}{b}\right)

But this is exactly the old value $V$ .

Hence the new potential difference remains unchanged:

\boxed{V}

Option check

A: $V$ ✅
B: $2V$ ❌
C: $-2V$ ❌
D: $4V$ ❌

Therefore, the correct answer is A.

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