Q

Solve this problem - Electrostatics - JEE Main-6

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 :

• Option 1)

$2\; V$

• Option 2)

$V$

• Option 3)

$-2\; V$

• Option 4)

$4\; V$

Views

Initially

$V_{1}=\frac{kQ}{a}+\frac{k(-Q+Q)}{b}$

$V_{2}=\frac{kQ}{b}+\frac{k(-Q+Q)}{b}$

and $\Delta V_{I} =V_{1}-V_{2}=KQ(\frac{1}{a}-\frac{1}{b})=V\; \; \; \; \; (1)$

After giving $-4Q$ to shell

$V_{1}=\frac{kQ}{a}+k\frac{(-Q-3Q)}{b}$

$V_{2}=\frac{kQ}{b}+k\frac{(-Q-3Q)}{b}$

$\Delta V _{II}=V_{1}-V_{2}=KQ(\frac{1}{a}-\frac{1}{b})=V$

$\Delta V_{II} =\Delta V _{I} =V$

Option 1)

$2\; V$

Option 2)

$V$

Option 3)

$-2\; V$

Option 4)

$4\; V$

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