Question 202 of 311

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 $–4Q,$ the new potential difference between the same two surfaces is:

A$–2V$B$2V$C$V$D$4V$

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Answer: (C) $V$

First condition:

Both conducting spheres are shown.
${V}_{in}-{V}_{out}=\left(\dfrac{KQ}{{r}_{1}}\right)-\left(\dfrac{KQ}{{r}_{2}}\right)$
$\Rightarrow V=KQ\left(\dfrac{1}{{r}_{1}}-\dfrac{1}{{r}_{2}}\right)$
Second condition:

Shell is now given charge $-4Q$ .
${V}_{in}-{V}_{out}=\left(\dfrac{KQ}{{r}_{1}}-\dfrac{4KQ}{{r}_{2}}\right)-\left(\dfrac{KQ}{{r}_{2}}-\dfrac{4KQ}{{r}_{2}}\right)$
$=\dfrac{KQ}{{r}_{1}}-\dfrac{KQ}{{r}_{2}}$
$=KQ\left(\dfrac{1}{{r}_{1}}-\dfrac{1}{{r}_{2}}\right)=V$