Question

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. $V$
• B. $2V$
• C. $-2V$
• D. $4V$

Answer 1

The correct option is A $V$


As given in the first condition,
both conducting spheres are shown.
$V_{in}-V_{out}=\left(\frac{kQ}{r_1}\right)-\left(\frac{kQ}{r_2}\right)$
$=kQ(\frac{1}{r_1}-\frac{1}{r_2})=V$
In the second condition,
Shell is now given charge $-4Q$.
$V_{in}-V_{out}=(\frac{kQ}{r_1}-\frac{4kQ}{r_2})-(\frac{kQ}{r_2}-\frac{4kQ}{r_2})$
$=\frac{kQ}{r_1}-\frac{kQ}{r_2}$
$=kQ(\frac{1}{r_1}-\frac{1}{r_2})=V$
Hence, we also obtain that potential difference does not depend on charge of outer sphere.
$\therefore$ potential difference remains same.
Note: The beauty of this type of system is that no matter how much charge you provide on outer surface, always the potential difference between surface of solid sphere and outer surface of shell remains constant.