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 o f 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

Answer 1

Answer: (A)

V

Answer 2

As given in the first condition :
Both conducting spheres are shown.
$V_{in } - V_{out } = \left(\frac{kQ}{r_{1}} \right) - \left(\frac{kQ}{r_{2}}\right)$
$=kQ \left(\frac{1}{r_{1}} - \frac{1}{r_{2}} \right)=V$
$V_{in } -V_{out } = \left(\frac{kQ}{r_{1}} - \frac{4kQ}{r_{2}}\right) - \left(\frac{kQ}{r_{2}} - \frac{4kQ}{r_{2}}\right)$
$= \frac{kQ}{r_{1}} - \frac{kQ}{r_{2}}$
$= kQ \left(\frac{1}{r_{1}} - \frac{1}{r_{2}}\right) = V$
Hence, we also obtain that potential difference does not depend on charge of outer sphere.
$\therefore$ P.d. remains same