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

• A. V
• B. 2V
• C. -2V
• D. 4V

Answer 1

Answer: (A)

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