Question

A solid conducting sphere having a charge Q is surrounded by an uncharged concentric 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 beV. If the shell is now given a charge of – 3Q, the new potential difference between the two surfaces is

• A. V
• B. 2V
• C. 4V
• D. –2V

Answer 1

Answer: (A)

V

Answer 2

If a and b are radii of spheres and spherical shell respectively, potential at their surfaces will be

$V_{\text{sphere}} = \frac{1}{4\pi\varepsilon_{0}}.\frac{Q}{a}$ and $V_{\text{shell}} = \frac{1}{4\pi\varepsilon_{0}}.\frac{Q}{b}$ and so according to the given problem.

$V = V_{\text{sphere}} - V_{\text{shell}}$ $= \frac{Q}{4\pi\varepsilon_{0}}\left\lbrack \frac{1}{a} - \frac{1}{b} \right\rbrack$ …. (i)

Now when the shell is given a charge –3Q the potential at its surface and also inside will change by

$V_{0} = \frac{1}{4\pi\varepsilon_{0}}\left\lbrack - \frac{3Q}{b} \right\rbrack$

So that now $V_{\text{sphere}} = \frac{1}{4\pi\varepsilon_{0}}\left\lbrack \frac{Q}{a} - \frac{3Q}{b} \right\rbrack$ and

$V_{\text{shell}} = \frac{1}{4\pi\varepsilon_{0}}\left\lbrack \frac{Q}{b} - \frac{3Q}{b} \right\rbrack$

hence $V_{\text{sphere}} - V_{\text{shell}} = \frac{Q}{4\pi\varepsilon_{0}}\left\lbrack \frac{1}{a} - \frac{1}{b} \right\rbrack = V$