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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 be $V$. If the shell is now given a charge of $-3Q$, the new potential difference between the same two surfaces is......$V$
$1$
$2$
$4$
$-2$
Solution
(a) In case of a charged conducting sphere
${V_{{\rm{inside}}}} = {V_{{\rm{centre }}}} = {V_{{\rm{surface}}}} = \frac{1}{{4\pi {\varepsilon _o}}}.\frac{q}{R}$, ${V_{{\rm{outside}}}} = \frac{1}{{4\pi {\varepsilon _0}}}.\frac{q}{r}$
If $a$ and $b$ are the radii of sphere and spherical shell respectively, then potential at their surface will be
${V_{{\rm{sphere }}}} = \frac{1}{{4\pi {\varepsilon _0}}}.\frac{Q}{a}$ and ${V_{{\rm{shell}}}} = \frac{1}{{4\pi {\varepsilon _0}}}.\frac{Q}{b}$
$\therefore $$V = {V_{{\rm{sphere }}}} – {V_{{\rm{shell}}}} = \frac{1}{{4\pi {\varepsilon _0}}}.\left[ {\frac{Q}{a} – \frac{Q}{b}} \right]$
Now when the shell is given charge $(-3Q)$, then the potential will be
$V{'_{{\rm{sphere}}}} = \frac{1}{{4\pi {\varepsilon _0}}}\left[ {\frac{Q}{a} + \frac{{( – 3Q)}}{b}} \right],$$V{'_{{\rm{shell}}}} = \frac{1}{{4\pi {\varepsilon _0}}}\left[ {\frac{Q}{b} + \frac{{( – 3Q)}}{b}} \right]$
$\therefore $$V{'_{{\rm{sphere }}}} – V{'_{{\rm{shell}}}} = \frac{1}{{4\pi {\varepsilon _0}}}\left[ {\frac{Q}{a} – \frac{Q}{b}} \right] = V$
