Question

Two concentric spherical conducting shells of radii $R$ and $2R$ carry charges $Q$ and $2Q$ respectively. Change in electric potential on the outer shell when both are connected by a conducting wire is:
(Where $K=\frac{1}{4\pi {ϵ}_0}$)

• A. Zero
• B. $\frac{3KQ}{2R}$
• C. $\frac{KQ}{R}$
• D. $\frac{2KQ}{R}$

Answer 1

The correct option is A Zero


Case $1:$ Net potential $\left(V_0\right)$ on the outer shell when both shells are not connected,

$V_0=V_1+V_2$
Here,
$V_1=$Potential on outer shell due to shell $1$,
$V_2=$Potential on outer shell due to shell $2$,

Substituting the values of $V_1$ and $V_2$ from figure we get,
$V_0=\frac{1}{4\pi {ϵ}_0}\left(\frac{Q}{2R}\right)+\frac{1}{4\pi {ϵ}_0}\left(\frac{2Q}{2R}\right)$

$\therefore V_0=\frac{1}{4\pi {ϵ}_0}\left(\frac{3Q}{2R}\right)\dots \dots \left(1\right)$

Case $2:$After connecting, all charges in the inner shell come to outer shell.

Net potential $\left(V_0^{\prime }\right)$ on the outer shell when both shells are connected by conducting wire,

$V_0^{{}^{\prime }}=\frac{1}{4\pi {ϵ}_0}\frac{(Q+2Q)}{2R}$

$\therefore V_0^{\prime }=\frac{1}{4\pi {ϵ}_0}\frac{3Q}{2R}\dots \dots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$,
The change in potential, $V_0^{{}^{\prime }}-V_0=0$

Hence option (a) is the correct answer.