Question

The surface charge densities of two thin concentric spherical shells are $\sigma$ and $-\sigma$ respectively as shown in the figure. Their radii are $R$ and $2R$. Now they are connected by a thin wire. Potential on either of the shell will be.

• A. $-\frac{3\sigma R}{2{\epsilon }_{\circ }}$
• B. $-\frac{2\sigma R}{3{\epsilon }_{\circ }}$
• C. $-\frac{\sigma R}{2{\epsilon }_{\circ }}$
• D. zero

Answer 1

The correct option is A $-\frac{3\sigma R}{2{\epsilon }_{\circ }}$

Total charge on inner shell is,
$q_i=\left(\sigma \right)\left(4\pi R^2\right)$

Total charge on outer shell is,
$q_o=(-\sigma )\left(4\pi \right)(2R{)}^2$

When they are connected by thin wire, the charge on the inner shell flows to outer shell till both the shells are of same potential.
Hence, net charge on outer shell is
$q=q_i+q_0$

$\Rightarrow q=\left(\sigma \right)\left(4\pi R^2\right)+(-\sigma )\left(4\pi \right)(2R{)}^2$

$\Rightarrow q=-12\sigma \pi R^2$

So, the potential on the either of the shells is,
$V=\frac{1}{4\pi {\epsilon }_o}\frac{q}{\left(2R\right)}$

$\Rightarrow V=\frac{-12\sigma \pi R^2}{8\pi {\epsilon }_{o}R}$

$\therefore V=-\frac{3\sigma R}{2{\epsilon }_o}$

Hence, option (a) is the correct answer.