Question

In the figure shown below the conducting shell $A$ having radius $R$ is given charge $Q$ initially. The shell $A$ is connected to shell $B$ of radius $2R$ via switch $S$ and a conducting wire. The switch $S$ is closed at $t=0,$ then charge density on the shell $B$ is
(Assuming separation between $A$ and $B$ is very very large)

• A. $\frac{Q}{48\pi R^2}$
• B. $\frac{Q}{24\pi R^2}$
• C. $\frac{Q}{12\pi R^2}$
• D. $\frac{Q}{6\pi R^2}$

Answer 1

The correct option is A $\frac{Q}{48\pi R^2}$

$V^{\prime }=V$

$\left(\frac{Q-q}{8\pi {\epsilon }_{0}R}\right)=\frac{q}{8\pi {\epsilon }_0\left(2R\right)}$

$Q=3q$

$q=\left(\frac{Q}{3}\right)$

Charge density $=\left(\frac{\text{Charge}}{\text{Area}}\right)$

$=\frac{Q}{3\left(4\pi {\left(2R\right)}^2\right)}$

Charge density $=\frac{Q}{48\pi R^2}$