Question

Two concentric thin conducting spherical shells having radius $R$ and $2R$ are shown in figure. A charge $+Q$ is given to shell $A$ and $-4Q$ is given to shell $B$. Now shell $A$ and $B$ are connected by a thin conducting wire, then the final charge on the sphere $B$ will be :

• A. $\frac{-3Q}{2}$
• B. $-4Q$
• C. $-3Q$
• D. $2Q$

Answer 1

The correct option is C $-3Q$

After connecting both the spherical shells by a conducting wire, they will have the same potential. Let $Q_1$​ and $Q_2$​ be the charges on the inner sphere and the outer sphere respectively after they are connected.



Potential of the spherical shell having radius $R$ :
$V_R=\frac{k{Q}_1}{R}+\frac{k{Q}_2}{2R}$​​

Potential of the spherical shell having radius $2R$ :
$V_{2R}=\frac{k{Q}_1}{2R}+\frac{k{Q}_2}{2R}$​​

Since, $V_R=V_{2R}$​

$\Rightarrow \frac{k{Q}_1}{R}+\frac{k{Q}_2}{2R}=\frac{k{Q}_1}{2R}+\frac{k{Q}_2}{2R}$

$\Rightarrow Q_1=0$

But, $Q_1+Q_2=Q+(-4Q)=-3Q$

$\Rightarrow 0+Q_2=-3Q$

Or, $Q_2=-3Q$