Question

Two concentric thin conducting spherical shells having radius R and 2 R are as 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 amount of heat produced after connection is:

• A. $\frac{3Q^2}{4\pi {ϵ}_{0}R}$
• B. $\frac{9Q^2}{16\pi {ϵ}_{0}R}$
• C. $\frac{5}{8}\frac{Q^2}{\pi {ϵ}_{0}R}$
• D. None of these.

Answer 1

The correct option is B $\frac{9Q^2}{16\pi {ϵ}_{0}R}$

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: $V_R=\frac{Q_1}{R}+\frac{Q_2}{2R}$
Potential of the spherical shell having radius 2R: $V_{2R}=\frac{Q_1}{2R}+\frac{Q_2}{2R}$
Since $V_R=V_{2R}$
$\frac{Q_1}{R}+\frac{Q_2}{2R}=\frac{Q_1}{2R}+\frac{Q_2}{2R}$
$\Rightarrow Q_1=0$
But $Q_1+Q_2=Q+(-4Q)=-3Q$
$\Rightarrow 0+Q_2=-3Q$
$\Rightarrow Q_2=-3Q$
Initial energy: $U_i=\frac{Q^2}{8\pi {ϵ}_{0}R}+\frac{(-4Q{)}^2}{8\pi {ϵ}_0\left(2R\right)}=\frac{9Q^2}{8\pi {ϵ}_{0}R}$
Final energy: $U_f=\frac{Q_1^2}{8\pi {ϵ}_{0}R}+\frac{Q_2^2}{8\pi {ϵ}_0\left(2R\right)}=0+\frac{(-3Q{)}^2}{8\pi {ϵ}_{0}2R}=\frac{9Q^2}{16\pi {ϵ}_{0}R}$
Heat produced after connection: $H=U_i-U_f=\frac{9Q^2}{16\pi {ϵ}_{0}R}$