Question

A conducting sphere of radius $R$ carrying charge $Q$ lies inside an uncharged conducting shell of radius $2R$. If they are joined by a metal wire, the amount of heat that will be produced is:

• A. $\frac{1}{4\pi {\in }_0}.\frac{Q^2}{4R}$
• B. $\frac{1}{4\pi {\in }_0}.\frac{Q^2}{2R}$
• C. $\frac{1}{4\pi {\in }_0}.\frac{Q^2}{R}$
• D. $\frac{2}{4\pi {\in }_0}.\frac{Q^2}{3R}$

Answer 1

The correct option is A $\frac{1}{4\pi {\in }_0}.\frac{Q^2}{4R}$

When connected by metal wire, whole of the charge would be transferred to the outer conducting shell.

Capacitance of the two system are $C_i=4\pi {ϵ}_0\left(R\right)$

and $C_f=4\pi {ϵ}_0\left(2R\right)$

Thus initial energy $=\frac{Q^2}{2C_i}$

Final energy $=\frac{Q^2}{2C_f}$

The energy amount of heat produced is equal to the change in potential energy of the system $=\frac{Q^2}{2C_i}-\frac{Q^2}{2C_f}=\frac{1}{4\pi {ϵ}_0}.\frac{Q^2}{4R}$