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

Answer 1

Answer: (A)

The capacitances of two are $C_1=4\pi {\epsilon }_{0}R$ and $C_2=4\pi {\epsilon }_0\left(2R\right)$.
Initial energy = $E_i=\frac{Q^2}{2C_1}$
Final energy = $E_f=\frac{Q^2}{2C_2}$
Heat produced = $E_i-E_f$
$=\frac{Q^2}{2}\left[\frac{1}{4\pi {\epsilon }_{0}R}-\frac{1}{2\times 4\pi {\epsilon }_{0}R}\right]$
$=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Q^2}{2R}\left[1-\frac{1}{2}\right]$
$=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Q^2}{4R}$