A conducting sphere of radius $R$ carrying charge $Q$ lies inside an uncharged conducting shell of radius $2…

A conducting sphere of radius $R$ carrying charge $Q$ lies inside an uncharged conducting shell of radius $2 R$. If they are joined by a metal wire, the amount of heat that will be produced is
  1. $\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q^2}{4 R}$
  2. $\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q^2}{2 R}$
  3. $\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q^2}{R}$
  4. $\frac{2}{4 \pi \varepsilon_0} \cdot \frac{Q^2}{3 R}$

Solution

The capacitances of two are $C_1=4 \pi \varepsilon_0 R$ and $C_2=4 \pi \varepsilon_0(2 R)$ Initial energy $=E_i=\frac{Q^2}{2 C_1}$ Final energy $=E_f=\frac{Q^2}{2 C_2}$ Heat produced $=E_i-E_f$ $\begin{aligned} & =\frac{Q^2}{2}\left[\frac{1}{4 \pi \varepsilon_0 R}-\frac{1}{2 \times 4 \pi \varepsilon_0 R}\right] \\ & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q^2}{2 R}\left[1-\frac{1}{2}\right] \\ & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q^2}{4 R}\end{aligned}$

Asked in: MHT CET Full Test 9

Practice more Electrostatics questions on Aicharya