Question

A solid conducting sphere of radius a having a charge $q$ is surrounded by a concentric conducting spherical shell of inner radius $2a$ and outer radius $3a$ as shown in figure. Find the amount of heat produced when switch is closed
$\left(K=\frac{1}{4\pi {\epsilon }_0}\right)$

• A. $K=\frac{k{q}^2}{2a}$
• B. $K=\frac{k{q}^2}{3a}$
• C. $K=\frac{k{q}^2}{4a}$
• D. $K=\frac{k{q}^2}{6a}$

Answer 1

The correct option is C $K=\frac{k{q}^2}{4a}$

Capacitance of a spherical capacitor having inner radius of sphere a and radius of outer sphere b is :

$C=\frac{ab}{K(b-a)}$

$b\to \infty C=\frac{a}{k}$

And energy of capacitor $=\frac{q^2}{2C}$

Before closing the switch charge distribution is given by

Total energy of system = Energy of capacitor (1) + Energy of capacitor (2)

Capacitance of 1st1st capacitor $C_1=\frac{a\times 2a}{K(2a-a)}=\frac{2a}{K}$

capacitance of 2nd capacitor is $C_2=\frac{3a}{K}$

$totalenergy=\frac{q^2}{2\left(\frac{2a}{K}\right)}+\frac{q^2}{2\left(\frac{3a}{K}\right)}$

$totalenergy=\frac{{Kq}^2}{4a}+\frac{K{q}^2}{6a}$

After closing the switch charge distribution is

$totalenergyfinal=\frac{q^2}{2\left(\frac{3a}{K}\right)}=\frac{K{q}^2}{6a}$

$\Delta TE={TE}_f-{TE}_i=\frac{K{q}^2}{6a}-(\frac{K{q}^2}{6a}+\frac{K{q}^2}{4a})=-\frac{K{q}^2}{4a}heatproduced=+\frac{K{q}^2}{4a}$

Option C is correct.

1t1capacitor