Question

A solid conducting sphere of radius ${}^{\prime }{a}^{\prime }$ having charge $q$ is surrounded by a concentric conducting spherical shell of inner radius $2a$ and outer radius $3a$ as shown in figure. If the amount of heat produced when switch is closed is $\alpha \left(\frac{K{q}^2}{24a}\right)$. Fill the value of $\alpha$.

Answer 1

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

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

As $b\to \infty$

$C=\frac{a}{K}$

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

Before closing the switch charge distribution is given by.

Total energy of system $=$ Energy of capacitor $\left(1\right)+$ Energy of capacitor $\left(2\right)$

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

Capacitance of $2^{nd}$ capacitor $C_2=\frac{3a}{K}$

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

$T.Ei$$=\frac{K{q}^2}{4a}+\frac{K{q}^2}{6a}$

After closing the switch charge distribution is as shown in Fig $3$

Total energy $=0+\frac{q^2}{2\left(\frac{3a}{K}\right)}$

$T.E_f=\frac{K{q}^2}{6a}$

$△T.E=\frac{-K{q}^2}{4a}=T.E_f-T.E_f$

Thus heat produced $=+\frac{K{q}^2}{4a}$

If the amount of heat produced when switch is closed is $6\left(\frac{K{q}^2}{24a}\right)$

Then $\alpha =6$