Question

A spherical shell of radius $R_1$ with a uniform charge $q$ has a point charge $q_0$ at its centre. Find the work performed by the electric forces during the shell expansion from radius $R_1$ to radius $R_2$.

Answer 1

Initially, energy of the system, $U_i=W_1+W_{12}$ where, $W_1$ is the self energy and $W_{12}$ is the mutual energy.
So, $U_i=\frac{1}{2}\frac{q^2}{4\pi {ϵ}_0{R}_1}+\frac{q{q}_0}{4\pi {ϵ}_0{R}_1}$
and on expansion, energy of the system,
$U_f=W_1^{\prime }+W_{12}^{\prime }$
$=\frac{1}{2}\frac{q^2}{4\pi {ϵ}_0{R}_2}+\frac{q{q}_0}{4\pi {ϵ}_0{R}_2}$
Now, work done by the field force, $A$ equals the decrement in the electrical energy,
i.e. $A=(U_i-U_f)=\frac{q(q_0+q/2)}{4\pi {ϵ}_0}(\frac{1}{R_1}-\frac{1}{R_2})$
Alternate : The work of electric forces is equal to the decrease in electric energy of the system,
$A=U_i-U_f$
In order to find the difference $U_i-U_f$, we note that upon expansion of the shell, the electric field and hence the energy localized in it, changed only in the hatched spherical layer consequently (Fig.).
$U_i-U_f={\int }_{R_1}^{R_2}\frac{{ϵ}_0}{2}(E_1^2-E_2^2)\cdot 4\pi r^{2}dr$
where $E_1$ and $E_2$ are the field intensities (in the hatched region at a distance $r$ from the centre of the system) before and after the expansion of the shell. By using Gauss' theorem, we find
$E_1=\frac{1}{4\pi {ϵ}_0}\frac{q+q_0}{r^2}$ and $E_2=\frac{1}{4\pi {ϵ}_0}\frac{q_0}{r^2}$
As a result of integration, we obtain
$A=\frac{q(q_0+q/2)}{4\pi {ϵ}_0}(\frac{1}{R_1}-\frac{1}{R_2})$.