A point charge $q$ is at the center of an uncharged conducting spherical shell of finite thickness, with inner and outer radii $a$ and $b$ respectively. Find the work done on the system when $q$ is removed from its original position to a very large distance from the conducting shell, through a small hole in it.
My attempt:
Idea was to calculate the energy of the initial configuration and negative of that would be the work required. Trying to use $ W = \frac {\epsilon_0}{2} \int E^2 d\tau $ didn't work as the integral goes to infinity at r=0. So I used $ W =\frac 12 \int \sigma V ds$ where $\sigma$ is the induced charge on the two surface at radii a and b.
$$ W= \frac {q}{8\pi \epsilon_0b} \left(\int _{r=a} \sigma_a da + \int_{r=b} \sigma_b da \right)
$$
which come out to be zero as each integral is equal to $q$ in magnitude and opposite in sign. I think this is wrong. How do I solve this problem?
EDIT:
I think I forgot the terms due to the potential of induced charges at the center.
$$ W= \frac {q}{4\pi \epsilon_0b} \left(\int _{r=a} \sigma_a da + \int_{r=b} \sigma_b da \right) + \frac {q^2}{4\pi \epsilon_0b} – \frac {q^2}{4\pi \epsilon_0a}
$$
This comes out as $$ W= \frac {q^2}{4\pi \epsilon_0} \left(\frac1b – \frac1a \right)
$$
External work done is negative of this so that a positive work is done in taking the charge to infinity. Is this correct?
Best Answer
Why do not you try this?
The shell is always an equipotential object. When the charge is inside the shell, the potential of the shell is $kq/R$ with $R$ the radius of the shell, so the energy of the charge is $kq^2/R$.
When the charge moves to infinity, the potential energy of it is zero. Therefore, the work is $kq^2/R$.