Riemann surface
matlab

(a) Show that the work done to remove the charge $q$ from a distance $r>a$ to infinity against the force, Eq. (2.6), of a grounded conducting sphere is
$$W=\frac{q^2 a}{8 \pi \epsilon_0\left(r^2-a^2\right)}$$
Relate this result to the electrostatic potential, Eq. (2.3), and the energy discussion of Section 1.11.

(a) The force is
$$|F|=\frac{q^2 a}{4 \pi \epsilon_0} \frac{1}{y^3\left(1-a^2 / y^2\right)^2}$$
directed radially inward. The work is
\begin{aligned} W & =-\int_r^{\infty} F d y \ & =\frac{q^2 a}{4 \pi \epsilon_0} \int_r^{\infty} \frac{d y}{y^3\left(1-a^2 / y^2\right)^2} \ & =\frac{q^2 a}{4 \pi \epsilon_0} \int_r^{\infty} \frac{y d y}{\left(y^2-a^2\right)^2} \ & =\frac{q^2 a}{4 \pi \epsilon_0} \int_{r^2-a^2}^{\infty} \frac{d u}{2 u^2} \ & =\frac{q^2 a}{4 \pi \epsilon_0}\left|-\frac{1}{2 u}\right|_{r^2-a^2}^{\infty} \ & =\frac{q^2 a}{8 \pi \epsilon_0\left(r^2-a^2\right)} \end{aligned}
To relate this to earlier results, note that the image charge $q^{\prime}=-(a / r) q$ is located at radius $r^{\prime}=a^2 / r$. The potential energy between the point charge and

its image is
\begin{aligned} P E & =\frac{1}{4 \pi \epsilon_0}\left(\frac{q q^{\prime}}{\left|r-r^{\prime}\right|}\right) \ & =\frac{1}{4 \pi \epsilon_0}\left(\frac{-q^2 a}{r\left(r-a^2 / r\right)}\right) \ & =\frac{1}{4 \pi \epsilon_0}\left(\frac{-q^2 a}{r^2-a^2}\right) \end{aligned}
Result (7) is only half of (8). This would seem to violate energy conservation. It would seem that we could start with the point charge at infinity and allow it to fall in to a distance $r$ from the sphere, liberating a quantity of energy (8), which we could store in a battery or something. Then we could expend an energy equal to (7) to remove the charge back to infinity, at which point we would be back where we started, but we would still have half of the energy saved in the battery. It would seem that we could keep doing this over and over again, storing up as much energy in the battery as we pleased.

I think the problem is with equation (8). The traditional expression $q_1 q_2 / 4 \pi \epsilon_0 r$ for the potential energy of two charges comes from calculating the work needed to bring one charge from infinity to a distance $r$ from the other charge, and it is assumed that the other charge does not move and keeps a constant charge during the process. But in this case one of the charges is a fictitious image charge, and as the point charge $q$ is brought in from infinity the image charge moves out from the center of the sphere, and its charge increases. So the simple expression doesn’t work to calculate the potential energy of the configuration, and we should take (7) to be the correct result.

(b) Repeat the calculation of the work done to remove the charge $q$ against the force, Eq. (2.9), of an isolated charged conducting sphere. Show that the work done is
$$W=\frac{1}{4 \pi \epsilon_0}\left[\frac{q^2 a}{2\left(r^2-a^2\right)}-\frac{q^2 a}{2 r^2}-\frac{q Q}{r}\right] .$$
Relate the work to the electrostatic potential, Eq. (2.8), and the energy discussion of Section 1.11.

(b) In this case there are two image charges: one of the same charge and location as in part a, and another of charge $Q-q^{\prime}$ at the origin. The work needed to remove the point charge $q$ to infinity is the work needed to remove the point charge from its image charge, plus the work needed to remove the point charge from the extra charge at the origin. We calculated the first contribution above. The second contribution is
\begin{aligned} -\int_r^{\infty} \frac{q\left(Q-q^{\prime}\right) d y}{4 \pi \epsilon_0 y^2} & =-\frac{1}{4 \pi \epsilon_0} \int_r^{\infty}\left[\frac{q Q}{y^2}+\frac{q^2 a}{y^3}\right] d y \ & =-\frac{1}{4 \pi \epsilon_0}\left|-\frac{q Q}{y}-\frac{q^2 a}{2 y^2}\right|_r^{\infty} \ & =-\frac{1}{4 \pi \epsilon_0}\left[\frac{q Q}{r}+\frac{q^2 a}{2 r^2}\right] \end{aligned}
so the total work done is
$$W=\frac{1}{4 \pi \epsilon_0}\left[\frac{q^2 a}{2\left(r^2-a^2\right)}-\frac{q^2 a}{2 r^2}-\frac{q Q}{r}\right] .$$

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