Riemann surface
matlab

A point charge $q$ is brought to a position a distance $d$ away from an infinite plane conductor held at zero potential. Using the method of images, find:
(a) the surface-charge density induced on the plane, and plot it;
(b) the force between the plane and the charge by using Coulomb’s law for the force between the charge and its image;
(c) the total force acting on the plane by integrating $\sigma^2 / 2 \epsilon_0$ over the whole plane;

(a) We’ll take $d$ to be in the $z$ direction, so the charge $q$ is at $(x, y, z)=(0,0, d)$. The image charge is $-q$ at $(0,0,-d)$. The potential at a point $\mathbf{r}$ is
$$\Phi(\mathbf{r})=\frac{q}{4 \pi \epsilon_0}\left[\frac{1}{|\mathbf{r}-d \mathbf{k}|}-\frac{1}{|\mathbf{r}+d \mathbf{k}|}\right]$$
The surface charge induced on the plane is found by differentiating this:
1
Homer Reid’s Solutions to Jackson Problems: Chapter 2
2
\begin{aligned} \sigma & =-\left.\epsilon_0 \frac{d \Phi}{d z}\right|{z=0} \ & =-\left.\frac{q}{4 \pi}\left[\frac{-(z-d)}{|\mathbf{r}+d \mathbf{k}|^3}+\frac{(z+d)}{|\mathbf{r}+d \mathbf{k}|^3}\right]\right|{z=0} \ & =-\frac{q d}{2 \pi\left(x^2+y^2+d^2\right)^{3 / 2}} \end{aligned}
We can check this by integrating this over the entire $x y$ plane and verifying that the total charge is just the value $-q$ of the image charge:
\begin{aligned} & \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \sigma(x, y) d x d y=-\frac{q d}{2 \pi} \int_0^{\infty} \int_0^{2 \pi} \frac{r d \psi d r}{\left(r^2+d^2\right)^{3 / 2}} \ & =-q d \int_0^{\infty} \frac{r d r}{\left(r^2+d^2\right)^{3 / 2}} \ & =-\frac{q d}{2} \int_{d^2}^{\infty} u^{-3 / 2} d u \ & =-\frac{q d}{2}\left|-2 u^{-1 / 2}\right|_{d^2}^{\infty} \ & =-q \ & \end{aligned}

(b) The point of this problem is that, for points above the $z$ axis, it doesn’t matter whether there is a charge $-q$ at $(0,0, d)$ or an infinite grounded sheet at $z=0$. Physics above the $z$ axis is exactly the same whether we have the charge or the sheet. In particular, the force on the original charge is the same whether we have the charge or the sheet. That means that, if we assume the sheet is present instead of the charge, it will feel a reaction force equal to what the image charge would feel if it were present instead of the sheet. The force on the image charge would be just $F=q^2 / 16 \pi \epsilon_0 d^2$, so this must be what the sheet feels.
(c) Total force on sheet
\begin{aligned} & =\frac{1}{2 \epsilon_0} \int_0^{\infty} \int_0^{2 \pi} \sigma^2 d A \ & =\frac{q^2 d^2}{4 \pi \epsilon_0} \int_0^{\infty} \frac{r d r}{\left(r^2+d^2\right)^3} \ & =\frac{q^2 d^2}{8 \pi \epsilon_0} \int_{d^2}^{\infty} u^{-3} d u \ & =\frac{q^2 d^2}{8 \pi \epsilon_0}\left|-\frac{1}{2} u^{-2}\right|_{d^2}^{\infty} \ & =\frac{q^2 d^2}{8 \pi \epsilon_0}\left[\frac{1}{2} d^{-4}\right] \end{aligned}

$$=\frac{q^2}{16 \pi \epsilon_0 d^2}$$
in accordance with the discussion and result of part b.

(d) the work necessary to remove the charge $q$ from its position to infinity;
(e) the potential energy between the charge $q$ and its image (compare the answer to part $d$ and discuss).
(f) Find the answer to part d in electron volts for an electron originally one angstrom from the surface.

(d) Work required to remove charge to infinity
\begin{aligned} & =\frac{q^2}{4 \pi \epsilon_0} \int_d^{\infty} \frac{d z}{(z+d)^2} \ & =\frac{q^2}{4 \pi \epsilon_0} \int_{2 d}^{\infty} u^{-2} d u \ & =\frac{q^2}{4 \pi \epsilon_0} \frac{1}{2 d} \ & =\frac{q^2}{8 \pi \epsilon_0 d} \end{aligned}
(e) Potential energy between charge and its image
$$=\frac{q^2}{8 \pi \epsilon_0 d}$$
equal to the result in part $d$.
(f)
\begin{aligned} \frac{q^2}{8 \pi \epsilon_0 d} & =\frac{\left(1.6 \cdot 10^{-19} \text { coulombs }\right)^2}{8 \pi\left(8.85 \cdot 10^{-12} \text { coulombs } \mathrm{V}^{-1} \mathrm{~m}^{-1}\right)\left(10^{-10} \mathrm{~m}\right)} \ & =7.2 \cdot\left(1.6 \cdot 10^{-19} \text { coulombs } \cdot 1 \mathrm{~V}\right) \ & =7.2 \mathrm{eV} . \end{aligned}

\end{prob}

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