# Hints and review to problem set 3

## 1. Electrostatic potential above the axis of a disk

### a) Expression for potential

We compute the electrostatic potential using the formula $\varphi = \frac{1}{4\pi\epsilon_0} \iint_S \frac{\sigma(\mb{r'})}{|\mb{r}-\mb{r}'|} \, \de{\mb{S'}}.$

We parameterize the disk using polar coordinates in the xy-plane, with $\mb{r}' = [r\cos\theta, r\sin\theta, 0]$, and we also choose a point along the axis of the disc, with $\mb{r} = [0, 0, z]$. Integrating gives $\varphi = \frac{\sigma}{2\epsilon_0} \left[ \sqrt{a^2 + z^2} - \sqrt{z^2} \right].$

It is crucial that you don't set $\sqrt{z^2} = z$ because this isn't true for negative values of $z$. Thus, we conclude that the potential function satisfies $\varphi = \frac{\sigma}{2\epsilon_0} \left[ \sqrt{a^2 + z^2} - |z| \right].$

### b) Jump across

It is always a good idea to check the limit when $|\mb{r}|$ is large. In this case, expanding the square root function gives $\varphi \sim \left[\frac{1}{4\pi \epsilon_0}\right] \frac{\sigma \pi a^2}{|z|},$

which we recognize in the form of $Q/(4 \pi \epsilon_0 |\mb{r}|)$, the electrostatic potential due to a point source of charge $Q$ at the origin. Seeking the electric field at the centre of the disc, note that from symmetry, $\varphi_x = 0 = \varphi_y$ at the centre, and we expect that the electric field only has a z component. The electric field along the axis is then given by $\mb{E} = -\nabla \varphi = -\pd{\varphi}{z} \mb{k} = -\mb{k} \frac{\sigma}{2\epsilon_0} \left[ -\frac{z}{|z|} + \frac{z}{\sqrt{a^2 + z^2}}\right].$

Thus as $z \to 0$, we see that $[\mb{E}]_{z=0^-}^{z=0^+} \sim \frac{\sigma z\mb{k}}{2\epsilon_0 |z|} \biggr\rvert_{z=0^-}^{z=0^+} = \mb{k} \frac{\sigma}{2\epsilon_0} \left[ 1 - (-1)\right] = \frac{\sigma}{\epsilon_0} \mb{k}.$

## 2. Gravitational potential for a cone

The analogous potential sends $\frac{1}{4\pi \epsilon_0} \to G$ and $\sigma \to m$. The potential above a disk of radius $a$ is $\phi = 2 \pi G \sigma\{ \sqrt{a^2 + b^2} - b\}.$

By summing over discs of mass $\sigma \de{z}$, we would obtain the expression $\phi_\text{total} = 2\pi G \sigma \int_0^h \left(s \sec\alpha - s\right) \, \de{s},$

of which the result follows after integration.

## 3. Gravitational potential for a straight rod

The potential is given by integrating $\phi = G\rho \int_{-a}^a \frac{\de{z'}}{\sqrt{R^2 + (z - z')^2}}.$

Let $z - z' = R\sinh(s)$ to significantly reduce the integral. The final substitution requires using $s = \log[\sqrt{A^2 + 1} + A]$

where $A = (z \mp a)/R$.