Find electric potential at a point at the edge of a charged disc of uniform surface charge density

Question

If I've made any mistakes, please point them out in the comments

Find electric potential at a point at the edge of a charged disc of uniform surface charge density

I started by setting up a curved charge element(an Arc), and took the point about which I want to find the potential as Origin

Let the angle the point at which the arc touches either end of the disc be $\theta$ and the distance between the point and the ends of each element vary from $r$ to $R$

First to find the area of the element $dQ=r \sigma dr d({\theta})$

where r varies from r to R and the angle varies from $0$ to $\theta $

that gives $\sigma \frac{(R^2-r^2)}{2}\theta $ as the charge on a given charge element

now $dV =\frac{kQ}{r}$

so here $$dV=\sigma \frac{(R^2-r^2)}{2}\theta $$

so not too sure how to proceed from here

Alternatively, starting by finding the Potential due to an arc at a distance $x$ from its midpoint

so that would be

$dV=\frac{\sigma xdxd\theta}{x}$

which would mean the potential would be

$$dV=\int_0^\theta \int_r^R\sigma dxd theta $$

which simply is $\sigma (R-r)\theta $

Edit:- If possible, I would like to stick to Cartesian coordinates and single integrals

Answer 1

Let's take your first approach. The problem is we need to know the distance between the point $r \times \theta$ (where $r \in [0, R]$ and $\theta \in [0, 2 \pi)$) and the location of the test charge, $R \times 0$. To find the distance, let's convert into Cartesian coordinates: the first point becomes $r \cos \theta \times r \sin \theta$, and the test charge becomes $R \times 0$. The (Euclidean) distance is therefore $d (r) = ((R - r \cos \theta)^2 + (r \sin \theta)^2)^\frac{1}{2}$.

You are correct that the differential charge is $dQ = \sigma r dr d\theta$. The contribution of the charge to the potential is $dV = k \frac{dQ}{d}$. Therefore, your desired potential is

$$V = k \sigma \int_0^R \int_0^{2 \pi} \frac{r}{(R^2 - 2 r R \cos \theta + r^2)^\frac{1}{2}} d\theta dr$$

You could non-dimensionalize a bit with $\hat{r} = \frac{r}{R}$ to get

$$V = [k \sigma R] \underbrace{\int_0^1 \int_0^{2 \pi} \hat{r} (1 - 2 \hat{r} \cos \theta + \hat{r}^2)^{-\frac{1}{2}} d\theta d\hat{r}}_I$$

At this point it just becomes a matter of evaluating integrals. The integral over $\theta$ gives

$$I = \int_0^1 \left( \frac{4 \hat{r}}{1 - \hat{r}} \right) K \left(- \frac{4 \hat{r}}{(1 - \hat{r})^2} \right) d\hat{r}$$

where $K(x)$ is an elliptic integral of the first kind. The result turns out to be $I = 4$. So your answer is

$$V = 4 k \sigma R$$

EDIT

The really simple answer suggests there might be a simpler approach. There is indeed for this problem. Consider arcs centered $R \times 0$. These arcs will intersect the disc at two points; let the separation angle be $2 \theta$. Note that $\theta$ ranges from $\frac{\pi}{2}$ (when the arc has $0$ radius) down to $0$ (when the arc has a radius of $2 R$). Through simple trigonometry, you can see that the relationship between the radius of the arc and $\theta$ is given by $r = 2 R \cos \theta$. The differential charge on an arc is $dQ = 2 \sigma r \theta dr$, so the contribution to the potential is $dV = k \frac{dQ}{r} = 2 k \sigma \theta dr = -4 k \sigma R \sin \theta d\theta$. Then

$$V = \int_{\frac{\pi}{2}}^0 -4 k \sigma R \sin \theta d\theta = 4 k \sigma R \underbrace{\int_0^{\frac{\pi}{2}} \sin \theta d\theta}_{1} = 4 k \sigma R$$

just the same as with the more complicated method.