Potential in the middle of a square [closed]

Question

Find the electric potential in the middle of a square with side length $a$ and charge $Q$.

If I put the origin in the middle of the square, for the potential I get: $$4\frac{1}{4\pi\epsilon_0}\int_0^\frac{a}{2}\int_0^\frac{a}{2}\frac{\sigma}{\sqrt{x^2 + y^2}}dxdy$$

,where $\sigma$ is surface charge density. Then I'd proceed to change to polar coordinates, so I get something like: $$\frac{\sigma}{\pi\epsilon_0}\int_0^{\frac{\pi}{2}} \int_0^{\frac{a}{2\cos\varphi}} dr d\varphi=\frac{\sigma}{\pi\epsilon_0}\frac{a}{2}\int_0^{\frac{\pi}{2}}\frac{d\varphi}{\cos \varphi}$$ which is a divergent integral.

What am I doing wrong?

I think I can imagine changing the approach in such a way that the limits of integration are different, but what's wrong with this one?

Answer 1

The Cartesian integral you've set up is correct. Now do the integrals: $$\begin{align*} V&=\frac{\sigma}{\pi\epsilon_0}\int_0^{a/2}dy\;\left.\sinh^{-1}\frac{x}{y}\right|_0^{a/2}\\ &=\frac{\sigma}{\pi\epsilon_0}\int_0^{a/2}dy\;\sinh^{-1}\frac{a}{2y}\\ &=\frac{\sigma}{\pi\epsilon_0}\int_{\infty}^{1}-\frac{a}{2u^2}du\;\sinh^{-1}u\\ &=\frac{\sigma a}{2\pi\epsilon_0}\int_{1}^{\infty}du\;\frac{\sinh^{-1}u}{u^2}\\ &=\frac{\sigma a}{\pi\epsilon_0}\sinh^{-1}(1) \end{align*}$$