I was teaching kids about how to find electric field using the superposition
principle for continuous charge distributions. I thought maybe I should derive
the formula for electric field due to a finite rectangular sheet of charge of charge on the surface $S$,
where
$$
S = \left\{(x,y,z)\in \mathbb{R}^3 \mid -a/2< x < +a/2; -b/2< y < +b/2 ; z = 0 \right\}
.$$
However, I got stuck at the following integration.
$$
E(0,0,r) = \frac{\sigma r}{4\pi\epsilon_o}
\int_{x=-a/2}^{x=+a/2}\int_{y=-b/2}^{y=+b/2} \frac{dx dy}{(x^2+y^2+r^2)^{3/2}},
$$
where $\sigma$ is the surface charge density.
Note: This integration can be done if $a$ or $b$ or both are very large i.e.
$\infty$ in which case we get usual result of $E=\frac{\sigma}{2\epsilon_o}$
So my question is, Can this integral be calculated? If not then what method
would I use to find the electric field in this case. Also It would be greate if
anyone can comment on how to find the electric field by directly solving the
poisson equation.
Consequently if we take case of finite disk the following is the resulting
integration.
$$
E = \frac{\sigma r}{2\epsilon_o}
\int_{\xi=0}^{\xi=R} \frac{\xi d\xi}{(\xi^2+r^2)^{3/2}}
$$
which can be solved as
$$
E = \frac{\sigma}{2\epsilon_o} \left(1- \frac{r}{\sqrt{r^2+R^2}}\right)
$$
Now by taking the limit $R \rightarrow \infty$ we can show that
$E \rightarrow \frac{\sigma}{2\epsilon_o}$.
Best Answer
The integrals are difficult but not impossible, unless I've made a mistake with WolframAlpha. The result is:
$$E = \frac{\sigma}{\pi \epsilon_0} \arctan\left( \frac{ab}{4r\sqrt{(a/2)^2+(b/2)^2+r^2}} \right)$$
When $a,b \to \infty$ the whole arctangent goes to $\pi/2$ and we recover $E=\frac{\sigma}{2\epsilon_0}$, which is definitely encouraging.
And I don't know what you mean by "directly solving Poisson's equation". As far as I know, the usual way to do that is to use Green's functions, i.e., this integral.