Electric field due to a uniformly charged FINITE rectangular plate

Question

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.

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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 $E \rightarrow \frac{\sigma}{2\epsilon_o}$.

Answer 1

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.

Answer 2

initially put $x^2 + r^2 = p^2$

then $y = p(\tan(A))$

solve it, it will be in terms in terms of $x$. and substitute $\frac{r(\tan(B))}{b} = \frac{x}{\sqrt{4x^2 + 4r^2 + b^2}}$ solve it you will get answer easily.

Answer 3

Actually this integral can be solved by the method of polar substitutions. x=rcos(A) and y=rsin(A) where r is the distance and A the angle in the polar plane. You can find further details in Thomas Calculus. Be sure to substitute the limits properly and multiply the integral by the Jacobian which in this case is r. Hope this answer helped you.

Answer 4

This integral cannot be solved in terms of elementary functions. You can easily do an expansion in $\frac{1}{r}$ in the integrand after doing on of the integrations, then doing the second integral after expanding you get $$ \frac{ab}{r^2}\left(1 - \frac{a^2+b^2}{12 r^2} + \mathcal{O}\left( \frac{1}{r^4}\right)\right) $$ If you want to solve the poisson equation, you have to use Green's function method because you have a charge distribution (unlike when you only have laplace equation with boundary conditions and you can just use separation of variables), this will bring you right back to this integral.

Note: this series is converging if you're interested in the region $r>\text{max }\left(a,b \right)$

Note: this is basically the multipole expansion, where the first term is the monopole contribution, the second is the quadrupole etc... (all odd multipole vanish because of symmetry)