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 the finite rectangular uniformly charged sheet on its axis (since the electric field is going to be along the axis due to symmetry) but I got stuck at the following integration.

$$ E = \frac{\sigma r}{4\pi\epsilon_o} \int_{x=-a/2}^{x=+a/2}\int_{y=-b/2}^{y=+b/2} \frac{dx dy}{(x^2+y^2+r^2)^{3/2}} $$

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


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.

  • $\begingroup$ thanks very much. I realised this integration is actually solvable and it is very straight forward as well. thanks again. $\endgroup$ – The Imp Apr 18 '15 at 2:06
  • $\begingroup$ @The Imp this was a while ago, but are you able to give a short description of how the integration was done in the end? At the moment there seem to be a few different methods floating in others answers to your original question $\endgroup$ – L. Maynard Oct 29 '17 at 23:53

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.

  • $\begingroup$ Hi, welcome to Physics SE! Please don't post formulae as pictures or plain text, but use MathJax instead. MathJax is easy for people on all devices to read, and can show up clearer on different screen sizes and resolutions. I've edited it here as an example. Look at this Math SE meta post for a quick tutorial. $\endgroup$ – user191954 Sep 9 '18 at 5:39

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)


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.

  • $\begingroup$ You realize that the integration is over a square? Polar coordinates would make this way more difficult than it has to be. Note also that in the case of a disc, OP used polar coordinates and was able to do the calculation. And finally, why do so many people reply to years-old questions which already have perfectly fine and accepted answers? $\endgroup$ – Noiralef Jan 28 '17 at 11:19

protected by Community Sep 16 '18 at 7:42

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