The electrical field for a charge density $\varrho(r)$ is $${\bf E}({\bf r})=\frac{1}{4\pi\varepsilon_0}\int \varrho({\bf r}')\frac{{\bf r}-{\bf r}'}{\hspace{.1cm }|{\bf r}-{\bf r}'|^3} \mathrm d^3r'.$$

Given a single particle of charge $q_C$ fixed in the centre of the coordinate system, a test charge $q_T$ feels a field which goes as $\propto\frac{1}{r^2}$. This field diverges as one gets closer (${\bf r}\to{\bf 0}$). On the other hand, if we have a big wall of charges with density $\varrho_W$, then its electrical field is constant and doesn't diverge as I get closer ($z\to0$) - for example imagine a capacitor place in the xy-plane.

Say I have a charge $q_T$ at position ${\bf x}(0):=(0,0,d)$ above charged plate in the xy-plane. It's eighter infinitely big of or radius $R$. Now I start moving the particle along the given trajectory ${\bf x}(t):=(0,0,d-v\ t)$. Here $d$ and $v$ are fixed constants. What is the value of ${\bf E}({\bf x}(t))$?

What is the electrical field in the case where I have a capacitor plane containing many charges, one of which is perfectly fixed at ${\bf 0}$?

  • $\begingroup$ Are you still interested in an answer to this question? I suppose the main question here is, why doesn't the field diverge when r = 0 for a charged sheet while it does diverge for a single point charge, am I right? $\endgroup$ – Gotaquestion Oct 20 '13 at 18:34
  • $\begingroup$ @Gotaquestion: Yes I think. I guess it has something to do with the mutual force the charges let each other feel. Physically, I think the divergence would vanish if the charges get repelled from the plate center, but if I just force the condition like I did above, I actually don't understand why the divergence isn't there anymore. Of course, it's known that classical electromagnetism has some energetic problems with divergences. $\endgroup$ – Nikolaj-K Oct 20 '13 at 19:35

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