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I am trying to find the potential at any point (x,y,z) due to a rectangular plate with a constant surface charge density. Let's assume the plate is centered on the X-Y plane and extends from -n to n in the x direction and from -m to m in the y direction. Here is what I can work out so far:

$$V(R)=\frac{1}{4 \pi \epsilon_0}\int_{s'}\frac{\rho_s(R')ds'}{|R-R'|}$$

With constant charge density, this simplifies to the following

$$V(R)=\frac{\rho_s}{4 \pi \epsilon_0}\int_{s'}\frac{1}{|R-R'|}\,ds'$$ $$V(x,y,z)=\frac{\rho_s}{4 \pi \epsilon_0}\int_{s'}\frac{1}{\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}}\,ds'$$ $$V(x,y,z)=\frac{\rho_s}{4 \pi \epsilon_0}\int_{-n}^{n}\int_{-m}^{m}\frac{1}{\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}}\,dy\, dx$$ $$V(x,y,z)=\frac{\rho_s}{4 \pi \epsilon_0}\int_{-n}^{n}\int_{-m}^{m}\frac{1}{\sqrt{(x-x')^2+(y-y')^2+z^2}}\,dy\, dx$$

Can the integral be evaluated from here? I have tried with mathematica, but the evaluation never seems to complete. Is there any better way to approach the problem?

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  • $\begingroup$ Hint: Instead of calculating the double integral over patches of area $dA$, try finding an equivalent single integral over a shape of known potential. $\endgroup$
    – Ultima
    Commented Jul 20, 2015 at 18:51
  • $\begingroup$ I am struggling to come up with a shape of known potential since I am solving for potential. Under what circumstances do I have an expression of known potential? $\endgroup$
    – BenP1192
    Commented Jul 20, 2015 at 19:19
  • $\begingroup$ Consider a rectangle of length $2L$ and width $2W$ with uniform linear charge density $\lambda$. $\endgroup$
    – Ultima
    Commented Jul 20, 2015 at 20:42
  • $\begingroup$ So, if I understand you correctly, I can say that $\lambda=2L\rho_s$ and then treat the problem as a finite line of charge of length $2W$? $\endgroup$
    – BenP1192
    Commented Jul 20, 2015 at 21:10
  • $\begingroup$ Actually, I think that only makes sense for points along the center of the rectangle $\endgroup$
    – BenP1192
    Commented Jul 20, 2015 at 21:27

1 Answer 1

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The general calculation to this potential is far from trivial and its not known in the present literature, since no symmetry can be invoked. However, in the particular case of point lying above the plate along the symmetry axis passing through the center of the plate, at distance z, the exact solution follows:

$$V(z)= \frac{\sigma}{\pi\epsilon_0}\left[a \sinh^{-1}\left( \frac{b}{\sqrt{z^2+a^2}}\right)+b \sinh^{-1}\left( \frac{a}{\sqrt{z^2+b^2}}\right) -z\tan^{-1}\left( \frac{ab}{z\sqrt{z^2+a^2+b^2}}\right)\right]$$

More details about the (long) calculation can be found in this paper: https://iopscience.iop.org/article/10.1088/1361-6404/ac362f

Cheers,

D.A. Fagundes

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