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Suppose i have some electrons stored in a empty shell container with a negative ion layer in the inner surface so the electrons keep bouncing inside without being able to leave the inner cavity.

I want to compute the electrostatic pressure that the electrons exert on the shell container. So i assume the electrons have a uniform volumetric charge density $\rho$, and the container has a radius $R_0$. My derivation is as follows:

electric field at radius $r$ is $$\frac{1}{3 \epsilon_0} \frac{\rho r^3}{r^2} = \frac{1}{3 \epsilon_0} \rho r$$

force on an element of charge $dQ = \rho dr dA$ is

$$F = p dA = \frac{1}{3 \epsilon_0} \rho^2 r dr dA $$

which means that the pressure satisfies

$$ \frac{dp}{dr} = \frac{1}{3 \epsilon_0} \rho^2 r $$

which by simple integration leads to

$$ p = \frac{1}{6 \epsilon_0} \rho^2 r^2 $$

evaluated at the shell this gives

$$ p(R_0) = \frac{1}{6 \epsilon_0} \rho^2 R_0^2 $$

Which looks reasonable. But the part that i'm not sure is if this result is sensitive to the assumption that the charge density is uniform inside the volume. The relaxed density of the electrons will tend to be maximum near $R_0$.

How can i estimate the relaxed density radial function and/or how will this change my result?

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btw, is there a better way in MathJax to render the differentials? they look too stacked –  user56771 Nov 11 '12 at 16:19
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