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I have founde the following problem. I am given a sphere of radius $R$, with charge density $\rho$ and the fact that the electrostatic field inside the sphere has constant magnitude $E_0$. The question is to find the charge distribution.

Gauss' Law states that $\nabla\cdot \vec{E} = \frac{\rho}{\epsilon_0}$. Thus, if I know a mathematical expression for the electrostatic field, I can calculate the charge density.

Obviously, the uniform field $\vec{E}=E_0 \cdot \hat{n} $ satisfies the constant magnitude constraint, but can there be another solution other than that trivial case? How can I prove that this is the only solution?

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1 Answer 1

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From Gauss' law, using a Gaussian surface of radius $R$ with $R$ inside your charged sphere: $$ \epsilon\int \vec E\cdot d\vec S=\oint \rho dV $$ Assuming $\rho$ is spherically symmetric we obtain $$ \epsilon \vert \vec E\vert\, 4\pi R^2 = 4\pi \int_0^R r^2 \rho dr $$ Thus, if $R_0$ is the radius of your charged sphere, any charge density $\rho$ of the form $$ \rho=\rho_0 R_0/r $$ for $\rho_0$ constant will produce $$ 4\pi \int_0^R r^2 \rho dr = 4\pi \int_0^R r \rho_0 R_0 = 2\pi \rho_0 R_0 R^2 $$ and thus $$ \vert \vec E \vert = \frac{2\pi \rho_0 R_0 R^2 }{4\pi \epsilon R^2}= \frac{\rho_0 R_0}{2\epsilon} $$ will be constant.

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  • $\begingroup$ Ok, there are two solutions to the problem. Are they the only ones? $\endgroup$ Sep 6, 2017 at 12:12
  • $\begingroup$ For the spherical geometry this is the only charge distribution that works. Note that the distribution is at $r=0$ so you might want to re-examine the situation if your sphere is hollow. $\endgroup$ Sep 6, 2017 at 12:34

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