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A ball with radius $R$ has a non-uniform charge density $\rho(r)$. The electric field of the ball as a function of position $E(r)$ is known. How would you then find $\rho(r)$? I was thinking that, by using Gauss' law to find the charge enclosed by concentric spherical surfaces: $$ \Phi_E(r)=E(r)\cdot4\pi r^2=\frac{q_{enc}(r)}{\varepsilon_0}\implies q_{enc}(r)=\frac{E(r)}{k_e}{r^2} $$ And then differentiating that would yield $\rho(r)\cdot 4\pi r^2$, but I wasn't really sure since it would make more sense if $\rho(r)$ was a surface charge density $\sigma(r)$.

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You might want to start by using the differential form: $$\nabla \cdot \vec{E} = \frac{\rho(r)}{\epsilon_0}.$$ Since you don't know the actual distribution, the charge enclosed is unknown and you can't calculate $\vec{E}$ using the integral version which relies on symmetries and knowing the enclosed charge to be useful. Therefore you would need $\vec{E} = - \nabla \phi$ and have to solve Poisson's equation altogether: $$\nabla^2 \phi = -\frac{\rho(r)}{\epsilon_0}.$$ More info on $\rho(r)$ could help you integrate this more easily (symmetries, for example).

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  • $\begingroup$ That sounds easier, but my professor didn't really cover the differential form of Gauss' law yet since it's a high-school class. This example relies on spherical symmetry, would it be rational to assume that the change in charge as the radius of the Gaussian surface increases, divided by the surface area of that surface, is the charge density at that specific radius? $\endgroup$ – Angelo Di Bella May 5 '20 at 20:29
  • $\begingroup$ In that case (spherical symmetry and only a dependence of $r,$) you can leave the charge enclosed as a function of the integral inside a certain circle, for example $$q_{enc}(R) \equiv 4\pi \int_o^{R} \mathrm{d}r \; r^2 \rho(r).$$ And then use Gauss' law as you suggested in your question, without knowing the form of $\rho(r)$ the exact value of $q_{enc}$ can't be calculated but you still have some idea of the way the field behaves, although not exact. $\endgroup$ – Nelson Vanegas A. May 6 '20 at 1:16

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