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In the problem that I'm working on, there is a spherical capacitor with charges $Q$ and $-Q$ over the inner and outer surfaces respectively and half of it is filled with a dielectric with permitivity $\epsilon$. And the task is to compute the electric field in the interior of the capacitor and the surface charge density at the inner surface.

Using Gauss's law I've found that the electric field is the same in the region with the dielectric and the empty one, and its value is: $$\vec{E(r)}=\frac{Q}{2\pi(\epsilon_0 +\epsilon)r^2}\hat e_r$$ I computed the surface charge density multiplying the electric field (evaluated at the inner radius of the capacitor) by the permitivity of each of the mediums, so I get a discontinuity at the joint.

On one hand it seems coherent to me to have a greater charge density over the hemisphere in contact with the dielectric but o the other hand I'm not sure how to interpret this jump when passing from one hemisphere to the other. What is worst, I think the problem might be that maybe it is not right to use Gauss's law in this case because the gaussian surface I've used is precisely a hemisphere surrounding each half of the inner conducting sphere and to make use of a symmetry argument (and be able to consider the electric field constant at each point of the hemisphere) I've assumed that the charge over each half of the conducting inner sphere was uniformly distributed and now this assumption seems dubious to me.

So, what's going on? Isn't it more realistic to assume the charge density over the conducting inner sphere is a smooth function varying with the angle $\theta$ (orientate the capacitor in such a way that the hemisphere filled with the dielectric is at the top) instead of a two-parts constant function with an abrupt jump at $\theta=\pi/2$? Does it sound coherent? If so, I guess there is no simple way to compute the electric field inside the capacitor and the charge density at the inner surface, or is it?

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Q: Is it reasonable to have a discontinuity on the surface charge density?

A:yes

Why: I think the best way to approach this to start with the boundary condition for an E field which is.

$(\vec{E}_{2}-\vec{E}_{1}) \cdot \hat{n}=\frac{\sigma}{\epsilon_0}$

Taken that $\vec{E}_{1}$ is the E Field inside the inner sphere, we know that it is

$\vec{E}_{1}=0$

Taken that $\vec{E}_{2}$ is the E Field inside the capacitor, we know that it is

$\vec{E}_{2}=E_{Upper}(r) \hat{r}$, For $0<\theta<\frac{\pi}{2}$

$\vec{E}_{2}=E_{Lower}(r) \hat{r}$, For $\frac{\pi}{2}<\theta<\pi$

Notice also that $\hat{n}=\hat{r}$ so we can write out the discontinuity of the surface charge at $\theta=\frac{\pi}{2}$ as a difference between E field boundary conditions from Eupper and Elower so

$\Delta\sigma|_{\theta=\frac{\pi}{2},r=R}=\sigma_{Upper}-\sigma_{Lower}= \epsilon_0 (E_{Upper}(R)-E_{Lower}(R))$

Where R is the radius of the inner sphere.

We can see that a surface charge density can have a discontinuity but if the E fields on both hemispheres have different E fields in their capacitance.

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