A shell with negligible thickness has a given charge density as a function of theta, and is filled with a dielectric material with a given electric susceptibility.
Using lovely legendre polynomials I was able to find the potentials inside and outside the shell, and then used $\mathbf{E} = -\nabla V$ to find the electric fields due to these potentials.
I am now tasked with finding the bound charge on the surface of the sphere inside the charged shell; the two have the same radius.
I must do this by finding the polarization, but I do not know how to find the polarization. This is because I am not sure if the total field inside the sphere is just the internal field due to the shell; my intuition tells me that the shell causes a field which polarizes the dielectric which causes a bound charge which causes a field which modifies the polarization ad infinitum.
How can I find the polarization, and subsequently the bound charge? my professor suggests "looking at the boundary conditions" but I do not know how this helps.
Thanks!
Best Answer
Looking at the boundary condition helps because the jump of the perpendicular component of electric field at a surface is determined by the surface charge density, as can easily be proven by using the Gauss theorem. $$ \int_S d\Sigma\, \sigma = \int_V d^3r\, \rho = \varepsilon_0 \int_{\partial V} d\vec \Sigma \cdot \vec E = \frac{A}{\varepsilon_0}(E_{\perp,1} - E_{\perp,2}).$$ Here $S$ is a small part of the surface, $V$ is a thin box around the surface $S$ and $\partial V$ is the surface of the box. If we now let the extents of $S$ go to zero (and make the heigh of $V$ smaller as well), we get $$ \sigma = \varepsilon_0(E_{\perp,1} - E_{\perp,2}).$$
Now in the case of the exercise the charge density has two parts: The known charge density of the charged shell and the bound charge density due to the polarization of the dielectric, so the latter can easily be calculated.