# Dielectric Problem Boundary Condition

I am given two concentric conductive spherical shells, one of radius $$a$$, and the other of radius $$b$$, with $$b>a$$. The space between these shells is filled with a dielectric of relative permittivity $$\kappa$$. The inner shell has a charge of $$+q$$ and the outer shell is grounded.

I would like to solve for the potential within the dielectric medium as a function of $$r$$, and use it to compute the surface charge density on the outer plate.

Considering Laplace Equation in Spherical Coordinates, the solution takes the form: $$V(r,\theta)=\sum_{l=0}^{\infty}{\bigg(A_l+\frac{B_l}{r^{l+1}}}\bigg)P_l(\cos\theta)$$

By spherical symmetry, we know that there exists no dependence on $$\theta$$, thus we need only to consider the first term in the summation: $$V(r)=A+\frac{B}{r}$$ We will need boundary conditions to determine $$A$$ and $$B$$. Consider the surface charge density on the inner plate: $$\sigma=\frac{+q}{4\pi a^2}$$ This surface charge is associated with a discontinuity of the Electric Field at $$r=a$$. Using Gauss Law considered in the vicinity of a surface charge: $$-\sigma=\epsilon_{above}\frac{\partial V_{above}}{\partial r}-\epsilon_{below}\frac{\partial V_{below}}{\partial r}\Biggr|_{r=a}$$ Since the shell is conductive, it is an equipotential i.e. $$V_{below} = constant$$. Thus the above formula reduces to: $$-\sigma=\kappa \epsilon_0 \frac{\partial V_{above}}{\partial r}\Biggr|_{r=a}$$ Substituting $$\sigma$$ and $$\partial_r V_{above}$$ we obtain: $$\frac{-q}{4\pi a^2}=\kappa \epsilon_0 \frac{-B}{a^2}$$ Simplifying: $$B=\frac{q}{4\pi \kappa \epsilon_0}$$ Applying the second boundary condition of the grounded potential $$V(b)=0$$: $$0=A+\frac{B}{b}$$ Rearranging and subsituting: $$A=\frac{-B}{b}=\frac{-q}{4\pi \kappa \epsilon_0 b}$$ Thus the potential in the bulk of the dielectric is given by: $$V(r)=\frac{q}{4\pi \kappa \epsilon_0}\biggr(\frac{1}{r}-\frac{1}{b}\biggr),\quad a < r < b$$ To compute the surface charge density at the outer shell, we need to consider the potential outside the outer shell $$r>b$$. We will repeat the process outside the shell.

Following a similar argument: $$V(r)=C+\frac{D}{r}$$ We have the boundary condition $$V(b)=0$$ thus: $$C+\frac{D}{b}=0$$ But we are missing a boundary condition...and now I'm stuck. We should probably get $$C=D=0$$, but I cannot justify it. Could someone justify it for me with another boundary condition, and some physical justification for it.

EDIT: I know that this is an overkill technique for a simple problem, the answer is $$-q/{4\pi b^2}$$ but I am trying to solve it as a boundary value problem for practice.

• ...why the close vote? This isn't a homework question, I already know the answer. Sep 8, 2020 at 12:30
• You can use Gauss law, picking a sphere that encloses both conductors as integration surface. Sep 8, 2020 at 12:55
• It doesn't have to be literally homework, hence the HW&E tag.
– Gert
Sep 8, 2020 at 13:11

You're missing the "trivial" boundary condition $$V \rightarrow0$$ as $$r\rightarrow\infty$$ for a charge distribution spread over a finite volume/area/length.
$$\therefore \hspace{2pt} V(\infty)=C=0$$ by using your first condition $$0+\frac{D}{b}=0$$ gives $$D=0$$.
Surface charge density of grounded shell $$-\sigma=\epsilon_{above}\frac{\partial V_{above}}{\partial r}\Biggr|_{r=b^+}-\epsilon_{below}\frac{\partial V_{below}}{\partial r}\Biggr|_{r=b^-}$$ $$\implies-\sigma=0-\kappa\epsilon_0\frac{q}{4\pi \kappa \epsilon_0}\bigg(-\frac{1}{b^2}\bigg)$$ $$\implies\sigma=-\frac{q}{4\pi b^2}$$
• My problem is, how can I ensure that $V(\infty)=0$, you see we set the reference potential arbitrarily, how can I be sure that our choice of the shell $r=b$ at zero potential is consistent with $V$ vanishing at $r\rightarrow \infty$, and not some other constant? Sep 8, 2020 at 18:16