Dielectric Problem Boundary Condition

Question

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.

Answer 1

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 $$-\sigma=\epsilon_{above}\frac{\partial V_{above}}{\partial r}\Biggr|_{r=b^+}-\epsilon_{below}\frac{\partial V_{below}}{\partial r}\Biggr|_{r=b^-}$$ $$\implies-\sigma=0-\kappa\epsilon_0\frac{q}{4\pi \kappa \epsilon_0}\bigg(-\frac{1}{b^2}\bigg)$$ $$\implies\sigma=-\frac{q}{4\pi b^2}$$