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:
$$V(r,theta)=sum_{l=0}^{infty}{bigg(A_l+frac{B_l}{r^{l+1}}}bigg)P_l(costheta)$$

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:
$$sigma=frac{+q}{4pi a^2}$$
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:
$$frac{-q}{4pi a^2}=kappa epsilon_0 frac{-B}{a^2}$$
Simplifying:
$$B=frac{q}{4pi kappa epsilon_0}$$
Applying the second boundary condition of the grounded potential $V(b)=0$:
$$0=A+frac{B}{b}$$
Rearranging and subsituting:
$$A=frac{-B}{b}=frac{-q}{4pi kappa epsilon_0 b}$$
Thus the potential in the bulk of the dielectric is given by:
$$V(r)=frac{q}{4pi kappa epsilon_0}biggr(frac{1}{r}-frac{1}{b}biggr),quad a < r < b$$
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 $-q/{4pi b^2}$ but I am trying to solve it as a boundary value problem for practice.