Laplace Equation of a Charged Conductive Sphere in an External Uniform Electric Field

I am trying to solve Problem 3.21 in Introduction to Electrodynamics, Griffiths where I am asked:

Find the potential outside a charged metal sphere of charge Q and radius R, placed in an otherwise uniform electric field $$\mathbf E_0$$.

Let us orient our coordinate system such that the electric field acts along the z-axis.

• BC 1: The sphere is conductive, thus set $$V(R, \theta)=0$$.
• BC 2: As $$r \rightarrow \infty$$, we notice that $$V \rightarrow -E_0r \cos \theta- \frac{Q}{4\pi\epsilon_0r}$$

Note the Laplace Equation solution in azimuthal-symmetric cases in spherical coordinates is given by:

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

I am currently stuck at trying to make the two boundary conditions work together, all I get is a limit form of what the coefficients should be, and even an incompatibility.

Applying BC 1: $$V(r,\theta)=\sum_{l=0}^{\infty}{A_l( r^l-\frac{R^{2l+1}}{r^{l+1}})P_l\cos(\theta)}$$

But clearly for significantly large $$r$$, the $$\frac{R^{2l+1}}{r^{l+1}}$$ terms vanish, and now we can't use the part of the second boundary condition that scales as $$\frac{Q}{4\pi\epsilon_0r}$$, which is not a surprise, but the problem is that the second boundary condition is incompatible with the the first, due to the $$\frac{Q}{4\pi\epsilon_0r}$$ and $$-E_0r \cos \theta$$ terms not fitting the form required when we first applied BC 1.

Please could someone clarify on the issue of this incompatibility (Though not actually solve the problem using a different method, I am trying to understand where I went wrong with this method.)

• Notice that $r = R$ only has to be an equipotential, but it does not necessarily mean $V$ must vanish. Also, in the limit $r \rightarrow \infty$ all the $\frac{1}{r^{l+1}}$ terms go away and so you get no information about these terms in this limit. Aug 20 '20 at 20:57
• We could set the sphere to have an arbitrary potential, so by convention, let us set it to zero. You are correct about the reciprocal terms vanishing. Aug 20 '20 at 23:21
• Sure, it's just that in this case you'll need a constant term $A_0$ to make the potential vanish at the surface of the sphere, which you won't need if you just let it be non-zero. Aug 21 '20 at 6:37

BC $$2$$ is wrong. For large $$r$$, the field due to the charge $$Q$$ on the sphere is negligible, and the only thing that's left is the uniform $$\mathbf{E_0}$$. So the condition at infinity is actually $$\mathbf{E}(r\rightarrow\infty,\theta)\rightarrow\mathbf{E_0}$$. Translate this to the potential. Hint: Try to impose $$V(r\rightarrow\infty,\theta)=-E_0r\cos\theta+C$$ (don't forget the constant $$C$$) to the result you got after applying BC $$1$$. You'll see that this implies $$A_{l\ge2}=0$$, and will allow you to determine $$A_1$$.
Finally, the result will depend on $$A_0$$ (equivalently $$C$$). Then you need a third BC, which is that the net charge on the sphere is $$Q$$. Hint: Try finding $$\mathbf{E}=-\mathbf{\nabla}V$$ and use Gauss' Law to find the link between $$A_0$$ and $$Q$$.