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.)

  • $\begingroup$ 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. $\endgroup$
    – secavara
    Aug 20, 2020 at 20:57
  • $\begingroup$ 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. $\endgroup$
    – Joeseph123
    Aug 20, 2020 at 23:21
  • $\begingroup$ 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. $\endgroup$
    – secavara
    Aug 21, 2020 at 6:37

1 Answer 1


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 $Q$.


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