Consider a neutral hollow spherical conducting shell. The shape of the shell is described with respect to its center $O$ in spherical coordinates as the locus $a<r<b$ (inner radius $a$, out radius $b$).

Suppose we place a charge $q$ at the center $O$ and we wish to calculate the electric field outside of the conducting sphere at the distance $r=b$. By symmetry, the electric field is spherical, i.e it only has a radial component, and thus taking a Gaussian surface centered at $O$ of radius $b$, we deduce that $E*4 \pi b^2 = q/\epsilon_0$.

Now suppose we displace the charge $q$, so that in Cartesian coordinates it is located at $(2a/3,0,0)$. Let's try to evaluate the electric field strength at the distance $r=b$ on the $x$ axis, i.e at the point $(b,0,0)$.

Apparently, we get the same answer because the field is apparently still spherically symmetric for $r \ge b$, which makes no sense to me.

What am I missing?

What I would really like to do is determine the field for all points along the $x$ axis, but I'm not sure if it's possible to describe nicely.


The field is the same. The simplest way to see this is to realize that the outer surface of your conductor is an equipotential irrespective of the position of the charge $q$ inside the hollowed part.

Thus, you can find the field outside your conductor by first solving Laplace's equation for the potential, with boundary conditions defined only on the surface of the conductor. Since $V$ is fixed on this boundary irrespective of the location of $q$, $\vec E=-\vec\nabla V$ will be the same everywhere outside the conductor, irrespective of the location of $q$.

Edit. We find the field by first finding the potential using Poisson's equation. This is simplest because the boundary condition is on the potential rather than the field. Here, $V$ is constant on the surface of a sphere (the outer surface of your hollowed conductor.) By symmetry, it must be that $V=V(r)$ only or otherwise the spherical surface would not be an equipotential. Thus, write $$ \nabla^2 V(r)=\frac{1}{r^2}\frac{d }{dr}\left(r^2 \frac{dV}{dr}\right)=0 $$ since $V(r)$ must be function of $r$ alone and there is no charge outside the conductor. Hence, $$ r^2\frac{dV(r)}{dr}= -C_0\, ,\qquad V(r)=\frac{C_0}{r}+C_1\, . $$ We can find the integration constants but this is inessential to the argument since $\vec E=E_r\hat r$ with $E_r=-\frac{dV}{dr}= \frac{C_0}{r^2}$, which is spherically symmetric and independent of the location of the charge $q$ inside the hollow sphere.

  • $\begingroup$ But why does the field remain spherically symmetric outside of the conductor? $\endgroup$ Jul 9 '17 at 3:16
  • $\begingroup$ The surface of a conductor is an equipotential so the outside surface of the spherical conducting shell is an equipotential. Thus $V=V(r)=A+B/r$ for $r>b$ by solving Poisson eqn. in spherical coordinates, from which $\vec E=\hat r B/r^2$ follows. $\endgroup$ Jul 9 '17 at 4:54
  • $\begingroup$ I understand that the conductor always an equipotential surface, but I don't understand how that gives information about what happens at point outside of the surface. Could you elaborate a bit more or write a formal proof that the field remains the same for $r>b$? $\endgroup$ Jul 9 '17 at 18:51
  • $\begingroup$ @JoshuaBenabou hopefully this will help. $\endgroup$ Jul 9 '17 at 19:19

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