Field outside hollow conducting shell with charge displaced from center

Question

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.

Answer 1

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 $q$.

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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.

Answer 2

Charge within the cavity attracts opposite charge to the inner shell in order to cancel out the electric field within the conductor. Consequently, on the inner surface, the charge density will be highest (in absolute value) where the cavity's field is highest. You can find the exact value solving the Laplacian when you apply Maxwell's Equations. The surface charge density on the inner surface is proportional to the field at the boundary.

Now that charge has to come from somewhere. If you have $q$ charge within the cavity, you have $-q$ induced on the inner surface of the cavity. However, that negative charge left a deficit of negative charge elsewhere, or in other words a net positive charge. Those positive charges (ions) can't "feel" any of the charge on the inner surface or within the cavity since those cancel out. They can feel their own field and must move in such a way as to minimize their potential. That requires the charges be even distributed on the outer surface. An evenly distributed charge on the surface of a sphere is identical to the same amount of charge concentrated at its center. So a conductor actually hides information about the charge distribution contained in a cavity within.

Is probably worth noting there is no net charge on the conductor.