Surface charge density for an off-centre charge in a spherical shell?

Question

How would I evaluate the surface charge density on the inner and outer surface of a neutral, spherical, conducting shell which has an off-centre charge $q$ inside? I believe that we can not use the method of image charges since even though we know the potential of the shell is constant we do not know its value; it is not even fixed (unlike a grounded shell).

Answer 1

In the case of a grounded conducting shell, it is well known that the method of images can be used to calculate how the total charge $-q$ on the inner surface is distributed. The solution is given in the wikipedia link above.

Now, you ask what happens if the the potential of the shell is fixed, but not necessarily zero. The electric field inside the conductor must still be zero. Using Gauss's law and a surface that is inside the conductor we know that there then must still be a charge $-q$ distributed over the inner surface in some way. To conserve charge there must now also be a charge $+q$ distributed over the outer surface.

We know that the electric field lines leaving the outer surface of a conductor must be perpendicular to that surface. But for a spherical outer surface, the only way this can be arranged and keep the outer surface as an equipotential is if the charge $+q$ is distributed uniformly over that surface. This spherically symmetric arrangement of charge contributes no net electric field inside the conducting shell or in its interior. i.e. the electric field outside the conducting shell will be exactly equivalent to that of a positive charge at the centre of the shell. Furthermore, as the contribution of this charge to the field inside the shell is zero, then it cannot alter the electric field deduced using the method of images for a grounded shell.

As the electric field at the inner surface of the shell is unchanged, then the surface distribution of charge must also be unchanged.

Therefore nothing changes about the inner shell charge surface distribution if the shell is not grounded.

Answer 2

Yes, you can use the method of images because uniqueness of the solution is guaranteed when you Know the total charge of an equipotential surface without knowing the value of the potential itself

I'll summarize a procedure to obtain the correct answer: Application of Gauss law tells us that there must be total charge -q on the inner surface then because of charge conservation there must be total charge q on the outer surface. How do we know the surface density of this charge? We solve the boundary value problem por the regions exterior to the outer shell and interior to the inner shell. For the exterior region we simply have $V_{ext}=\frac{1}{4\pi\epsilon_0}\frac{q}{r}$ which gives a constant surface density

For the interior region we use the method of images. Place a charge outside the inner shell of magnitud $q'=-\frac{d}{a}q$ at a distance from the center of $d=\frac{a^2}{\delta}$ where $\delta$ is the distance from center to the charge placed inside and "a" is the shell radius.

The physical reason for the internal shell charge density not being uniform is that a uniform surface density can never compensate the potential of an off-centre point charge

Answer 3

I guess you can still use the method of image charges - the image charge will be in the inversion point - http://en.wikipedia.org/wiki/Method_of_image_charges#Reflection_in_a_conducting_sphere

Answer 4

Rob Jeffries's answer is correct, but here's another way to see it. Consider two solutions to Laplace's/Poisson's equation:

1. $V_1(\vec{r})$ is the solution (in all space) for a spherical conducting shell at potential $V_0$, with $V \to 0$ as $r \to \infty$.
2. $V_2(\vec{r})$ is the solution (in all space) for a grounded spherical conducting shell with an off-center charge $+q$ inside.

Obviously $V_1(\vec{r})$ is just $V_0 R/r$ outside the sphere and $V_0$ inside; $V_2(\vec{r}) = 0$ outside the shell, and can be found inside the shell via the method of images as noted above.

Now consider the solution $V(\vec{r}) = V_1 + V_2$. This has the property that the potential is a constant $V_0$ on the shell, goes to zero at infinity, and has the charge distribution corresponding to an off-center charge inside. But the boundary conditions (and sources) of the original problem are just a specific case of these boundary conditions. To relate the constant potential $V_0$ to the charge magnitude $q$, we can just use Gauss's Law, with the usual result that $V_0 = q/(4 \pi \epsilon_0 R)$.