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).

  • $\begingroup$ Gauss law should help... $\endgroup$ – danimal May 10 '15 at 11:08
  • $\begingroup$ @danimal I think I can only use Gauss's to determine the net charge on each surface rather then the charge density at a given point. $\endgroup$ – Quantum spaghettification May 10 '15 at 11:17

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.

  • $\begingroup$ How did you conclude this "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." ? $\endgroup$ – Sidd May 20 '15 at 7:53
  • 1
    $\begingroup$ @Sidd The electric field parallel to the surface must be zero, otherwise the charges would move to make it so. Any anisotropy in the charge distribution on a spherical surface would give rise to an electric field. And there can't be a contribution from charge inside because the electric field in the conductor is zero. $\endgroup$ – Rob Jeffries May 20 '15 at 10:53

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

  • $\begingroup$ Can you please elaborate on the boundary value problem to be used to determine $V_ext$? $\endgroup$ – Sidd May 20 '15 at 12:50
  • $\begingroup$ You have to find a solution of laplace eq Knowing the total charge on the suface and also knowing that the surface is equipotencial.Under this conditions you know there exist only one solution(unless a constant) $\endgroup$ – facenian May 20 '15 at 13:51
  • 1
    $\begingroup$ It is quite easy because outside the conductor the equation for the electric potential is $\Delta V =0$ because there are no charges. More than this you know that over a circle centered in the origin the potential is constant (the boundary of the conductor). Thus the problem has spherical symmetry and the Laplace equation becomes an ordinary, homogenous, differential equation that you can solve easily. $\endgroup$ – falematte May 20 '15 at 19:11
  • $\begingroup$ $V_{ext}=\frac q {4πϵ_0r}$ will give a constant charge density only if the charges inside the outer surface, i.e., the off-center charge and inner shell charge density, contribute nothing to the potential. I can't understand why won't they contribute anything. $\endgroup$ – Sidd May 21 '15 at 11:10
  • $\begingroup$ You are right may be it is not easy to understand why they don't contribute, I will let you think about it however I'll note that the solution is mathematically correct, and that is precisely the beauty and utility of the unicity theorem $\endgroup$ – facenian May 21 '15 at 11:46

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

  • $\begingroup$ I am convinced that method of image charges will NOT work in this situation. The method of image charges requires the potential on the boundary to be known, a requirement for uniqueness theorem for which this method is based. The sphere they use on the wiki page is grounded so we know the potential is 0 and thus uniqueness theorem holds. This is not we we have here. $\endgroup$ – Quantum spaghettification May 10 '15 at 11:27
  • $\begingroup$ You can just define one shell to be at zero potential. But then the other shell might still be problematic. $\endgroup$ – Sebastian Riese May 10 '15 at 11:38
  • $\begingroup$ @Joseph: "Alternatively, application of this corollary to the differential form of Gauss' Law shows that in a volume V surrounded by conductors and containing a specified charge density ρ, the electric field is uniquely determined if the total charge on each conductor is given." -same source as in my answer. $\endgroup$ – akhmeteli May 10 '15 at 11:50
  • $\begingroup$ @Joseph: Maybe you should first solve the problem for the interior of the shell, then for the exterior of the shell, but I believe in both cases you can use the image charge method with inversion. $\endgroup$ – akhmeteli May 10 '15 at 11:53
  • $\begingroup$ @Sebastian Riese: Please see my second comment. $\endgroup$ – akhmeteli May 10 '15 at 12:01

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 $V_0 = q/(4 \pi \epsilon_0 R)$.

  • $\begingroup$ Why $V_2(r)=0$ outside the shell? $\endgroup$ – falematte May 20 '15 at 19:05
  • $\begingroup$ For the second setup, the sphere is grounded, i.e., $V_2(R) = 0$. This then implies that $V_2 = 0$ for all $r > R$. $\endgroup$ – Michael Seifert May 20 '15 at 19:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.