This is question from Bangladesh Physics olympiad:

Find the potential of an uncharged solid metal sphere provided that a point charge $+q$ is located at a distance $r$ from its center.

I think question is ambiguous as it doesn't mention whether the charge is inside or outside the sphere. Let's suppose, it's outside. Then how would be the charge distribution?

I first thought the whole outer surface of the hemisphere facing the point charge would build up negative charge uniformly and the opposite hemisphere would built up the opposite. But, after diacussing with my friend, I have got a feeling that charge buildup won't be uniform. Rather, it would be dense where the distance from the point charge is small and less dense where the distance is large.

But, if so, then can we derive any parametric equation for the charge distribution on the sphere or not?


closed as off-topic by Jon Custer, Kyle Kanos, David Hammen, John Rennie, ZeroTheHero Aug 1 '17 at 16:54

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    $\begingroup$ The potential at every point on the sphere must be the same (as it is a conductor). Does that help? $\endgroup$ – Floris Jul 31 '17 at 14:31
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    $\begingroup$ Have a look at "method of images" for how to solve these types of problems. $\endgroup$ – Mikael Fremling Jul 31 '17 at 14:35
  • $\begingroup$ There is flag calling my question off-topic. I don't how it is off-topic. $\endgroup$ – Mockingbird Jul 31 '17 at 14:43
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    $\begingroup$ @Mockingbird The reasoning is this: Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better. I think your question fulfills this strict criterium, so I vote it to "leave open" - hopefully there will be 2 other voter doing the same. Over 300 reputation, also you can cast close/reopen votes to your own posts. $\endgroup$ – user259412 Jul 31 '17 at 15:09

You're right, the charge build-up won't be uniform over the hemisphere.

This is a rather standard example of using the method of images where space is separated into two regions by a conducting surface (inside and outside a sphere or above and below an infinite plane are some classic examples).

You may then replace the net effect of the charge distribution on the (grounded or uncharged) conducting surface with an equivalent "image charge" which satisfies the same boundary conditions (usually the value of the potential on the conductor), and solve that problem which is generally considerably simpler. The uniqueness theorem in Electrostatics then implies that the solutions to both these problems should be the same, as they both satisfy the same boundary conditions.

I could write out a solution, but there's already one given on the Wikipedia page, as well as a rather good one here, which seems quite detailed.

There is a very good answer to a similar question that could also help you.

EDIT: I'm including a hint at the solution for an ungrounded sphere.

Finding the potential:

The method of images works well when we assume a grounded sphere (so that the potential $V=0$ on the surface). However, with a small modification, the same basic model can also handle a sphere at an arbitrary potential $V_0$. We do this by introducing a second charge.

As shown on this page, the solution for the potential of a grounded sphere of radius $R$ and a charge $+q$ at a distance $z=a$ ($a>R$) from its center is given by replacing the entire sphere with an induced charge $$q' = -\frac{q R}{a}$$

at a position $$z = \frac{R}{a^2} < R$$

As I explained, both these problems are guaranteed to have the same solution by the uniqueness theorem. How do we extend this to a sphere at some arbitrary potential? We need to increase the potential on the surface of the sphere while keeping it an equipotential surface! It should be quite obvious that the way to do this is in fact to introduce a second image charge (say $q''$) at the center of the sphere $z=0$. Since potentials are additive, this merely changes the potential on the surface of the sphere from being $V=0$ to a constant $V = V_0 = \frac{1}{4\pi \epsilon_0}\frac{q''}{R}$, where $q''$ can be chosen depending on $V_0$.

If the sphere is neutral (as is the case in your problem) then we simply require that $q' + q'' = 0$. Thus, you would now need to find the potential for the following problem 3 charges

which is a trivial extension that you should be able to do if you've understood how to do the grounded case, but with 3 terms this time, the third being the potential due to the charge $q'' = -q'$ at the origin.

Finding the surface charge density:

The field within a conductor is zero, and the field outside infinitesimally close to it is given by

$$\mathbf{E} = -\mathbf{\vec{\nabla}}V = \frac{\sigma}{\epsilon_0}\mathbf{\hat{n}}$$

(If you're having trouble with this part, I would very strongly advise you to read Chapter 2 in Griffith's Electrodynamics. I haven't seen it better explained anywhere else.)

Thus if we know the potential $V$ using this formula, we can calculate the surface charge density. Defining

$$\frac{\partial V}{\partial n} = \vec{\nabla}V \cdot \mathbf{\hat{n}}, \quad \quad \quad \quad \implies \sigma = -\epsilon_0 \frac{\partial V}{\partial n}\Big|_\text{on the surface}$$

So, the case of the sphere is pretty simple, since on the surface of the sphere the normal direction is $r$. Once you've found $V(r,\theta)$, you can easily find

$$\sigma(\theta) = -\epsilon_0 \frac{\partial V(r,\theta)}{\partial r}\Big|_{r=R}$$

which you should be able to do. If you'd like to check your results, here are two solutions:

1) For a grounded sphere, the induced surface charge density is given by

$$\sigma_0 (\theta) = \frac{q}{4\pi R}\left( R^2 - a^2 \right) \left(R^2 + a^2 - 2 R a \cos\theta \right)^{-3/2}$$

(You could check this by integrating over all $\theta$. What should the result be?)

2) For an ungrounded sphere,

$$\sigma(\theta) = \sigma_0 (\theta) + \frac{q}{4 \pi R a}$$

  • $\begingroup$ What would be the distribution of positive charge? $\endgroup$ – Mockingbird Aug 1 '17 at 1:03
  • $\begingroup$ On the conductor? Nothing obvious, as far as I know. It seems a rather difficult problem to solve and I can't think of a way to do it off-hand. I believe it'd look something like this. The strength of the method of images is that you don't need to calculate this complicated distribution to find the solution for the potential, all you need is to solve a much simpler problem, and uniqueness guarantees that the solutions will be the same. $\endgroup$ – Philip Cherian Aug 1 '17 at 1:10
  • $\begingroup$ The sphere you mentioned in the link is grounded. So, obviously no positive charge would build up on the sphere. $\endgroup$ – Mockingbird Aug 1 '17 at 1:16
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    $\begingroup$ Oh of course, you're right. To move from the 'grounded' to the 'ungrounded' cases is actually quite simple: imagine first that you have a grounded sphere. As described in the link, this induces some negative charge $q'$ on the sphere. Now imagine that you cut the ground wire and add an additional positive charge $q''=-q'$ to the center whose sole purpose is to make the (now ungrounded) sphere neutral. This brings us to the problem you want to solve. I'm sorry if it's not clear, I'm a bit sleepy, I'll edit my answer tomorrow including this along with my formula for the surface charge density. $\endgroup$ – Philip Cherian Aug 1 '17 at 2:04
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    $\begingroup$ I'm not really sure what you mean. Which positive charge are you talking about? $q$ or $q''$, the second image charge? I suggest editing your question (which is currently "on hold") with your calculations including what you think the potential should be, and explaining clearly where you're stuck and what you're having a problem understanding. $\endgroup$ – Philip Cherian Aug 3 '17 at 1:49

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