Electric Potential of a Sphere: A Puzzling Problem

[Trisztan]

Homework Statement

Consider an infinite conducting plane with uniform charge density $\eta$. Now suppose we place a neutral conducting sphere of radius $r$ a perpendicular distance $L$ above the conducting plane. What is the electric potential of the sphere?

Relevant Equations

Gauss' Law: $$\oint{\mathbf{E}\cdot d\mathbf{A}}=\frac{q}{\varepsilon_0}.$$
Electric field and potential: $$E_s = -\frac{dV}{ds}.$$

I can calculate the electric field strength at any point above the plane with Gauss' Law ($E = \frac{\eta}{\varepsilon_0}$) and so the electric potential at any point a perpendicular distance $z$ above the conducting plane ($V=−\frac{\eta}{\varepsilon_0}z$).

But I'm having trouble taking into account the sphere. Since it's neutral, does it have any impact on the potential at that point? Also, does "of the sphere" imply I need to find the electric potential generated by the sphere? And if so, wouldn't that be 0, since the sphere is neutral?

Any help or advice would be appreciated.

 

[BvU]

Hello @Trisztan ,

$\qquad$ !​

taking into account the sphere

The sphere is neutral, but conducting: the total charge on the sphere is and remains zero, but the charge carriers are free to move. And they will move until they no longer experience an electric field (a potential difference). [edit]Or end up at the boundary which they can't get off .

I can calculate the electric field strength at any point above the plane

Sure, but that is in the absence of the sphere ! In that case equipotential surfaces are planes parallel with the plate. The sphere, however, changes that ! As explained above, the sphere surface is an equipotential surface...
And since the plane is conducting also, the charges on its surface will also move until they no longer experience a potential difference in a direction parallel to the surface...

I suspect you are going to need to look into the method of mirror charges (or: method of images), which google (I like Errede, Illinois).

And I think this exercise is not all that trivial. But perhaps there is an elegant answer.

$\$

 

[Trisztan]

@BvU Thank you for your answer. I've looked around prior to this and come across the method of mirror charges, but the course I'm currently taking doesn't teach that, so there should be some other way to do the question.

And I think this exercise is not all that trivial. But perhaps there is an elegant answer.

As for its difficulty, the question is actually one part of one of the questions on a past paper. So, I would expect it to take, at maximum, around 5 minutes to solve.

But I've asked others about this question as well and am yet to receive an answer, so maybe the professor aimed a bit too high with this one.

 

[BvU]

Unless there is an elegant path that avoids the need to know the field.

@Orodruin ?

 

[Orodruin]

Just thinking out loud (just gave a lecture so a bit tired): You can compute the potential on the sphere due to a point charge using the method of images for the spherical case. Then use the result to express the case at hand as an integral over the surface charge.

This integral may or may not be divergent depending on the choice of zero level. I have not checked. The problem should specify the relevant zero level of the potential, which should make it possible to renormalize the integral if divergent.

 

[Trisztan]

@Orodruin The choice of zero level is specified, it's the plane of charge. My bad for not including that.

However, this method would still require the method of image charges, which we are not taught in this course. But thank you for your time regardless.