# Homework Help: Conducting Sphere and Grounded Conducting Plate

1. Dec 9, 2008

### source.decay

1. The problem statement, all variables and given/known data
A conducting sphere of radius R carries a total charge of +Q. Its center is placed a distance 2R to the right of a grounded conducting plate. What is the potential at a general point in space outside the sphere and to the right of the plane? (Carry out to at least four terms)

2. Relevant equations
La Place's Equation, solved using separation of variables.
Legendre Polynomials

3. The attempt at a solution
I'm trying to do this as an image charge problem. I placed a sphere of charge -Q on the opposite side of the conducting plate. My boundary conditions state that at 2R, potential goes to zero and as r goes to infinity, potential goes to zero. I set up an equation for the charge distribution of the -Q sphere: Q = integral(sigma * da). My first attempt was to approximate both spheres as point charges, but the boundary conditions do not allow for that. The directions say to include the first four terms, thus I know think that I must use the infinite sum solution to la Place's equation.

Thank you

2. Dec 9, 2008

### gabbagabbahey

The method of images looks like the easiest way to do this to me. It will give you an exact answer which you can expand in spherical harmonics (Legendre Polynomials) to get the same solution that solving Laplace;s equation would give.

(1) Method of images:
You should know that the potential outside of a conducting sphere with charge Q is the same a a point charge Q at the center of the sphere. Use that to obtain the exact solution, and check that it gives a potential of zero at the location of the plane.

Then, expand your solution in terms of Legendre polynomials and calculate the first 4 coefficients.

(2)Laplace's equation--SoV:

The boundary condition that the potential on the plane is zero is fine, although you need to be careful to express that condition properly. (Saying that $V(2R,\theta)=0$ is not the right condition)

Your other boundary condition; that $V\to 0$ as $r\to \infty$ is true, but(!) way too restrictive! It contains no information about the charged sphere.. Instead, try using a more informative boundary condition: what is the potential far from the conducting plane, but not so far that the potential due to the charged sphere is zero? (i.e. what is V at $r\gg 2R$?)

If you post your work for either method (or both) I will be happy to assist you further.

3. Dec 9, 2008

### turin

I agree with using the method of images, and I don't understand why the problem talks about "the first four terms", because the solution can be found exactly. Hint: what is the shape of the equipotential surfaces due to two opposite point charges?

4. Dec 10, 2008

### source.decay

I thought that I could do the point charge at the center of the sphere for each one and then the problem become trivial, but don't you have to take into account the radius the sphere?

So with the idea that the sphere becomes a point charge, you have two point charges, one positive, one negative a distance 4 r away from each other. Giving a potential of

V(x,y,z) = 1/(4*pi*epsilon naught)*[q/(sqrt(x^2+y^2+(z-2R)^2))-q/(sqrt(x^2+y^2+(z+2R)^2))]

When z=0, at the plate, then V=0 and when x^2+y^2+z^2 >>(2R)^2, V goes to 0.

Then I could convert it over to the spherical and put in the Legendre polynomials?

5. Dec 10, 2008

### gabbagabbahey

Hmmm... maybe there is more than meets the eye to this problem: For a spherical shell of uniform charge density, and total charge Q, the potential outside the shell will be the same as the potential due to a point charge at the center of the sphere (you can see this by using Guss' law to quickly find the field outside the sphere).

But what if the charge on the shell isn't uniform? Does the same thing apply?

Can you be sure that the charge on the shell is uniform in this case?

IF the charge on the spherical shell is uniform, then this would be a correct approach.

So, the burning question is whether or not the charge is uniform on the conducting shell. Can you reason why it isn't?