E-field of cylindrical conductor above infinite ground plane

1. Mar 21, 2015

bleach2015

1. The problem statement, all variables and given/known data
Find the electric field between the conductor and ground. The conductor is at:
• Potential = +V0,
• distance d from the ground plane.
2. Relevant equations
I used image theory to create a conductor at -V0 at distance -d from the ground plane.

Laplace's equation: ∇2V = 0

3. The attempt at a solution
My main road block on this problem is that the cylindrical conductor is NOT at the origin. This means the symmetry in φ and z do not apply, and I cannot easily solve Laplace's equation since V = V(r,φ,z) instead of V = V(r) for a cylinder at the origin. Note that I am in second year electrical engineering, so I don't think we are supposed to solve PDEs... Also, if I set my positive charged conductor at the origin, then my image conductor is still not symmetric with respect to φ and z, and I face the same concern.

Can someone give me a hint on what to do when the conductor is not centered at the origin? Right now, I have the ground plane set as the y = 0 plane and the conductor lying on x = 0.

Thanks a lot!!

Last edited: Mar 21, 2015
2. Mar 22, 2015

BvU

Hi bleach, a belated welcome to PF !

I'm having trouble to imagine the situation; the problem statement seems incomplete to me .
From your attempt at solution I gather the ground plane is conducting and at zero potential (perhaps a standard convention in EE - but you're posting in a forum where physicists roam free...).

But I can't make out why the cylinder can't be at the origin of the coordinate system, because you don't say much about that system.
(This is not to say that a coordinate system with the cylinder at the origin would be a smart choice, for sure !)

So if you tell me the cylinder is with its center of mass at (0,0,d) and is oriented either along the z axis, or parallel to the x-axis, then at least I know what problem you are facing.

And either way, I do think your mirror charge approach is the way to go.

For a complete expression of $\vec E$ all over the place you will also need the length of the cylinder. But it could well be that in EE jargon, "between the conductor and ground" means " means "in some specific area or on some specific line"...

3. Mar 22, 2015

bleach2015

Hi BvU, thanks for the reply!

You're correct about the ground plane being a conductor at zero potential. The problem setup is shown in the picture below:

The cylinder and ground can both be considered to be infinitely long.

I just need the Electric field on the y axis between the cylinder and ground. Can that simplify my calculations somewhere?

4. Mar 22, 2015

BvU

Aha! so I can use the pile of scratch paper I reserved for the whole exercise for something else ! Good for me and for the environment !

Infinitely long means the end effects aren't there and this is a 2d problem (from mirror symmetry wrt the xy plane there's no z component). Phew !

And you only need the field on the y-axis. Piece of cake.

The whole mirror charge thingy is based on the uniqueness of solutions to the Laplace equation. Find one, then that's it!

Mirror disk with charge -q at -d is good.

The problem statement still is incomplete to me : we don't have something for the radius or the diameter, and somehow I don't think I can hope that it's zero .
But if it is, we're done.

If it isn't zero, then we can still have the bright idea that for small r/d a (2d) dipole field is a good step towards the solution. Turns out to be the way to go.
Check out the shape of the equipotential planes for two parallel line charges !