Electrical Field of a Sphere

1. Oct 5, 2008

eintagsfliege

Hi together

I want to calculate an electrical field between a sphere and a plate, where the sphere has a certain voltage and the plate is on ground.
Has someone a suggestion to begin?

2. Oct 5, 2008

Ben Niehoff

Put an image sphere held at -V on the other side of the plate.

3. Oct 5, 2008

eintagsfliege

The more little the radius of the sphere is, the higher the electric field grows.
But how should the formula look like for the electric field?

4. Oct 5, 2008

granpa

plate=infinite plane? (i.e. much larger than the sphere and much larger than the distance between the sphere and the plane)

are you familiar with lines of flux?

Last edited: Oct 5, 2008
5. Oct 5, 2008

clem

Ben's suggestion of an image sphere is correct, but there is more to the problem.
You should know how to do a point charge and a conducing sphere at V.
You have to keep adding image charges in a series to get the pot for two spheres.

6. Oct 5, 2008

ZapperZ

Staff Emeritus
Did you not read what Ben Niehoff suggested?

The problem with your question here is that you neglected to let the rest of us know what YOU know, i.e. are you able to solve the typical advanced undergraduate E&M problem? Can you solve Poisson's equation? Are you familiar with the method of images?

Someone familiar with those would have automatically understood what is meant by putting an image sphere on the other side of the plate.

Zz.

7. Oct 5, 2008

eintagsfliege

I am sorry.
Some years ago, I had lectures about Electrodynamics. Unfortunately, I don't have my books here. But so far I remember these Image Method.
I just can't remember how to calculate the electrical field between the inifinite plate and the charged sphere with a radius a.
Maybe someone has a little time to explain.

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8. Oct 5, 2008

granpa

I misunderstood the question.

post 2 is pretty much all there is to it.

9. Oct 5, 2008

Antenna Guy

Draw rings of constant field strength about the sphere at uniform intervals. When the rings reach the plate, connect the ends (at ground) using an arc of the same radius - but from the center of a mirror image of the sphere (i.e. mirrored across across the ground plane).

Field strength falls off with with 1/r, so each succesive ring represents 1/n of the field at the first ring (i.e. 1/1, 1/2, 1/3, ...). Vector add the direct and reflected fields where the rings cross over.

Regards,

Bill

10. Oct 5, 2008

granpa

and there is no field at all on the other side of the plate?

11. Oct 5, 2008

eintagsfliege

No field, just the one of the sphere and the plate

12. Oct 5, 2008

granpa

I mean due to the sphere

13. Oct 5, 2008

ZapperZ

Staff Emeritus
This is NOT a simple problem. You have to set up the Green's function for the geometry using the image sphere at an opposite potential as your real sphere. This is the only geometry that will give you the same boundary condition on your infinite plate. Setting and solving this isn't trivial, especially if you haven't done this in a while or haven't done Jackson.

Zz.

14. Oct 5, 2008

granpa

ah yes the plate is an equipotential (its at ground) so the field lines must enter it at right angles.

15. Oct 5, 2008

eintagsfliege

True!
Could you give me a hint to set up the green function for a sphere?

16. Oct 5, 2008

Ben Niehoff

The problem of two spheres is quite non-trivial. Do you know the Green function for a point charge near a conducting sphere, in spherical polar coordinates? You will need to take that Green function, and integrate it over the second sphere.

Alternatively, you can consider an infinite series of image charges, derived by finding the images of each sphere in the other, recursively, ad infinitum.

Yet a third way is to take the Green function for a point charge above an infinite plane, and integrate that over your original sphere.

Any way you cut it, the problem is not simple.

17. Oct 5, 2008

granpa

I've never heard of this before. Is this common with other geometries or is it particular to spheres? are the image charges inside the spheres? if so then how is the net charge of the sphere conserved?

18. Oct 5, 2008

Ben Niehoff

It works with parallel cylinders, too, although in that case it is probably easier to use conformal mapping.

The image charges are placed inside the spheres, yes. Charge is conserved by placing equal and opposite image charges at the centers of the spheres, as well. I.e., if the sphere has charge Q and a particular image charge is q, then you place q at the image location, and put -q at the center of the sphere.

However, if the spheres are held at constant potentials, then charge isn't conserved anyway (because it can flow through whatever apparatus holds the spheres at constant potential).