offsetted concentric spheres

Ah.....

So the *radius* of the outer sphere is 64mm. Before you said the *diameter* was 64mm. Now *that* can actually be built ;-)

If you calculate the field formed by two parallel plates 10 mm apart with

100kV difference, that at least is the upper bound for the field intensity. On the radial going outward from the center of the large sphere, through the center of the small sphere, the field will be strongest in the 10 mm space between the two. Everywhere else it will drop off since the distance between the two surfaces increases.

The weakest part will be on the opposite side of the inner sphere. It will be stronger than field between two parallel plates that are 99mm apart (the distance between the inner sphere and the opposite wall of the outer sphere).

Not a lot of help, but it at least gives you a couple of 'boundary' values.

daestrom

Smythe. Smythe. Smythe.

-- Ferme le Bush

Thank you Deastrom, for your help indeed it was a mistake (diameter vs radius). I 'll try if i can solve the problem now. Maby also combined with mirror charges.

The problem of a cable with a central conductor which is not co-axial with the sheath is similar- only with a line charge- not a point charge (and is more practical in that it is possible). In any case, the problem comes down to locating a charge and its image. Note that the potential of either sphere, with respect to a reference is an equipotential surface and is described by a constant ratio of the distances to the charge and its image. This reduces to the equation of a circle offset from the origin . Note that for a charged conductor above ground, using images, the location of the charge is not at the center of the conductor. Solve this problem. Same thing for two equipotentials. You then can find the charge location to give the desired equipotentials and from this and the voltage difference, find the charge. knowing charge and its location, fid the field in the in-between space.

I've done it for cables and the change to spheres doesn't seem to be onerous. starting hint based on a cylinder or line charge case: based on the line charge case where V=charge /(2*pi*eps0)ln R2/R1 where r2 is the distance from the image and R1 is the distance from the primary charge. For an equipotential R2/R1 is constant and from this the equation of the equipotential circle can be found. This is the conductor surface and this leads to: Conductor of radius b with center at distance h from a reference plane. charge and image located at +a and -a from this reference plane (which in your case is not the surface of the outer conductor). Then a^2=h^2-b^2 .

P.S. don't take me on trust-I could be lying to you- do it for yourself.

--

Don Kelly @shawcross.ca remove the X to answer

In most cases, it is not possible to morph two dimensional potential distributions into valid three dimensional potential distribution. The best bet is to use the reference Smythe that I mentioned in previous posts.

For spheres, geometric inversion may be the way to go. Many images may be necessary. Smythe, not an easy read, is the way to go. These days, with computers, the art of Smythe's methods is probably a lost art.

Bill

-- Ferme le Bush

------------- I agree that, in general cases, morphing will not work and if there are more than two spheres (or circles as it may be) then one has to consider multiple images.

In this case, with the offset in one direction only, it appears solvable by extending from the 2 dimensional case (which in turn can be solved by eventually considering only the one dimension). The object is to locate the equivalent point charges with respect to the centers of the spheres. The final solution for fields is a bit more complex in the point charge case than the line charge case simply because of the added dimension.

With the line charge case one looks at an expression of the form (x-a)^2

+y^2 =R^2 and evaluates R and a for point on the equipotential where y=0. With a sphere there is a z component but again consider that y=z=0 and the rest follows. At least it appears so to me.

Just to refresh my memory, I looked at my copy of Smythe. There is a section that covers the exact problem of unequal spheres, both internal and external. To solve that problem using images, an infinite set of images is required.

If the problem is really important, GO TO THE LIBRARY! BUY THE BOOK.

Bill

-- Ferme le Bush

Are there cases where this is not necessary or does he deal with a general case? Just curiosity. Certainly the multi image method is tedious with images of images. The problem may reduce to one of locating the charge within a single sphere so that the sphere is an equipotential. I was considering only an offset on one axis.

The problem with spheres is not y important to me except for curiosity and it is not convenient for me to buy Smythe at my present location as a retiree living in the boondocks. If I get the chance, I will try to look at the book. Thank you

Many of the problems described by Smythe require use of infinite series. The exposition regarding spheres starts at Section 5.08 in the book. For a single sphere, you get a spherical equipotential by placing a charge at the center of the sphere. The presence of the second sphere distorts that equipotential and you need to add appropriate image charges. These image charges are not equal to the central charge.

Smythe realized that infinite series gave computational difficulty. With modern computation that is not much of a problem. Nevertheless, he describes how to solve such problems using three-dimensional inversion. If you have a solution to Laplace's eq1uation, a geometrical inversion of the field leaves to another field that is a solution as well.

His exposition of three-dimensional inversion starts with Section 5.09. He also describes inversion for two-dimensional problems.

I cannot overemphasize that numerical answers are readily achieved using computer methods. Because static temperature distributions obey Laplace's equation, many electrostatic problems can be solved by adapting thermal codes.

In the end, the curse of computation often is that you get answers without understanding.

Bill

-- Ferme le Bush