**Required math: calculus**

**Required physics: electrostatics**

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 4.24.

As an exercise in applying Laplace’s equation to a problem with dielectrics, suppose we have a conducting sphere of radius surrounded by a spherical shell of dielectric of outer radius and susceptibility , with the whole system in a uniform electric field .

The general solution to Laplace’s equation for the potential in spherical coordinates is

where is the degree- Legendre polynomial.

Inside the sphere, the potential is constant (since the field is zero inside a conductor), so we might as well take it to be zero.

In the region , the solution is the general one given above. For , in order to avoid an infinite field for large , we have to drop the terms except for since we need a potential that gives a constant field. If we take the field to lie in the direction, then since we have for this region

Note that we’ve used a different set of coefficients here, since the solution in this region is distinct from that for the region . That is, the coefficients .

We can obtain conditions on the coefficients by equating terms of the various Legendre polynomials in the boundary conditions. Continuity of the potential at gives the condition

Continuity at gives

Finally, we can use the condition at the boundary of two dielectrics to get, since there is no free charge at the boundary:

Consider first the terms for . We get

Since both and are proportional to , the only way the third boundary condition (at the dielectric/vacuum boundary) above can be satisfied is if for . We are therefore left with the terms. For these we get

Plugging these into the third boundary condition gives

In terms of the dielectric constant we get

The potential inside the dielectric shell is therefore

The field can be found from the gradient

This reduces to the situation of a conducting sphere in a uniform field if we set (effectively replacing the dielectric by a vacuum). In that case, the potential reduces to

The first term is just the applied field, and the second term arises from the induced charge on the conductor.

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georgianajackalHad been working on this problem since last night and finally I just put down the books at 1:30 am, started working again at 4:00 am in the morning. Finally, done it now but I had to take help from here. it was 6:00 am then.

so with help I could do it in a quarter to thee hours. Goodness knows how much longer would have it taken without all this help.

Thanks Growescience.

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Forrest ShriverI believe I caught an error. In equation 7, in the second term on the LHS, should the term in the denominator not be , instead of ? Sorry if I’m mistaken.

gwrowePost authorFixed now. Thanks.

GeorgiaHow are you able to take the potential inside the sphere to be zero, whilst also defining the potential in the limit that r-> infinity as E_0 rcos(theta)?

I would have thought that would result in the potential at large r being E_0 rcos(theta) – some constant that is required to make the potential inside the sphere zero?

Many thanks- This page is amazing!!!!

gwrowePost authorThere was an error in my equation 2, in that the solution for should have a different set of coefficients (instead of ), which I’ve now fixed. I think the source of your confusion is that we have a

differentsolution function in each of the three zones, and we make them all consistent with each other by matching them up at the boundaries. Thus we are free to set for and also for large , provided that we impose the boundary conditions. The solution is adifferentfunction from , so we shouldn’t include any information about in the series for .DustinFor the boundary condition at r = b how did you get E_o *b for the first term?

gwrowePost authorIf you mean equation 4 (please use equation numbers when asking about an equation!), that’s the term at , which comes from equating the corresponding terms in equations 1 and 2.