I had been asked by some of my friends to post a solution for this problem. Personally, I feel this is an extremely important problem to understand as a lot of other similar problems emerge which requires a similar approach. Understanding this problem also helps us appreciate how we could know so many different things about a system by just knowing its polarisation. So do follow through and let me know if some parts are unclear.

We essentially have a spherical shell which is made up of some dielectric material we don't care of. What we do care about is how the electric field changes as we go from the centre of the sphere to the inner walls and the outer walls. So if we were to imagine the centre as the origin then we are interested in finding how the electric field changes when 0<r<a , a<r<b and r>b.

So lets first start with finding how the electric field changes when we go from centre to the inner walls. For this, we could use our good old Gauss's law. Gauss's law gives us this really beautiful and crucial relationship:

But we don't know what Q is or do we? Total charge is essentially made up of free charges and bound charges. In this system, we don't have any free charges. This makes our job much easier. So all we are interested in is finding the bound charges. Lets recall the relationship between the bound charges and polarisation.

Lets first start with finding the volume bound charge density. Its important to notice that the polarisation is only dependant on r so we are only interested in the first term of the divergence in spherical coordinates formula.

After doing the necessary computation, we get the bound volume charge density as -k/r^2. Now lets do something similar with surface bound charges. Because its a sphere, the normal vector will be perpendicular to the surface which is just the vector r. The normal vector r will be pointing inwards for the inner sphere.

So after completing the necessary computation, the surface charge density for the inner sphere comes out to be -k/a.

All right, so we are done with the hard part. So now we could simply use our gauss's law to find the electric field. We will have to repeat this method when handling with a<r<b

and r>b parts of the problem. Below is the complete computation of the problem.

Do let me know if you are still unclear of some part.

## Comments