Solving Polarization Question for Electric Field in Spherical Shell

[rbnphlp]

A thick spherical shell (inner radius a, outer radius b) is made of dielectric materials with a "frozen-in"polarization
P(r) =k/r rhat;
where k is a constant and r is the distance from the centre. There is no free charge.
Find the electric field E in all three regions by two different methods:
(a) Locate all the bound charge, and use Gauss’ law to calculate the field E produced.
Im stuck for a<r<b
so using gauss's law for a<r<b ,

I need to find the Q enc= ∫-k/r^2dr (4πr^2) between r and a , k/r^2 comes from usinfg ∇.P=-k/r^2 , which I get to be -4∏(r-a)k ..which is not the answer its -4∏(r)k , which means Qencl is actually=-4∏(a)+4∏(r-a)k, so I have missed the charge at a ..But isn't the integral take into account all the charges from r to a, do I need to an additional -4∏(a)k just at r=a?

can someone explain why?

Thanks :)

 

[eljose79]

Thank you for your question. I can understand why you are confused about the charge at r=a. Let me try to explain it to you in a simpler way.

Firstly, let's think about what Gauss' law tells us. It states that the electric flux through a closed surface is equal to the enclosed charge divided by the permittivity constant. In this case, we have a spherical shell, so we can use a spherical Gaussian surface to apply Gauss' law. This surface will have a radius r, where a<r<b.

Now, let's think about the charge enclosed within this surface. As you correctly mentioned, the bound charge is given by ∫-k/r^2dr (4πr^2) between r and a. This means that the charge enclosed is actually the difference between the charge at r and the charge at a. So, the total charge enclosed is (-4∏(r)k) - (-4∏(a)k) = 4∏(a)k - 4∏(r)k = 4∏(a-r)k.

I hope this helps clarify your confusion. Please let me know if you have any further questions. Good luck with your calculations!