# Variable Dielectric

1. May 18, 2013

### sazerterus

1. The problem statement, all variables and given/known data

Calculate the capacitance of a spherical capacitor of inner radius A and outer radius B which is filled with a di-electric varying as e=c+dcos^2θ . c and d are constants and θ is the angle made with the z axis.(also known as the polar angle)

2. Relevant equations
$C=εA/d$

3. The attempt at a solution
At first,i thought maybe i should take the potential by keeping θ constant,and then taking the average of the potential to get avg V and then find the potential.Again,I realized that we could also consider the spheres to be thin parallel plates.But,both the process give different answer.Is either of them right?and if one is,can someone explain the process.

2. May 19, 2013

### Thaakisfox

Use the fact that the polar angle dependence of the permittivity arises only in the displacement field not the electric field, hence the electric field has only radial dependence.
To find the capacity you need to find the potential difference between the two plates. For this use the integral form of Gauss's law (take care of the displacement field) to find the radial dependence of the electric field. From here finding the potential difference should be easy.

3. May 19, 2013

### sazerterus

Could you please show how to start the calculations?Could you provide setup of any similar problem?Thanks for the help.

4. May 19, 2013

### Thaakisfox

Suppose the charge on the inner sphere is Q. Now surround the sphere by an imaginary sphere S of radius r. By Gauss's law:

$$\oint_{S}\textbf{D}\;d\textbf{f}=Q$$

Where df is the element vector of the spherical surface. Consider a linear medium in this case the displacement field proportional to the first power of the electric field:

$$\textbf{D}(r,\theta)=\varepsilon(\theta)\textbf{E}(r)$$

(convince yourself the electric field only depends on r).

Now then we have:

$$E(r)r^2\oint_{S}\varepsilon(\theta)d\Omega=Q$$

Now just calculate the integral of the permitivity over the solid angle, and from here express the electric field.

When you have the electric field, integrate it between A and B to get the potential difference and from there the capacitance is self evident.

5. May 19, 2013

### sazerterus

Finally I got it!! Thnx a lot.