# Electric Field in a cavity of a dielectric medium

1. Jul 24, 2011

### XCBRA

1. The problem statement, all variables and given/known data
The polarization charge in the surface of a spherical cavity is $-\sigma_e \cos\theta ,$at a point whose radius from the centre makes and angle $\theta$ with a give axis Oz. Prove that the field strength at the centre is $\frac{ \sigma_e}{3 \sigma_e}$ parallel to Oz.

If the cavity is in a uniform dielectric subject to a field strength $E_0$ parallel to the direction $\theta = 0$, show that
$$\sigma_e = 2 E_0 \epsilon_0 \frac{(\epsilon_r - 1)}{(1+\epsilon_r)}$$,
where $\epsilon_r$ is the relative permittivity of the dielectric.

2. Relevant equations

3. The attempt at a solution

First to find the field stength i have used Gauss' law:

$$\epsilon_0\int E.ds = \int \sigma da$$

$$4\pi R^2 \epsilon_0 E = \int \int \sigma R^2 d\theta d\phi$$

Then for the second part,

I have let $V_1$ for the potential inside the cavity and $V_2$ for the potential outside the cavity.

I assume that:

$$V_1 = B_1 r\cos\theta + \frac{B_2 \cos\theta}{r^2}$$
$$V_2 = -E_0 r\cos\theta + \frac{A_2 \cos\theta}{r^2}$$

then since $V_2 \neq \infty, B_2 = 0$.

Then since $V_1 = V_2 at r=R$

$$(B_1 + E_0) R^3 = A_2$$

then find the radial electric field components:

$$E^r_1 = - \frac{\partial V_1}{\partial r} = -B_1 \cos\theta$$

$$E^r_2 = - \frac{\partial V_1}{\partial r} = E_0\cos\theta+\frac{A_2 \cos\theta}{r^3}$$

then taking the boundary condition

$$D^r_1= D^r_2 \epsilon_0 \epsilon_r E^r_2 = \epsilon_0 E^r_1$$

to give

$$B_1 = - \frac{3 \epsilon_r E_0}{2 \epsilon_r + 1}$$

therefore inside the cavity:

$$P = (\epsilon_r - 1) \epsilon_0 E_1$$

$$P.n = P \cos\theta = \sigma_e cos\theta$$

$$P = \sigma_e = \frac{3(\epsilon_r-1) \epsilon_0 \epsilon_r E_0}{(2\epsilon_r +1)}$$

I am not sure why i have the extra $\epsilon_r$ in the numerator, and I am not sure if there is a simpler method to do this?

Thank you for any help.

2. Jul 25, 2011

### mathfeel

Write down the spherical harmonic expansion on either side of the boundary and match the appropriate boundary condition. The potential outside is particularly simple. Because at large distance from the cavity, it must reduces to $\vec{E} = \frac{E_0}{\epsilon_r} \hat{z}$, which corresponds to potential: $\phi = - \frac{E_0}{\epsilon_r} z = - \frac{E_0}{\epsilon_r} r \cos\theta$.

3. Jul 25, 2011

### XCBRA

The question I had asked earlier was indeed the first part of the question, but I am now having trouble with he second part and felt necesary to include the first part answer and soltuion to give the context to the second part.

I just realsied that there is typo in

and it should indeed be:

$$\sigma_e = 3 E_0 \epsilon_0 \frac{\epsilon_r -1}{2\epsilon_r +1}$$

with regards to the expansion. Wouldnt the field outside the cavity need to have an extra term other thant just the term at infinity as the cavity will distort the field around it?

With the expansion i did above,

I get for inside the sphere $V_1 = - (\frac {3\epsilon_r}{1+2\epsilon_r}) E_0 r \cos\theta$

then for outside: $V_2 = - (1-\frac {R^3(\epsilon_r - 1)}{r^3(1+2\epsilon_r)}) E_0 r \cos\theta$,

however i am not entirly sure i understand how to then go about finding the surface charge density of the cavity?

Thnak you for you time.