# Dipole in a dielectric medium

1. Jul 25, 2011

### XCBRA

1. The problem statement, all variables and given/known data
A dipole p is situated at thecentre of a spherical cavity of radius a in an infiite medium of relative permitivity $\epsilon_r$. show that the potential in the dielectric medium is the same as would be produced by a dipole p' immersed in a continuous dielectric, where

$$p'=p\frac{3\epsilon_r}{2\epsilon_r +1}$$

and that the field strength inside the cavity is equal to that which the dipole would produce in the absence of the dielectric, plus a uniform field E

$$E=\frac{2(\epsilon_r-1)}{2\epsilon_r + 1}\frac{p}{4\pi\epsilon_0a^3}.$$

2. Relevant equations

3. The attempt at a solution
I am not sure at all how to approach this question. I would like to say that I would use spherical harmonics but i am not sure how to apply them in this case.

Would it be possible to say that at large distances

$$V_2= -\frac{p\cos\theta}{4\pi\epsilon_0\epsilon_r r^2}$$

then to add then assume that outside the sphere that

$$V_2= -\frac{p\cos\theta}{4\pi\epsilon_0\epsilon_r r^2} + \frac{A_2\cos\theta}{r^2}$$

and inisde the sphere that

$$V_1= B_1 r \cos\theta + \frac{B_2\cos\theta}{r^2}$$

and then solve the problem using the boundary conditions for tangential E and perpendicular D?

I am really unsure of how to solve this and any help will be greatly appreciated.

2. Jul 25, 2011

### mathfeel

Kind of. Since $A_2$ is unknown, you might as well just try the potential: $V_2 = - \frac{p^{\prime} \cos\theta}{4\pi \epsilon_0 \epsilon_r r^2}$, where $p^{\prime}$ is unknown and to be solved.

The whole point of the problem is that at large $r$, the field looks like something due to some effective dipole moment $p^{\prime}$, whose value you are to find.

$D$ is okay, but instead of using $E$, it is easier to use the condition that $V$ is continuous.

So now you have two equations but three unknown: $B_1$, $B_2$, and $p^{\prime}$. But one of them can be found by consider the limit $r\to 0$.

Last edited: Jul 25, 2011
3. Jul 26, 2011

### XCBRA

Ahh ok that makes a lot more sense, thank you for the help.