**Required math: calculus**

**Required physics: electrostatics**

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 4.34.

Another example of solving a boundary value problem in a system with linear dielectrics. Suppose we have an ideal dipole at the centre of a sphere of dielectric (dielectric constant ) and radius . What is the potential at any point (inside or outside the sphere)?

We can use a similar approach to that of the problem of a dielectric cylinder in an electric field. We first specify the boundary conditions. At the surface of the sphere, the potential must be continuous, so we have

As we saw in the cylinder problem, the condition on the normal derivative of the potential is, since on the outside of the sphere, :

Since there is no external field, we must have as . Finally, as , the potential must behave like that of an ideal dipole, so, assuming that points in the direction:

Note that we’re using rather than in this formula, since the dipole is inside the dielectric and the potential is reduced by a factor of , where .

With these conditions, we must solve Laplace’s equation in spherical coordinates, so we can quote the general form of the solution:

Inside the sphere, we have

Outside, we have

Applying the continuity condition 1 and equating coefficients of the , we get

We can therefore express the in terms of :

Using the condition on the derivatives 2 and equating coefficients as before, we get

Substituting for into the second equation gives

The only way this can be satisfied is if for , so we get for .

For the case, we can solve the two equations in and to get (using ):

The potential is

### Like this:

Like Loading...

*Related*

jerrywhile solving the equations for A1 and B1, why do we not take into account the cos(theta)

growescienceThe shouldn’t have appeared in the equation for and since we are equating coefficients of . I’ve fixed the equation now.

Pingback: Dipole in dielectric sphere | physics

TimWhy is it that is ok? This will blow up as approaches 0.

growescienceI think it’s because the dipole is ideal, in the sense that it’s a point object. In any real system, we’d need to take account of the non-zero size of the charged object. You get the same problem with the potential and electric field of point charges.

LauraFor the potential for an ideal dipole, why is the permittivity epsilon instead of epsilon naught or epsilon_r?

growescienceIn a dielectric, we have to use the permittivity of the dielectric rather than which is the permittivity of free space (that is, vacuum). See here for more details.

Pingback: Dipole in a spherical cavity in an infinite dielectric – Internet and Tecnnology Answers for Geeks

drabussShouldn’t B_1 equal

B_1 = (3p)/(4 \pi \epsilon)*(\epsilon_r / ( \epsilon_{r}+2))?

gwrowePost authorI assume you’re referring to eqn 17 (please use equation numbers in comments!). It is equal to what you say, but if you then use , it simplifies to eqn 17.