Point charge inside Dielectric Sphere embedded in Dielectric Slab

OsCiLL8

I've been working on this for a little while now (in CGS units), and am not really sure where I've gone wrong at in calculating the potential, so I've come here! Here is the problem:

What is the potential caused by placing a point charge Q at the center of a dielectric sphere ([tex]\epsilon[/tex]₂), radius R, that is embedded inside some other infinite slab of dielectric ([tex]\epsilon[/tex]₁)?

Here's what I've determined so far:
D(r) = Q/r²
E(r<R) = Q/[tex]\epsilon[/tex]₂*r²
E(r>R) = Q/[tex]\epsilon[/tex]₁*r²

So, letting P = (D-E)/4[tex]\pi[/tex] , I've found

[tex]\Phi[/tex](r<R) = Q/r + ([tex]\epsilon[/tex]₂-1)*Q/(3*[tex]\epsilon[/tex]₂*r)

[tex]\Phi[/tex](r>R) = Q/r + ([tex]\epsilon[/tex]₁-1)*Q/(3*[tex]\epsilon[/tex]₁*r)


My question is, shouldn't I have the option of allowing the potential to be continuous at the interface?? Have I left out some surface charge polarization or something?
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution

gabbagabbahey

Okay, so far so good...

Why are you calculating the polarization, and how did you go from the polarization to the potential?


"Option" is a poor choice of words, the potential must be continuous everywhere....the fact that yours is not should be a dead giveaway that you've done something wrong.

OsCiLL8

Calculating the polarization allows me to determine the contribution to the potential from the polarization surface charge density and polarization volume charge density.

I think your incorrect about the need for a continuous potential. The parallel E component and the perpendicular D componenent have to be continuous, implying that the potentials at the boundary be equal UP TO a constant. Setting the constant equal to zero makes the potential continuous, while setting the constant equal to a nonzero number implies that there is some work function required to go from one dielectric to the other.

gabbagabbahey

Okay, but you will also need to account for the potential due to the point charge, and you have made an error somewhere. If you post your calculations, I can point it out to you.

Alternatively, you can save yourself from the hassle of this method altogether by just using the definition of potential:

[tex]\textbf{E}=-\mathbf{\nabla}V\Longleftrightarrow V(\textbf{r})=\int_{\mathcal{O}}^{\textbf{r}}\textbf{E}\cdot d\mathbf{l}[/tex]

No, the potential must be continuous everywhere....What would the force be on a point charge located at a discontinuity in potential?

ehild

The integral is a continuous function of the upper limit. Integrate E(r) from the place where the potential is chosen 0 (that is infinity) to r.

ehild