# Point charge inside Dielectric Sphere embedded in Dielectric Slab

1. Oct 4, 2009

### 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 ($$\epsilon$$2), radius R, that is embedded inside some other infinite slab of dielectric ($$\epsilon$$1)?

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

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

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

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

2. Oct 5, 2009

### 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.

3. Oct 5, 2009

### 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.

4. Oct 5, 2009

### 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:

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

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

5. Oct 5, 2009

### 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