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Homework Help: Point charge inside Dielectric Sphere embedded in Dielectric Slab

  1. Oct 4, 2009 #1
    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
  2. jcsd
  3. Oct 5, 2009 #2


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    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.
  4. Oct 5, 2009 #3
    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.
  5. Oct 5, 2009 #4


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    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?
  6. Oct 5, 2009 #5


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    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.

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