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Homework Help: Electric Potential inside insulating solid sphere.

  1. Sep 25, 2010 #1
    1. The problem statement, all variables and given/known data

    Hi everyone,
    I'm supposed to find an expression for the electric potential as a function of r, the radial distance inside a solid and non conducting sphere of radius R. A total charge of q is uniformly distributed throughout its volume. The annoying part is that I'm supposed to do that using a relation between the potential and [tex] \rho[/tex] , the charge density.

    2. Relevant equations

    The equation the problem wants me to use is:

    [tex] V( \textbf{r} ) = \frac{1}{4 \pi \epsilon_{0}} \int \frac{\rho ( \textbf{r} )}{r} d \tau ' [/tex]

    [tex]r[/tex] being the distance between source & field (observation) point.

    I already have obtained a result using a different method (integrating the electric field) and got:

    This is also what I found on different sources so I'm pretty sure the result is correct.

    3. The attempt at a solution

    Yet I can't set up my integration is such a way so as to retrieve this result.
    Using the law of cosines on the following figure:

    I have: [tex] r^{2} = r'^{2} + z^{2} - 2 r' z cos ( \theta )[/tex]

    Using spherical coordinates, I make the integral range from 0 to 2pi and 0 to pi for phi and theta.

    For [tex]r'[/tex] I am more doubtful. Indeed, when we integrate with respect to the two angles first, we obtain an expression that involves the square root of the following difference: [tex](r' - z)^{2}[/tex].
    Thus depending on where the observation point is on the sphere, the positive or negative root is to be taken. So I would guess I could split the integration into two parts, from R (the radius of the sphere) to r, and then from r to 0. Since [tex]\rho[/tex] is zero everywhere outside the sphere, there is no need to integrate from infinity, where the potential is assumed to go to zero.

    Yet I do not get the expected result, and this using different integration limits. I am assuming I am doing something wrong initially, and if someone has any suggestion I would be very grateful for that.
    Last edited: Sep 25, 2010
  2. jcsd
  3. Sep 25, 2010 #2


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    Homework Helper

    In that element [tex](r' - z)^{2}[/tex], [tex]r'[/tex] and [tex]z[/tex] are simple distances, not vectors, right?
    Don't you mean from R to z, then z to 0? (or rather, 0 to z and then z to R, I guess)

    Subject to the couple of notes above, your setup seems fine. I did the problem setting it up the same way you've described, and I got the expected answer, so perhaps you have a math error somewhere. If you post the details of your work we can help you identify it.
  4. Sep 26, 2010 #3
    Hi diazona,

    yes just distances, and yes again for z instead of r, that's what I meant indeed.

    I'm going to post here the details of my calculations by Sunday afternoon, thanks a lot for taking a look at my problem.
  5. Sep 26, 2010 #4
    hi again,

    I miraculously realized that I had a math mistake indeed when I was carrying my integration.
    I obtained the exact same result, except that I got a minus sign in front of it.
    You mentioned that the integration limits should be from 0 to z, then z to R. I did the opposite, which thus gave me the negative of my expected result - yet I'm a bit confused.

    Isn't the integration assuming that our reference point is at infinity, thus we're "coming from" infinity, integrating from +∞ to z (or simply R to z since the charge density is zero outside the sphere) and then z to 0. That's what sounds logical to me even though it obviously leads the wrong answer. Do you know why?
  6. Sep 26, 2010 #5


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    Homework Helper

    You're mixing up two completely different equations. The one that requires a reference point at infinity is the definition of V as the path integral of E. When you calculate
    [tex]\int_A^B\vec{E}\cdot\mathrm{d}\vec{s} = V(A) - V(B)[/tex]
    all you really get is a difference - that is, you only know that the potential at point A is such-and-such relative to the potential at point B. In order to define "the potential at A" you have to have a standard reference point B.

    In this problem, though, the integral defines V as the sum of contributions from each of the individual charge elements that make up the sphere. Conceptually you could think of it as a limit of
    [tex]V = \sum_i V_i = \sum_i \frac{kq_i}{r_i}[/tex]
    as the individual charge elements [itex]q_i[/itex] go to zero. There is no path involved here, and what you get isn't a difference between two values, so there's no need to have a standard reference point to compare everything to. The result you get is already absolute.
  7. Sep 26, 2010 #6

    Oh goodness, I got it. In the second equation we're simply integrating the charge density over the whole sphere while there is no such idea of reference point I was getting myself confused with - we're just adding the "potential" contribution of each individual charge point. Thanks a lot for your help, I really appreciate it.
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