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Homework Help: Problem with charge distribution and potential

  1. Oct 4, 2005 #1
    Hi, I have a problem with finding charge distribution. If anyone could provide help or any tips as to how to solve it, I would greatly appreciate it.

    I have already solved half of the problem.

    The potential V(r) of a spherically symmetric charge distribution is given by V(r) = (Q/4*Pi*e0*R)[-2 + 3(r/R)^2] for r<R and V(r) = Q/4*Pi*e0*r for r>R. Use Gauss's law applied to Gaussian surfaces at various radii to calculate the charge distribution that gives rise to the potential given.

    First I found the Electric field. -dV/ds=E

    E(r<R) = (Q/4*Pi*e0)(6r/R^3)
    E(r>R) = (Q/4*Pi*e0*r^2)

    How would I find the charge distribution?

    Thanks for any help! =)
  2. jcsd
  3. Oct 4, 2005 #2
    gauss's law (in differential form!) is

    \nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0}.

    now have at it!
  4. Oct 4, 2005 #3
    thanks for the quick response! but, we haven't learned that in class...

    can you explain to me how [tex]\nabla \[/tex] works?

    how would I use it to solve the problem?
  5. Oct 4, 2005 #4

    hm. that's interesting.

    well, this problem is in spherical coordinates. so far, i haven't had a need to use any but the first term in the full expression for the divergence in spherical coordinates, so for the sake of my latex typing, i'll just give you the first part (which is all you'll need, anyway):


    \nabla \cdot \mathbf{E} (\mathbf{r}) = \frac{1}{r^2}\frac{\partial}{\partial r} (r^2 E_r).

    word of caution!

    \nabla \cdot \frac{\mathbf{r}}{r^2} = 4\pi\delta^3 (\mathbf{r}).

    please note that the bold "r" above on the left hand side should be an r-hat!
    Last edited: Oct 4, 2005
  6. Oct 4, 2005 #5
    oh, just take the divergence of the "inside" electric field.

    the outside one makes the sphere look like a point charge. now, a point charge would actually have a charge density with a delta function playing a role, but as far as the spherical charge is concerned, this is just an illusion.

    so just worry about that field inside--the field outside is kind of deceiving!

    i'm beginning to think that you're in a first-year physics course... in which case it would make sense you haven't seen divs and curls and the like.

    but... hell, it makes the problem pretty straightforward.

    (i guess the way you were supposed to solve the problem was work backwards from the value of the field inside the sphere to gauss's law in integral form.)
  7. Oct 4, 2005 #6
    in fact, here's how to do it this way:

    we have the fields

    E(r<R) = (Q/4*Pi*e0)(6r/R^3)
    E(r>R) = (Q/4*Pi*e0*r^2).

    like i said above, the second field isn't of our concern, since it makes us think we have a point charge on our hands.


    E = Q/4*pi*e0 * (6r/R^3)
    =Q/4*pi*e0 * r^2/r^2 (6r/R^3)
    =Q/4*pi*e0 * 6r^3/R^3 * 1/r^2


    4pi r^2 * E = Q/e0 6r^3/R^3.

    the left hand side is clearly the left hand side of gauss's law in integral form for this particular case.

    and the right hand side gives q_enclosed as a function of r.

    then i guess you make some sort of leap of faith and get rho.

    (or you could use the fancy math in the above posts!)
  8. Oct 4, 2005 #7
    thanks for all the help!
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