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Dipole in a sphere

  1. Jun 12, 2004 #1

    I have problems solving this:
    Given: A sphere(Radius R) with a mathematical dipole in its center. I have to find the charge distribution on the sphere´s surface,

    [tex] \sigma(r,\phi,\theta) [/tex]

    so that the resulting potential is zero for r>R. I think that
    [tex] \sigma(r,\phi,\theta)=\sigma(\theta) [/tex]
    ,but I have no idea how to find this special charge distribution(I suppose it is unique). I know the electric field of a dipole, but that doesn´t help much.
    If anybody has an idea, please tell me!
  2. jcsd
  3. Jun 14, 2004 #2


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    Homework Helper

    Superposition. There is a relationship between surface charge density and the normal component of the electric field that it generates. The dipole will also generate an electric field with a component normal to the sphere. You want the field of your surface charge to cancel the field for the dipole.

    It is as you say: the surface charge density depends only on the polar angle.

    To find the field due to the dipole, you can either look it up, or do a limit as a -> 0 and qa = constant, where a is the separation between two equal and opposite charges, q and -q.
  4. Jun 14, 2004 #3

    thanks for the answer.
    The normal component of the dipole field is
    [tex] E_r=\frac{qd 2 cos(\theta)}{4\pi\epsilon_0 r^3} [/tex]
    So I thought of choosing

    [tex] \sigma(\theta)=-\frac{qd 2 cos(\theta)}{4\pi R^3} [/tex]

    then the normal component should be cancelled out according to the following law:

    [tex] n(E_o-E_i)=\frac{\sigma}{\epsilon_0} [/tex]

    where E_o is the normal component outside(=0) and E_i the normal component inside. n is the unit normal vector.

    But one professor of our university solved this problem too and he has the following result:

    [tex] \sigma(\theta)=-\frac{qd 3 cos(\theta)}{4\pi R^3} [/tex]

    He has a 3 instead of a 2, so my solution seems to be wrong.
    So I´m still confused:(
    Last edited by a moderator: Sep 7, 2017
  5. Jun 16, 2004 #4


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    The dipole moment of a sphere with radius R and a dipole charge distribution of C0=D0 * cos(theta) is (4pi/3)(R**3)D0. (this follows from the standard integral for a dipole moment) If this is to be equal and opposite to the original dipole moment, p0, then

    D0 = p0 (3/4pi)/R**3

    So the 3 is correct.
    Reilly Atkinson
  6. Jun 16, 2004 #5
    Ok, thanks!

    I see now that my first way to solve the problem was wrong.
  7. Jun 16, 2004 #6


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    Science Advisor

    Jonas -- Sorry , I fogot the minus sign in the sphere's dipole moment. RA
  8. Jun 16, 2004 #7
    No prob,
    I got the idea of your solution, so I noticed that on my own.
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