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Electric Dipole - Oppositely charged hemispheres

  1. Sep 24, 2011 #1
    1. The problem statement, all variables and given/known data
    A sphere of radius [itex] a[/itex] carries a uniform charge density [itex] \rho_0 [/itex] in one hemisphere and the opposite charge density [itex] -\rho_0 [/itex] in the other hemisphere. The charged sphere produces an electric potential [itex]\Phi (r,\theta,\phi)[/itex]. For [itex] r>>a[/itex], the potential to leading order is of the form,

    \Phi(r,\theta,\phi) \approx \frac{g_n(\theta,\phi)}{r^n}

    Find [itex] g_n(\theta,\phi)[/itex] and the numerical value for n.

    2. Relevant equations

    I presume the multiple expansion could be used, but this was a test question so I think I could do it without having the expansion memorized.

    Any ideas?

  2. jcsd
  3. Sep 25, 2011 #2

    rude man

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    If you had to get an expression for gravitational force exerted by this body at large distances, except that one-half would consist of "negative" matter, what would you do?
  4. Sep 25, 2011 #3
    If I wanted the gravitational force, I would set up the integral over the region, taylor expand it and then integrate the leading term. Does that sound right?

    Plus, since I need potential and not force, it will be easier since its a scalar.
  5. Sep 25, 2011 #4

    rude man

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    That sounds good overall. You want to find the c.m., pretend all the charge is there, now you have a standard dipole. Potentials may be the way to go, but I'm not sure. There are probably formulas floating around out there in cyberspace giving the c.m. of a hemisphere .... I would forget about Taylor series expansions. The formula for the field at large distances from a dipole is very standard textbook stuff.
  6. Sep 25, 2011 #5
    The question clearly asks for the leading order term of the potential though.
  7. Sep 25, 2011 #6

    rude man

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    I interpreted that to mean the far-field formula. The far-field formula IS a first-order ("leading") approximation. That's why it's called far-field. Does not apply when near the dipole, gets too inaccurate. It's 100% accurate only at infinity.

    See http://hyperphysics.phy-astr.gsu.edu/hbase/electric/dipole.html

    Look at "dipole moment" part. Not that z has to be >> r, the distance between the spehers' c.m. in your case.

    However, I just noticed they want an expression for everywhere around the dipole (albeit at far distances). The formula in the link only givs the field along the line z, perpendicular to r.

    This is a bigger deal than I thought. May come back to it, might not be too bad.

    Here's the best I've found on your problem:


    If the last equation is expanded in spherical coordinates, that should be the answer. I would set up the dipole along the z axis, with the origin half-way between the two effective point charges.
    Last edited: Sep 26, 2011
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