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General solution to Laplace's equation where V depends only on r

  1. Sep 30, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the general solution to Laplace's equation in spherical coordinates, for the case where V depends on on r. Do the same for cylindrical coordinnates assuming V depends only on r.


    2. Relevant equations
    Laplace's Eq (spherical): 1/r^2 (d/dr)(r^2(dV/dr)) + 1/(r^2sin(theta))(d/dtheta)(sin(theta)(dV/dtheta)) + 1/(r^2sin^2(theta))(d^2V/dphi^2))


    3. The attempt at a solution
    Having it only depend on r should I just use the first term of the eq. Everything before the first plus.

    and, should I use this EQ for V:

    V = kq/r?
     
  2. jcsd
  3. Sep 30, 2008 #2

    gabbagabbahey

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    Just use [itex]V(r,\theta,\phi)=V(r)[/itex] (that is [itex]V[/itex] depends only on [itex]r[/itex])

    What are [itex]\frac{\partial V}{\partial \theta}[/itex] and [itex]\frac{\partial V}{\partial \phi}[/itex] then?

    What does Laplace's equation look like now?
     
  4. Oct 1, 2008 #3
    wouldnt it just be

    1/r^2 (d/dr)(r^2(dV/dr)) dV/dr = (-1/4*pi*e0)(q/r^2) and the two r^2 cancel and you get

    1/r^2(d/dr)((-1/4*pi*e0)(q)) and that goes to 0 because there are not any r's inside the partial.
     
  5. Oct 1, 2008 #4

    gabbagabbahey

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    No, Laplace's equation is [itex]\nabla^2V=0[/itex] NOT [itex]\nabla^2V=V_{pointcharge}[/itex].

    [tex]\Rightarrow \frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{dV(r)}{dr} \right) = 0[/tex]

    Which gives you a second order, ordinary differential equation to solve for [itex]V(r)[/itex].

    Why would you think that the Laplacian of V was equal to the potential of a point charge?:confused:
     
  6. Oct 1, 2008 #5
    I was saying if you use the point charge formula for V(r) when you do the partial the r^2 that is being multiplied by the partial cancels the r^2 from V'(r) and you get a constant, so when you take the second partial with respect to r, you get 0 and you have that (dell^2)V = 0
     
  7. Oct 1, 2008 #6

    gabbagabbahey

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    The potential due to a point charge at the origin will satisfy Laplace's equation (except at r=0) but it is not the general solution to Laplace's equation! Can you solve the above ODE I posted?
     
  8. Oct 1, 2008 #7
    not really sure how
     
  9. Oct 1, 2008 #8

    gabbagabbahey

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    Okay, then I think you need to review the basics of ODE's.
     
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