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Poisson and Laplace's equation

  1. Nov 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Use Poisson's equation and Laplace's equation to determine the scalar potential inside and outside a sphere of constant charge density po. Use Coulomb's law to give the limit at very large r, and an argument from symmetry to give the value of E at r=0.


    2. Relevant equations
    del2V = - p/epsilono
    del2V = 0

    3. The attempt at a solution
    Not sure how to start this just using Poisson's/Laplace's equation.
     
  2. jcsd
  3. Nov 12, 2009 #2

    gabbagabbahey

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    What is [itex]\nabla^2V[/itex] in spherical coordinates?...Are there any symmetries that allow you to simplify it?
     
  4. Nov 12, 2009 #3
    Hmm okay. I'm not sure what symmetries you would need to consider to simplify [itex]\nabla^2V[/itex] in spherical coordinates. Because the charge density is constant, V does not depend on r inside the sphere...I'm not sure about anything else.
     
  5. Nov 12, 2009 #4

    gabbagabbahey

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    What makes you think [itex]V[/itex] doesn't depend on [itex]r[/itex]?:confused:

    Surely you can say that it doesn't depend on either the azimuthal or polar angles though right?....After all, a test charged placed at any given value of [itex]r[/itex], will "see" the same charge distribution at say [itex]\theta=\phi=\pi/4[/itex] as it would at say [itex]\theta=3\pi[/itex], [itex]\phi=\pi/7[/itex] or any other angle wouldn't it? (Draw a picture to convince yourself of this!)...This type of symmetry is called spherical symmetry.
     
  6. Nov 12, 2009 #5
    Oops sorry, I meant to say that. Okay so using the simplified form of the equation for [itex]\nabla^2V[/itex] in spherical coordinates, and Poisson's equation, I was able to get the equation for V inside the sphere. For outside the sphere, the charge density is 0 obviously, so you have to use Laplace's equation. I tried to do this but I can't anywhere near the form of the equation I'm meant to be getting :frown:
     
  7. Nov 12, 2009 #6

    gabbagabbahey

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    How about you show me your steps (for both inside and outside) and I'll see if I can spot where you are going wrong...
     
  8. Nov 12, 2009 #7
    Inside the sphere:
    [tex]\nabla^2V = \frac{1}{r}[/tex][tex]\frac{d}{dr}(r^2\frac{dV}{dr}) = \frac{\rho_{o}}{\epsilon_{o}}[/tex]
    [tex]\Rightarrow V = - \frac{\rho_{o} r^2}{6\epsilon_{o}}[/tex]

    Outside the sphere:
    [tex]\nabla^2V = \frac{1}{r}[/tex][tex]\frac{d}{dr}(r^2\frac{dV}{dr}) = 0[/tex]
    [tex]\Rightarrow r^2\frac{dV}{dr} = constant = a[/tex]
    [tex]\Rightarrow V = \frac{-a}{r} + c[/tex]
    I'm assuming that you now have to use some sort of boundary conditions...
     
    Last edited: Nov 12, 2009
  9. Nov 12, 2009 #8

    gabbagabbahey

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    You're missing a minus sign in your first equation (although it looks like that's just a typo, since your answer has the correct sign)....More importantly, shouldn't you have two constants of integration for the solution to this 2nd order ODE?:wink:
     
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