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Finding the potential of a charged, solid sphere using the charge density

  1. Aug 13, 2012 #1
    1. The problem statement, all variables and given/known data

    Solid ball of charge with radius R and volume charge density ρ(r) = ρ0r2, centred at the origin.

    I have already found the electric field for r<R and r>R and also the potential at the origin by using the formula:
    V = -∫E.dl

    Now i want to find the potential at the origin using the charge density but am at a loss for the first step

    2. Relevant equations

    V = (1/4∏ε)∫(ρ/r)d[itex]\tau[/itex]


    3. The attempt at a solution

    I don't need the full solution, just the first step would be much appreciated.
     
  2. jcsd
  3. Aug 13, 2012 #2
    Find the potential at the center due to a thin shell of charge and then integrate over all the shells in the ball?
     
  4. Aug 13, 2012 #3

    vela

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    Just evaluate the integral. Or is there something specific about the integral you don't understand?
     
  5. Aug 13, 2012 #4
    Yes i'm unsure of what my limits are in this case and what d[itex]\tau[/itex] is equal to
     
  6. Aug 13, 2012 #5

    vela

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    The volume element ##d\tau## will depend on which coordinate system you choose to use. To calculate the potential at a point, you integrate over all space, but really you only need to worry about where the charge density ##\rho## doesn't vanish because everywhere else won't contribute to the integral.
     
  7. Aug 14, 2012 #6
    So would the integral from 0 to R work if d[itex]\tau[/itex] = r2sinθdrdd[itex]\phi[/itex] ?
     
  8. Aug 14, 2012 #7
    Is V = ρ0R4 / 4ε0 the answer to this problem?
     
  9. Aug 14, 2012 #8
    V = [itex]\frac{1}{4\pi\epsilon_{0}}[/itex][itex]∫^{2\pi}_{0}[/itex]d[itex]\phi[/itex][itex]∫^{\pi}_{0}[/itex]sin[itex]\theta[/itex]d[itex]\theta[/itex][itex]∫^{R}_{0}[/itex] [itex]\frac{ρ_{0}r^{2}}{r}[/itex] r[itex]^{2}[/itex]dr

    Solving this:

    V= [itex]\frac{ρ_{0}R^{4}}{4\epsilon_{0}}[/itex]
    I'm quite happy with this answer, but if it is incorrect please let me know.
    Thanks for the help
     
  10. Aug 14, 2012 #9

    vela

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    I haven't worked it out, but your method looks fine. Does it match the answer you got using the previous method?
     
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