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Electrostatic Potential over all space

  1. Aug 20, 2012 #1
    If I have a sphere with radius R which has a charge distribution given by

    [itex]\rho(r)=\frac{5Q}{\pi R^{5}}r(r-R)[/itex]

    and [itex]\rho = 0 [/itex] at r bigger or equal to R, how do I find the electrostatic potential of this overall space? There is a charge Q, in addition, at the origin.

    My original thought was to just do the usual and use

    [itex]V(r)=\frac{1}{4\pi\epsilon_{0}}\int \frac{\rho(r')}{r}dt'[/itex],

    which if I am correct the integration goes from 0 to R, correct. Or does it extend from infinity in to R? This has never made much sense to me. Somebody help me out with this idea. Thanks!
     
  2. jcsd
  3. Aug 20, 2012 #2
    Or do I integrate from 0 to R, plus integrate from R to r? That seems a lot more in line with "over all-space". . . let me know your thoughts.
     
  4. Aug 21, 2012 #3

    mfb

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    For the potential at a given radius r, you can neglect all charges outside (r'>r) and assume that all charges inside are at r'=0. This is similar to gravity, and follows from the spherical symmetry.

    Therefore, for radius r, [itex]\frac{dV(r)}{dr}=\frac{Q(r)}{r^2}[/itex] with prefactors depending on your units. Q(r) is the total charge up to radius r: [itex]Q(r)=Q_0 + \int_0^r 4 \pi r'^2 \rho(r') dr'[/itex].
    You can find an analytic expression for Q(r), this can be used in the first equation, and another integration will give you the potential.
     
  5. Aug 21, 2012 #4

    vanhees71

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    I'd rather solve the differential equation (written in Heaviside-Lorentz units)
    [tex]\Delta \Phi=-\rho.[/tex]
    Since the charge distribution is radially symmetric, you can make the ansatz in spherical coordinates,
    [tex]\Phi(\vec{x})=\Phi(r).[/tex]
    Then from the Laplacian in spherical coordinates you get
    [tex]\Delta \Phi=\frac{1}{r^2} [r^2 \Phi'(r)]',[/tex]
    and the equation becomes an ordinary differential equation, which you have to solve with the appropriate boundary conditions. This leads to mfb's solution (modulo a sign and prefactors depending on the system of units used).
     
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