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Electric Potential of a Spherical Charge Distribution

  1. Oct 21, 2012 #1
    1. The problem statement, all variables and given/known data
    A total charge q is uniformly distributed throughout a sphere of radius a.
    Find the electric potential in the region where r1<a and r2>a.
    The potential is defined anywhere inside the sphere.

    2. Relevant equations
    letting ρ = volume charge density and ε = permittivity constant,
    [tex]
    \nabla^2\phi = \frac{\rho}{\epsilon}
    [/tex]


    3. The attempt at a solution
    for r1 <a,
    [tex]
    \rho = \frac{q}{\frac{4}{3}\pi a^3} = \frac{3q}{4 \pi a^3}
    [/tex]
    there is only an r term, so
    [tex]
    \frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{d \phi}{dr} \right) =\frac{3q}{4 \pi a^3}\\
    \frac{d \phi}{dr} = \frac{q r}{4 \pi a^3} + \frac{C_1}{r^2}\\
    \phi = \frac{q r^2}{8 \pi a^3} - \frac{C_1}{r} + C_2\\
    [/tex]
    C1 = 0, so that the potential is defined at the origin (r = 0)
    so,
    [tex]
    \phi(r_1) = \frac{q r_1^2}{8 \pi a^3} + C_2
    [/tex]
    -------------------
    for r2 <a,
    [tex]
    \phi(r_2) = -\frac{C_3}{r_2} + C_4\\
    [/tex]
    ----------------------
    how do I find C2, C3, and C4?
    please help :/
     
  2. jcsd
  3. Oct 21, 2012 #2
    What are the boundary conditions? There should be one for 0, one for infinity, and one for the edge of the sphere.
     
  4. Oct 21, 2012 #3
    thank you for the quick reply

    the potential vanishes as r approaches infinity, so I guess that means C4 =0

    but there was no boundary condition given for the edge of the sphere. that is why I'm having trouble... :/
     
  5. Oct 21, 2012 #4
    If you've learned Gauss' Law, you should be able to determine the boundary conditions that way fairly quickly.
     
  6. Oct 21, 2012 #5
    I think I know Gauss' Law for electric fields, but I still do not see how it can be used to find the boundary conditions.

    I know that Gauss's Law states that the total charge enclosed by any closed surface equals the sum of components of the electric field perpendicular to the surface. I doubt it says how to find the charge ON THE SURFACE itself. :/

    Please educate me.
     
  7. Oct 21, 2012 #6
    Draw a Gaussian surface in the shape of a prism with the faces parallel to the surface, at the boundary between the surface. Let the area be A, and the height be h. What is the enclosed charge in the volume as h goes to zero? What does that say about the electric field, and in turn, the derivative of the electric potential?
     
  8. Oct 21, 2012 #7
    One more thing, the potential doesn't necessarily have to vanish as r approaches infinity. The electric field has to vanish, which means the potential just has to have a zero derivative. If you have C4=0, then you need C2=0 as well.
     
  9. Oct 22, 2012 #8
    Scratch that last. I was wrong, if C4=0, then C2 can't be zero. There is no way to make the equations work out. What I should have said is that you will get an expression for C2 that depends on C4. That comes from the fact that the potential is always continuous. C4 still doesn't have to equal zero, but then you would increase or decrease C2 by that same amount. It's just convenient and typical to set V=0 at infinity.
     
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