Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electric potential of a point charge inside a spheric conductor

  1. Aug 30, 2011 #1
    I'm trying to find the electric potential inside a thin (not charged) spheric conductor of radius R, containing a point charge of charge +q exactly in the center.

    To do this, I'm trying to solve the differential equation [itex]\Delta V = 0[/itex] (harmonic). Together with the boundary condition [itex]V(R) = c[/itex].

    From the null laplacian, one can obtain [itex]V(r) = -\dfrac{\alpha}{r} + \beta[/itex], but I'm lacking one boundary condition. Can this method be applied even when there's a singularity (i.e. [itex]V(0) = \infty[/itex])?

    How does one solve this problem? What additional boundary conditions are there?
     
  2. jcsd
  3. Aug 30, 2011 #2

    Meir Achuz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    V(R)=q/R.
     
  4. Aug 30, 2011 #3
    Care to explain that conclusion that's not trivial to me, Meir Achuz?
     
  5. Aug 31, 2011 #4

    clem

    User Avatar
    Science Advisor

    By Gauss's law, the field outside the conductor is E=q/r^2, so phi=q/r.
    If you want SI, divide by 4piepsilonzero.
     
  6. Aug 31, 2011 #5

    clem

    User Avatar
    Science Advisor

    By Gauss's law, the field outside the conductor is E=q/r^2, so phi=q/r.
    If you want SI, divide by 4piepsilonzero.
     
  7. Aug 31, 2011 #6
    Draw a spherical Gaussian surface around the whole system and use Gauss' law in integral form, after figuring out the total enclosed charge to get the field outside the conductor. Do this again with the Gaussian surface around the central charge but inside the conductor to get the field inside. You can't use the Laplace equation directly because it assumes you have no charge in the region of interest, and you do have a charge inside.
     
  8. Aug 31, 2011 #7
    Poisson's equation shouldn't, however, be yielding results? I tried it but the left-hand side of the equation is zero (unless I messed up a calculation, but I can't find an error). I'm a little confused :confused:
     
  9. Aug 31, 2011 #8
    You have to use the mathematical identity:

    2(1/r) = -4πδ(r) where δ(r) is the three-dimensional Dirac delta in the separation distance r.

    Poisson's equation states in SI units in free space: ∇2V = ρ/ε0

    For a charge density of only a point charge +q, this becomes: ∇2V = q0 δ(r)

    Try the a solution of the form V = a/r + b to find: a2(1/r) = q0 δ(r)

    Use the identity to find: a (-4πδ(r)) = q0 δ(r)

    This leads to: a = - q/(4π ε0 ) and:

    V = - q/(4π ε0 r) + b

    As you realized earlier, the other constant b is determined by applying the boundary condition on the conductor.
     
  10. Sep 4, 2011 #9
    hmmm, I'm having a little trouble with this.

    In spherical coordinates, the r component of the Laplacian is:

    [tex]\frac{1}{r^2}\frac{\partial}{\partial r}(r^2\frac{\partial}{\partial r}V)[/tex]

    then

    [tex]\frac{1}{r^2}\frac{\partial}{\partial r}(r^2\frac{\partial}{\partial r}(\frac{a}{r}+b))=\frac{1}{r^2}\frac{\partial}{ \partial r}(r^2(-\frac{a}{r^2}))=\frac{1}{r^2}\frac{\partial}{ \partial r}(-a)=0[/tex]

    I'm probably missing something obvious, but I can't derive your solution:

    [tex]a ∇^2(\frac{1}{r}) = \frac{q}{ε_0} δ(r)[/tex]

    Also, please correct me if I'm wrong, but shouldn't we use

    [tex]V=a ln(r)+b[/tex]

    in spherical coordinates?
     
  11. Sep 4, 2011 #10
    An even simpler analysis. The potential of a point charge assuming NO conductors is just q/R. But this thin spherical conductor is symmetric around the charge, so it won't disturb the potential field... there's no induced charge on the conductor at all (by symmerty). So the solution with the conductor is just the same as the solution with no conductor: q/R.

    The problem is more interesting when the charge is inside the sphere but offset from the center. This can be solved by the method of images for the potential inside the sphere. Outside the sphere, the potential remains q/R, even with the offset internal charge. (The conductor's induced charge compensates.)
     
  12. Sep 5, 2011 #11

    clem

    User Avatar
    Science Advisor

    "So the solution with the conductor is just the same as the solution with no conductor: q/R."

    This thread is so long that nobody noticed posts #2 and 4.
     
  13. Sep 5, 2011 #12
    It is because your math is not correct at r=0. d/dr 1/r is not -1/r^2 at r=0
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook