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Compute the volume charge density of a charged spherical conductor

  1. Aug 21, 2009 #1
    In this problem we assume that the photon mass is m > 0. We take a conducting sphere of radius R and put a charge of Q on it. Some fraction of the charge will then reside at the surface and a fraction will move into the bulk. Evaluate the volume charge density in the bulk.

    I was shocked to learn in another thread that some people believe that the fact that there are no charges inside a conductor has nothing to do with Coulomb's law (apart from the mere fact that it leads to a repulsion between like charges). So, I invented this problem, solved it, and because it turned out to be quite easy, posted it here.

    B.t.w., what is even easier is to simply see that not all of the charge can reside at the surface.
  2. jcsd
  3. Aug 21, 2009 #2
    I forgot a bit of electrodynamics, so I'll ask:
    For a conductor, any charge applied to it can by uniformly distributed across the volume of the conductor, right?
  4. Aug 26, 2009 #3
    It will stay on the surface, unless Coulomb's law is invalid.
  5. Aug 26, 2009 #4
    If no one solves the problem by Monday, I'll write up the answer then.
  6. Aug 26, 2009 #5
    Are you assuming idealized point-like charge carriers and a perfect conductor, etc.? If so, then doesn't Gauss's Law ensure that the charges will all accumulate a the surface? I'm just imagining a spherical Gaussian surface concentric with the conducting sphere and with a radius less than but arbitrarily close to the radius of the conductor.
  7. Aug 26, 2009 #6
    Yes, that is correct. So, if Coulomb's law is valid, then we can derive that Gauss' law is valid and then you find this result. The question is what happens if Coulomb's law is not valid. You can e.g. consider the case of a massive photon and explicitely work out the case of a charged sphere. You then need to consider how the Maxwell equations (particularly Gauss' law) changes. The case of the massive photon is simpler than any other changes of Coulomb's law.
  8. Aug 30, 2009 #7
    Deadline is 30 hours from now. An F minus grade will be awarded to PF if there is no response by then.
  9. Aug 30, 2009 #8
    Do you mean proton? And since you mentioned mass, are you trying to involve gravity into the equations?
  10. Aug 30, 2009 #9

    Vanadium 50

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    No, he means photon. This is leftover from another thread.

    What I suspect is going to happen is that Count Iblis will note that [itex]\nabla \cdot E = -\rho/\epsilon_0 - m^2 \phi [/itex], and then argue that with a surface charge the field isn't zero because of the [itex]m^2 \phi [/itex]. There will then be a long drawn-out discussion about what a conductor is: is it a material where the electric field is zero everywhere, or is it a material where charges are free to move? In Maxwell electrodynamics, the two are the same. In Proca electrodynamics, they are not.
    Last edited by a moderator: Aug 30, 2009
  11. Aug 30, 2009 #10
    What is Proca electrodynamics?
  12. Aug 30, 2009 #11
    A conductor is an object in which charges are free to move and that implies (also in Proca theory) that the electric field is zero in the interior of the conductor.
  13. Aug 30, 2009 #12
    Where do photons come into play?
  14. Aug 30, 2009 #13
    I can't say that the assuptions of your set-up are clear. There can be electric fields within a conductor--or is this about electrostatics, only? If so, why the massive photon?
  15. Aug 30, 2009 #14
    This is about electrostatics. To answer also Gear's question, The massive photon causes the Coulomb potential of a point charge q to be modified. It becomes the Yukawa potential:

    V(r) = q/(4 pi epsilon_0) exp(-m c/hbar r)/r
  16. Aug 30, 2009 #15
    Looks like there is more involved with this problem than I thought there was.
  17. Aug 31, 2009 #16
    It is in fact pretty easy to solve. You can e.g. simply start with the Yukawa potential and derive from that how to generalize Gauss' law and proceed from that.
  18. Sep 2, 2009 #17

    [tex]\rho = \frac{Q}{4\pi R \lambda^{2}}\frac{1}{1+\frac{R}{\lambda}+\frac{R^2}{3\lambda^2}}[/tex]



    is the photon Compton wavelength
  19. Sep 19, 2009 #18
    Last edited by a moderator: Apr 24, 2017
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