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I Charge on conductors

  1. Apr 14, 2016 #1
    upload_2016-4-14_23-42-59.png
    - David J Griffiths

    But how s it valid for some conductor like →
     
  2. jcsd
  3. Apr 14, 2016 #2
  4. Apr 14, 2016 #3

    Dale

    Staff: Mentor

    Given:
    1) Maxwells equations.
    2) Electrostatic: ##J=0##, ##d\rho/dt=0##, ##dE/dt=0##, ##dB/dt=0##
    3) Conductor: ##J=\sigma E##

    Assume there is charge in the interior. Then by Gauss law there is an E field in the interior. Then by assumption 3) there is a current. This contradicts assumption 2). Therefore there can be no charge in the interior.
     
  5. Apr 14, 2016 #4
    So does that mean -
    if some charge is given to an arbitrarily shaped conductor, it will surface and regardless of everything; the field inside will be zero and the surface will be equipotential
     
  6. Apr 17, 2016 #5
    please reply
    somebody !!
     
  7. Apr 17, 2016 #6

    vanhees71

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    The reasoning is very simple. I don't know, why Griffiths likes to obscure things in his textbooks. In electrostatics by definition we have all fields and sources time-independent and also no current density (see #3). Then within a conductor you have
    $$\sigma \vec{E}=\vec{j}=0 \; \Rightarrow \; \vec{E}=0.$$
    The electric charge density at every point is given by Gauss's Law (Heaviside-Lorentz units)
    $$\vec{\nabla} \cdot \vec{E}=\rho.$$
    Since ##\vec{E}=0## inside the conductor also ##\rho=\vec{\nabla} \cdot \vec{E}=0## inside the conductor. So if the conducting body is overall charged, the charge must be on its surface.

    Caveat: To be more accurate, of course microscopically the entire body consists of charged particles (atomic nuclei and electrons). So there are charges inside the body but they are compensating each other precisely in the static case, when looked from a macroscopic (coarse-grained) point of view. So within macroscopic electrodynamics the macroscopic charge distribution vanishes in conductors for electrostatic situations.
     
  8. Apr 17, 2016 #7
    so #4 is right ?
     
  9. Apr 17, 2016 #8
    So does that mean -
    if some charge is given to an arbitrarily shaped conductor, it will surface and regardless of everything; the field inside will be zero and the surface will be equipotential
     
  10. Apr 17, 2016 #9
    by the way can you put this physically , I don't know this math
     
  11. Apr 17, 2016 #10

    Dale

    Staff: Mentor

    Only in the electrostatic case, yes. This does not hold in magnetostatics or electrodynamics.
     
  12. Apr 19, 2016 #11

    SammyS

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    Yes. The surface is equipotential.

    That does not mean that the surface charge is uniformly distributed.
     
  13. Apr 20, 2016 #12
    Then we should have current over the surface, as its made of conductor, does that happen in steady state ?
    It won't be electrostatics then !
     
  14. Apr 20, 2016 #13

    SammyS

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    I didn't mention current, nor did I imply it.

    The surface charge density is not one uniform value over the whole surface. It varies from location to location, but it is static.
     
  15. Apr 22, 2016 #14
    that means it is equipotential
     
  16. Apr 22, 2016 #15

    Dale

    Staff: Mentor

    Yes.
     
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