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I Is it possible to charge a conducting sphere from the inside

  1. Apr 23, 2016 #1
    i just saw this amazing Forum and wanted to contribute, so here is my first question and i hope i get an answer x.d
    I was told that at electrostatic equilibrium, the electric field inside a conducting sphere will be ZERO,
    How can i charge thissphere from the inside, so the Electric field in side it wont equal zero.?
  2. jcsd
  3. Apr 23, 2016 #2
    Electric field inside a conductor's cavity (in the classical realm atleast) will be zero, no matter what the shape of the cavity is. You will have an electric field only if you physically place a charge in the cavity (here inside the sphere).

    This directly follows from Gauss's Law.
  4. Apr 23, 2016 #3
    I dont really get what you are talking about ...
    Why would you charge the sphere from the inside ?
    Then sure, you would have an electric field in there (assuming it is a hollow sphere and you somehow got your charge in there),but if the charge somehow
    gets the chance to distribute itself along the sphere,it will do so in a way that results in 0 electirc field inside.

    This principle isn't just exclusive to spheres.
    ANY conductive object however charged has zero electric field inside.
    It is a bit like saying there is no pressure difference in a gas filled container (in the absence of gravity)
    This follows from the fact that if there was a difference in pressure / (electrostatic potential for your problem) inside it, this differnce would cause the gas / (charges)
    to redistribute in a way that is "stable" and therefore does not cause any current to flow.
    In other words some arrangement that has no pressure gradient /(electric field) inside it.
  5. Apr 23, 2016 #4
    Then how about.
  6. Apr 23, 2016 #5
    That's why I was careful to mention "in the classical realm". In reality, electric forces on an electron far outweigh any other kind of forces.
    Doing some rough order of magnitude calculations, gives you that the gravitational field will be at least 10^20 order of magnitude lesser than electric field strength. This is impossible to detect in a lab.

    Even if you make the field strengths same (maybe around a black hole/neutron star or something),
    the gravitaional force/electrical force ~ m/q ratio of the electron ~ 10^-11. which is again small but might be detectable.

    Now if you can to go in that domain, let's see: Considering the continous generation/annihilation of virtual photons (electrostatic force carriers) in free space might give some interesting results. What if we place a black hole inside conducting cavity. Considering the net charge on the black hole to zero, there will still be some detectable photons coming from the event horizon, which might interact with the electrons. This might lead to a net charge inside the cavity surface.
  7. Apr 24, 2016 #6
    I actually came up with an easier example.
    Just fling your object really fast through a really strong magnetic field.
    It will act like some kind of mechanical hall effect sensor where the electric field is not created by
    the electric current but the bulk movement of the object.
    However you could argue that the lorentz force is fundimentally an electric effect... :I

    Actually now that i think about it every generator and transformer defies the principle of
    0 E-field inside it, but those are not in a stable state...
    The inductive currents slow down the relative movement.

    The Law should be more like: "In any conductor on wich no external forces act and wich is in a stable state there is
    no electric field.(If you approximate charge to be continuous)"

    Because if you imagine a low number of charge carriers on the object,
    it becomes apparent that it is impossible to arrange them in a way that would result in exactly 0 electric field.
    But since the "charge quanta" are so tiny the approximation of a continuous charge distribution is valid
  8. Apr 24, 2016 #7
    I agree with you but then I was going more "complicated" cause I was still trying to preserve the conditions of electrostatics. You're right in restating the law but then the OP already mentions the E field being zero in electrostatic conditions. And in electrostatics the electric force on an electron far exceeds any other kind of force.

    P.s. A more 'simpler' (not electrostatic) example: just make current flow through the conductor which has the cavity!!
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