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Force between Conductor Plates

  1. Nov 12, 2005 #1
    How does one prove that the force between two conductor plates is F=q^2/2eA, where e=epsilon.

    If I use F=qE. Where the electric field is generated by one of the conducting plates: E=q/eA, then F=q^2/eA.

    I think what the answer assumes is that the E-field is from a uniform plate (insulator): E=q/2eA. In that case, F=q^2/2eA, which is the correct answer. But how can you make that assumption when obviously two conductor plates will have all their charge on the surface facing the other plate?
  2. jcsd
  3. Nov 12, 2005 #2


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  4. Nov 12, 2005 #3

    I must have some concept wrong, because they use that equation again for the conductor when they are caculating the electric field inside a conductor with a dielectric, i.e. E=qf/2eA+qf/2eA-qi/2eA-qi/2ea (where qf=free charge on conducting plate, and qi=charge bound to dielectric).

    I am a little unsure how they are coming up with that 1/2, because in my book they derived the Electric field using surface charge density (with Gauss' Law, of course), and the result is E=o/e (surface charge density/epsilon)--no 1/2. I suppose if you are dealing with charge and not charge density, then an excess charge 'q' will distribute q/2 on each surface.

    So, I am now completly lost. Because I thought the influence of another charged plate would draw all of the 'q' to one surface, even with a dielectric.
  5. Nov 12, 2005 #4
    I understand now, thanks. I think Halliday, Resnik could have done a better job, though.
  6. Nov 12, 2005 #5


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    Yeah, when I first studied EM, it was not so intuitively obvious to me, despite the fact that the textbooks and professors claimed it so.

    Anyway, it comes down to the charge distribution and Gauss's law.

    See if this helps.


    In the case, where there are equal and opposite charge densities, the contribution to the electric field must be equal, so that each charge distribution contributes

    E = [itex]\frac{\sigma}{2\epsilon_o}[/itex], and the combined field is

    E = [itex]\frac{\sigma}{\epsilon_o}[/itex]

    All textbooks seem to use the same example, and I think they could all do a better job.
  7. Nov 12, 2005 #6
    I looked at the links, but i'm still having a little trouble..

    (1) There seems to be an inconsistency when applying Gauss's Law. When they calculate the electric field between two insulated sheets they use superposition to find the net electric field. But for two conducting sheets, the derivation does not require superposition. In both cases the electric field is the same (E=q/eA), but if superposition was used for the conductor then E would be twice as large (two sheets), E=2q/eA. EDIT: This could be because they keep interchanging the charge density based on the situation--this isn't helping.

    (2) Does a dielectric block the external electric field between two capacitor plates? Because the book solution for the electric field between a capacitor with a dielectric is: E=qf/2eA+qf/2eA-qi/2eA-qi/2ea (where qf=free charge on a conducting plate, and qi=charge bound to dielectric). The qf/2 implies the charge distribution is divided on the two halves of the conducting plate, whereas typically all the charge is on the surface facing the other plate.
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