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Discontinuity of electric field at surface of a conductor

  1. Aug 29, 2013 #1
    Hi all,

    So I've been reviewing for the PGRE at the end of next month (wish me luck, I'm going to need it) and I stumbled across something that confused me in my old textbook. I was reading about the discontinuity of the electric field at the surface of a conductor, and also about the strength of the electric field immediately outside of a conductor. Here is what puzzles me:

    So as I understand it the discontinuity of the electric field (which has magnitude σ/ε) is due to the fact that the electric field outside the conductor is is equal to σ/2ε on both sides of the surface of a conductor in opposite directions. This is just what you get when you assume your point is very close to the surface and thus get to treat the surface of the conductor as a plane of charge. That's all fine and good, but then the next section argues that since the electric field must be 0 inside (once the conductor has reached electrostatic equilibrium) and the discontinuity of the electric field is σ/ε as just proven, the electric field immediately outside the conductor must actually be σ/ε. Wait a minute! We just said it was σ/2ε outside! And we said that just inside in the opposite direction it was σ/2ε, talking as though there were actually a field immediately within.

    What's going on here?
     
  2. jcsd
  3. Aug 30, 2013 #2

    mfb

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    You are mixing two different cases.
    1) A plane where you don't care about its (small) thickness, and charge density σ.
    2) A plane with finite width, where you consider the charge density on both sides separately. If you want to compare this to case 1, let σ/2 be on both sides, and the results are consistent again.
     
  4. Aug 30, 2013 #3
    Turns out that the field produced by the local charge density is indeed σ/2ε pointing outwards just outside of the conductor surface and σ/2ε pointing inwards just inside of the conductor surface. But that's not the whole story. There is also the field produce by all the other charges present in the conductor. Those far charges produce a field σ/2ε pointing outwards on both sides of the surface. On the inside the two fields cancel each other out while on the inside they add up to σ/ε.
     
  5. Aug 30, 2013 #4
    Thank you for the replies. Dauto's response makes a bit more sense to me. mfb, in both cases I was referring to the charge found on the surface of a conductor. The distribution of charge itself was assumed to be a surface having no width. Do you disagree with Dauto's explanation?

    Dauto, if your response was correct it is a nice way to think about these things and a more intuitive explanation for "no electric field in a conductor at electrostatic equilibrium."

    Thanks both of you.
     
  6. Aug 30, 2013 #5

    ehild

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    σ is the charge per unit area of a planar array of charges here. σ/ε field lines emerge from charge σ, they go symmetrically up and down, so the density of field lines (the electric field strength) is σ/2ε on both sides of the plane.

    Here you have a plate of conductor of finite thickness. Giving charge to it, the charge appears on both surfaces, but there is no charge inside. The surface charge density is σ on both surfaces. Imagine that you remove the metal, and keep the charges only. You have two planes, separating three regions - one above, the second between the planes and the third below the plate. The electric field is the sum of the fields due to both planes, the red ones are field lines from the top plane and the blue ones are from the bottom plane. You see, there are 2(σ/2ε) field lines both above and below the metal plate and the field lines from the top plane are opposite to those from the bottom plate between the planes, so they cancel: the electric field is zero inside.

    ehild
     

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    Last edited: Aug 30, 2013
  7. Aug 31, 2013 #6

    mfb

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    All three explanations here just use different words to describe the same thing, I agree with all of them.
     
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