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Homework Help: E between parallel plates

  1. Mar 7, 2010 #1
    I'm not sure if this qualifies as a 'homework question'. There is no specific problem...I have a question about something in the text. I posted in 'Classical physics' with no luck. Maybe someone can help me here?

    It gives the situation of a conducting plate with charge density sigma 1 on each side. The E field on each side due to the conducting plate is = sigma 1/epsilon.

    When a second conducting plate (plate 2) is brought near, the charge on the opposite sides of each of the plates is attracted to the interior side. So now one side of each plate has charge density 0 and one side has charge density 2 * sigma 1 (or sigma 2).

    Then it says that the E field between the plates is equal to 2 *sigma 1/ epsilon or twice what it was before the second plate was brought over.

    But why wouldn't the second plate generate a field with equal magnitude, making the total E field in between the plates = 4 * sigma 1 / epsilon? Why is the total E field in between the two plates only calculated from one plate, and not considering both?

    Then this website: http://www.ac.wwu.edu/~vawter/Physic...ParallCap.html [Broken]

    Says that the E fields are equal to (sigma 1)/2 epsilon and that the E fields add. So the book says that the charge densities double but the E fields don't add, and this website says that the charge densities stay the same but the E fields add.

    Also, the E field is 0 on the outside of the plates. The website says it is because the E fields cancel. The book says it is because there is no excess charge on the outside of the plates since it all went to the inside. but why the E field be zero just because there is no charge near it? There's still charge on the inside of the plate, just a little farther away.

    What's really going on? I'm confused! Help!
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 8, 2010 #2


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    The field between the parallel plates is the sum of the fields arising from the charges on both plates. You get it by applying Coulomb's law for point charges arranged uniformly along a plane of infinite extension and integrating for the whole plane and then adding the contributions of both plates.

    Note the distinction between "plate" which has two sides and the plane of charges that contains all the charges on the plate.

    For one plane with sigma positive charge density, the integration results in an electric field of sigma/(2epsilon) on both sides of the plane, with direction pointing away from the plane. It is the same, only pointing towards the plane for the other plane with negative charge. Inside the plates this results of E=sigma/epsilon. Outside the plates, the electric fields are opposite, so their resultant is zero.

    In case of a metal plate, sigma surface charge density means 2sigma charge density in the plane of the plate, as there are charges on both sides. So the electric field of those parallel plates is 2sigma/epsilon between the plates.

    You would get the same result by applying Gauss' law for the resultant electric field.

  4. Mar 8, 2010 #3
    So you're treating the two metal plates as four planes of charge, each with surface charge density of sigma? If so that makes sense to me, I just want to clarify.
  5. Mar 9, 2010 #4


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    Yes, you can take one plate as two planes apart at zero distance or one plane with twice the surface charge of the metal sheet.

  6. Mar 9, 2010 #5
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