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Oppositely charged parallel plates I don't get why

  1. Apr 24, 2010 #1
    "the sum of the force from the top plate and the force from the bottom plate always equals the same amount"...as shown here:


    I mean, if a charge was placed at different places (ie distances) between 2 point charges, the net force would not be the same???

    Am I using Coloumb's law incorrectly?

    For example, let the charge be q and the distance between them be 4r, then if the test charge is placed half way between them, the net force would be

    kq²/(2r)² + kq²/(2r)²

    and if the test charge is placed at 1/4 of the distance (ie at r), then the net force would be...

    kq²/(r)² + kq²/(3r)²

    Any/all help/enlightenment would be appreciated. :)
    Last edited by a moderator: Apr 25, 2017
  2. jcsd
  3. Apr 24, 2010 #2


    Staff: Mentor

    The charge on a plate is not a point charge. If you want to use Coulomb's law you will have to integrate it over the plate. Of course, for a very large (infinite) plate you can use Gauss' law much easier.
  4. Apr 24, 2010 #3
    oop.s. I meant Coulomb's law.
  5. Apr 24, 2010 #4
    Thanks...are you saying that if we treat each plate as an infinite number of point charges and use Coulomb's law an infinite number of times to find the sum(s) (as in an intergral) they will balance each other? Could you show/tell me the intergral(s) please?

    btw...How would you explain this without knowledge of intergrals? :)
  6. Apr 24, 2010 #5


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    Homework Helper

    I'm not sure this can be done without integrals. Also instead of an infinite number of point chargers, which would be infinite charge, you have a finite amount of charge per unit area from a infinite large disk or plane.

    The math for field strength from a solid disc is explained here:


    as [tex]R^2 \ \rightarrow \infty [/tex]

    then [tex] \frac{z} {\sqrt{z^2 + R^2}} \ \rightarrow \ 0 [/tex]

    and you end up with [tex]E_z = k \ \sigma \ 2 \ \pi [/tex]

    An alterative approach is to consider the field from an infintely long line (= 1/z), then integrate an infinitely large plane composed of infintely long rectangles that approach infinitely long lines as their width approaches zero.

    For an infinite line charge, E = 2 k λ / r, where λ is charge per unit length and r is distance from the line. Here is the derivation:
    http://hyperphysics.phy-astr.gsu.edu/%E2%80%8Chbase/electric/elelin.html#c1 [Broken]

    For an infinite plane case, infinitely long strips (rectangles) approximate a line as their width -> 0. The area of a strip of lengh L is L dx, and the charge dq = σ L dx. The charge per unit length dλ = dq/L = σ dx. Assume the plane exists on the x-y plane, then the magnitude of the field at any point in space from a strip is dE = 2 k σ / r, where r is the distance from a strip to that point in space. In the case of the entire plane, the x components cancel because of symmetry, with only a net force in the z direction, and for each strip of the plane, dEz = dE (z / r) = dE sin(θ).

    [tex]E = 2 k \sigma \int_{-\infty}^{+\infty} \frac{{sin}(\theta)dx}{r}[/tex]

    [tex]{sin}(\theta) = z / r[/tex]

    [tex]E = 2 k \sigma \int_{-\infty}^{+\infty} \frac{z {dx}}{r^2}[/tex]

    [tex]r^2 = z^2 + x^2[/tex]

    [tex]E = 2 k \sigma z \int_{-\infty}^{+\infty} \frac{dx}{z^2 + x^2}[/tex]

    [tex]E = 2 k \sigma z \left [ \frac{1}{z} tan^{-1}\left (\frac{x}{z}\right )\right ]_{-\infty}^{+\infty} [/tex]

    [tex]E = 2 k \sigma z \left ( \frac{1}{z} \right) \left ( \frac{+ \pi}{2} - \frac{- \pi}{2} \right ) [/tex]

    [tex]E = 2 \pi k \sigma [/tex]
    Last edited by a moderator: May 4, 2017
  7. Apr 26, 2010 #6
    thanks...you've (both) given me a lot to go on...will do :)
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