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Uniform plane of charge

  1. Feb 24, 2015 #1
    1. The problem statement, all variables and given/known data
    GP4JG.jpg

    Charge is distributed uniformly over a large square plane of side l, as shown in the figure. The charge per unit area (C/m^2) is [itex]\sigma [/itex]. Determine the electric field at a point P a distance z above the center of the plane, in the limit [itex]l \to \infty[/itex].
    [Hint: Divide the plane into long narrow strips of width dy, and use the result of Example 11]


    2. Relevant equations

    Result of Example 11: [itex]\frac{2k\lambda }{x}[/itex] (electric field at a distance x due to an infinitely long wire)(that point is symmetric about the x-axis, so there is no y component of the electric field.)

    [tex]k = \frac{1}{4\pi\epsilon_0}[/tex]

    3. The attempt at a solution

    Charge densities:
    [tex]\sigma = \frac{dq}{dy*l}[/tex] (an infinitely small q over an infinitely small surface)
    [tex]\lambda = \frac{dq}{l}[/tex] (total charge of the strip / total length)

    [tex]dE = \frac{2k\lambda}{\sqrt{y^2+z^2}}[/tex]
    (electric field due to a long strip)
    [tex] dE_z = dE sin\theta = \frac{2k\lambda y}{{(y^2+z^2)}^{3/2}}[/tex]
    (its z component is what we need)
    [tex] dE_z = dE sin\theta = \frac{2k\sigma y}{{(y^2+z^2)}^{3/2}}dy[/tex]
    (dy is necessary, so replace lambda with sigma)

    The following is what I get after the integration,

    [tex]{-2\sigma k} \frac{1}{\sqrt {y^2+z^2}}[/tex]

    The limits are zero and infinity, so I end up with;
    [tex]\frac{2\sigma k}{z}[/tex]

    There is an example of uniformly charged disk in my textbook. The formula for electric field for that disk does not depend on the distance. That's why I believe I've done this question wrong. What do you think about my solution? I am not sure if I wrote charge densities correct, so that may be the mistake.
     
    Last edited: Feb 24, 2015
  2. jcsd
  3. Feb 24, 2015 #2

    TSny

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    When getting the z-component you used a factor of sinθ. Is that correct?
     
  4. Feb 24, 2015 #3
    My bad... It should be cosine, shouldn't it?
     
  5. Feb 24, 2015 #4

    TSny

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    Yes, that's right.
     
  6. Feb 24, 2015 #5

    TSny

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    Also, should you end up with a power of 3/2 in the denominator of the integrand?
     
  7. Feb 24, 2015 #6
    It must be raised to the first power I suppose, right?
     
  8. Feb 24, 2015 #7

    TSny

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    Yes. What should be the limits on the integral in order to cover the entire plate?
     
  9. Feb 24, 2015 #8
    I know we are doing good right now but I am not good at identifying the limits. Probably wrong, but I would say 0 and l.
     
  10. Feb 24, 2015 #9

    TSny

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    You are integrating over strips. Where is the strip corresponding to y = 0? As y varies between 0 and ##l##, what strips are included?
     
  11. Feb 24, 2015 #10
    Then I guess -l/2 & l/2 would be the limits, or simply 2 * [0 to l/2]?
     
  12. Feb 24, 2015 #11

    TSny

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    Yes. Good. ( I should have said as y varies from 0 to ##l/2## above.)

    So, as ##l## goes to infinity, you can see what the limits on the integral should be.
     
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