Uniform plane of charge

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Homework Statement


GP4JG.jpg

[/B]
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]


Homework Equations


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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]

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.
 
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Answers and Replies

  • #2
TSny
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When getting the z-component you used a factor of sinθ. Is that correct?
 
  • #3
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When getting the z-component you used a factor of sinθ. Is that correct?
My bad... It should be cosine, shouldn't it?
 
  • #4
TSny
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Yes, that's right.
 
  • #5
TSny
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Also, should you end up with a power of 3/2 in the denominator of the integrand?
 
  • #6
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Also, should you end up with a power of 3/2 in the denominator of the integrand?

It must be raised to the first power I suppose, right?
 
  • #7
TSny
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Yes. What should be the limits on the integral in order to cover the entire plate?
 
  • #8
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Yes. What should be the limits on the integral in order to cover the entire plate?
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.
 
  • #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?
 
  • #10
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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?

Then I guess -l/2 & l/2 would be the limits, or simply 2 * [0 to l/2]?
 
  • #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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