How Does Charge Distribution Affect Electric Field Above an Infinite Plane?

[hitemup]

Homework Statement

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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 $\sigma$. Determine the electric field at a point P a distance z above the center of the plane, in the limit $l \to \infty$.
[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: $\frac{2k\lambda }{x}$ (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.)

$$k = \frac{1}{4\pi\epsilon_0}$$

The Attempt at a Solution

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

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

The following is what I get after the integration,

$${-2\sigma k} \frac{1}{\sqrt {y^2+z^2}}$$

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

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.

 

[TSny]

When getting the z-component you used a factor of sinθ. Is that correct?

 

[hitemup]

When getting the z-component you used a factor of sinθ. Is that correct?

My bad... It should be cosine, shouldn't it?

 

[TSny]

Yes, that's right.

 

[TSny]

Also, should you end up with a power of 3/2 in the denominator of the integrand?

 

[hitemup]

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?

 

[TSny]

Yes. What should be the limits on the integral in order to cover the entire plate?

 

[hitemup]

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.

 

[TSny]

You are integrating over strips. Where is the strip corresponding to y = 0? As y varies between 0 and $l$, what strips are included?

 

[hitemup]

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]?

 

[TSny]

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.