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So the problem is asking me to find the Electric field a height z above the center of a square sheet of side a

I approach the problem a different way than the book, I derive the electric field due to a line of charge of side $a$ a height z above the center of a square loop, and I verified it to be $\frac{1}{4\pi\epsilon_0}$ $\frac{\lambda a z}{(z^2+\frac{a^2}{4}) (z^2+\frac{a^2}{2})^\frac{1}{2}}$ $\hat z$

Now the way I do it is that I let that line have a thickness $da$ where da is a width element not an area element (as the side of the square is a), so now the linear charge density $\lambda$ is equal to the surface charge density multiplied by that small thickness $da$ , that is

$\lambda = \sigma da$

So the Electric field $dE$ due to a line of small thickness $da$ is

$dE$ = $\frac{1}{4\pi\epsilon_0}$ $\frac{\sigma da z a}{(z^2+\frac{a^2}{4}) (z^2+\frac{a^2}{2})^\frac{1}{2}}$ $\hat z$

I integrate this field from $0$ to $a$ then,

$E$ = $\frac{\sigma z}{4\pi\epsilon_0}$ $\int_0^a$ $\frac{ada}{(z^2+\frac{a^2}{4}) (z^2+\frac{a^2}{2})^\frac{1}{2}}$ $\hat z$

This integral yields $\frac{4}{z}$ $\tan^{-1}(\sqrt{1+\frac{a^2}{2z^2}}$ $|^{a}_{0}$

= $\frac{4}{z}$ $[\tan^{-1} (\sqrt{1+\frac{a^2}{2z^2}} -\frac{\pi}{4}]$

That is the value of the integral, now multiply it by $\frac{\sigma z}{4\pi\epsilon_0}$

Then $E$=$\frac{\sigma}{\pi\epsilon_0}$ $[\tan^{-1} (\sqrt{1+\frac{a^2}{2z^2}} -\frac{\pi}{4}]$

I'm missing it by a factor of 2, the answer should be $\frac{2\sigma}{\pi\epsilon_0}$ $[\tan^{-1} (\sqrt{1+\frac{a^2}{2z^2}}) -\frac{\pi}{4}]$

I'm pretty sure about the mathematical steps, I'm assuming I made a false assumption at the beginning, but its been more than 20 hours and I still haven't figured out what it is, any help would be appreciated.

Here's a picture to show you how I think I can do it

enter image description here

This red line is of width $da$ and I want to integrate $dE$ from $0$ to $a$

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  • $\begingroup$ Two things that jump out to me. First, $a$ is the length of the segment, your integration boundary and the variable of integration. It's easy to mix those up. Secondly, shouldn't the integral run from $-\frac{1}{2}a$ to $\frac{1}{2}a$? $\endgroup$ – Aaganrmu Nov 28 '18 at 16:21
  • $\begingroup$ Yes, I didn't bother putting primes on the variable, a is the side of the square, $da$ is a width element that is a small segment of the side $a$ (the left one), second, I ran the integral from 0 to a because originally, the electric field due to the line at point p was taken by assuming that the origin is at the center of the line [I'll edit the thread to show you the coordinate system] $\endgroup$ – khaled014z Nov 28 '18 at 16:26
  • $\begingroup$ Just to make things clear, I want to integrate line by line on that sheet, not along the red line, the red line was just an illustration of its width $da$, thus I need to integrate from 0 to a to cover the whole sheet with lines of infinitesimal widths $\endgroup$ – khaled014z Nov 28 '18 at 16:33
  • $\begingroup$ Did you.only.take the z conponent for dE $\endgroup$ – lalala Nov 28 '18 at 17:14
  • $\begingroup$ The equation $dE$ = $\frac{1}{4\pi\epsilon_0}$ $\frac{\sigma da z a}{(z^2+\frac{a^2}{4}) (z^2+\frac{a^2}{2})^\frac{1}{2}}$ $\hat z$ is the z component of the Electric field of the line of thickness $da$ ,yes, since all x and y components cancel $\endgroup$ – khaled014z Nov 28 '18 at 17:18
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Your math is correct as far as the calculations are concerned, but you made an error in your choice of variables. The contribution of a single line of charge at horizontal position $y$ is $$E_{z}(y) = \frac{1}{4\pi\epsilon_0} \frac{\lambda a z}{\left( z^2 + y^2 \right) \left( z^2 + y^2 + \frac{a^2}{4} \right)^\frac{1}{2}} .$$ At position $y = a/2$, which is the segment you evaluated, this reduces to your result (your first formula). But as you integrate over a range of $y$-values, the difference between $y$ and $a/2$ becomes significant. You are keeping the integration boundary $a/2$ equal to the horizontal coordinate $y$ (since they are both called $a/2$ in your calculation), so you are actually integrating over the yellow section of the plane in this picture: Integration area

That happens to be half the plane, and therefore you got half the correct result.

To find the total field strength, you would integrate the expression above: $$E = \int_{-\frac a2}^{\frac a2} E_z(y)\, dy = \frac{\sigma}{4\pi\epsilon_0} a z \int_{-\frac a2}^{\frac a2} \frac{dy}{\left( z^2 + y^2 \right) \left( z^2 + y^2 + \frac{a^2}{4} \right)^\frac{1}{2}}$$ which according to an engine works out to $$E = \frac{\sigma}{4\pi\epsilon_0} \left. 2 \arctan{\left( \frac{a^2}{z \left( a^2 + 4 (y^2 + z^2) \right)^\frac12} \right) } \right|_{-\frac a2}^{\frac a2} = \frac{\sigma}{\pi\epsilon_0} \arctan{\left( \frac{a^2}{2\,z\,\left( 2 a^2 + 4 z^2 \right)^\frac12} \right)}\,.$$

A few checks to see if the extreme cases turn out correct. From a large distance, with $z \gg a$, the plane looks like a point - and indeed, since $\arctan x \approx x$ for small $x$, the equation reduces to $$E_{z \to \infty} = \frac{\sigma}{\pi\epsilon_0} \frac{a^2}{2\,z\,\left(4z^2\right)^\frac12} = \frac{Q}{4\pi\epsilon_0 z^2} $$ which is the field of a point charge.

Conversely, if $a$ is very large, we should have the field of an infinite plane, which does not depend on the distance $z$: $$E_{a \to \infty} = \frac{\sigma}{\pi\epsilon_0} \lim_{a \to \infty} \arctan{\left( \frac{a^2}{2\,z\,\left( 2 a^2 + 4 z^2 \right)^\frac12}\right)} = \frac{\sigma}{\pi\epsilon_0} \frac{\pi}{2} = \frac{\sigma}{2\epsilon_0}\, .$$

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  • $\begingroup$ Is this what you're trying to tell me, look at the post edit, that I should consider the y positions of each line and that is the variable I should integrate on? However when I try finding the electric field due to a line on a position y I get a different result than yours, is my coordinate system valid? In other words is the the direction I'm integrating through positive? $\endgroup$ – khaled014z Nov 29 '18 at 13:44
  • $\begingroup$ @khaled You say: "when I try finding the electric field due to a line on a position y I get a different result than yours". What do you get? With $y=a/2$, my result and your first formula are the same. $\endgroup$ – Jim Danner Nov 29 '18 at 13:54
  • $\begingroup$ Actually, I think I got your point now, I had the axes flipped before, now I understand it, thank you $\endgroup$ – khaled014z Nov 29 '18 at 13:55
  • $\begingroup$ Ok so when I now tried integrating for the whole sheet, I'm integrating $dE_{z}(y) = \frac{\sigma az}{4\pi\epsilon_0} \int\frac{dy}{\left( z^2 + y^2 \right) \left( z^2 + y^2 + \frac{a^2}{4} \right)^\frac{1}{2}} .$ From $y=\frac{-a}{2}$ to $y=\frac{a}{2}$ , I get, $E_z(y)=\frac{\sigma}{2\pi\epsilon_0} [\tan^{-1}(\frac{a^2}{2z \sqrt{\frac{5a^2}{4}+4z^2}}) - \tan^{-1}(\frac{-a^2}{2z \sqrt{\frac{5a^2}{4}+4z^2}})]$ It's not identical to the answer, is it mathematically correct? However when I take the limit as $a \rightarrow\infty$ , it is the correct electric field for an infinite sheet. $\endgroup$ – khaled014z Nov 29 '18 at 16:00
  • $\begingroup$ I've worked out the final result (though a computer did the heavy lifting), and updated my answer. Your result looks a little different, but I'm not sure why. $\endgroup$ – Jim Danner Nov 30 '18 at 12:55

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