# Field from non-conducting plate?

For a non-conducting sheet, the electric field is given by:

$$E = \frac{\sigma}{2\epsilon_0}$$ where $\sigma$ is the surface charge density.

This equation holds well for a finite nonconducting sheet as long as we are dealing with points close to the sheet and not too near its edges.

Why does the equation hold better with points closer to the sheet? I understand why the approximation worsens near the edges (because symmetry fails and causes fringe effects) but why is the approximation better near the sheet?

• If you are close to the sheet, the edge effects are negligible. Imagine sitting very close to the sheet. As far as you are concerned, the sheet is infinite because you can't see the edges. A similar thing happens here. – Prahar Mitra Mar 11 '14 at 18:57
• @Prahar Could you please give me a more formal explanation? Maybe one that uses symmetry? I don't really get the analogy you gave above. Just because I'm closer, it doesn't mean the sheet is any bigger. Just because it looks bigger doesn't mean it has grown bigger. – dfg Mar 12 '14 at 17:50
• The question you must always ask when you use the word "big" is "big with respect to what?" The Earth is big w.r.t. to us, but not w.r.t. the sun. When discussing the electric field due to a sheet, the size of sheet is compared to our distance from the sheet. It is then definitely true, that when we are closer to the sheet, in comparison, the sheet has "grown bigger" and therefore can essentially be considered as an infinite sheet and the edge effects can be ignored. – Prahar Mitra Mar 12 '14 at 23:52
• However, since you are asking for a more formal answer, I will write one. – Prahar Mitra Mar 12 '14 at 23:53

Consider a square sheet with edges located at $(a,0)$, $(-a,0)$, $(0,a)$ and $(0,-a)$. Suppose, we wish to find the electric field at a point $(0,0,z)$. By symmetry, this electric field will point solely in the $z$-direction. To find the electric field, consider a small element on the sheet located at $(x,y)$ of area $dx dy$. The charge of this element is $\sigma dx dy$. The magnitude of the electric field at $(0,0,z)$ due to this element is then (treating the element as a point charge) $$dE = \frac{1}{4\pi \epsilon_0} \frac{\sigma dx dy}{x^2 + y^2 + z^2 }$$ The $z$-component of this electric field is $$dE_z = \frac{1}{4\pi \epsilon_0} \frac{\sigma z dx dy}{\left( x^2 + y^2 + z^2 \right)^{3/2}}$$ Integrating this over the sheet, we find the total electric field at $(0,0,z)$ as $$E_z = \frac{\sigma}{ \pi \epsilon_0} \tan^{-1} \left[ \frac{a^2}{z \sqrt{ 2a^2 + z^2 } } \right]$$ Let us now take the limit of small $z$. However, $z$ is a dimensionfull quantity, and you can't discuss the largeness or smallness of dimensionfull quantities, only dimensionless numbers. The only dimensionless number that I can construct using $z$ is $\frac{z}{a}$. So, when I say, $z$ is small, I really mean $\frac{z}{a}$ is small.

In this limit, we find $$E_z = \frac{\sigma}{ \pi \epsilon_0} \tan^{-1} \left[ \frac{1}{(z/a)\sqrt{ 2 + (z/a)^2 } } \right] = \frac{\sigma}{2\epsilon_0} + {\cal O}(z/a)$$ Thus, when we are sitting close to the sheet, the field takes the form you described above.

But, here's the important thing. We didn't really care if $z$ itself is small (that sentence doesn't even make sense). What we really care about is if $z/a$ is small. Now, there are two ways to make this small -

1. Make $z$ small compared to $a$, i.e. move in very close to the sheet.

2. Make $a$ large compared to $z$, i.e. make the sheet very very large.

Both the statements above are completely equivalent.

• Thank you! Your proof shows that in the limit, the magnitude of the field approaches the formula I gave. However, how do we know the direction of the field approaches 90 degrees? (with respect to the surface) In other words, you showed that $E_z$ approaches my formula, but how do we know $E_y$ and $E_x$ approach zero? – user41086 Mar 13 '14 at 1:01
• Why don't you do the computation? Compute the electric field at a general point $(x,y,z)$ in space-time. This electric field will have in general all 3 - components $(E_x, E_y, E_z)$. Work them all out and show that in the small $z/a$ limit, $E_x$ and $E_y$ vanish, while $E_z$ goes to $\frac{\sigma}{2\epsilon_0}$. – Prahar Mitra Mar 13 '14 at 1:16
• Fair enough. One more thing - your proof calculates the field at $(0,0,z)$ - does this work for other points too? – user41086 Mar 13 '14 at 1:21
• I computed the field at $(0,0,z)$ so that I have enough symmetry to say $E_x = E_y = 0$ even for a finite plate. At a different point, there is no symmetry, so $E_x , E_y \neq 0$ which only makes the computation more complicated. I only to described the simplest possible case to explain my point. – Prahar Mitra Mar 13 '14 at 4:38
• That's totally fine, but if you had picked another point would have gotten the same result? – user41086 Mar 13 '14 at 4:48