I'm confused as to how this expression $$E= \frac{\sigma}{\epsilon_0}$$ can't be used to calculate the electric field of a perfectly flat part of a surface even farther from just above the surface. If you just extend the same cylindrical Gaussian surface used for this proof, wouldn't the field stay the same no matter how far out you go? The charge enclosed on the surface of the conductor wouldn't change and neither would the area of the circular end of the cylinder sticking out of the conductor since it sticks out of a flat surface. Obviously this simply cannot be so as the field should grow weaker with distance from its charged source, so I wonder what about this proof I'm misunderstanding. On a related note, what about the derivation for $$E= \frac{\sigma}{2\epsilon_0}$$ for an infinitely charged plane takes into account its infinite size, as it seems to me that it could just as easily be used to calculate the electric field for a finite sheet of charge?
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$\begingroup$ More on capacitors and factors of 2: physics.stackexchange.com/q/110480/2451 and links therein. $\endgroup$– Qmechanic ♦Sep 1, 2017 at 12:55
3 Answers
The difference between the finite and the infinite sheet, is that with the infinite sheet, we can assume by symmetry that $\vec{E}$ is everywhere perpendicular to the surface, which makes taking the surface integral for Gauss's law trivial.
When the sheet is finite, we can no longer make that assumption, and would need to consider edge effects. Indeed as you move further away from the finite plane, the smaller and smaller it will look, eventually if you were far enough away, you could imagine it would look almost like a point charge, and you'd expect a $\frac{\vec{r}}{r^3}$ type of field in the far limit. Intuitively, with a finite sheet, you could imagine looking at the size of the sheet to say how far you are away from the sheet. But with the infinite sheet, no matter how far you are away from it, you only ever see an infinite sheet. So the distance away from the sheet has no meaning, hence the field must be constant at all distances.
I'm confused as to how this expression $E= \frac{\sigma}{\epsilon_0}$ can't be used to calculate the electric field of a perfectly flat part of a surface even farther from just above the surface.
What does one mean by just above the surface?
For an infinite flat surface the distance of any point which is a finite distance from the surface is infinitely small when compared with the size of the surface ie the point is "just above the surface".
Once you are dealing with finite surfaces then you do have a scale which will enable you to quantify what is meant by "just above the surface".
In such a situation you can then ascertain how close the electric field at a point is to that if the surface was infinite e.g. the electric field between the plates of a parallel plate capacitor.
In the first case ($\sigma / \varepsilon_0$), if you extend the gaussian surface of the cylinder to very far away from the surface of the conductor a central point will not be true in the calculation: The flux on this lateral surface will not be zero.
This is a central point in the case of a conductor: it is reasonable suppose that there are no tangential electric field just upside the conductor surface, or else there will be currents. So the point is that the tangential component of the field must be zero, adding no flux to the lateral area of the cylindrical gaussian surface...
The second case is trickier. It's totally non-intuitive to obtain as a result of a charge distribution a constant field as far as in the flux expression you have an electric field that should decay as $1/ r^2$ as you pointed.
I would pinpoint some arguments that I think can give you some hints.
(i) The situation of an infinite charged plane is quite artificial. It will off course never occur in a laboratory. That said, the calculation of a infinite plane is usefull if you think of a finite plane, for which you want to identify the field in the central regions.
(ii) This situation is a very good approximation for parallel plate capacitor, if the distance between the plates is very small and the deviation of the field from that occurs more noticeably in the ends of the plates.
(iii) About the dilema of a non-intuitive result I can think of a physical argument, why it must occur, which is current in the textbooks... Take a central point $P$ that is distant a quantity $x$ from the plane. If you take for instance a small $dQ$ in the plane distant $x\sqrt{2}$ of this point, you will have a right triangle, $x,x,x\sqrt{2}$. Now, take the opposed $dQ$ that forms the same right triangle and sum the electric field of both charges in $P$. You will get an electric field perpendicular to the plane as a result that depends on the sine of the distance $dQ$-$P$ and the plane. If you take another point distant $2x$ first as the distance increase by 2, the field diminishes by 4. But this sine term increases if you go from $x$ to $2x$. So, all in all if you integrate over all the sheet the influences equalize each other... (I will try to develop a little bit better point (iii) soon).
