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Assume there is an infinitely large plane with a charge density $\sigma$. I understand how to derive, using Gauss' Law, that $E = \frac{\sigma}{2\epsilon_0}$ is the electric field at a distance $r$ from the plane.

However, intuitively, I don't understand how this could be true. Shouldn't the electric field be more powerful closer to the plane, since the electric field falls of as $\frac{1}{r^2}$? The larger I make $r$, the further it is from every point, meaning the electric field should decrease - shouldn't it?

Any help understanding this would be appreciated.

Qmechanic
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Cisplatin
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1 Answers1

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Here's a simple way of looking at it:

If you are close to an infinite plane, you may be feeling stronger attraction by every individual part of it, but "more" of those parts are pulling you at a significant angle. This way, a lot of the attraction is canceling out. As it happens (this is anything but coincidence though), these two opposite effects exactly cancel eachother out when you move away from (or closer to) the plane.

overactor
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