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The electric field of an uniformly charged infinite surface plane is totally independent of how far away the test charge is. But isn't electric field depends upon distance by the relation $1/r^2$? I understand that if the plane is truly infinite then it does'nt matter how far away you are. But there is no such thing as infinite plane. So, is the independency of distance works in real or practical life?

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    $\begingroup$ The $1/r^2$ law works for point charges, and a plane or segment of a plane is not a point. If you get very far away from a segment of a plane it will "look like" a point, and the behavior of the electric field closely resembles $1/r^2$. It will never be exactly $1/r^2$ but it you can make it as close to it as you would like by moving farther away. $\endgroup$ – garyp Dec 18 '16 at 14:58
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1/r² (inverse square) is valid for a point charge. Infinite wires have 1/r dependency.

It is all consistent.

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  • $\begingroup$ Why is it consistent? Do add that. $\endgroup$ – user36790 Dec 18 '16 at 10:54
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In fact, the electric field of an infinite plane cannot vary with with the inverse square law. If you apply Gauss theorem, you find that the electric field of that plane depends only on the surface charge distribution

The independency of distance works in real life, if you have a charge or a charged conductor that is small enough and close enough to the plane, it sees the electric field lines parallel to each other.

On the other hand, if you get close to the edge, you will get edge effects and the electric field lines will not be parallel.

In a real life example: when you calculate the electric acceleration of a charged particle through a synchrotron, like the LHC, you usually assume the particle to be small enough and the plates to be big enough so that no edge effects apply. It makes calculations easier, and it gives a good first approximation before getting in the real equations.

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