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In this video

At $6:00$ Khan gives an example problem with electric potential energy, electric field and electric force. He finds the electric force $F = E Q$ and then calculates the work $W = F d$ needed to be done to move this charge $3 \text{ m}$ with the already found electric force $(10\text{ N})$ and indeed that will be the electric potential energy. My question is, as the charge gets closer, shouldn't the electric force from the electric field increase, since we are getting closer to the electric field? And from there shouldn't the work be more since we must apply more than $10 \text{ N}$ to "beat" the electric field's generated force? Am I understanding something wrong, or Khan did it that way for simplicity?

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    $\begingroup$ Not listening the whole video I would assume that the example used a constant electric field; it doesn't get stronger as you move along it. $\endgroup$
    – Communisty
    Feb 26 '18 at 8:20
  • $\begingroup$ Oh, yeah... I think he said something like that but I didn't think it meant that. So a constant electric field means that the electric force due to the field is the same, everywhere? $\endgroup$ Feb 26 '18 at 8:28
  • $\begingroup$ Yes, the field strength and direction are the same everywhere within the field. $\endgroup$
    – Communisty
    Feb 26 '18 at 9:05
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The analysis done in the video is correct as the electric field strength due to a uniformly charged infinite conducting plane at a given position is the same irrespective of the chosen position.

You are not alone in thinking that as you get closer to a uniformly charged infinite plane conductor the electric field should should increase.
It is mentioned in the video that you can prove that the electric field is constant using calculus.

Perhaps a non-mathematical approach can also be used to convince you that the field is uniform which very much relies on the uniformly charged conducting plane being infinite.

Imagine in your mind a positive test charge in the vicinity of a the uniformly charged infinite conducting plane.
Will the picture differ if the test charge was one millimetre or one metre away from the charged plane?

The answer is "no" because you cannot gauge the distance that the charge is away from the infinite plane with reference to any other distance as the plane itself is infinite in length.

So no matter how far the test charge is away from the plane the arrangement that you visualise always looks exactly the same.
If you cannot decide relative to the size of the plane how far the charge is away from the plane then the force on the test charge must be the same everywhere as must be the electric field strength due to the charged plane.

Here a slightly more mathematical approach but again remember that the plane is infinite.

The test charges are at distance $a$ and $b$ from the plane.

enter image description here

The charge at a distance $b$ feels a smaller force from each of the charges "beneath" it in the cone described by the $25^\circ$ angle than the test charge at a distance $a$ because it is further away from the plane, force $\propto \dfrac{1}{\rm distance ^2}$, but there are more charges in that cone $\propto \text {base radius of cone} ^2$ (remeber that it is a two dimensional plane) which exactly compensates for the effect of the greater distance.

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    $\begingroup$ > If you cannot decide relative to the size of the plane how far the charge is away from the plane then the force on the test charge must be the same everywhere as must be the electric field strength due to the charged plane. -- This is an invalid argument, on two accounts: 1) it does not give the right answer in other scale-free situations, like two charged points in empty space 2) it gives the right answer for infinite plane only if the field is falling off as inverse square of distance. The actual reason for the result is the inverse square of distance character of electric field. $\endgroup$ Feb 26 '18 at 17:13

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