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A very common question about Gauss law is being asked in schools or Internet:

Why the electric field of an infinite plane is independent on distance?

Actually I was one of who asked this question and I found an answer, and I will explain it in the next paragraph.

A very important thing we have to remember when think about this kind of problems is that the electric field is a vector, and the $Ex$ vector only effect on it because $Ey$ is cancelled by another charge.

Since we go further. the electric field is being weaker, also the angle $θ$ is being less too So $\cos θ$ is getting larger so that compensated the lack of electric field. enter image description here

Now, why we cant apply this argument on an infinite line, as you see in the picture?

enter image description here

Please don't explain it using Gauss law because it's hard to visualize it, because the number of imaginary concepts like : (Flux, Electric field, epsilon ... etc.) So it's hard to realize.

Finally, please don't mark my question as a duplicate because I didn't understand Eeko explanation.

Why does the electric field of an infinite line depend on the distance, but not on an infinite plane?

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    $\begingroup$ Not understanding an answer doesn't make it not a duplicate. You should either explain why your question is not a duplicate, or specifically talk about what in the duplicate does not make sense to you $\endgroup$
    – BioPhysicist
    Commented Apr 8, 2019 at 14:47
  • $\begingroup$ You need to revise the title from "Infinity line" to "infinite plane". $\endgroup$
    – Bob D
    Commented Apr 8, 2019 at 16:10

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One way to think about this is through symmetry. If your plane is infinite then as you move away from it, nothing appears to change. Exactly half of your field of view is filled with plane.

Here's another way to look at it:

Take your cone of view. If I double the distance from the plane, the charge in that cone is twice as far away. Inverse square -- you get 1/4 of the contribution to the electric field.

But your cone is now grabbing a circle of plane that is twice the diameter => 4 times the area, so the contribution is 4 times as much. Two factors balance out.

When you calculate this using a simple integral, you figure it out usually integrating over your angle from 0 to 90. Part of this you figure out the differential amount of charge in the plane for $\delta\theta$ and when you do the distance from the plane cancels out.

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  • $\begingroup$ Your geometrical argument is good, but the argument in your first paragraph, I'm not sure. If I move away from an infinite line of charge, nothing appears to change. $\endgroup$
    – garyp
    Commented Apr 8, 2019 at 15:16
  • $\begingroup$ @garyp It works as long as you treat the line as the limiting case of a cylinder of finite size. This is the same argument I gave in the linked answer that the question asker doesn't understand. (Although stated a bit more succinctly). $\endgroup$
    – Eeko
    Commented Apr 8, 2019 at 15:26
  • $\begingroup$ @Eeko I am sorry about this controversy , But I really want to understand this , I am very annoyed about this question . Let us using another way of thinking , What is the mistake I made in the second picture mathematically or physically ? $\endgroup$
    – Mohammad Alshareef
    Commented Apr 8, 2019 at 19:14
  • $\begingroup$ So it does. The limiting cylinder had a natural length scale (the radius), while the plane does not. With the cylinder we can compare the distance of the observation point to the size of the cylinder. No such comparison can be made for the plane. The fact that the scaling argument works in the limit is somewhat interesting. $\endgroup$
    – garyp
    Commented Apr 8, 2019 at 19:48
  • $\begingroup$ @garyp It's an interesting case where the order you take the limits in matters. $\endgroup$
    – Eeko
    Commented Apr 8, 2019 at 19:56
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You're correct that there is no electric field parallel to the line, since the line is symmetric across any point you pick. (A benefit of being infinite in length). But that isn't enough to say that the electric field doesn't vary with distance perpendicular to the line.

If you look at the diagram for the infinite plane, you'll see that as you look at larger and larger angles, the circumference of the ring of charges you consider grows in proportion to the sin($\theta$). The line on the other hand, only has two points of charge for any angle $\theta$, so rather than growing with sin($\theta$), the contribution of charges to the electric field is independent of $\theta$.

Since, as you noted, the electric field is constant for an infinite plane, it makes sense that the electric field of an infinite line should decrease with distance, as the amount of contributing charge in the plane is more than in the line.

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  • $\begingroup$ I have another explanation of this before asking the question , when we say $q$ in the infinity line : $q = λL$ , But in infinity plane $q = σA$ , So when we pick a charge from the line it does not like a charge picked from plane , so the compensating does not happen perfectly ? I am trying to find a mistake from the second picture , Because I believe in logical mathematical proof more than anything . IS THAT RIGHT ? $\endgroup$
    – Mohammad Alshareef
    Commented Apr 8, 2019 at 19:37
  • $\begingroup$ @MohammadAlshareef The mathematical proof is given by Gauss' law, so by excluding that explanation there's not much I can do in the way of mathematical rigor. Remember, since the line and plane are infinite, we can't talk about total charge, but instead differentials. $dq = \lambda dL$. In this sense, the area has more "$dq$'s" per angle than the line does. So the strength of the plane's field doesn't fall off with distance, while the line's does. $\endgroup$
    – Eeko
    Commented Apr 8, 2019 at 20:04
  • $\begingroup$ That's it . Thank you very very much , Now really I am feeling comfortable by understanding this . I LOVE YOU ^_^ $\endgroup$
    – Mohammad Alshareef
    Commented Apr 8, 2019 at 20:12

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