Electric field of infinite plane

AI Thread Summary
The electric field of an infinite plane does not depend on distance (r) because the field lines are straight and uniform, unlike those from a point charge, which spread out and weaken with distance. As the plane is infinite, the field remains constant regardless of the observer's position. This concept is supported by Gauss's Law, which mathematically shows that the electric field is invariant in two dimensions. An intuitive analogy compares the situation to hovering above an infinite plane, where the view remains unchanged at any height. Thus, the electric field strength is consistently equal to 2πσ, regardless of distance from the plane.
omri3012
Messages
60
Reaction score
0
Hallo,

Why does the field of a infinite plane does not depend on r? i know it's equal to

two \pi\sigma but why does his "infinite" makes it independent on r?

thanks

Omri
 
Physics news on Phys.org
Hallo Omri! :smile:

(have a pi: π and a sigma: σ :wink:)
omri3012 said:
Why does the field of a infinite plane does not depend on r? i know it's equal to

two \pi\sigma but why does his "infinite" makes it independent on r?

The field lines from a point charge spread out, so the field decreases with increasing distance.

But the field lines from an infinite plane are straight, and don't spread out, so the field is the same at any distance. :smile:
 
The math for field strength from a solid disc is explained here:

http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/elelin.html#c3

as R^2 \ \rightarrow \infty

then \frac{z} {\sqrt{z^2 + R^2}} \ \rightarrow \ 0

and you end up with E_z = k \ \sigma \ 2 \ \pi

An alterative approach is to consider the field from an infinitely long line (= 1/z), then integrate an infinitely large plane composed of infinitely long rectangles that approach infinitely long lines as their width approaches zero.
 
Last edited:
thanks everyone,

it was very helpful.
 
You can do a web search for "guass law infinite plane" and see a few various approaches for this.
 
One thing to note is that the problem is invariant in two-dimensions. If the plane lies in the x-y plane, then you know right away that the field cannot rely on x or y since the source is infinite and invariant along those directions. So all you need to do is convince yourself that the resulting fields will also be invariant with regards to z, the remaining dimension. Jeff's post is good for showing this part.
 
The 'math' answer: The derivation of the field strength, using Gauss's Law, shows that the Electric Field is independent of z. Not a very satisfying answer, but a truthful one.

The Intuitive Answer: Imagine you are a point charge hovering in a balloon over an infinite plane of charge. The plane extends infinitely, as far as the eye can see, in every direction below you. The view will be exactly the same, no matter how close or far you are from the plane. In fact, for a blank, empty, infinite plane with no points of reference on it, you will have no way of telling how far from the plane you are from where you sit in your balloon.
 
about the ituitive answer, i can claim the same argument over aninfinite wire but is field does depend on location.
 
omri3012 said:
about the ituitive answer, i can claim the same argument over aninfinite wire but is field does depend on location.

he he :biggrin:

(the field lines from a line charge also spread out)

omri3012 1

JazzFusion 0 :wink:
 
  • #10
tiny-tim said:
(the field lines from a line charge also spread out)
...so if you move laterally away from the line charge, the 'view' changes.
 
Back
Top