Electric field of infinite plane

[omri3012]

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

 

[tiny-tim]

Hallo Omri!

(have a pi: π and a sigma: σ )

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.

 

[rcgldr]

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.

 

[omri3012]

thanks everyone,

it was very helpful.

 

[rcgldr]

You can do a web search for "guass law infinite plane" and see a few various approaches for this.

 

[Born2bwire]

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.

 

[JazzFusion]

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.

 

[omri3012]

about the ituitive answer, i can claim the same argument over aninfinite wire but is field does depend on location.

 

[tiny-tim]

about the ituitive answer, i can claim the same argument over aninfinite wire but is field does depend on location.

he he

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

omri3012 1

JazzFusion 0

 

[JazzFusion]

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

...so if you move laterally away from the line charge, the 'view' changes.