# Infinite plane of charge

1. Dec 23, 2011

### tomwilliam

I'm a little confused by the textbook I'm working from, which asserts:

"the electrical field from an infinite plane of charge does not decrease with distance"

Why doesn't it decrease as the square of distance from the plane, from Coulomb's law?

2. Dec 23, 2011

### Morgoth

there are several ways to look at it.
From Gauss's law you can extract that the potential is independent of the distance.

It seems strange that the field does not depend from the distance from the plane because of coulombs law. One way to understand it is to think that AS FAR THE OBJECT MOVES FROM THE PLANE the more "effective" charge it can feel from it. So of course it would fall because of its distance, but on the other hand it raises because the "charge it sees" changes.

I got that explanation from Griffith's book in electrodynamics.

3. Dec 23, 2011

### Antiphon

When something involving the inverse square law moves twice as far away, it looks half as large.

The infinite plane looks exactly the same no matter how far you get from it so the net fields are not a function of distance.

4. Dec 23, 2011

### FireStorm000

The basic thing to see is that coulomb's law is for point charges; other arrangements have to be looked at as a collection of point charges if you want to use this law.

For example, if you look at a line of infinite charge, the field decreases by the inverse of the distance rather than the inverse square. By extending the charge from existing in dimention 0, a point, to dimension 1, a line, we have also reduced the reliance on r. If we further extend the charge into a plane, dimension 2, then the denominator disappears entirely, and you're left with a constant field.

If you are in a calculus based class, then you can take the sheet of charge, and count up every infinitesimal contribution (integrate), and see that the distance term disappears entirely for an infinite charged plane(It's a rather cool problem, so if you have the math background, you should definitely try it).

5. Dec 23, 2011

### rcgldr

It's assumed that the plane has a uniform amount of charge per unit area. The components of force parallel to the plane cancel out leaving only the components of force perpendicular to the plane as a net force. For a point near a plane most of the net force is due to the area near the point. The force on that point from any point on the plane is relative to sin(θ) where θ is the angle between a line along the plane to a line going towards that point near the plane. Assume you only consider the net force related to a cone originating from the point near the plane back to the plane, with the cone having a fixed angle of width. If the point is located futher away from the plane, the force from each point decreases by distance2, but the area of the base of the cone increases by distance2, so the force remains constant.

With similar logic, the force from an infinite line will be equal to 1 / distance instead of 1 / distance2.

This can be demonstrated using calculus.

The math for the electric field (force per unit charge) for a line, ring, and disc of charge is shown here:

http://hyperphysics.phy-astr.gsu.edu/%E2%80%8Chbase/electric/elelin.html [Broken]

For the disc of charge, as R → ∞, Ez → kσ2π.

The math for an infinite plane is shown in post #2 of this thread:

Last edited by a moderator: May 5, 2017
6. Dec 23, 2011

### Foofur

Explanation 1: Integrate it.

Explanation 2: Imagine you're looking at an infinite TV screen that's just glowing white. If you step away, does it look any different?

7. Dec 24, 2011

### granpa

the solution is obvious when you picture the lines of force. (thats why people use them)

8. Dec 25, 2011

### tomwilliam

Thanks for all those replies...that's much clearer.

9. Dec 25, 2011

### Morgoth

granpa, without solving the equations you can draw no lines of force.

mathematically it is always easier to use Gauss law and see what happens with the electric field, and physically i think the posts #2 and 3 are the right ones.