Is the Electric Field from an Infinite Plane of Charge Constant with Distance?

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SUMMARY

The electric field generated by an infinite plane of charge remains constant regardless of distance, contrary to the expectations set by Coulomb's law, which applies to point charges. This phenomenon can be understood through Gauss's law, which indicates that the electric field is independent of distance when considering an infinite plane. As one moves away from the plane, the effective charge perceived does not diminish due to the uniform distribution of charge across the plane, leading to a constant electric field. This concept is further illustrated by comparing the electric fields from different charge distributions, such as lines and planes, where the field from an infinite line decreases inversely with distance.

PREREQUISITES
  • Understanding of Gauss's law in electrostatics
  • Familiarity with Coulomb's law and its application to point charges
  • Basic calculus for integrating charge distributions
  • Knowledge of electric field concepts and vector fields
NEXT STEPS
  • Study the application of Gauss's law to various charge distributions, including infinite lines and planes
  • Learn how to derive electric fields using calculus for different geometries
  • Explore the implications of electric field lines and their visualization in electrostatics
  • Investigate the differences between point charge fields and continuous charge distributions
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Students of physics, particularly those studying electromagnetism, educators explaining electric fields, and anyone seeking to deepen their understanding of electrostatic principles and charge distributions.

tomwilliam
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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?

Thanks in advance
 
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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.
 
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.
 
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 dimension 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).
 
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

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

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

https://www.physicsforums.com/showthread.php?t=471647
 
Last edited by a moderator:
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?
 
tomwilliam said:
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?

Thanks in advance

the solution is obvious when you picture the lines of force. (thats why people use them)
 
Thanks for all those replies...that's much clearer.
 
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.
 

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