Electric field is constant around charged infinite plane. Why?

[zenterix]

Homework Statement

Consider an infinitely large non-conducting plane in the xy-plane with uniform surface charge density $\sigma$. Determine the electric field everywhere in space.

Relevant Equations

I know of two ways to answer this question. The first way uses Gauss's Law, the second way uses Coulomb's Law. I will use Gauss's Law below.

Using Gauss's Law

By using a symmetry argument, we expect the magnitude of the electric field to be constant on planes parallel to the non-conducting plane.

We need to choose a Gaussian surface. A straightforward one is a cylinder, ie a "Gaussian pillbox".

The charge enclosed is $q_{enc}=\sigma A$ where $A=A_1=A_2$ is the area of the cylinder end-caps.

The total flux through the pillbox is

$$\Phi_E=\iint_S \vec{E} \cdot d\vec{A}=E_1A_1+E_2A_2 + 0=A(E_1+E_2)$$

where the double integral is a surface integral over a closed surface, the cylinder. We've also chosen the cylinder such that the two end-caps (where there is non-zero flux) are the same distance from the plane so $E_1=E_2$.

$$\Phi_E=2EA$$

Now we apply Gauss's Law

$$2EA=\frac{q_{enc}}{\epsilon_0}=\frac{\sigma A}{\epsilon_0}$$

The magnitude of the electric field is thus

$$E=\frac{\sigma}{2\epsilon_0}$$

The magnitude of the electric field does not seem to depend on distance from the uniformly charged plane. I can't understand why this result occurs. I would have imagined that close the the plane the field would be stronger. Does anyone have an intuitive explanation of this result?

 

[rude man]

Homework Statement:: Consider an infinitely large non-conducting plane in the xy-plane with uniform surface charge density $\sigma$. Determine the electric field everywhere in space.
Relevant Equations:: I know of two ways to answer this question. The first way uses Gauss's Law, the second way uses Coulomb's Law. I will use Gauss's Law below.

View attachment 292638
Using Gauss's Law

By using a symmetry argument, we expect the magnitude of the electric field to be constant on planes parallel to the non-conducting plane.

We need to choose a Gaussian surface. A straightforward one is a cylinder, ie a "Gaussian pillbox".

The charge enclosed is $q_{enc}=\sigma A$ where $A=A_1=A_2$ is the area of the cylinder end-caps.

The total flux through the pillbox is

$$\Phi_E=\iint_S \vec{E} \cdot d\vec{A}=E_1A_1+E_2A_2 + 0=A(E_1+E_2)$$

where the double integral is a surface integral over a closed surface, the cylinder. We've also chosen the cylinder such that the two end-caps (where there is non-zero flux) are the same distance from the plane so $E_1=E_2$.

$$\Phi_E=2EA$$

Now we apply Gauss's Law

$$2EA=\frac{q_{enc}}{\epsilon_0}=\frac{\sigma A}{\epsilon_0}$$

The magnitude of the electric field is thus

$$E=\frac{\sigma}{2\epsilon_0}$$

The magnitude of the electric field does not seem to depend on distance from the uniformly charged plane. I can't understand why this result occurs. I would have imagined that close the the plane the field would be stronger. Does anyone have an intuitive explanation of this result?

If you have an infiite plane you can't tell what is "close" vs. what is "far". Everything's infinite distance!

 

[PeroK]

The magnitude of the electric field does not seem to depend on distance from the uniformly charged plane. I can't understand why this result occurs. I would have imagined that close the the plane the field would be stronger. Does anyone have an intuitive explanation of this result?

If you take a point close to the plane, then very little of the electric field from the charges in the plane points in the direction away from the plane. Most of the field from each charge is parallel to the plane. By symmetry, the net field is that pointing away from the plane.

If we take a point further away from the plane, then more of the electric field from each charge points away from the plane.

So, we have two factors: as distance from the plane increases and the field strength from a typical ring of charge decreases, so the proportion of the field pointing away from the plane increases. These two factors cancel out for a plane, if you do the maths, and leave the field constant as distance from the plane increases.

Note that, more generally:

The field from a point charge drops of as $1/r^2$; the field from an infinite line of charge drops off as $1/r$ and the field from an infinite sheet of charge is constant.

 

[anuttarasammyak]

The magnitude of the electric field does not seem to depend on distance from the uniformly charged plane. I can't understand why this result occurs. I would have imagined that close the the plane the field would be stronger. Does anyone have an intuitive explanation of this result?

Let us see the case of charged disk of radius R

Re: post #3 of https://www.physicsforums.com/threa...arged-disk-how-to-do-the-integration.1009062/

\phi(z)=\frac{\sigma}{2\epsilon_0}(\sqrt{z^2+R^2}-|z|)
E(z)=-\nabla\phi(z)=\frac{\sigma}{2\epsilon_0}(\frac{-z}{\sqrt{z^2+R^2}}+sgn(z))
where $\sigma$ is charge area density and sgn(z) is signature of z.
We observe the first term decrease to zero as we enlarge R. In this limit E(z) does not depend on |z| but signature of z.

 

[rude man]

Consider also your Gaussian surface, a right cylinder with major axis aligned with the normal to the dielectric sheet.. Gauss' principle is useful in this case if all the electric field flux pokes out of the ends only. If a portion pokes out the sides all bets are off. Your surface integral must include that sideways flux - Gauss' is always correct but only if all the flux piercing the entire surface is taken into account.

If the sheet is of finite dimensions then the flux will not be perpendicular to the sheet except right at its surface. At any finite distance above the sheet surfaces the flux lines will bend away from the sheet normal (called "fringing"). As you approach the sheet's sides and/or move farther away from the sheet's surfaces the bending increases and your Gaussian surface flux calculation will be more and more in error.

If on the other hand the dielectric sheet is infinitely large then the flux lines have to be perpendicular at any finite distance above the sheet's surfaces. Then no flux escapes thru the Gaussian cylinder's sides and your E calculations are accurate.