# How Does Gauss' Law Apply to Infinite Planes with Different Charge Densities?

• tony873004
In summary, the conversation is discussing the calculation of the electric field between two parallel, infinite planes with surface charge densities. The electric field is perpendicular to the planes and can be calculated using Gauss' Law by placing a cylinder on the plane. The electric field above the top plane is half of the total electric field, while the electric field below the bottom plane is the same. Between the planes, the electric field lines cancel out.
tony873004
Science Advisor
Gold Member
I'm still not confident with these kinds of problems. Hopefully I got it right, but can someone double check my work? Thanks!

Two parallel, infinite planes are separated by a distance d. Find the electric field everywhere (a) if both planes carry a surface charge density $$\sigma$$ and (b) if one plane has a charge density $$\sigma$$ and the other - $$\sigma$$.

Here's my attempt. I'm only trying part (a) here.
The electric field is perpendicular to the planes because of symmetry. The components of the electric field parallel to the planes add to zero because of symmetry. Therefore, the electric field is perpendicular to the planes. So we can place a cylinder anywhere on the plane and use Gauss’ Law to compute the flux passing through the cylinder. The flux passing through the top of the cylinder will be half of the total flux.

$$\begin{array}{l} \Phi _{total} = \Phi _{side} + \Phi _{top} + \Phi _{bottom} \\ \\ \Phi _{side} = 0{\rm{ because }}\overrightarrow E \bot {\rm{\hat n}} \\ \\ \Phi _{{\rm{top}}} = \overrightarrow {E_z } \cdot {\rm{\hat k}} \cdot {\rm{\hat n}} \cdot A,\,\,\,\,{\rm{\hat k}} \cdot {\rm{\hat n = 1 because the normal vector, \hat n, is in the direction \hat k}}{\rm{.}} \\ \Phi _{{\rm{top}}} = \overrightarrow {E_z } \cdot A \\ \\ \Phi _{{\rm{bottom}}} = - \overrightarrow {E_z } \cdot - {\rm{\hat k}} \cdot {\rm{\hat n}} \cdot A,\,\,\,\, - {\rm{\hat k}} \cdot {\rm{\hat n = }} - {\rm{1 because the normal vector, \hat n, is in direction }} - {\rm{\hat k}}{\rm{.}} \\ \Phi _{{\rm{bottom}}} = \overrightarrow {E_z } \cdot A \\ \\ \Phi _{total} = 2\overrightarrow {E_z } \cdot A = \frac{{Q_{enclosed} }}{{\varepsilon _0 }},\,{\rm{ }}Q_{enclosed} = \sigma A \\ \overrightarrow {E_z } \cdot = \frac{{\sigma }}{{2\varepsilon _0 }} \\ \overrightarrow {E_z } = \frac{\sigma }{{2\varepsilon _0 }} \\ \end{array}$$

The electric field coming out of the top of the cylinder is half of the electric field coming out of the total cylinder. Therefore, the electric field above the top infinite plane is
$$\overrightarrow {E_z } = \frac{{\frac{\sigma }{{2\varepsilon _0 }}}}{2} =$$ $${\frac{\sigma }{{4\varepsilon _0 }}{\rm{,\hat k}}}$$

Similarly, the electric field below the bottom infinite plane is
$${\frac{\sigma }{{4\varepsilon _0 }}, - {\rm{\hat k}}}$$

Between the planes, the electric field lines will cancel. Therefore there are no electric field lines between the planes.

I change my guess for the top and bottom. It should be twice as much as the field lines from the plates combine.

This is because the electric field from one plane is equal and opposite to the electric field from the other plane, resulting in a net electric field of zero.

Overall, your calculations are correct and your reasoning is sound. However, just to clarify, in part (a) both planes have the same surface charge density, while in part (b) one plane has a positive charge density and the other has a negative charge density. This results in a slightly different calculation for the electric field above and below the planes, as shown below:

Above the top plane: \overrightarrow {E_z } = \frac{{\sigma }}{{2\varepsilon _0 }}, {\rm{\hat k}}

Below the bottom plane: - \overrightarrow {E_z } = - \frac{{\sigma }}{{2\varepsilon _0 }}, - {\rm{\hat k}}

Between the planes: \overrightarrow {E_z } = 0, {\rm{\hat k}}

Overall, your understanding of Gauss' Law and its application to infinite planes is correct. Keep practicing and you will become more confident with these types of problems. Good work!

## What is Gauss' Law for infinite planes?

Gauss' Law for infinite planes is a fundamental law in electrostatics that states the electric flux through a closed surface surrounding a point charge is directly proportional to the enclosed charge. In the case of an infinite plane, the electric field is considered to be uniform and perpendicular to the surface.

## How is Gauss' Law applied to infinite planes?

Gauss' Law is applied to infinite planes by using a Gaussian surface, which is a hypothetical surface that encloses the infinite plane. The electric flux through this surface is calculated, and it is equal to the charge enclosed by the surface divided by the permittivity of free space.

## What is the electric field equation for an infinite plane using Gauss' Law?

The electric field equation for an infinite plane using Gauss' Law is E = σ/2ε₀, where E is the electric field, σ is the surface charge density of the infinite plane, and ε₀ is the permittivity of free space. This equation only applies to the region directly in front of or behind the infinite plane, and the electric field is assumed to be uniform and perpendicular to the surface.

## How does the electric field vary in different regions around an infinite plane?

The electric field varies in different regions around an infinite plane. In the region directly in front of or behind the plane, the electric field is uniform and perpendicular to the surface. In the region above or below the plane, the electric field decreases with distance from the plane and becomes zero at an infinite distance. In the region parallel to the plane, the electric field is zero.

## What are some real-life examples of infinite planes in which Gauss' Law can be applied?

Gauss' Law can be applied to real-life examples such as parallel plate capacitors, which are made up of two infinite conducting plates separated by a small distance. It can also be applied to the electric field created by a charged conducting sheet, such as a metal sheet with a uniform surface charge. Additionally, the electric field between two large parallel conducting plates can also be approximated as an infinite plane for the purpose of applying Gauss' Law.

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