# Field of an infinite plane sheet of charge

• fishturtle1
In summary: In other words, the only component of the electric field that contributes to the flux is the one that is perpendicular to the surface of the cylinder. Therefore, only the electric field that is perpendicular to the sheet contributes to the flux, and that is why the diagram shows a single electric field line going straight through the center of the cylinder. In summary, when using Gauss's Law to find the electric field caused by a thin, flat, infinite sheet with a uniform positive surface charge density σ, it is only necessary to consider the flux through the ends of a Gaussian cylinder, as the flux through the sides is zero due to the cancellation of electric field components. This leads to the equation E = σ/2ε0 for the electric field strength. The
fishturtle1

## Homework Statement

Use Gauss's law to find the electric field caused by a thin, flat, infinite sheet with a uniform positive surface charge density σ.

## Homework Equations

φ(flux) = ∫E*dA = qencl / ε0

## The Attempt at a Solution

I set up a diagram like this: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesht.html#c1

1) I'm sorry if this is a really vague question but I don't get why the cylinder is smaller than the sheet? I thought the cylinder would have to enclose the sheet? If we don't enclose all the charge of the plane, how can we use Gauss's Law?

then EA = qencl / ε0, because I assumed E and A are constants in this situation.

A = 2πrL, where L is the length of the gaussian cylinder.

then E = qencl / (ε0 * A)

E = qencl / ε02πrL

which is so wrong..

I read the answer in the book but these are what makes no sense to me:

1) The flux though the cylindrical part of the cylinder is zero because E * n = 0 everywhere.

2) The flux through each flat end of the surface is +EA because E * n = E everyone, so the total flux through both ends is +2EA.

If I assume these two things are true I understand what they do after which is

2EA = σA / ε0

E = σ/2ε0

Please help on how I can understand what the book is telling me in 1) and 2) and also how do they arrive at the diagram that they do?

Last edited:
So you have an infinite, infinitesimally thin sheet of charge with a uniform charge density. If you were to treat every point on that sheet as a point charge of some uniform magnitude, and then drew a vector field for each one, you would see that everything except the vertical components cancel from the charges next to it. So, if you draw a Gaussian cylinder, there is no net flux through the sides of the cylinder and therefore you only need to worry about the caps.

Now, take Gauss's Law. We know the charge enclosed is equal to the integral of the charge density over the area, but since its uniform, we can simplify that to the surface charge density sigma multiplied by the area. On the left side of the equation, we can assume E will have constant magnitude and will always be parallel to the caps of the cylinder, so we can pull that out and simplify the equation to the magnitude of the electric fields multiplied by the area. However, since we have two caps, this will actually be the area of both circles. Divide, the areas cancel, and you're left with your answer.

So there is only one electric field line that goes straight though the center of the cylinder? Because for every other electric field, there will exist another that is equal in magnitude but opposite in sign?

Well, it's better to think of it as a vector, where all of the x and y components cancel, and all the z components add.

## 1. What is the concept of a field of an infinite plane sheet of charge?

The field of an infinite plane sheet of charge is a physical concept in which a two-dimensional surface is uniformly charged. This creates an electric field that is perpendicular to the plane and decreases in strength as the distance from the plane increases.

## 2. How is the electric field intensity calculated for an infinite plane sheet of charge?

The electric field intensity is calculated by dividing the surface charge density by 2 times the permittivity of free space. This gives the magnitude of the electric field at any point above or below the plane.

## 3. What is the direction of the electric field for an infinite plane sheet of charge?

The electric field for an infinite plane sheet of charge is always perpendicular to the surface of the plane. This means that the field lines are parallel and evenly spaced, extending infinitely in both directions.

## 4. How does the electric field change as you move away from the infinite plane sheet of charge?

The electric field strength decreases as you move away from the plane. This is because the electric field is inversely proportional to the distance from the charged surface. As the distance increases, the field strength decreases.

## 5. Can the electric field for an infinite plane sheet of charge ever be zero?

Yes, the electric field can be zero in two specific locations - directly above and directly below the plane. In these locations, the electric field vectors cancel each other out, resulting in a net electric field of zero.

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