# Field of an infinite plane sheet of charge

1. Mar 12, 2017

### fishturtle1

1. The problem statement, all variables and given/known data
Use Gauss's law to find the electric field caused by a thin, flat, infinite sheet with a uniform positive surface charge density σ.

2. Relevant equations
φ(flux) = ∫E*dA = qencl / ε0

3. 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: Mar 12, 2017
2. Mar 12, 2017

### TJGilb

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.

3. Mar 12, 2017

### fishturtle1

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?

4. Mar 12, 2017

### TJGilb

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.