Field of an infinite plane sheet of charge

  • #1
334
44

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:

Answers and Replies

  • #2
156
38
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
334
44
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
156
38
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.
 

Related Threads on Field of an infinite plane sheet of charge

Replies
2
Views
117
Replies
12
Views
7K
Replies
1
Views
4K
Replies
8
Views
1K
Replies
3
Views
3K
Replies
7
Views
1K
Replies
3
Views
36K
Replies
8
Views
17K
Top