# Infintite plane of charge guass law

1. Jan 31, 2016

### kiwileaf

when deriving the electric field through using gauss's law, I do not completely understand why when
calculating for the charge enclosed the answer key says
"The surface charge distribution on is uniform. The area of the intersection of he non-conducting plane with the Gaussian cylinder is equal to the area of the end-caps, . Therefore the charge enclosed in the cylinder is q=σA"

I thought that the total charge would be the integral of sigma with respect to the volume of a cylinder (dv), which would make the total charge enclosed q=σV

However, even if it is q=σA I do not understand why we would not include the entire area of the cylinder because wouldn't the charge be over that entire Gaussian surface? why do we disregard the sides and only include the end caps for area?

2. Jan 31, 2016

### kiwileaf

(this is for an infinite plane of charge with a uniformly distributed positive charge density σ) and using Gaussian surface of a cylinder where the infinite plan is at x=0 and the two sides of the cylnider extend to include x<0 and x>0

3. Jan 31, 2016

### Aniruddha@94

We disregard the area of the sides because the normal vectors to the sides (, i.e. the curved part) is perpendicular to the electric field vector... It means no flux passes through the curved part of the cylinder.

4. Jan 31, 2016

### kiwileaf

I understand that when computing the integral of E dot dA the E dot dA of the sides would be zero, but I was talking about simply calculating and integrating the amount of charge enclosed (but I think I understand that the charge enclosed would be the σ times the area of the circle of the cylinder that intersects the plane not a three dimensional volume where the sides are included)

5. Jan 31, 2016

### Suraj M

So that clears it out for you?

6. Feb 1, 2016

### Aniruddha@94

Yes, you've got it. We take ( sigma)X(end area) because that's the only part of the sheet inside the cylinder.