Question

Question using Gauss Law to determine the surface charge density of a charged conducting sheet. I will attach a picture of it. I'm not sure which Gaussian Surface to use in this case. I was thinking a Cylinder, but it's a Surface and not a Line

Answer 1

Gauss Law Problem

Answer 2

You use something called a "Gaussian pillbox". See attached diagram. So the pill box magically goes through the sheet like a ghost. But it's okay because only the top and bottom cross-section of the cylinder contribute to the flux. That is, the only E-fields that do not cancel out are the ones normal to the surface. Of course with almost all Gauss's Law problems one already knows the answer before you start. What Gauss's Law does for us is give us a mathematical proof of our intuition. In this case, for an infinite sheet, the E-field is constant everywhere! I call this the "Flat Earth" solution because is equivalent to a gravitational field near the surface of where curvature of the Earth is negligible. The "other side" of the gravitational sheet in that case is straight through the Earth to the other side where someone is standing upside down - the gravitational field reverses direction. This is different the Electricity (Coulomb's Law basically) where the E-field is either repulsive or attractive using the convention of a positive charge to define the field. By the way, there is one other solution to the charged flat sheet, a neutrally charged sheet - zero E-field everywhere. Analogously, if one sits inside a cavity at the center of the Earth the gravitational field is zero. Which means the curvature of space is truly zero in that situation!

Answer 3

Hey, I'm the one that posted that question, I didn't realize i was using my GF's account. So using the "PillBox" I get that \[\sigma = E2\epsilon_0\]

Answer 4

Then E=(mgtan(30)+q)2\epsilon_0

Answer 5

\[E=(mgtan \theta + q)2\epsilon_0\]

Answer 6

@Paul_Kent let me know if im in the right track

Answer 7

\[\sigma=(mgtan \theta + q)2\epsilon_0\]
made a mistake above.