# Application of Gauss's Law to Charge Distribution

1. Sep 2, 2013

### zachfoltz

1. The problem statement, all variables and given/known data
A 10.0 g piece of Styrofoam carries a net charge of -0.700 $\mu$C and is suspended in equilibrium above the center of a large, horizontal sheet of plastic that has a uniform charge density on its surface. What is the charge per unit area on the plastic sheet?

2. Relevant equations

Fe = $\frac{KqQ}{r^2}$ (q is the charge of the styrofoam- given, and Q is the total charge of the plastic sheet. K is Coulumbs Constant = 8.99x10^9)
Fg = mg
σ = $\frac{Q}{A}$ (Q is the total charge of the sheet, A is its area)
$\Phi$ = $\oint$ E$\cdot$dA = qin0 (E is the electric field. dA is an infinitesimal area VECTOR hence the dot product. qin is the total charge inside a Gausian surface. ε0 is the permittivity of free space which is equal to 1/(K4$\pi$) which means 1/ε0 = K4$\pi$

3. The attempt at a solution
Obviously since the Styrofoam is floating at equilibrium Fg = Fe and I set those equal to each other, but since I don't know "r" - the height that the Styrofoam is suspended at, or "A" - the area of the plastic sheet, or "Q" (the total charge of the sheet) I don't know how to solve for σ. I couldn't come up with a Gaussian surface that would make E and A vectors parallel to simplify the surface integral, so I don't know if I can do anything with that equation. Thank you for taking the time to read and hopefully help.

2. Sep 2, 2013

### Staff: Mentor

That gives the force between two point charges. Not relevant here.

What's the field from a charged sheet? What field is required to put that Styrofoam piece in equilibrium? (Treat it as a particle.)

3. Sep 2, 2013

### zachfoltz

Oh, I didn't know that only applied to particles. The formula for a force from a uniformly distributed charge on a charge q would be: F= q$\oint$ E$\cdot$dA. Unfortunately I don't know how to solve that integral because I can't think of a Gaussian surface that would make E$\cdot$da = E*dA (because the vectors are parallel), or a surface on which E is always constant (the same distance away at all points) therefore able to take out of the integral.

4. Sep 2, 2013

### Staff: Mentor

The force would be F = Eq. The first job is to find the field from a uniform sheet of charge.

Treat the charged sheet as being infinitely big. How would the field lines look? Choose a Gaussian surface that has its surfaces parallel and perpendicular to the charged surface. (Many shapes will do.)

This is a standard exercise, so your textbook may have a discussion worth reviewing.

5. Sep 2, 2013

### zachfoltz

THANK YOU. It just clicked. Much gratitude brother!