Application of Gauss's Law to Charge Distribution

  • Thread starter zachfoltz
  • Start date
  • #1
zachfoltz
27
0

Homework Statement


A 10.0 g piece of Styrofoam carries a net charge of -0.700 [itex]\mu[/itex]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?


Homework Equations



Fe = [itex]\frac{KqQ}{r^2}[/itex] (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 = [itex]\frac{Q}{A}[/itex] (Q is the total charge of the sheet, A is its area)
[itex]\Phi[/itex] = [itex]\oint[/itex] E[itex]\cdot[/itex]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[itex]\pi[/itex]) which means 1/ε0 = K4[itex]\pi[/itex]


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.
 

Answers and Replies

  • #2
Doc Al
Mentor
45,462
1,950
Fe = [itex]\frac{KqQ}{r^2}[/itex] (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)
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
zachfoltz
27
0
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[itex]\oint[/itex] E[itex]\cdot[/itex]dA. Unfortunately I don't know how to solve that integral because I can't think of a Gaussian surface that would make E[itex]\cdot[/itex]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
Doc Al
Mentor
45,462
1,950
The formula for a force from a uniformly distributed charge on a charge q would be: F= q[itex]\oint[/itex] E[itex]\cdot[/itex]dA.
The force would be F = Eq. The first job is to find the field from a uniform sheet of charge.

Unfortunately I don't know how to solve that integral because I can't think of a Gaussian surface that would make E[itex]\cdot[/itex]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.
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
zachfoltz
27
0
THANK YOU. It just clicked. Much gratitude brother!
 

Suggested for: Application of Gauss's Law to Charge Distribution

Replies
1
Views
398
Replies
10
Views
674
  • Last Post
Replies
1
Views
324
Replies
9
Views
561
Replies
1
Views
320
Top