Application of Gauss's Law to Charge Distribution

AI Thread Summary
The discussion revolves around applying Gauss's Law to determine the charge per unit area on a charged plastic sheet that balances a negatively charged piece of Styrofoam. The participant struggles with finding a suitable Gaussian surface to simplify the integral for the electric field. Key points include recognizing that the electric field from a uniformly charged sheet can be treated as constant and that the force on the Styrofoam can be expressed as F = Eq. Suggestions are made to consider the charged sheet as infinitely large to visualize the field lines and choose an appropriate Gaussian surface. The participant expresses gratitude upon gaining clarity on the problem.
zachfoltz
Messages
27
Reaction score
0

Homework Statement


A 10.0 g piece of Styrofoam carries a net charge of -0.700 \muC 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 = \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\cdotdA = 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


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.
 
Physics news on Phys.org
zachfoltz said:
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)
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.)
 
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\cdotdA. Unfortunately I don't know how to solve that integral because I can't think of a Gaussian surface that would make E\cdotda = 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.
 
zachfoltz said:
The formula for a force from a uniformly distributed charge on a charge q would be: F= q\oint E\cdotdA.
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\cdotda = 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.
 
  • Like
Likes 1 person
THANK YOU. It just clicked. Much gratitude brother!
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top