Electric Field inside a nonconducting sheet

Click For Summary
SUMMARY

The discussion focuses on calculating the electric field within a nonconducting sheet that is 10.0 cm thick, with a uniform charge distribution of +95.0 nC/m² on the upper face and -25.0 nC/m² on the lower face. The electric field at a point 2.00 cm below the upper face is determined using Gauss's law and the principle of superposition. Participants clarify that the varying charge densities on each side must be considered separately, treating the sheet as two parallel sheets of charge. The relevant equation for electric flux is also highlighted as electric flux = ∫E*da = q(enc)/ε_0.

PREREQUISITES
  • Understanding of Gauss's law and electric flux
  • Familiarity with charge density concepts
  • Knowledge of the principle of superposition in electrostatics
  • Basic proficiency in calculating electric fields from charge distributions
NEXT STEPS
  • Study the application of Gauss's law in different geometries
  • Learn about electric fields due to infinite sheets of charge
  • Explore the principle of superposition in electrostatics
  • Investigate the effects of varying charge densities on electric fields
USEFUL FOR

Students in physics, particularly those studying electromagnetism, as well as educators and anyone seeking to understand electric fields in nonconducting materials.

forestmine
Messages
201
Reaction score
0

Homework Statement



A huge (essentially infinite) horizontal nonconducting sheet 10.0 cm thick has charge uniformly spread over both faces. The upper face carries +95.0 nC/m2 while the lower face carries -25.0 nC/ m2. What is the magnitude of the electric field at a point within the sheet 2.00 cm below the upper face? (\varepsilon_0 = 8.85 × 10-12 C2/N · m2)

Homework Equations



electric flux = ∫E*da = q(enc)/ε_0


The Attempt at a Solution



I figured this problem should be relatively easy, but the fact that the charge densities are different on each side is throwing me off...

First, I thought to find the E field at some particular point, I ought to use a Gaussian surface, and my thought was to use a cylinder, sticking in as far as .02m below the upper face. Doing so, gives me E=σA/ε_0*A, and the A's cancel, of course.

What I'm really confused about is how to take into account the fact that it's only .02m into the sheet. And for that matter, because the charge density is different on each side, I'm not sure how to go about finding the charge density at that particular point of .02m inside the sheet.

If anyone could help clear some of this up, I'd really appreciate it.

Thanks!
 
Physics news on Phys.org
Principle of superposition.
You can do it OK as a single sheet of charge right?
 
Thanks for the reply.

That's right, just confused about the varying charge density. So if I understand you correctly, I should find the field of the one side, the other side, and then again for 2cm in? How do I determine a correct charge density for the point that's 2 cm in, however?
 
Your description says the charge is spread uniformly over the faces only - not penetrating into the sheet. So what you have is two parallel 2D sheets of charge.

so put the +Q sheet at z=+5cm and the -Q sheet at z=-5cm and work out the field at z=+3cm.
 

Similar threads

  • · Replies 26 ·
Replies
26
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
969
Replies
9
Views
855
Replies
23
Views
4K
Replies
9
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
4
Views
1K
Replies
10
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K