Infinite sheets of charge w/ conducting slab

Click For Summary
SUMMARY

The discussion centers on calculating the surface charge density on each side of an infinite conducting slab placed between two infinite sheets with surface charge densities σ1 = 8.85 μC/m² and σ2 = 1.5 μC/m², and a net charge per unit area of σC = -3 μC/m². The conducting slab has a thickness of a = 2 cm. The user successfully calculated the x-component of the net electric field at x = 0 but seeks assistance in determining the surface charge densities on either side of the slab, confirming that these must equal the total surface charge density of the slab.

PREREQUISITES
  • Understanding of electric fields and surface charge density
  • Familiarity with Gauss's Law and its applications
  • Knowledge of electrostatics, particularly in relation to conductors
  • Basic calculus for solving electric field equations
NEXT STEPS
  • Study the application of Gauss's Law in electrostatics
  • Learn how to calculate electric fields due to surface charge densities
  • Explore the concept of electric field discontinuity at the surface of conductors
  • Review problems involving multiple charged plates and their interactions
USEFUL FOR

Students preparing for exams in electrostatics, physics educators, and anyone studying electric fields in the context of conductors and charge distributions.

rgalvan2
Messages
26
Reaction score
0

Homework Statement


Two infinite sheets with surface charge density σ1 and σ2, respectively, are oriented perpendicular to the x-axis. An infinite, conducting slab of thickness a is placed between the charged sheets as shown in the figure. The conducting plate has a net charge per unit area of σC.

σ1=8.85\muC/m^2
σ2=1.5\muC/m^2
σC=-3\muC/m^2
a=2cm

Homework Equations


E=\sigma/2\epsilon0

The Attempt at a Solution



I figured since the magnitude of the electric field is not determined by displacement, I could just use the above equation. The only problem it that the conducting slab has different surface charge density on each side and I'm not sure how to start this. I have an exam tomorrow so any help is greatly appreciated!
 
Physics news on Phys.org
What are you asked to find?
 
Oops sorry:
Calculate the x-component of the net electric field at x = 0.
I figured that part out though.
The next part asks to calculate the surface charge density on each side of the slab. I'm not sure how to do that. I do know that both sides need to equal the total surface charge density of the slab. Is that right?
 

Similar threads

  • · Replies 26 ·
Replies
26
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
1
Views
1K
Replies
6
Views
1K
Replies
9
Views
4K
  • · Replies 17 ·
Replies
17
Views
14K
  • · Replies 3 ·
Replies
3
Views
21K
Replies
12
Views
2K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K