E Field caused by infinite plate

Click For Summary
SUMMARY

The electric field (E field) caused by an infinite plate of charge is defined by the equation E = σ/(2ε₀), where σ represents the surface charge density and ε₀ is the permittivity of free space. This E field remains constant and uniform regardless of the distance from the plate. When considering a solid conducting slab with a thickness of 2 cm, the charges distribute evenly on both sides, effectively treating it as two infinite plates, which also maintains a uniform E field within the slab.

PREREQUISITES
  • Understanding of electric fields and charge distributions
  • Familiarity with the concepts of surface charge density (σ)
  • Knowledge of permittivity of free space (ε₀)
  • Basic principles of electrostatics and conductors
NEXT STEPS
  • Study the derivation of the electric field due to an infinite plane sheet of charge
  • Explore the effects of thickness on the electric field of conducting slabs
  • Learn about the superposition principle in electrostatics
  • Investigate the differences between electric fields from thin plates versus thick plates
USEFUL FOR

Students and professionals in physics, electrical engineering, and anyone interested in understanding electrostatics and electric fields generated by charged objects.

WarpDrive
Messages
6
Reaction score
0
What is the equation for the electric field caused by an infinite plate of charge? Is it sigma/2epsilon0?

So does that mean that the distance from a point to the plate doesn't matter in computing the E field? How does this change if a solid conducting slab with significant thickness (2 cm) is used? I know that the charges should build up on each side, with half of the charges on each side, so is that treated then as 2 plate charges?
 
Physics news on Phys.org
WarpDrive said:
So does that mean that the distance from a point to the plate doesn't matter in computing the E field?

That's right. An infinite plate of charge has a constant, uniform E field.

How does this change if a solid conducting slab with significant thickness (2 cm) is used? I know that the charges should build up on each side, with half of the charges on each side, so is that treated then as 2 plate charges?

See the following website:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesht.html

The first segment, entitled Electric Field: Sheet of Charge covers the thin plate, and the second segment, entitled Electric Field: Parallel Plates covers thick plates.
 

Similar threads

  • · Replies 31 ·
2
Replies
31
Views
2K
  • · Replies 26 ·
Replies
26
Views
3K
Replies
1
Views
1K
Replies
11
Views
3K
Replies
6
Views
1K
  • · Replies 20 ·
Replies
20
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
9
Views
824
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K