E Field caused by infinite plate

[WarpDrive]

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?

 

[quantumdude]

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.

 

[quicknote]

The equation for the electric field caused by an infinite plate of charge is indeed sigma/2epsilon0, where sigma represents the surface charge density of the plate and epsilon0 is the permittivity of free space. This equation assumes that the plate is infinitely large and has a constant charge density throughout.

The distance from a point to the plate does not affect the calculation of the electric field because the plate is assumed to be infinite in size, so the distance to the edge of the plate is essentially infinite. This means that the electric field will remain constant at any distance from the plate.

If a solid conducting slab with significant thickness is used, the electric field will still be constant at any distance from the plate, but the charges will distribute differently. In this case, the electric field will be created by the charges on both sides of the slab, so it can be treated as two separate plates with opposite charges. However, the overall electric field will still be the same as that of an infinite plate with the same surface charge density.