Electric Field/ infinite charged sheet

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 7K views
Andy13
Messages
16
Reaction score
0

Homework Statement


There are three charged sheets with various charge densities. The following link shows a diagram, and solutions (not mine; these are some professor's solutions but I don't understand them yet):

http://www.phy.syr.edu/courses/PHY212.08spring/HW/WHW-3.pdf

Draw vectors at each of the points A-D to show the direction and relative magnitude at the point due to a) sheet 1 b) sheet 2 c) sheet 3.

Homework Equations



E field due to infinite sheet = σ/2ε
(where ε = permittivity constant and σ = charge density)

Flux = ∫ EdA = E*A = |E||A|cos(θ)

Gauss' Law: Flux = q(enclosed)/εo

The Attempt at a Solution



I know which direction the vectors go, but I don't know what their magnitudes are. Surfing the internet gave me this (scroll to last page):

http://www.phy.syr.edu/courses/PHY212.08spring/HW/WHW-3.pdf

I know that the magnitudes of each vector of the electric field from each individual sheet from points A-D should be equal, because the electric field from a charged sheet doesn't depend on radius (for example, sheet 1's field could give vectors of magnitude |a| at every point).

However, I don't understand why sheet 2 (σ = -1), according to the above solutions, has vectors of twice the magnitude of sheets 1 and 3 (σ= +1). What am I missing?
 
Last edited by a moderator:
Physics news on Phys.org
SammyS said:
I agree with you.

His solution is consistent with the center sheet having a surface charge density of -2σ0 .

(I have taught this subject for more than 20 years.)

Thanks SammyS-- I'm glad I wasn't misunderstanding something.