Electric Field Problem: Solving for E in a System of Oppositely Charged Plates

AI Thread Summary
In a system of two large, parallel metal plates with opposite surface charge densities of 17.0*10^-22, the electric field (E) in the outer regions of both plates is zero due to the cancellation of their electric fields. This cancellation occurs because the electric fields from the plates are uniform and equal in magnitude, despite the distances not being equidistant. The electric field from a single large sheet of charge is given by E = sigma/2×vacuum permittivity, and since the plates have opposite charges, their fields only add together in the space between them. The importance of the plates being "large" lies in the uniformity of the electric field they produce, which is valid except near the edges. Thus, the net electric field outside the plates remains zero.
Prashasti
Messages
63
Reaction score
2
Question : Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0*10^-22. What is E (Electric field) : (a) in the outer region of the first plate, (b) in the outer region of the second plate?

Answer: According to the book, answers are (a) 0 (b) 0 And the reason they've given is - "Net electric field is Zero because electric fields due to both the plates will cancel out each other".

My objection : How? How do they cancel each other's effect when that particular region is NOT EQUIDISTANT from each other, and as Electric field varies inversely with the square of distance, HOW ARE THEY EQUAL IN MAGNITUDE?
 
Physics news on Phys.org
...as Electric field varies inversely with the square of distance
... this is only correct for sherical charge distributions - i.e. a sphere or a point charge.

Why is it important that the plates are "large"?

The electric field from two sheets of charge like that is the sum of the fields due to the single sheets alone.
What is the electric field due to a single large sheet of charge?
 
I got it. Thank you so much. I think it is E = sigma/2×vacuum permittivity?!
 
Well done - yeah: the field due to a large plate (except near the edges) is uniform.
Since the plates have opposite signs, their fields cancel everywhere except in between them (and by the edges) where they add together.
 
It may be shown from the equations of electromagnetism, by James Clerk Maxwell in the 1860’s, that the speed of light in the vacuum of free space is related to electric permittivity (ϵ) and magnetic permeability (μ) by the equation: c=1/√( μ ϵ ) . This value is a constant for the vacuum of free space and is independent of the motion of the observer. It was this fact, in part, that led Albert Einstein to Special Relativity.
Back
Top