Finding the electric field at a point in between 3 electrodes

[kwozz48]

Homework Statement

The figure is an edge view of three charged metal electrodes. Let the left electrode be the zero point of the electric po*tential. Assume that Q=100 nC and q=50 nC.

The left electrode has a charge of -q, the middle electrode has a charge of +Q, and the right electrode has a charge of -q. There is 1 cm in between the left electrode, the middle electrode is 1 cm wide, and there is 1 cm between the middle electrode and the right electrode. All of the electrodes have 2.0 cm x 2.0 cm cross section. I apologize for the lack of a visual, I realize this would make it easier.

a.) What are E at 0.5 cm
b.) What are V at 0.5 cm
c.) What are E at 1.5 cm
d.) What are V at 1.5 cm
e.) What are E at 2.5 cm
f.) What are V at 2.5 cm

Homework Equations

E=Q/(e_0*A)
V=Ed

The Attempt at a Solution

If I have the equations that are supposed to be used correct, then getting parts b, d, and f are very simple. You just plug the answer from parts a, c, and e in correct? I can't figure out how to use the first equation to the get the E field at those points. Any explanation of this problem you can give me would be much appreciated. I need to understand it and be able to work it, not just know the answer. Thanks a lot.

 

[Mraaz4mu]

You got the first equation wrong!
It is:
E = Q / (A*2e_0)
you forgot to multiply 2 with e_0.
Hope that helps!
SOURCE:
https://answers.yahoo.com/question/index?qid=20080129214944AAQ5iJv

 

[rude man]

No, your 1st equation is OK. The D field close to a surface charge σ is D = σ.
And D = ε₀E and σ = Q/A where Q is the total charge on the surface of area A.

For a dielectric plate the D field close to the plate (but in vacuum) would be σ/2.

Your problem is to determine the surface charge density on all 6 surfaces, then E is immediately available and so as you point out is V.
EDIT: OK, you only need the inner 4 surfaces.

BTW the fact that the plates are only 2 x 2 cm. makes your answers necessarily very approximate since there would be appreciable field fringing taking place.