Find charge density on two parallel planes

[Thorn]

Homework Statement

Two planes of charge with no thickness, A and B, are parallel and vertical. The electric field in region 1 to the left of plane A has magnitude 3σ/(2*ε0) and points to the left. The electric field in the region to the right of B has magnitude 3σ/(2*ε0) and points to the right. The electric field in the region between the the two planes has magnitude σ/(2*ε0) and points to the right. What is the surface charge density on planes A and B respectively?

Homework Equations

E = σ/(2*ε0)

The Attempt at a Solution

I have no idea where to begin. I was thinking that the charges were of magnitude 3σ, but that wouldn't explain the inside field...I just don't know how to think of this problem correctly...

 

[Doc Al]

Homework Equations

E = σ/(2*ε0)

That's the correct formula for the field from one plane of charge. When you have more than one, you'll add up the field contributions from each. (The principle of superposition.)

Hint: Don't forget that the direction (sign) of the field from a plane of charge is different on each side.

 

[Thorn]

adding up the field contributing from each? you mean that plate B is contributing to the field to the left of plate A, "region 1"? Well, I suppose it would have to be, I just don't see how. As for the superposition principle, I got that, since all these fields are on the same axis, I will just add or subtract them accordingly. But still...I have no clue how that formula works. Like I said, Using it on the left side of plane A tells you that the charge density is 3σ, which just isn't right...you get the same answer if you use it on plane B in the region to the right. But these charge densities clearly wouldn't get the E field we see in between the two planes...

 

[Thorn]

hah! Nevermind...I figured it out...you can get 2 systems of equations and solve for the 2 charge densities...

Answer.. Plate A has charge density 2σ and B has charge density σ.

Thanks much!