Two infinite sheets of different charge density

[purplegreen]

Homework Statement

From the picture attached it can be seen that there are two plates with charge density σa and -σb. We will chose → as the n direction.
What is that electric field everywhere for this system?

Homework Equations

∫∫E.dS = ∫∫σ/ε0 dA

The Attempt at a Solution

Using Gauss' law we can take three gaussian surfaces. One containing both of the plates and two more containing each individual plate.

Using the surface containing both plates, we have:
ESn + ESn = σ(a-b)/ε0 n

hence,
E = σ(a-b)/2ε0 n outside of the plates

Inside I think I should use superposition of the two other surfaces. However I'm not sure which direction the electric field goes in for these two surfaces?

Generally I'm not convinced by my answer and lack some understanding which I would like some help with. Thanks in advance!

 

[Dickfore]

Using Gauss's Lw, you should be able to show that an infinite charged sheet creates a homogeneous electric field directed perpendicularly to the sheet:
$$\vec{E} = \frac{\sigma}{2 \epsilon_0} \, \hat{n}$$
where $\hat{n}$ is a unit normal directed from the sheet outwards (on each side).

Now you can use the principle of superposition to find the electric field due to two sheets of charge.

 

[purplegreen]

Using Gauss's Lw, you should be able to show that an infinite charged sheet creates a homogeneous electric field directed perpendicularly to the sheet:
$$\vec{E} = \frac{\sigma}{2 \epsilon_0} \, \hat{n}$$
where $\hat{n}$ is a unit normal directed from the sheet outwards (on each side).

Now you can use the principle of superposition to find the electric field due to two sheets of charge.

Ok so would the answers for this question be:

on the left hand side of the 2 plates:
E= -σ(a-b)/2ε n

In the middle of the plates:
E= σ(a+b)/2ε n

Finally, on the right hand side of the plates:

E= σ(a-b)/2ε n

Thanks for the help.