Electric field inside at points on a conductive box

[Caiti]

Homework Statement

I'm having trouble understanding a concept.
Given a conducting box with point a located outside the box, point b located within the (thick) wall of the box and point c located within the cavity of the box, I have to calculate the electric field at each of these points.
There is excess negative charge on the box and a surface density of 2.10×10^10 e/m^2.

I managed to figure out all parts, but I struggle to understand why this is the case.

Homework Equations

Gauss' Law: ∫ E.dA = Q/ε_0
Q = Ne

The Attempt at a Solution

For point a:
Q = Ne ∴ Q = 1.6*10^(-19)*-2.10*10^(10) = 3.36*10^-9

∫E.dA = Q/ε_0 = (3.36*10^-9)/(8.85*10^-12) ≈ -379.7 N/C

I know the answer is meant to be positive, but I don't understand how. I believed it would be negative as the charge on the box is negative, and so would 'emit' a negative field.

For point b:
0 N/C because it is inside the walls of the box.

For point c:
I initially thought this would be (+ or -) 379.7 N/C to balance out point a, but found the answer to be 0 N/C. I can't find a reason why this would be, other than that the net charge must be negative; however, I don't see why this couldn't contribute a negative charge in order to increase the magnitude of the net negative charge.

 

[Gil]

For point a, I don't understand exactly what is your question.

For point c think the following. Where are the charges? Are there charges INSIDE the box? The electrical field inside the box is related to the electric charge that is inside the box. Another way to see this is the following. In point b there is no field, so the outside fields become zero in this point, what do you expect to happen as you go to the interior?

 

[Caiti]

Here's a copy of an image explaining where the points are. It has them labelled as 1, 2 and 3, but I used a, b and c respectively.

So with point c, inside, it is determined by point b? I'm sorry, but I don't fully understand

Thank you (=