# Gauss' law giving zero field where field is not zero?

Two plastic sheets with charged densities as shown:

I'm trying to find the field at $B$. I obtained the correct answer by adding up the fields created by each charge density. But I realized that since the field is uniform in the region between the two sheets, I should be able to make a Gaussian surface between the sheets with the shape of a box and with one edge at $B$. Thus, the flux would be

$$\int{E \cdot dA} ~=~ EA ~=~ \frac{q_{encl}}{\epsilon_{0}}~=~0.$$

Because there would be no charged enclosed inside the surface. However, that means that $EA$ is $0$ (Note the integral reduced to $EA$ because $E$ is uniform). Since $A$ is not zero (it is the area of two sides of the box), this means that $E$ must be zero. However, $E$ is not zero there, as you can see by adding $\frac{\sigma}{2\epsilon_0}$ for each charge density.

What am I doing wrong when using Gauss' law?

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Draw your little box and draw the E field vector. The vector points into the box on one side, out of the box on the other side. When you use Gauss's law you need to dot the field with the outward pointing surface normal vector. The total flux is EA - EA = 0, that does not imply EA=0. – user1631 Nov 3 '11 at 17:40
Hall Vladimir, I guess You could not beare that Guass? :=) – Georg Nov 4 '11 at 14:48

You tell us that one surface of the box is at $B$, but you're a little vague on where the opposite face is. You do say that your surface is "between the two sheets", so I think you may mean that the surface is entirely contained in the space between the two sheets. The box does not intersect any charged surface. With that, and a uniform electric field in that region, the flux on the opposite face will have a sign opposite to the flux at the surface at $B$. The total flux is zero. The field and flux at any portion of the surface can be non-zero. Gauss speaks only to the total flux.
If you draw a gaussian box between the plates, the flux in the surface near to $\sigma_3$ is equal to that of the surface $\sigma_2$. In this case $\vec{E}\cdot\vec{A}$ is the same as $\vec{E}\times\vec{A}$.
If you apply the superposition principle and for each separate plate and calculate the field by Gauss law then adding the fields gives: $E= \frac{\sigma_3-\sigma_2}{\epsilon_0}$