## Electric Field Due to Infinite Uniform Positive Charged Plate

We can use Gauss's Law to find the electric field due an an infinite flat plate with a uniform positive surface charge
$\alpha$
. We enclose a section of the plate within a closed cuboid and apply Gauss's Law to the surface.

Gauss's Law states
$\int_S \mathbf{E} \cdot \mathbf{n} dS = \frac{q}{\epsilon_0}$

where
$q$
is the charge enclosed by the surface. By summetry
$\mathbf{E}$
is constant and perpendicular to the surface of the plate. This means that at the sides of the surface the normal
$\mathbf{n}$
is at right angles to the electric field
$\mathbf{E}$
. The only contributions to the surface integral comes from the surfaces of the cuboid parallel to the plate.
$\int_S \mathbf{E} \cdot \mathbf{n} dS = \frac{q}{\epsilon_0} \rightarrow 2 AE = \alpha A \rightarrow E =\frac{A \alpha}{\epsilon_0} \rightarrow E = \frac{\alpha}{2 \epsilon_0}$