## Electric Field Due to Infinite Uniform Positive Charged Plate

\[\alpha\]

. 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.\[\mathbf{E}\]

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. \[\int_S \mathbf{E} \cdot \mathbf{n} dS = \frac{q}{\epsilon_0} \rightarrow 2 AE = \alpha A \rightarrow E =\frac{A \alpha}{\epsilon_0} \rightarrow E = \frac{\alpha}{2 \epsilon_0}\]