Electric field from infinitely large metal

The fields from both the plates add up in the same direction to get the required field =sigma/epsilon

Taking the following formulae for capacitance:

$C = \frac{\epsilon_{0} A}{d}$ and $C = \frac{q}{V}$

equating these with other yields $\frac{\epsilon_{0} A}{d} = \frac{q}{V}$ . Now the capacitor plates have a surface charge of $\sigma$ coulombs/sq. unit, or $q = \sigma A$ coulombs of charge. The electric field between these plates is expressible as $E = \frac{V}{d}$ volts/unit. Substitution of these values now yields:

$\frac{\epsilon_{0} A}{d} = \frac{q}{V} \Rightarrow \frac{\epsilon_{0} A}{d} = \frac{\sigma A}{V} \Rightarrow \frac{V}{d} = \frac{\sigma}{\epsilon_{0}} \Rightarrow \boxed{E = \frac{\sigma}{\epsilon_{0}}}.$