By Gauss’s law
$ \oint \overrightarrow{E}.\overrightarrow{ds}=\frac{q}{\in_0}$$\therefore{2{\text{EA}}}=\frac{\sigma{\text{A}}}{{\in_0}}$
$\therefore\text{E}=\frac{\sigma}{2\in_0}\text{ }\text{or}\frac{\sigma}{2\in_0}\text{A}$
Electric field between two identical charged sheets
$\because$ Both the sheets have same charge density, their electric fields will be equal and opposite in the region between the two sheets.
Hence the net field is zero.Alternate Answer
$\text{E}_1=\frac{\sigma}{2\in_0}$
$\text{E}_2=-\frac{\sigma}{2\in_0}$
Resultant electric field between the plates = E1 + E2$\frac{\sigma}{2\in_0}-\frac{\sigma}{2\in_0}=0$]
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