Answer
$E=(-7.91\times 10^{-11}\frac{N}{C})(i^{\wedge})$
Work Step by Step
The electric field between the two plates is given as:
$E=(\frac{\sigma}{2\epsilon_{\circ}})(-i^{\wedge})+(\frac{\sigma}{2\epsilon_{\circ}})(-i^{\wedge})$
$E=(\frac{\sigma}{\epsilon_{\circ}})(-i^{\wedge})$
We plug in the known values to obtain:
$E=\frac{7.00\times 10^{-22}}{8.85\times 10^{-12}}(-i^{\wedge})=(-7.91\times 10^{-11}\frac{N}{C})(i^{\wedge})$
