Answer
$E=5.65\times 10^4\frac{N}{C}$
Work Step by Step
In the given scenario, we can find electric field as:
$E=\frac{\sigma_1}{2\epsilon_{\circ}}+\frac{\sigma_2}{2\epsilon_{\circ}}+\frac{\sigma_3}{2\epsilon_{\circ}}$
$E=\frac{\sigma_1+\sigma_2+\sigma_3}{2\epsilon_{\circ}}$
We plug in the known values to obtain:
$E=\frac{(2.00+4.00-5.00)\times 10^{-6}}{2(8.85\times 10^{-12})}$
$E=5.65\times 10^4\frac{N}{C}$
