Answer
$E_{(r)}=1.35\times 10^4\frac{N}{C}$
Work Step by Step
We know that
$E_{(r)}=\frac{1}{4\pi \epsilon_{\circ}}\frac{q_1+q_2}{r^2}$
$E_{(r)}=K\frac{q_1+q_2}{r^2}$ (where $K=\frac{1}{4\pi \epsilon_{\circ}}$)
We plug in the known values to obtain:
$E_{(r)}=8.99\times 10^9\times \frac{(4.00+2.00)\times 1\times 10^{-8}}{(0.200)^2}=1.35\times 10^4\frac{N}{C}$
