Answer
$E_{(r)}=2.50\times 10^4\frac{N}{C}$
Work Step by Step
We know that:
$E_{(r)}=\frac{1}{4\pi \epsilon_{\circ}}\frac{q_1}{r^2}$
$E_{(r)}=K\frac{q_1}{r^2}$ Z(where $K=\frac{1}{4\pi \epsilon_{\circ}}$)
We plug in the known values to obtain:
$E_{(r)}=8.99\times 10^9\times \frac{4.00\times 10^{-8}}{(0.120)^2}=2.50\times 10^4\frac{N}{C}$
