## Fundamentals of Physics Extended (10th Edition)

$V=2.9KV$
We know that $V=V_1+V_2$ $\implies V=\frac{Kq_1}{r_1}+\frac{Kq_2}{r_2}$ We plug in the known values to obtain: $V=\frac{9\times 10^9\times 1\times 10^{-8}}{0.03}+\frac{9\times 10^9\times (-3\times 10^{-8})}{2.0-0.03}=2.9\times 10^3=2.9KV$