Answer
0.957V
Work Step by Step
The total electric potential is given as:
$V=K\frac{q_1}{R}+K\frac{q_2}{R}+K\frac{q_3}{R}$
We plug in the known values to obtain:
$V=9\times 10^9\times \frac{4.52\times 10^{-12}}{8.50\times 10^{-2}}+9\times 10^9\times \frac{-2\times4.52\times 10^{-12}}{8.50\times 10^{-2}}+9\times 10^9\times \frac{3\times4.52\times 10^{-12}}{8.50\times 10^{-2}}$
Solving the equation, we find that:
$V=0.957V$