Answer
At the point B, the electric potential is $9.0~V$
Work Step by Step
We can find the electric potential at the point B:
$V = \frac{k~q_1}{r_1}+ \frac{k~q_2}{r_2}$
$V = k~(\frac{q_1}{r_1}+ \frac{q_2}{r_2})$
$V = (9.0\times 10^9~N~m^2/C^2) (\frac{2.0\times 10^{-9}~C}{1.0~m} + \frac{-1.0\times 10^{-9}~C}{1.0~m})$
$V = 9.0~V$
At the point B, the electric potential is $9.0~V$.
