Answer
$V_{BA}=\frac{2kq(2b-d)}{b(d-b)} $.
Work Step by Step
Use equation 17–5, the voltage due to a point charge.
$$V_{BA}=V_B-V_A=(\frac{kq}{d-b}+\frac{k(-q)}{b})-(\frac{kq}{ b}+\frac{k(-q)}{d-b})$$
$$V_{BA}=kq (\frac{1}{d-b}-\frac{1}{b}-\frac{1}{ b}+\frac{1}{d-b})$$
$$V_{BA}=2kq (\frac{1}{d-b}-\frac{1}{b})$$
$$V_{BA}=\frac{2kq(2b-d)}{b(d-b)} $$
