Answer
The correct answer is:
(a) It is the same as that at D.
Work Step by Step
We can find an expression for the potential at point B:
$V_B = \frac{k~q_1}{s}+\frac{k~q_2}{s} = \frac{k~(q_1+q_2)}{s}$
We can find an expression for the potential at point D:
$V_D = \frac{k~q_1}{s}+\frac{k~q_2}{s} = \frac{k~(q_1+q_2)}{s}$
The correct answer is:
(a) It is the same as that at D.
