## College Physics (4th Edition)

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.