Answer
a) $ q_2=\frac{-Q}{\sqrt 2 }$.
b) positive
Work Step by Step
(a) We know that point B is closer to $ q_2$ whereas it is at a distance of $\sqrt 2$ times greater from $ q_1$. The potential at B will be zero, so the charge at $ q_2$ must be less than the magnitude of $ q_1$ by a factor $\sqrt 2$ and also $ q_2$ must be negative, thus $ q_2=\frac{-Q}{\sqrt 2 }$.
(b) We know that the point A is closer to $ q_1$ than to $ q_2$ and also the magnitude of the negative charge is smaller, hence the electric potential at point A must be positive.
