Answer
(a) $-\sqrt{2}Q$
(b) negative
Work Step by Step
(a) We know that
$r_2=\sqrt{r^2+r^2}$
$\implies r_2=\sqrt{2}r$
The net potential at point A is given as
$V_A=\frac{Kq_1}{r_1}+\frac{Kq_2}{r_2}$
Given that the net potential at point A is zero, we have:
$\implies \frac{Kq_1}{r_1}+\frac{Kq_2}{r_2}=0$
$\implies \frac{q_1}{r_1}+\frac{q_2}{r_2}=0$
$\implies q_2=-(\frac{q_1}{r_1})r_2$
We substitute $+Q$ for $q_1$, r for $r_1$ and $\sqrt{2}r$ for $r_2$
$\implies q_2=-(\frac{+Q}{r})\sqrt{2}r$
$\implies q_2=-\sqrt{2} Q$
(b) We know that
$V_B=K(\frac{q_1}{r_1}+\frac{q_2}{r_2})$
$\implies V_B=K(\frac{+Q}{\sqrt{2}r}+(\frac{-\sqrt{2}Q}{r}))$
$\implies V_B=\frac{-KQ}{\sqrt{2}r}$
Thus, the negative electric potential exists at point B.
