(a) Potential at the mid point of the line joining the charges $Q_1$ and $-Q_2$ is $V = k \; \frac{Q_1}{\frac{r}{2}} + k \; \frac{Q_2}{\frac{r}{2}}$ ( $r$ is the distance between $Q_1$ and $-Q_2$ ) Or, $V = \; \frac{2 k Q_1}{r} - \; \frac{2 k Q_2}{r}$ Work done to place $Q_3$ is $W = V \times Q_3$ Or, $W = \left( \; \frac{2 k Q_1}{r} - \; \frac{2 k Q_2}{r} \right) \times Q_3$ $\therefore \;\; W = \; \frac{2k}{r} (Q_1-Q_2) Q_3$(b) Work done will be zero when $V=0$Let us consider that a point at a distance $x$ from the charge $Q_1$ where the potential will be zero.So, $\; \frac{k Q_1}{x} - \; \frac{k Q_2}{r-x} = 0$Or, $\; \frac{Q_1}{x} - \; \frac{Q_2}{r-x} = 0$Or, $\;\; \frac{Q_1}{x} = \; \frac{Q_2}{r-x}$Or, $Q_1(r-x)=Q_2 x$$\therefore \;\; x = \; \frac{Q_1 r}{Q_1+Q_2}$So, at a distance $\; \frac{Q_1 r}{Q_1+Q_2}$ from charge $Q_1$ on the line joining the two charges, the work done will be zero.