CBSE Class 12 · 2011 · All India · ElectrostaticsShort Answer

Coulomb's law gives $F=\frac{1}{4\pi {\epsilon }_0}{q}_1{q}_2\cdot \frac{1}{r^2}$, so a plot of $F$ against $\frac{1}{r^2}$ is a straight line through the origin whose slope is $\frac{q_1{q}_2}{4\pi {\epsilon }_0}$. For $(1\mu \text{C},2\mu \text{C})$ the product is positive, so the force is repulsive and the slope is $\frac{2}{4\pi {\epsilon }_0}{\text{N m}}^2$ (positive). For $(2\mu \text{C},-3\mu \text{C})$ the product is negative, so the force is attractive (negative slope), and the magnitude of the slope is larger ($6$ against $2$ units), making the line steeper.