Answer
c) i=iii$\lt$ii$\lt$iv
Work Step by Step
Electrostatic potential energy
$E_{el}=k\frac{Q_{1}Q_{2}}{d}$, $k$ is a constant.
For (i), $E_{el}=k\frac{(+1)(-3)}{d}=-3\times\frac{k}{d}$
For (ii), $E_{el}=k\frac{(+2)(-2)}{d}=-4\times\frac{k}{d}$
For (iii), $E_{el}=k\frac{(+2)(-3)}{2d}=-3\times\frac{k}{d}$
For (iv), $E_{el}=k\frac{(+4)(-4)}{3d}=-\frac{16}{3}\times\frac{k}{d}$
Comparing the electrostatic potential energies, we can arrange them in increasing order as:
i=iii$\lt$ii$\lt$iv
The correct option is (c).
