Question:
Two charged particles placed at a separation of $20 \mathrm{~cm}$ exert $20 \mathrm{~N}$ of Coulomb force on each other. What will be the force if the separation is increased to $25 \mathrm{~cm}$ ?
Solution:
$F=\frac{k q_{1} q_{2}}{r^{2}}$
$F \propto \frac{1}{r^{2}}$
$\frac{F}{F^{\prime}}=\frac{r^{\prime 2}}{r^{2}}$
$F^{\prime}=\frac{20 \mathrm{x}(0.20)^{2}}{(0.25)^{2}}$
$F^{\prime}=12.8 \mathrm{~N}$