KEAM 2021 Physics and Chemistry Solved Paper

Electrostatic force between two charges $q_{1}$ and $q_{2}$ is given as $F = \frac{K q_{1} q_{2}}{r^{2}}$ where, $K$ is the Coulomb's constant and $r$ is the distance between the charges. According to the question, for two protons with charge $q$ (say) and at a distance $r$ apart, the electrostatic force is given as $F_{p} = \frac{K q q}{r^{2}} = \frac{K q^{2}}{r^{2}}$.....(i) Similarly, for two electrons with charge $e$ and at a distance $r$ apart, the electrostatic force is given as $F_{e} = \frac{K e e}{r^{2}} = \frac{K e^{2}}{r^{2}}$.....(ii) On dividing Eq. (i) by Eq. (ii), we get $\frac{F_{p}}{F_{e}} = \frac{ \frac{K q^{2}}{r^{2}} }{ \frac{K e^{2}}{r^{2}} }$ $\Rightarrow$ $\frac{F_{p}}{F_{p}} = \frac{q^{2}}{e^{2}}$ Since, the magnitude of charge on proton $=$ the magnitude of charge on electron i.e. $|q| = |e| = 1.6 \times 10^{-19} \, \mathrm{C}$ $\Rightarrow$ $\left| \frac{F_{p}}{F_{e}} \right| = \left| \frac{q^{2}}{e^{2}} \right| = \left| \frac{q^{2}}{q^{2}} \right| = \frac{1}{1}$ Or $|F_{p} : F_{e}| = 1 : 1$