JEE Main 2024 — Electrostatics Question with Solution

To find the ratio of the Coulombic force to the gravitational force between an electron and a proton in a hydrogen-like system, we use the formulae for both forces and then divide them.

The Coulombic (electrostatic) force, $F_C$, between two charges is given by Coulomb's law:

$F_C=k\frac{|q_1{q}_2|}{r^2}$

where $k$ is Coulomb's constant ($8.987\times 10^9{\text{Nm}}^2{\text{C}}^{-2}$), $q_1$ and $q_2$ are the magnitudes of the charges, and $r$ is the distance between the charges. For a proton and an electron, $q_1=q_2=e$, where $e$ is the elementary charge ($1.602\times 10^{-19}\text{C}$).

The gravitational force, $F_G$, between two masses is given by Newton's law of universal gravitation:

$F_G=G\frac{m_1{m}_2}{r^2}$

where $G$ is the gravitational constant ($6.674\times 10^{-11}{\text{Nm}}^2{\text{kg}}^{-2}$), and $m_1$ and $m_2$ are the masses of the two objects. For a proton and an electron, $m_p\approx 1.673\times 10^{-27}\text{kg}$ and $m_e\approx 9.109\times 10^{-31}\text{kg}$, respectively.

The ratio of the Coulombic to the gravitational force is therefore:

$\frac{F_C}{F_G}=\frac{k\frac{|e^2|}{r^2}}{G\frac{m_p{m}_e}{r^2}}=\frac{k{e}^2}{G{m}_p{m}_e}$

Plugging in the values:

$\frac{F_C}{F_G}=\frac{(8.987\times 10^9)(1.602\times 10^{-19}{)}^2}{(6.674\times 10^{-11})(1.673\times 10^{-27})(9.109\times 10^{-31})}$

$\frac{F_C}{F_G}=\frac{(8.987\times 10^9)\times (2.568\times 10^{-38})}{(6.674\times 10^{-11})\times (1.523\times 10^{-57})}$

$\frac{F_C}{F_G}\approx 2.3\times 10^{39}$

Therefore, the ratio of the Coulombic force to the gravitational force between an electron and a proton in a hydrogen-like system is on the order of $10^{39}$, making Option B the correct answer.