Question

The ratio of the magnitude of electrostatic force and gravitational force for an electron and a proton is:
A) $6.6 \times {10^{39}}$
B) $2.4 \times {10^{39}}$
C) $6.6 \times {10^{29}}$
D) $19.2 \times {10^{29}}$

Answer 1

The electrostatic force of an electron is directly proportional to the charge of an electron. Gravitational force is directly proportional to the product of the masses and indirectly proportional to the square of the distance between them. Using the above statement determines the ratio of the electrostatic and gravitational force of an electron.

Formula used:
Gravitational force,
${F_g} = \dfrac{{G{m_e}{m_p}}}{{{r^2}}}$
Electrostatic force,
${F_e} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{e^2}}}{{{r^2}}}$
$e$is the charge of an electron, ${m_e}$ is the mass of the electron, ${m_p}$ is the mass of the proton.

Complete step by step solution:
Gravitational force is a force present in every particle in the universe that attracts every other particle that is directly proportional to the product of the masses and indirectly proportional to the square of the distance between them. This force is along the line joining particles. Gravitational force is the weakest force in nature. It plays a very important role in controlling the structure of the universe.
The gravitational constant is equal to $6.67 \times {10^{ - 11}}N{m^2}/k{g^2}$.
${F_g} = \dfrac{{G{m_e}{m_p}}}{{{r^2}}}$
There will be a force of attraction or repulsion between the charges. It also exists in a vacuum.
The electrostatic force of an electron is directly proportional to the charge of an electron. The square of the distance between them is inversely proportional to the gravitational force. The value of $\dfrac{1}{{4\pi {\varepsilon _0}}}$ is equal to $9 \times {10^9} units$. It is a central force and a conservative force. Gravitational force is attractive but the electrostatic force is both attractive and repulsive.
${F_e} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{e^2}}}{{{r^2}}}$
Now the ratio between the two forces,
$\dfrac{{{F_e}}}{{{F_g}}} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{e^2}}}{{G{m_e}{m_p}}}$
$\dfrac{{{F_e}}}{{{F_g}}} = \dfrac{{9 \times {{10}^9} \times {{\left( {1.6 \times {{10}^{ - 19}}} \right)}^2}}}{{6.67 \times {{10}^{ - 11}} \times 9 \times {{10}^{ - 31}} \times 1.66 \times {{10}^{ - 27}}}} = 2.4 \times {10^{39}}$

Hence, the correct answer is in option $\left( B \right) \Rightarrow 2.4 \times {10^{39}}$.

Note: Electrostatic force depends on the medium between them. Gravitational force acts along the line joining particles. It is independent of the medium present between the gravitational force is a force of attraction or repulsion between the charges. It also exists in a vacuum. The electrostatic force is a central force and a conservative force.