# Find the ratio of the electric and gravitational forces between two protons.

## Question:

Find the ratio of the electric and gravitational forces between two protons.

## Gravitational Force, Electrostatic Force:

{eq}\\ {/eq}

The gravitational force and the electrostatic force are two fundamental forces in nature.

The gravitational force exists between two objects because of their mass. It is always attractive in nature and is characterized by Newton's Law of Gravitation.

The electrostatic force is the force that exists between two objects because of their charge. It is either attractive or repulsive depending on the nature of the charge. The electrostatic force between two charged objects is characterized by Coulomb's Law.

{eq}\\ {/eq}

• The mass of a proton is: {eq}m=1.67\times 10^{-27}\;\rm kg {/eq}
• The charge on a proton is: {eq}q=1.6\times 10^{-19}\;\rm C {/eq}

The gravitational force between two objects of mass, {eq}m_1 {/eq}, and {eq}m_2 {/eq}, separated by a distance, {eq}r {/eq}, is given by the following equation:

{eq}F_g=\dfrac{Gm_1m_2}{r^2} {/eq}

Here,

• {eq}G=6.67\times 10^{-11}\;\rm m^3/kg.s^2 {/eq} is the gravitational constant.

The magnitude of the electrostatic force between two point charges {eq}q_1 {/eq} and {eq}q_2 {/eq} is given by the equation:

{eq}\begin{align*} F_e&=\dfrac{q_1q_2}{4\pi \epsilon_0 r^2}\\ &=\dfrac{Kq_1q_2}{r^2} \end{align*} {/eq}

Here,

• {eq}r {/eq} is the distance between the charges.
• {eq}\epsilon_0 {/eq} is the permittivity of free space.
• {eq}K=9\times 10^{9}\;\rm N.m^2/C^2 {/eq}

Therefore, the ratio of the two forces is:

{eq}\begin{align*} \dfrac{F_e}{F_g}&=\dfrac{\dfrac{Kq^2}{r^2}}{\dfrac{Gm^2}{r^2}}\\ &=\dfrac{Kq^2}{Gm^2}\\ &=\dfrac{9\times 10^{9}\times \left ( 1.6\times 10^{-19} \right )^2}{6.67\times 10^{-11}\times \left ( 1.67\times 10^{-27} \right )^2}\\ &=\boxed{1.24\times 10^{36}} \end{align*} {/eq} 