Find the ratio of the magnitude of the electric force to the gravitational force acting between two protons.

#### Solution

Charge of the proton, q = \[1 . 6 \times {10}^{- 19} C\]

Mass of the proton = \[1 . 67 \times {10}^{- 27} \text{ kg }\]

Let the distance between two protons be r.

Coulomb force (electric force) between the protons is given by

\[f_e = \frac{1}{4\pi \in_0} \times \frac{q^2}{r^2}\]

\[ = \frac{9 \times {10}^9 \times (1 . 6 )^2 \times {10}^{- 38}}{r^2}\]

Gravitational force between the protons is given by

\[f_g = \frac{\text{ G m}^2}{r^2}\]

\[ = \frac{6 . 67 \times {10}^{- 11} \times (1 . 67 \times {10}^{- 27} )^2}{r^2}\]

On dividing , \[f_e \text{ by } f_g\] , We get :

\[\frac{f_e}{f_g} = \frac{1}{4\pi \in_0} \times \frac{q^2}{r^2} \times \frac{r^2}{\text{ G m} ^2}\]

\[ = \frac{9 \times {10}^9 \times 1 . 6 \times 1 . 6 \times {10}^{- 38}}{6 . 67 \times {10}^{- 11} \times 1 . 67 \times 1 . 67 \times {10}^{- 54}}\]

\[ = \frac{9 \times (1 . 6 )^2 \times {10}^{- 29}}{6 . 67 \times (1 . 67 )^2 \times {10}^{- 65}}\]

\[ = 1 . 24 \times {10}^{36}\]