Question

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

Answer 1

Charge of the proton, q = $1.6\times {10}^{-19}\mathrm{C}$
Mass of the proton = $1.67\times {10}^{-27}\mathrm{kg}$
Let the distance between two protons be r.
Coulomb force (electric force) between the protons is given by
$$\begin{gathered} f_e=\frac{1}{4\mathrm{\pi }{\in }_0}\times \frac{q^2}{r^2} \\ =\frac{9\times {10}^9\times (1.6{)}^2\times {10}^{-38}}{r^2} \end{gathered}$$
Gravitational force between the protons is given by
$$\begin{gathered} f_g=\frac{\mathrm{G}{m}^2}{r^2} \\ =\frac{6.67\times {10}^{-11}\times (1.67\times {10}^{-27}{)}^2}{r^2} \end{gathered}$$

On dividing $f_e\mathrm{by}{f}_g$, we get:
$$\begin{gathered} \frac{f_e}{f_g}=\frac{1}{4\mathrm{\pi }{\in }_0}\times \frac{q^2}{r^2}\times \frac{r^2}{\mathrm{G}{m}^2} \\ =\frac{9\times {10}^9\times 1.6\times 1.6\times {10}^{-38}}{6.67\times {10}^{-11}\times 1.67\times 1.67\times {10}^{-54}} \\ =\frac{9\times (1.6{)}^2\times {10}^{-29}}{6.67\times (1.67{)}^2\times {10}^{-65}} \\ =1.24\times {10}^{36} \end{gathered}$$