Coulomb's law for electrostatic force between two point charges and Newton's law for gravitational force between two stationary point masses,both have inverse-square dependence on the distance between the charges and masses respectively.
$(a)$ Compare the strength of these forces by determining the ratio of their magnitudes $(i)$ for an electron and a proton and $(ii)$ for two protons.
$(b)$ Estimate the accelerations of electron and proton due to the electrical force of their mutual attraction when they are $1 \mathring A \left( = 10^{-10} \, m \right)$ apart? $\left( m_{p} = 1.67 \times 10^{-27} \, kg, m_{e} = 9.11 \times 10^{-31} \, kg \right)$

(N/A) $(a) (i)$ The electric force between an electron and a proton at a distance $r$ is $F_{e} = \frac{1}{4 \pi \varepsilon_{0}} \frac{e^{2}}{r^{2}}$. The gravitational force is $F_{G} = G \frac{m_{p} m_{e}}{r^{2}}$. The ratio is $\left| \frac{F_{e}}{F_{G}} \right| = \frac{e^{2}}{4 \pi \varepsilon_{0} G m_{p} m_{e}} \approx 2.4 \times 10^{39}$.
$(ii)$ For two protons,the ratio is $\left| \frac{F_{e}}{F_{G}} \right| = \frac{e^{2}}{4 \pi \varepsilon_{0} G m_{p}^{2}} \approx 1.3 \times 10^{36}$.
$(b)$ The magnitude of the electric force is $F = \frac{1}{4 \pi \varepsilon_{0}} \frac{e^{2}}{r^{2}} = (8.99 \times 10^{9}) \frac{(1.6 \times 10^{-19})^{2}}{(10^{-10})^{2}} \approx 2.3 \times 10^{-8} \, N$.
Acceleration of electron: $a_{e} = \frac{F}{m_{e}} = \frac{2.3 \times 10^{-8}}{9.11 \times 10^{-31}} \approx 2.5 \times 10^{22} \, m/s^{2}$.
Acceleration of proton: $a_{p} = \frac{F}{m_{p}} = \frac{2.3 \times 10^{-8}}{1.67 \times 10^{-27}} \approx 1.4 \times 10^{19} \, m/s^{2}$.