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 Å (= $10^{-10}$ m) apart? ($m_p$ = 1.67 × $10^{–27}$ kg, $m_e$ = 9.11 × $10^{–31}$ kg)

NCERT Example 1.3 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 Å (= $10^{-10}$ m) apart? ($m_p$ = 1.67 × $10^{–27}$ kg, $m_e$ = 9.11 × $10^{–31}$ kg)

(a) 

(i) Between Proton and Electron,

\[ \frac{F_{electric}}{F_{gravity}}=\frac{\frac{1}{4\pi\epsilon_o}\frac{q_1 q_2}{r^2}}{G\frac{m_1 m_2}{r^2}}=\frac{\left( 9 \times 10^9 \right) \times \frac{\left(1.6 \times 10^{-19}\right) \left(1.6 \times 10^{-19}\right)}{r^2}}{\left( 6.67 \times 10^{-11}\right) \times \frac{\left( 1.67 \times 10^{-27}\right) \left( 9.11 \times 10^{-31} \right) }{r^2}}=2.4 \times 10^{39}\]

We can see that the electric force is much larger than the gravitational force, 39 orders of magnitude larger.

(ii) Between Proton and Proton,

\[ \frac{F_{electric}}{F_{gravity}}=\frac{\frac{1}{4\pi\epsilon_o}\frac{q_1 q_2}{r^2}}{G\frac{m_1 m_2}{r^2}}=\frac{\left( 9 \times 10^9 \right) \times \frac{\left(1.6 \times 10^{-19}\right) \left(1.6 \times 10^{-19}\right)}{r^2}}{\left( 6.67 \times 10^{-11}\right) \times \frac{\left( 1.67 \times 10^{-27}\right) \left( 1.67 \times 10^{-27} \right) }{r^2}}=1.3 \times 10^{36}\]

(b) 

(i) Acceleration of Proton,

\[a_{proton} =\frac{F_e}{m_p}=\frac{\left( 9 \times 10^9 \right) \times \frac{\left(1.6 \times 10^{-19}\right) \left(1.6 \times 10^{-19}\right)}{\left(10^{-10}\right)^2}}{\left( 1.67 \times 10^{-27}\right)} =1.4 \times 10^{19} \; m/s^2 \] 

(ii) Acceleration of Electron,

\[a_{electron} =\frac{F_e}{m_e}=\frac{\left( 9 \times 10^9 \right) \times \frac{\left(1.6 \times 10^{-19}\right) \left(1.6 \times 10^{-19}\right)}{\left(10^{-10}\right)^2}}{\left( 9.11 \times 10^{-31}\right)} =2.5 \times 10^{22} \; m/s^2 \]

We can see that mass of electron is approximately 2000 times or 1800 times lighter than that of proton. So the acceleration of electron is so many times higher than proton.

Note that intermediate calculations such as $F_e, F_g$ are not encouraged as many of the terms may get cancelled out in the following steps.