Example

Question:

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 (\( = 10^{-10} \, \text{m} \)) apart. (\(m_p = 1.67 \times 10^{-27} \, \text{kg}, \, m_e = 9.11 \times 10^{-31} \, \text{kg}\))

Solution:

(a) (i) The electric force between an electron and a proton at a distance \( r \) apart is:
\[ F_e = -\frac{1}{4 \pi \varepsilon_0} \frac{e^2}{r^2} \] where the negative sign indicates that the force is attractive.
The corresponding gravitational force (always attractive) is:
\[ F_G = -G \frac{m_p m_e}{r^2} \] where \( m_p \) and \( m_e \) are the masses of a proton and an electron respectively.
Thus, \[ \left| \frac{F_e}{F_G} \right| = \frac{e^2}{4 \pi \varepsilon_0 G m_p m_e} = 2.4 \times 10^{39} \] (ii) The ratio of the magnitudes of electric force to the gravitational force between two protons at a distance \( r \) apart is: \[ \left| \frac{F_e}{F_G} \right| = \frac{e^2}{4 \pi \varepsilon_0 G m_p^2} = 1.3 \times 10^{36} \] For two protons inside a nucleus (distance \(\sim 10^{-15}\) m), \( F_e \sim 230 \) N and \( F_G \sim 1.9 \times 10^{-34} \) N.
The ratio of the two forces shows electrical forces are enormously stronger than gravitational forces.

(b) The electric force \( \mathbf{F} \) exerted by a proton on an electron is: \[ |\mathbf{F}| = \frac{1}{4 \pi \varepsilon_0} \frac{e^2}{r^2} = 8.987 \times 10^9 \, \text{Nm}^2/\text{C}^2 \times \left(1.6 \times 10^{-19} \, \text{C}\right)^2 / \left(10^{-10} \, \text{m}\right)^2 = 2.3 \times 10^{-8} \, \text{N} \] Using \( F = ma \), the acceleration for electron: \[ a = \frac{2.3 \times 10^{-8} \, \text{N}}{9.11 \times 10^{-31} \, \text{kg}} = 2.5 \times 10^{22} \, \text{m}/\text{s}^2 \] For the proton: \[ a = \frac{2.3 \times 10^{-8} \, \text{N}}{1.67 \times 10^{-27} \, \text{kg}} = 1.4 \times 10^{19} \, \text{m}/\text{s}^2 \] Thus, the effect of gravitational field is negligible compared to Coulomb force for both particles.