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/masses. (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 $1\mathring{A}(={10}^{-10}m)$ apart? $(m_p=1.67\times {10}^{-27}kg,m_e=9.11\times {10}^{-31}kg)$

Answer 1

(a)

The ratio of Coulomb force to gravitational force for an electron and a proton is given as,

$r_1=\frac{\frac{1}{4\pi {\epsilon }_0}\times \frac{q_1{q}_2}{R^2}}{\frac{G{m}_1{m}_2}{R^2}}$

$=\frac{9\times {10}^9\times \left(1.6\times {10}^{-19}\right)\times \left(1.6\times {10}^{-19}\right)}{6.67\times {10}^{-11}\times 9.1\times {10}^{-31}\times 1.67\times {10}^{-27}}$

$=\frac{2.27\times {10}^{39}}{1}$

The ratio of Coulomb force to gravitational force for two protons is given as,

$r_2=\frac{\frac{1}{4\pi {\epsilon }_0}\times \frac{q_1{q}_2}{R^2}}{\frac{G{m}^2}{R^2}}$

$=\frac{9\times {10}^9\times \left(1.6\times {10}^{-19}\right)\times \left(1.6\times {10}^{-19}\right)}{6.67\times {10}^{-11}\times {\left(1.67\times {10}^{-27}\right)}^2}$

$=\frac{1.24\times {10}^{36}}{1}$

(b)

The magnitude of Coulomb force between electron and proton is given as,

$F=\frac{1}{4\pi {\epsilon }_0}\times \frac{q_1{q}_2}{R^2}$

$=\frac{9\times {10}^9\times \left(1.6\times {10}^{-19}\right)\times \left(1.6\times {10}^{-19}\right)}{{\left({10}^{-10}\right)}^2}$

$=2.304\times {10}^{-8}N$

The acceleration of electron is given as,

$a_1=\frac{F}{m_1}$

$=\frac{2.304\times {10}^{-8}}{9.1\times {10}^{-31}}$

$=2.5\times {10}^{22}m/s^2$

The acceleration of proton is given as,

$a_2=\frac{F}{m_2}$

$=\frac{2.304\times {10}^{-8}}{1.67\times {10}^{-27}}$

$=1.4\times {10}^{19}m/s^2$