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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.
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?  $$(m_p=1.67\times 10^{-27}\mathit{kg},m_e=9.11\times 10^{-31}\mathit{kg}).$$  Speed
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04:29 ## QuestionPhysicsClass 12

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.
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?  $$(m_p=1.67\times 10^{-27}\mathit{kg},m_e=9.11\times 10^{-31}\mathit{kg}).$$

4.6     4.6     ## Solution

Given:
Distance between proton and electron,$$r=1Å=10^{-10} \ m$$
The magnitude of electrostatic force is given by,
$$\mid F_e\mid =\dfrac 1{4\pi \varepsilon _0}\dfrac{e^2}{r^2}$$
The magnitude of gravitational force is given by :
$$\mid F_g\mid =G\dfrac{m_pm_e}{r^2}$$
$$m_p$$  and  $$m_e$$  are the masses of a proton and electron respectively.
The ratio of electric force between an electron and a proton to the gravitational force is :
$$\dfrac{F_e}{F_g}=\dfrac{e^2}{4\pi \epsilon _0Gm_pm_e}$$
$$\dfrac{F_e}{F_g}=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{6.67\times 10^{-11}\times 1.67\times 10^{-27}\times 9.1\times 10^{-31}}$$
$$\dfrac{F_e}{F_g}=2.4\times 10^{39}$$
The ratio of electric force between two protons to the gravitational force is :
$$\dfrac{F_e}{F_g}=\dfrac{e^2}{4\pi \epsilon _0Gm_pm_p}$$
$$\dfrac{F_e}{F_g}=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{6.67\times 10^{-11}\times \left(1.67\times 10^{-27}\right)^2}$$
$$\dfrac{F_e}{F_g}=1.3\times 10^{36}$$          