Solved Problem on Coulomb's Law

The distance between the proton and the electron in a hydrogen atom is 5.3 × 10⁻¹¹ m. Determine:
a) The magnitude of gravitational force between proton and electron;
b) The magnitude of electric force between proton and electron;
c) Compare the two forces.
Consider the following values:
mass of proton:   \( m_p=1.7\times 10^{-27}\;\mathrm{kg} \) ;
mass of electron:   \( m_e=9.1\times 10^{-31}\;\mathrm{kg} \) ;
Universal Gravitational Constant:    \( G=6.67\times 10^{-11}\;\mathrm{\frac{N.m^2}{kg^2}} \) ;
charge of a proton:   \( q_p=1.6\times 10^{-19}\;\mathrm C \) ;
charge of an electron:   \( q_e=-1.6\times 10^{-19}\;\mathrm C \) ;
Coulomb Constant:   \( k_e=9\times 10^9\;\mathrm{\frac{N.m^2}{C^2}} \) .

Problem data:

Distance between proton and electron: \( r=5.3\times 10^{-11}\;\mathrm m \) ;
Mass of proton: \( m_p=1.7\times 10^{-27}\;\mathrm{kg} \) ;
Mass of electron: \( m_e=9.1\times 10^{-31}\;\mathrm{kg} \) ;
Universal Gravitational Constant: \( G=6.67\times 10^{-11}\;\mathrm{\frac{N.m^2}{kg^2}} \) ;
Charge of a proton: \( q_p=1.6\times 10^{-19}\;\mathrm{C} \) ;
Charge of an electron: \( q_{e}=-1.6\times 10^{-19}\;\text{C} \) ;
Coulomb Constant: \( k_e=9\times 10^9\;\mathrm{\frac{N.m^2}{C^2}} \) .

Solution

a) Applying Newton’s Law of Universal Gravitation, the magnitude of gravitational force is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {F_{\small G}=G\frac{M\;m}{r^2}} \end{gather} \]

substituting the data for the proton and the electron, the gravitational force between them will be

\[ \begin{gather} F_{\small G}=G\frac{m_p m_e}{r^2}\\[5pt] F_{\small G}=\left(6.67\times 10^{-11}\;\mathrm{\frac{N.m^2}{kg^2}}\right)\times \frac{\left(1.7\times 10^{-27}\;\mathrm{kg}\right)\times\left(9.1\times 10^{-31}\;\mathrm{kg}\right)}{\left(5.3\times 10^{-11}\;\mathrm m\right)^{2}}\\[5pt] F_{\small G}=\left(6.67\times 10^{-11}\;\mathrm{\frac{N.\cancel{m^2}}{\cancel{kg^2}}}\right)\left(\frac{15.5\times 10^{-58}\;\mathrm{\cancel{kg^2}}}{28.1\times 10^{-22}\;\mathrm{\cancel{m^2}}}\right)\\[5pt] F_{\small G}=\frac{103.4\times 10^{-69}\times 10^{22}}{28.1}\;\mathrm{N} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {F_{\small G}=3.7\times 10^{-47}\;\mathrm N} \end{gather} \]

b) Applying Coulomb's Law, the magnitude of electric force is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {F_{\small E}=k_e\frac{|\;Q\;||\;q\;|}{r^2}} \end{gather} \]

substituting the data to the proton and the electron, the electric force between them will

\[ \begin{gather} F_{\small E}=k_e\;\frac{|\;q_p\;||\;q_e\;|}{r^2}\\[5pt] F_{\small E}=\left(9\times 10^9\;\mathrm{\frac{N.m^2}{C^2}}\right)\times \frac{|\;1.6\times 10^{-19}\;\mathrm C\;|\times|\;-1.6\times 10^{-19}\;\mathrm C\;|}{\left(5.3\times 10^{-11}\;\mathrm m\right)^{2}}\\[5pt] F_{\small E}=\left(9\times 10^9\;\mathrm{\frac{N.\cancel{m^2}}{\cancel{C^2}}}\right)\left(\frac{2.6\times 10^{-38}\;\mathrm{\cancel{C^2}}}{28.1\times 10^{-22}\;\mathrm{\cancel{m^2}}}\right)\\[5pt] F_{\small E}=\frac{23.4\times 10^{-29}\times 10^{22}}{28.1}\;\mathrm{N} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {F_{\small E}=8.3\times 10^{-8}\;\mathrm{N}} \end{gather} \]

c) Comparing the two forces

\[ \begin{gather} \frac{F_{\small E}}{F_{\small G}}=\frac{8.3\times 10^{-8}\;\mathrm{\cancel N}}{3.7\times 10^{-47}\;\mathrm{\cancel N}}\\[5pt] \frac{F_{\small E}}{F_{\small G}}=2.2\times 10^{-8}\times 10^{47}\\[5pt] F_{\small E}=2.2\times 10^{39}F_{\small G} \end{gather} \]

This result means that the electric force has is 2.2×10³⁹ times greater than the gravitational force between a proton and an electron in the hydrogen atom.

Note: Imagine a force

2,200,000,000,000,000,000,000,000,000,000,000,000,000

greater than another. In many practical situations, the gravitational force between particles can be neglected in the calculations.

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .