Answer :
To compare the magnitude of the electromagnetic and gravitational forces between two electrons separated by a distance of [tex]\(2.00 \, \text{m}\)[/tex], we can follow these steps:
1. Calculate the electrostatic force ([tex]\(F_e\)[/tex]) using Coulomb's law:
[tex]\[ F_e = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q_1 q_2}{r^2} \][/tex]
Where:
- [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] are the charges of the electrons, [tex]\(1.61 \times 10^{-19} \, \text{C}\)[/tex],
- [tex]\( r \)[/tex] is the distance between the electrons, [tex]\(2.00 \, \text{m}\)[/tex],
- [tex]\( \epsilon_0 \)[/tex] is the vacuum permittivity, [tex]\(8.854 \times 10^{-12} \, \text{F/m}\)[/tex].
2. Calculate the gravitational force ([tex]\(F_g\)[/tex]) using Newton's law of universal gravitation:
[tex]\[ F_g = G \cdot \frac{m_1 m_2}{r^2} \][/tex]
Where:
- [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] are the masses of the electrons, [tex]\(9.11 \times 10^{-31} \, \text{kg}\)[/tex],
- [tex]\( G \)[/tex] is the gravitational constant, [tex]\(6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2}\)[/tex],
- [tex]\( r \)[/tex] is the distance between the electrons, [tex]\(2.00 \, \text{m}\)[/tex].
3. Determine the ratio of the electrostatic force to the gravitational force:
[tex]\[ \frac{F_e}{F_g} \][/tex]
After performing the calculations, we find the following results:
- The electrostatic force ([tex]\(F_e\)[/tex]) is:
[tex]\[ F_e = 5.82 \times 10^{-29} \, \text{N} \][/tex]
- The gravitational force ([tex]\(F_g\)[/tex]) is:
[tex]\[ F_g = 1.38 \times 10^{-71} \, \text{N} \][/tex]
- The ratio of the electrostatic force to the gravitational force is:
[tex]\[ \frac{F_e}{F_g} = 4.21 \times 10^{42} \][/tex]
Thus, the magnitudes are:
[tex]\[ \begin{array}{l} F_e=5.82 \times 10^{-29} \, \text{N} \\ F_g=1.38 \times 10^{-71} \, \text{N} \\ \frac{F_e}{F_g}=4.21 \times 10^{42} \end{array} \][/tex]
1. Calculate the electrostatic force ([tex]\(F_e\)[/tex]) using Coulomb's law:
[tex]\[ F_e = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q_1 q_2}{r^2} \][/tex]
Where:
- [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] are the charges of the electrons, [tex]\(1.61 \times 10^{-19} \, \text{C}\)[/tex],
- [tex]\( r \)[/tex] is the distance between the electrons, [tex]\(2.00 \, \text{m}\)[/tex],
- [tex]\( \epsilon_0 \)[/tex] is the vacuum permittivity, [tex]\(8.854 \times 10^{-12} \, \text{F/m}\)[/tex].
2. Calculate the gravitational force ([tex]\(F_g\)[/tex]) using Newton's law of universal gravitation:
[tex]\[ F_g = G \cdot \frac{m_1 m_2}{r^2} \][/tex]
Where:
- [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] are the masses of the electrons, [tex]\(9.11 \times 10^{-31} \, \text{kg}\)[/tex],
- [tex]\( G \)[/tex] is the gravitational constant, [tex]\(6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2}\)[/tex],
- [tex]\( r \)[/tex] is the distance between the electrons, [tex]\(2.00 \, \text{m}\)[/tex].
3. Determine the ratio of the electrostatic force to the gravitational force:
[tex]\[ \frac{F_e}{F_g} \][/tex]
After performing the calculations, we find the following results:
- The electrostatic force ([tex]\(F_e\)[/tex]) is:
[tex]\[ F_e = 5.82 \times 10^{-29} \, \text{N} \][/tex]
- The gravitational force ([tex]\(F_g\)[/tex]) is:
[tex]\[ F_g = 1.38 \times 10^{-71} \, \text{N} \][/tex]
- The ratio of the electrostatic force to the gravitational force is:
[tex]\[ \frac{F_e}{F_g} = 4.21 \times 10^{42} \][/tex]
Thus, the magnitudes are:
[tex]\[ \begin{array}{l} F_e=5.82 \times 10^{-29} \, \text{N} \\ F_g=1.38 \times 10^{-71} \, \text{N} \\ \frac{F_e}{F_g}=4.21 \times 10^{42} \end{array} \][/tex]
