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. Compare the strength of these forces by determining the ratio of their magnitudes for an electron and a proton and for two protons. \((m_p=1.67\times 10^{-27}\;\mathit{kg},m_e=9.11\times 10^{-31}\;\mathit{kg})\)
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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. Compare the strength of these forces by determining the ratio of their magnitudes for an electron and a proton and for two protons. \((m_p=1.67\times 10^{-27}\;\mathit{kg},m_e=9.11\times 10^{-31}\;\mathit{kg})\)
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Solution
We know that the force between an electron and a proton separated at a distance \(r\) can be given as,
\(F_e=\frac{-1}{4\pi \varepsilon _0}\frac{e^2}{r^2}\)
where as here the negative sign indicates that the force is attractive. The corresponding gravitational force which is always attractive can be given by,
\(F_G=-G\frac{m_pm_e}{r^2}\)
where as \(m_p\) is the mass of a proton and \(m_e\) is the mass of an electron.
Known data:
\(e=1.6\times 10^{-19}\mathit{C.}\)
\(G=6.67408\times 10^{-11}m^3\mathit{kg}^{-1}s^{-2}.\)
mass of a proton, \(m_p=1.67\times 10^{-27}\;\mathit{kg}\)
mass of an electron, \(m_e=9.11\times 10^{-31}\;\mathit{kg}\)
\(\varepsilon _0\)- the permittivity of free space .
\(\varepsilon _0=8.854\times 10^{-12}\;C^2N^{-1}m^{-2}\)
Therefore, the ratio of the magnitudes of electric force \(F_e\) to the gravitational force \(F_G\) between a proton and an electron separated at a distance\(r\)is,
\(\frac{F_e}{F_G}=\frac{\frac{-1}{4\pi \varepsilon _0}\frac{e^2}{r^2}}{-G\frac{m_pm_e}{r^2}}=\frac{e^2}{4\pi \varepsilon _0Gm_pm_e}=\frac{9\times 10^9 \times (1.6\times 10^{-19})^2}{6.67\times 10^{-11}\times 1.67\times 10^{-27} \times 9.11\times 10^{-31}}=2.4\times 10^{39}\)
The ratio of the magnitudes of electric force \(F_e\) to the gravitational force \(F_G\) between two protons separated at a distance \(r\) is,
Using all known data,
\(\frac{F_e}{F_G}=\frac{e^2}{4\pi \varepsilon _0Gm_pm_p}=\frac{9\times 10^9 \times (1.6\times 10^{-19})^2}{6.67\times 10^{-11}\times 1.67\times 10^{-27} \times 1.67\times 10^{-27}}=1.3\times 10^{36}\)
\(F_e=\frac{-1}{4\pi \varepsilon _0}\frac{e^2}{r^2}\)
where as here the negative sign indicates that the force is attractive. The corresponding gravitational force which is always attractive can be given by,
\(F_G=-G\frac{m_pm_e}{r^2}\)
where as \(m_p\) is the mass of a proton and \(m_e\) is the mass of an electron.
Known data:
\(e=1.6\times 10^{-19}\mathit{C.}\)
\(G=6.67408\times 10^{-11}m^3\mathit{kg}^{-1}s^{-2}.\)
mass of a proton, \(m_p=1.67\times 10^{-27}\;\mathit{kg}\)
mass of an electron, \(m_e=9.11\times 10^{-31}\;\mathit{kg}\)
\(\varepsilon _0\)- the permittivity of free space .
\(\varepsilon _0=8.854\times 10^{-12}\;C^2N^{-1}m^{-2}\)
Therefore, the ratio of the magnitudes of electric force \(F_e\) to the gravitational force \(F_G\) between a proton and an electron separated at a distance\(r\)is,
\(\frac{F_e}{F_G}=\frac{\frac{-1}{4\pi \varepsilon _0}\frac{e^2}{r^2}}{-G\frac{m_pm_e}{r^2}}=\frac{e^2}{4\pi \varepsilon _0Gm_pm_e}=\frac{9\times 10^9 \times (1.6\times 10^{-19})^2}{6.67\times 10^{-11}\times 1.67\times 10^{-27} \times 9.11\times 10^{-31}}=2.4\times 10^{39}\)
The ratio of the magnitudes of electric force \(F_e\) to the gravitational force \(F_G\) between two protons separated at a distance \(r\) is,
Using all known data,
\(\frac{F_e}{F_G}=\frac{e^2}{4\pi \varepsilon _0Gm_pm_p}=\frac{9\times 10^9 \times (1.6\times 10^{-19})^2}{6.67\times 10^{-11}\times 1.67\times 10^{-27} \times 1.67\times 10^{-27}}=1.3\times 10^{36}\)
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