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(D) The electrostatic force between two protons is given by $F_e = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}$.The gravitational force between two proto

Vedclass · Accepted Answer

$36$

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(D) The electrostatic force between two protons is given by $F_e = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}$.The gravitational force between two protons is given by $F_g = G \frac{m_p^2}{r^2}$.The ratio of these forces is $\frac{F_e}{F_g} = \frac{1}{4\pi\epsilon_0} \frac{e^2}{G m_p^2}$.Substituting the values: $e = 1.6 	imes 10^{-19} \ C$,$m_p = 1.67 	imes 10^{-27} \ kg$,$G = 6.67 	imes 10^{-11} \ N \cdot m^2/kg^2$,and $\frac{1}{4\pi\epsilon_0} = 9 	imes 10^9 \ N \cdot m^2/C^2$.$\frac{F_e}{F_g} = \frac{(9 	imes 10^9) 	imes (1.6 	imes 10^{-19})^2}{(6.67 	imes 10^{-11}) 	imes (1.67 	imes 10^{-27})^2} \approx 1.24 	imes 10^{36}$.Comparing this with $10^n$,we get $n \approx 36$.

For any fixed distance,the electromagnetic force between two protons is $10^n$ times the gravitational force between them. Then $n=$

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