If varepsilon_0 is the permittivity of free s | vedclass

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(B) The gravitational force between two protons separated by a distance $r$ is given by $F_g = \frac{G m_p^2}{r^2}$.The electrostatic force between tw

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$[M^0 L^0 T^0 A^0]$

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(B) The gravitational force between two protons separated by a distance $r$ is given by $F_g = \frac{G m_p^2}{r^2}$.The electrostatic force between two protons separated by the same distance $r$ is given by $F_e = \frac{1}{4 \pi \varepsilon_0} \frac{e^2}{r^2}$.Taking the ratio of the two forces,we get $\frac{F_e}{F_g} = \frac{\frac{1}{4 \pi \varepsilon_0} \frac{e^2}{r^2}}{\frac{G m_p^2}{r^2}} = \frac{e^2}{4 \pi \varepsilon_0 G m_p^2}$.Since the ratio of two forces is a dimensionless quantity,the dimensional formula for $\frac{e^2}{4 \pi \varepsilon_0 G m_p^2}$ is $[M^0 L^0 T^0 A^0]$.

If $\varepsilon_0$ is the permittivity of free space,$e$ is the charge of a proton,$G$ is the universal gravitational constant,and $m_p$ is the mass of a proton,then the dimensional formula for $\frac{e^2}{4 \pi \varepsilon_0 G m_p^2}$ is:

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