The ratio of electrostatic and gravitational | vedclass

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Question

(a) Gravitational force ${F_G} = \frac{{G{m_e}{m_p}}}{{{r^2}}}$
${F_G} = \frac{{6.7 	imes {{10}^{ - 11}} 	imes 9.1 	imes {{10}^{ - 31}} 	imes 1.6

Vedclass · Accepted Answer

$2.36 	imes 10^{39}$

Vedclass · Answer

(a) Gravitational force ${F_G} = \frac{{G{m_e}{m_p}}}{{{r^2}}}$
${F_G} = \frac{{6.7 	imes {{10}^{ - 11}} 	imes 9.1 	imes {{10}^{ - 31}} 	imes 1.6 	imes {{10}^{ - 27}}}}{{{{(5 	imes {{10}^{ - 11}})}^2}}}= 3.9 	imes 10^{-47} \,N$
Electrostatic force ${F_e} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{e^2}}}{{{r^2}}}$
${F_e} = \frac{{9 	imes {{10}^9} 	imes 1.6 	imes {{10}^{ - 19}} 	imes 1.6 	imes {{10}^{ - 19}}}}{{{{(5 	imes {{10}^{ - 11}})}^2}}} = 9.22 	imes 10^{-8}\, N$
So, $\frac{{{F_e}}}{{{F_G}}} = \frac{{9.22 	imes {{10}^{ - 8}}}}{{3.9 	imes {{10}^{ - 47}}}} = 2.36 	imes {10^{39}}$

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