{F_g} and {F_e} represents gravitational and | vedclass

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Question

(d) Gravitational force between electrons ${F_G} = \frac{{G{{({m_e})}^2}}}{{{r^2}}}$
Electrostatics force between electrons ${F_e} = k.\frac{{{e^2}}}

Vedclass · Accepted Answer

${10^{ - 43}}$

Vedclass · Answer

(d) Gravitational force between electrons ${F_G} = \frac{{G{{({m_e})}^2}}}{{{r^2}}}$
Electrostatics force between electrons ${F_e} = k.\frac{{{e^2}}}{{{r^2}}}$
$\frac{{{F_G}}}{{{F_e}}} = \frac{{G{{({m_e})}^2}}}{{k.{e^2}}} = \frac{{6.67 	imes {{10}^{ - 11}} 	imes {{(9.1 	imes {{10}^{ - 31}})}^2}}}{{9 	imes {{10}^9} 	imes {{(1.6 	imes {{10}^{ - 19}})}^2}}}$$ = 2.39 	imes {10^{ - 43}}$

${F_g}$ and ${F_e}$ represents gravitational and electrostatic force respectively between electrons situated at a distance $10\, cm$. The ratio of ${F_g}/{F_e}$ is of the order of

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