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Physics Question on Gravitation

If FgF_g and FeF_e are gravitational and electrostatic forces between two electrons at a distance 0.1 m then Fg/FeF_g/F_e is in the order of

A

104310^{43}

B

10−4310^{-43}

C

103510^{35}

D

10−3510^{-35}

Answer

104310^{43}

Explanation

Solution

FeFg=q1q24πϵ0r2/Gm1m2r2\frac{ F _{ e }}{ F _{ g }}=\frac{ q _{1} q _{2}}{4 \pi \epsilon_{0} r ^{2}} / \frac{ Gm _{1} m _{2}}{ r ^{2}}
FeFg=q24πϵ0Gm2\frac{ F _{ e }}{ F _{ g }}=\frac{ q ^{2}}{4 \pi \epsilon_{0} Gm ^{2}}
FeFg=9×109×(1.6×10−19)26.67×10−11×(9.1×10−31)2=0.417×1043\frac{ F _{ e }}{ F _{ g }}=\frac{9 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{6.67 \times 10^{-11} \times\left(9.1 \times 10^{-31}\right)^{2}}=0.417 \times 10^{43}