Question

Let the electrostatic force between two electrons $\left(F_e\right)$ be x times the gravitational force $\left(F_g\right)$ between them. Then, x is of the order of?

• A. ${10}^{40}$
• B. ${10}^{42}$
• C. ${10}^{38}$
• D. ${10}^{37}$

Answer 1

Answer: (B)

${10}^{42}$

Electric force between two electrons given by coloumb law

$F_e=\frac{1}{4\pi {\epsilon }_{\circ }}\frac{q_1{q}_2}{r^2}$

$F_e=\frac{1}{4\pi {\epsilon }_{\circ }}\frac{q^2}{d^2}$

Gravitational force given by Newtons universal law

$F_g=G\frac{m_1{m}_2}{r^2}$

$F_g=G\frac{m^2}{d^2}$

$\frac{F_e}{F_g}=\frac{\frac{1}{4\pi {\epsilon }_{\circ }}\frac{q^2}{d^2}}{G\frac{m^2}{d^2}}$

$\frac{F_e}{F_g}=\frac{1}{4\pi {\epsilon }_{\circ }}\times \frac{1}{G}\times \frac{q^2}{m^2}$

$\frac{F_e}{F_g}=(9\times {10}^9)\times \frac{1}{6.67\times {10}^{-11}}\times \frac{(1.6\times {10}^{-19}{)}^2}{(9.1\times {10}^{-31}{)}^2}$

$\frac{F_e}{F_g}\approx 4.17\times {10}^{42}$

$\frac{F_e}{F_g}$ is the order of ${10}^{42}$