Question

The ratio of electromagnetic and gravitational forces between two electrons is nearly, $($ Charge of the electron $e=1.6\times {10}^{-19}$C, mass of the electron $m=9.1\times {10}^{-31}$kg, permittivity of free space $\frac{1}{4\pi {\epsilon }_0}=9\times {10}^9$ Nm${}^2$C${}^{-2}$, universal gravitational constant $G=6.67\times {10}^{-11}$Nm${}^2$kg${}^{-2}$ $)$

• A. $4\times {10}^{42}$
• B. $2\times {10}^{31}$
• C. $3\times {10}^{21}$
• D. $6\times {10}^{72}$

Answer 1

The correct option is A $4\times {10}^{42}$

Electromagnetic force between 2 charged particles is: $F_E=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r^2}$
Gravitational force between the 2 electrons is: $F_G=\frac{G{m}_1{m}_2}{r^2}$
Here, $q_1=q_2=e=1.6\times {10}^{-19}$ C, $\frac{1}{4\pi {ϵ}_0}=9\times {10}^9$ $N{m}^2$C${}^{-2}$
$m_1=m_2=9.1\times {10}^{-31}$ kg, $G=6.67\times {10}^{-11}$$N{m}^2$kg${}^{-2}$

$\frac{F_E}{F_G}=\frac{\frac{1}{4\pi {ϵ}_0}\frac{e^2}{r^2}}{\frac{G{m}^2}{r^2}}$

$\frac{F_E}{F_G}=\frac{1}{4\pi {ϵ}_0}\frac{e^2}{G{m}^2}$

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

$\frac{F_E}{F_G}\approx 4\times {10}^{42}$