Q. The ratio of electrostatic and gravitational forces acting between electron and proton separated by a distance $5 \times 10^{- 11}m,$ will be (Charge on electron $= 1.6 \times 10^{- 19}C,$ mass of electron $= 9.1 \times 10^{- 31}kg,$ mass of proton $= 1.6 \times 10^{- 27}kg,\ G = 6.7 \times 10^{- 11}Nm^{2}/kg^{2}$)
A) $2.36 \times 10^{39}$
B) $2.36 \times 10^{40}$
C) $2.34 \times 10^{41}$
D) $2.34 \times 10^{42}$
Explanation
(a) Gravitational force $F_{G} = \frac{Gm_{e}m_{p}}{r^{2}}$
$F_{G} = \frac{6.7 \times 10^{- 11} \times 9.1 \times 10^{- 31} \times 1.6 \times 10^{- 27}}{\left( 5 \times 10^{- 11} \right)^{2}} = 3.9 \times 10^{- 47}N$
Electrostatic force $F_{e} = \frac{1}{4\pi\varepsilon_{0}}\frac{e^{2}}{r^{2}}$
$F_{e} = \frac{9 \times 10^{9} \times 1.6 \times 10^{- 19} \times 1.6 \times 10^{- 19}}{\left( 5 \times 10^{- 11} \right)^{2}} = 9.22 \times 10^{- 8}N$
So, $\frac{F_{e}}{F_{G}} = \frac{9.22 \times 10^{- 8}}{3.9 \times 10^{- 47}} = 2.36 \times 10^{39}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.