Solve the following :


Find the ratio of the magnitude of the electric force to the gravitational force acting between two protons.


$63 \times R$ SAQ - 10 Find the ratio.

$F_{e}=\frac{k q_{p} q_{p}}{r^{2}}$

$F_{e}=\frac{9 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{r^{2}}$

$F_{G}=\frac{6.67 \times 10^{11} \times\left(1.732 \times 10^{-27}\right)^{2}}{r^{2}}$

Ratio is

$\frac{F_{e}}{F_{G}}=\frac{\frac{9 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{r^{2}}}{\frac{6.67 \times 10^{11} \times\left(1.732 \times 10^{-27}\right)^{2}}{r^{2}}}$

$\frac{F_{\text {e }}}{F_{G}}=1.24 \times 10^{36}$


Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now