Question

under mutual gravitational and electric force then calculate the order of $\frac{Q}{M}$ in SI units.
The force between two point charges is 100 N in air. Calculate the force if the distance between them is increased by 50%.

Answer 1

Since $F_{electrostatic}=F_{gravity}$

$\frac{K{Q}^2}{r^2}=\frac{G{M}^2}{r^2}$

We get $\frac{Q}{M}=\sqrt{\frac{G}{K}}=\sqrt{\frac{6.67\times {10}^{-11}}{9\times {10}^9}}=8.6\times {10}^{-11}$

Let the initial distance between the charges be $d$.

New distance between them $d_2=1.5d$

Force $F=\frac{K{Q}^2}{d^2}$

$⟹$ $\frac{F_2}{F_1}=(\frac{d_1}{d_2}{)}^2$

Or $\frac{F_2}{100}=(\frac{d}{1.5d}{)}^2=\frac{1}{2.25}$

$⟹F_2=44.44N$