Question

The force between two electrons when placed in air is equal to $0.5$ times the weight of an electron. Find the distance between two electrons (mass of electron $=9.1\times {10}^{-31}kg$)

Answer 1

Electrostatic force $F_e=\frac{k{e}^2}{r^2}$

Weight of electron $W=mg$

So, $F_e=0.5W$

Or $\frac{k{e}^2}{r^2}=0.5mg$

Or $\frac{9\times {10}^9(1.6\times {10}^{-19}{)}^2}{r^2}=0.5\times 9.1\times {10}^{-31}\times 10$

$⟹r=7.2m$