Question

Calculate the force of repulsion between two charge of 10 and 20 separated by a distance of $20cm\frac{1}{\left(4n{ϵ}_0\right)}=9\times {10}^{9}N{m}^2/c^2)$

Answer 1

$q_1=10$

$q_1=20$

$d=20$cm

$F=\frac{1}{4\pi {\in }_o}\frac{q_1{q}_2}{d^2}$

$=9\times {10}^9\times \frac{10\times 10}{(0.2{)}^2}$

$=\frac{9\times {10}^{11}}{0.04}$

$=225\times {10}^{11}$.