Question

Calculate the electric force between charges of 2.5 nC and 6 x 10^8C if they are 8.0 cm apart.

Answer 1

Gives

$Q_1=2.5\times10^{_9}C$

$Q_2=6\times10^{-8}C$

r=0.08m

$F=\frac{KQ_1Q_2}{r^2}$

Put value

$F=\frac{9\times10^9\times2.5\times10^{-9}\times6\times10^{-8}}{0.08^2}$

$F=1.56\times10^{-6}N$