Question

the electronic force between two-points charge with a magnitude of-800nC and is 15N.how far apart are the two charges from each other in centimeters? with solution.

Answer 1

Force

$F=\frac{kq_1q_2}{r^2}$

$r^2=\frac{kq_1q_2}{F}$

$$r^2=\frac{9\times10^9\times800\times10^{-9}\times800\times10^{-9}}{15}=3.84\times10^{-04} m$$

$r=0.0195m$

$r=1.95cm$