Question

A point charge of +6.4μ C is located 0.15 m from a second point charge of -5.3μ C. What is the absolute magnitude of the force on each charge?

Answer 1

Given is:-

First charge  $q_1 = 6.4 \mu C$

Second charge  $q_2 = -5.3 \mu C$

distance between the charges r = 0.15m

Now,

Electrostatic force between two charges is given by

$F = \frac{Kq_1q_2}{r^2}$   

By putting all the values in above equation, we get

$F = \frac{8.99 \times 10^{9} \times 6.4 \times 10^{-6} \times (-5.3 \times 10^{-6})}{(0.15)^2}$

Thus by solving, we get

$F = -13.5529 \hspace{4} Newtons$

Negative sign represents that the force is of attractive nature

thus the absolute magnitude of the electrostatic force between the given two charge is

$|F|= 13.5529 \hspace{4} Newtons$