Answer to Question #253414 in Electricity and Magnetism for shamma

Question

A positive and a negative charge, each of magnitude 2.6 x 10^–5 C, are separated by a distance of 25 cm. Find the force on each of the particles

Answer 1

Given:

$q=2.6*10^{-5}\:\rm C$

$r=0.25\:\rm m$

$k=9*10^9\:\rm N\cdot m^2/C^2$

The Coulomb law says

$$F=k\frac{q_1q_2}{r^2}=k\frac{q^2}{r^2}$$

$$F=9*10^9*\frac{(2.6*10^{-5})^2}{0.25^2}=97\:\rm N$$

The force is attraction.