Question

A particle of charge of +3.40 ✕ 10−6 C is 17.5 cm distant from a second particle of charge of −2.00 ✕ 10−6 C. Calculate the magnitude of the electrostatic force between the particles.

Answer 1

Answer:

0.47 N

Explanation:

q1 = 3.4 x 10^-6 C

q2 = 2 x 10^-6 C

d = 17.5 cm = 0.175 m

The electrostatic force is given by

$F=\frac{Kq_{1}q_{2}}{d^{2}}$

$F=\frac{9\times10^{9}\times3.4\times 10^{-6}\times 2\times 10^{-6}}{0.175\times 0.175}$

F = 0.47 N

Thus, the force is 0.47 N.