Question

Two particles, one with charge -6.29 × 10^-6 C and one with charge 5.23 × 10^-6 C, are 0.0359 meters apart. What is the magnitude of the force that one particle exerts on the other?

Answer 1

Answer:

Force, F = −229.72 N

Explanation:

Given that,

First charge particle, $q_1=-6.29\times 10^{-6}\ C$

Second charged particle, $q_2=5.23\times 10^{-6}\ C$

Distance between charges, d = 0.0359 m

The electric force between the two charged particles is given by :

$F=k\dfrac{q_1q_2}{d^2}$

$F=9\times 10^9\times \dfrac{-6.29\times 10^{-6}\times 5.23\times 10^{-6}}{(0.0359)^2}$

F = −229.72 N

So, the magnitude of force that one particle exerts on the other is 229.72 N. Hence, this is the required solution.