Question

In a vacuum, two particles have charges of qs and q, where 91-+4.0 C. They are separated by a distance of 0.23 m, and particle 1 experiences an attractive force of3.1 N. What is q (magnitude and sign)?

Answer 1

`q_2 = - 1.66* 10^(-13) c`

Step-by-step explanation:

Given data:

if two charge are opposite in charge then force will be attractive between then or vice versa

if one charge is positive then other charge will be negative

from coulomb's law

`f = (1* q_1* q_2)/(4\pi \epsilon_o * r^2)`

`(1)/(4\pi \epsilon_o) = 9* 10^(9)`

`3.1 N = 9* 10^(9) (91.4* q_2)/(0.23^2)`

`q_2 = (3.1* 0.21^2)/(9* 10^(9)* 91.4)`

`q_2 = - 1.66* 10^(-13) c`