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

Answer:

$q_2 = - 1.66\times 10^{-13} c$

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

$f = \frac{1\times q_1\times q_2}{4\pi \epsilon_o \times r^2}$

$\frac{1}{4\pi \epsilon_o} = 9\times 10^{9}$

$3.1 N = 9\times 10^{9} \frac{91.4\times q_2}{0.23^2}$

$q_2 = \frac{3.1\times 0.21^2}{9\times 10^{9}\times 91.4}$

$q_2 = - 1.66\times 10^{-13} c$