Question

In a vacuum, two particles have charges of q1 and q2, where q1 = +3.11 μC. They are separated by a distance of 0.241 m, and particle 1 experiences an attractive force of 3.44 N.

What is the magnitude and sign of q2?

Answer 1

Charge of particle 2,
`q_2=-7.13\ \mu C`

Step-by-step explanation:

Given that,

Charge 1,
`q_1=3.11\ \mu C=3.11* 10^(-6)\ C`

The distance between charges, r = 0.241 m

Force experienced by particle 1, F = 3.44 N

We need to find the magnitude of electric charge 2. It can be calculated using formula of electrostatic force. It is given by :

`F=k(q_1q_2)/(r^2)`

`q_2=(Fr^2)/(kq_1)`

`q_2=(3.44* (0.241)^2)/(9* 10^9* 3.11* 10^(-6))`

`q_2=7.13* 10^(-6)\ C`

or

`q_2=7.13\ \mu C`

So, the magnitude of electric charge 2 is
`q_2=7.13\ \mu C`. Since, the force is attractive then the magnitude of charge 2 must be negative.