Answer:
Charge of particle 2, [tex]q_2=-7.13\ \mu C[/tex]
Explanation:
Given that,
Charge 1, [tex]q_1=3.11\ \mu C=3.11\times 10^{-6}\ C[/tex]
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 :
[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]
[tex]q_2=\dfrac{Fr^2}{kq_1}[/tex]
[tex]q_2=\dfrac{3.44\times (0.241)^2}{9\times 10^9\times 3.11\times 10^{-6}}[/tex]
[tex]q_2=7.13\times 10^{-6}\ C[/tex]
or
[tex]q_2=7.13\ \mu C[/tex]
So, the magnitude of electric charge 2 is [tex]q_2=7.13\ \mu C[/tex]. Since, the force is attractive then the magnitude of charge 2 must be negative.
