The net force on q₃ is 0.46 N vertically upwards along the +Y axis. The product of the two charges is directly proportional to the force.
The electrostatic force is directly proportional to the product of the two charges and inversly proportional to the product of the two charges.
The given data in the problem is;
F is the electric force =?
Q₁ is the charged particle 1 = 2 C
Q₂ is the charged particle 2 = 2 C
d₁₂ is the distance between particles = 0.5 m
The electric force is given by;
[tex]\rm F=\frac{ Kq_1q_2}{r^2} \\\\[/tex]
[tex]\rm F=\frac{ 9\times 10^9 (2\times 10^{-6} \times 4\times10^{-6}}{(0.5)^2} \\\\\rm F=0.288\ N[/tex]
This force is applied on both charges
So the result of the net force is given by;
[tex]\rm F_{net}= 2Fcos \theta \\\\ \rm F_{net}==2 \times 0.288 cos 37^0 \\\\ \rm F_{net}=0.46 \ N[/tex]
Hence the net force on q₃ is 0.46 N vertically upwards along the +Y axis.
To learn more about Coulomb's law refer to the link;
https://brainly.com/question/506926
