Answer:
3.825*10^-6 N
Explanation:
As particle 1 and particle 3 has opposite types of charge, particle 3 will be attracted to particle 1. And as particle 2 and 3 has the same sign, they will repel each other. Due to the position of the particles, both the Force that 1 exerts on 3 and the force that 2 exerts on 3, will have the same direction. Now, you need the magnitude. You can use the following expression:
[tex]F_e = K\frac{q_1*q_3}{r^2}[/tex]
K is the Coulomb constant equal to 9*10^9 N*m^2/C^2, q1 and q2 is the charge of the particles, and r is the distance.
Force 1 on 3 is equal to:
[tex]F_e = K\frac{q_1*q_3}{r^2} = 9*10^9 \frac{Nm^2}{C^2} * \frac{2.3*10^{-9}C * 7.5*10^{-9}C}{(-0.6m - (-0.3m))^2} = 1.725 * 10^{-6} N[/tex]
Force 2 on 3:
[tex]F_e = K\frac{q_2*q_3}{r^2} = 9*10^9 \frac{Nm^2}{C^2} * \frac{2.8*10^{-9}C * 7.5*10^{-9}C}{(0m - (-0.3m))^2} = 2.1 * 10^{-6} N[/tex]
The magnitude of the resultant force is the addition of both forces:
[tex]F_e = 1.725*10^{-6} N + 2.1*10^{-6} N= 3.825 * 10^{-6} N[/tex]
