Two point charges are located on the y axis as follows: charge q1 = -2.30 nc at y1 = -0.600 m , and charge q2 = 2.80 nc at the origin (y = 0). what is the magnitude of the net force exerted by these two charges on a third charge q3 = 7.50 nc located at y3 = -0.300 m ?

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:

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:

Force 2 on 3:

The magnitude of the resultant force is the addition of both forces:

the entire statement is true and correct.  

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