Two point charges are fixed on the y axis: a negative point charge q1 = -27 µc at y1 = +0.21 m and a positive point charge q2 at y2 = +0.35 m. a third point charge q = +9.0 µc is fixed at the origin. the net electrostatic force exerted on the charge q by the other two charges has a magnitude of 23 n and points in the +y direction. determine the magnitude of q2.

+40.21 μC

Explanation:

As the charge q1 is negative and charge q is positive, the force that charge q experiments will try to put the particles together, therefore will have a positive direction in the y-axis. It's magnitude is given by the equation:

K is the Coulomb constant equal to 9*10^9 N*m^2/C^2, q1 and q is the charge of the particles, in Coulombs, and r is the distance between the particles, in meters.

With this, we determine the force that the other particle must apply to achieve the net force of 23 N:

The negative sign means the force has to be a repulsive force, therefore, q2 is a positive charge. We use the first equation to find q2:

or 40.21 μC

d is the correct answer