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.

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:

[tex]F_q_1 = K\frac{q_1q}{r^2}[/tex]

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.

[tex]F_q_1 = K\frac{q_1q}{r^2} = 9*10^9 \frac{Nm^2}{C^2} * \frac{27*10^{-6}C*9*10^{-6}C}{(0.21m)^2} = 49.6 N[/tex]

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

[tex]E_y: 49.6N + F_q_2 =23N\\F_q_2 = 23N - 49.6N = -26.6N[/tex]

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:

[tex]F_q_2 = K\frac{q_2q}{r^2}\\q_2 = F_q_2\frac{r^2}{Kq} = 26.6N\frac{(0.35m)^2}{9*10^9 \frac{Nm^2}{C^2}*9*10^{-6}C} = 40.21 *10^{-6}C[/tex]

or 40.21 μC