Question

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.

Answer 1

+40.21 μC

Step-by-step 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:

`F_q_1 = K(q_1q)/(r^2)`

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.

`F_q_1 = K(q_1q)/(r^2) = 9*10^9 (Nm^2)/(C^2) * (27*10^(-6)C*9*10^(-6)C)/((0.21m)^2) = 49.6 N`

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

`E_y: 49.6N + F_q_2 =23N\\F_q_2 = 23N - 49.6N = -26.6N`

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:

`F_q_2 = K(q_2q)/(r^2)\\q_2 = F_q_2(r^2)/(Kq) = 26.6N((0.35m)^2)/(9*10^9 (Nm^2)/(C^2)*9*10^(-6)C) = 40.21 *10^(-6)C`

or 40.21 μC