Question

Two point charges are fixed on the y axis: a negative point charge q1 = -25 μC at y1 = +0.16 m and a positive point charge q2 at y2 = +0.38 m. A third point charge q = +8.3 μC is fixed at the origin. The net electrostatic force exerted on the charge q by the other two charges has a magnitude of 20 N and points in the +y direction. Determine the magnitude of q2.

Answer 1

force on charge q due to q1 is attractive and is upwards.

force on charge q due to q2 is repulsion and is downwards.

so,

net force = kqq1 / r² - kqq2 / r²

20 = 9e9 * 8.3e-6 * 25e-6 / 0.16² - 9e9 * 8.3e-6 * q2 / 0.38²

20 = 72.95 - 9e9 * 8.3e-6 * q2 / 0.38²

-52.95 = - 9e9 * 8.3e-6 * q2 / 0.38²

q2 = 1.023e-4 C

or

q2 = 102.3 uC