Answer to Question #132408 in Mechanics | Relativity for Jessica

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 = +0.37 m. A third point charge q = +8.8 μC is fixed at the origin. The net electrostatic force exerted on the charge q by the other two charges has a magnitude of 25 N and points in the +y direction. Determine the magnitude of q2.

Answer 1

Let us determine the net force exerted by two charges:

${F}_y = k\cdot\dfrac{q|q_1|}{y_1^2} - k\cdot\dfrac{qq_2}{y_2^2}\,.$

$q_2 = -\dfrac{y_2^2F_y}{kq} + |q_1|\cdot\dfrac{y_2^2}{y_1^2} = -\dfrac{0.37^2\cdot25}{9\cdot10^9\cdot8.8\cdot10^{-6}} + 27\cdot10^{-6}\cdot\dfrac{0.37^2}{0.21^2} = 40.6\mu\text{C}.$