Question

Two point charges are fixed on the y axis: a negative point charge q1 = -24 µC at y1 = +0.19 m and a positive point charge q2 at y2 = +0.33 m. A third point charge q = +8.0 µC is fixed at the origin. The net electrostatic force exerted on the charge q by the other two charges has a magnitude of 18 N and points in the +y direction. Determine the magnitude of q2.

Answer 1

Answer:

4.51 * 10^{-5} C

Explanation:

The force between two charge  q and q1 is given as

$F = \frac{k*q*q1}{r^2}$

$= \frac{(9.0 * 10^9)(24 * 10^{-6})(8 * 10^{-6} C)}{(0.19m)^2}$

= 47.86 N in the +y direction

We need the force between q and q2 to be (47.86 - 18) =  29.86 N in the other direction to get the desired result.

solving for q2,

$q2 = \frac{Fr^2}{(kq)}$

$= \frac{(29.86)(0.33 m)^2}{(9.0 * 10^9*8*10^{-6} C)}$

$= 4.51 * 10^{-5} C$