Two point charges are fixed on the y axis: a negative point charge q1 = -22 µC at y1 = +0.24 m and a positive point charge q2 at y2 = +0.37 m. A third point charge q = +7.7 µC is fixed at the origin. The net electrostatic force exerted on the charge q by the other two charges has a magnitude of 24 N and points in the +y direction. Determine the magnitude of q2.
physics - Marth, Thursday, January 6, 2011 at 4:52pm
q1 exerts a force on q in the positive direction, and q2 exerts a force on q in the negative direction.
F = k(q1q2)/r^2 is the force exerted on one point charge by another.
ΣF = k(q*q1)/(0.24)^2 - k(q*q2)/(0.37)^2 = 24.
q and q1 are given. q2=?
physics - michelle, Saturday, January 8, 2011 at 5:17pm
im getting the 26-k(q*q2)/(.37)^2=24
so 24-26 gets -2.0. -k(q*q2)/(.27)2=-2.0. and then i solve for q2 but im not geting the right answer.