Coulomb's Law magnitude problem

[Bradman]

Homework Statement

Two point charges are fixed on the y axis: a negative point charge q1 = -24 µC at y1 = +0.20 m and a positive point charge q2 at y2 = +0.37 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 26 N and points in the +y direction. Determine the magnitude of q2.

Homework Equations

F = K(q1q2)/r^2

The Attempt at a Solution

First I found the magnitude of force between q and q1 and set up the equation for q1 and q2:

[(8.99*10^9)(9.0*10^-6)(2.4*10-5)]/(0.20^2) = 48.546N
[(8.99*10^9)(2.4*10^-5)(q2)]/(0.17^2) = F

Then I did my equation to find the force for the 2nd equation (-48.546 since it's an attractive force):
F - 48.546N = 26N
F = 74.546

Then solved for q2:

q2 = [(74.546)(0.17^2)]/[(8.99*10^9)(2.4*10^-5)] = 9.985*10^-6

WebAssign tells me I'm incorrect, and I can't tell where I've gone wrong...

 

[rl.bhat]

Hi Bradman, welcome to PF.
What is the distance between q2 and q?

 

[Bradman]

0.37m

 

[rl.bhat]

0.37m

q2 = [(74.546)(0.17^2)]/[(8.99*10^9)(2.4*10^-5)] = 9.985*10^-6

Then check the force between q2 and q.

 

[Bradman]

Well, the sum of the forces on q is +26N, so since the Force of q1 on q is an attractive force of 46.546N in one direction, then the force of q2 on q should be a repulsive force 74.546N in the other direction. That makes the magnitude of the q2 charge:

q2 = [(74.546)(0.37^2)]/[(8.99*10^9)(9.0*10^-6)] = 1.261*10^-4 C

Which still tells me I'm wrong (it was my final submission).

 

[rl.bhat]

Force between q and q1 is 48.546 N along +y direction.

Net force is 26 N along the + y direction.

Hence force F between q and q2 must be along the - y direction.

So 26 N = 48.546 N - F.

Now proceed.