Net Force Exerted on 4.0*10-9 C Charge at Origin

[undrcvrbro]

Homework Statement

A 1.4*10-9 C charge is on the x-axis at x = -2.0 m, a 7.0*10-9 C charge is on the x-axis at x = 2.5 m. Find the net force exerted on a 4.0*10-9 C charge located at the origin.

Homework Equations

F=k(q1*q2) / (r^2)
k= 9*10^9

The Attempt at a Solution

I simply took the difference of the two Forces existing between the central charge and each end of the axis.
I took:
F= (9*10^9)(1.4*10^-9)(4.0*10^-9) / (2)^2
and then subtracted it from:
F= (9*10^9)(7.0*10^-9)(4.0*10^-9) / (2.5)^2

But that obviously isn't correct, because Web assign gave me a big red X when I plugged the answer in. Any help is greatly appreciated!

 

[Kurdt]

Force is a vector and the total force on the central charge can be found using vector addition. Work out what direction each force is acting and the answer should become apparent.

 

[undrcvrbro]

Force is a vector and the total force on the central charge can be found using vector addition. Work out what direction each force is acting and the answer should become apparent.

Yes, that's what I thought, but I got the wrong answer...I'm missing something here.

Because all vectors are on x axis, and its asking for the net force on the center charge, wouldn't I just take the force imposed on the center from the left, and subtract it from the force between the center and the right charge? It seems too simple.

 

[Kurdt]

Yes, that's what I thought, but I got the wrong answer...I'm missing something here.

Because all vectors are on x axis, and its asking for the net force on the center charge, wouldn't I just take the force imposed on the center from the left, and subtract it from the force between the center and the right charge? It seems too simple.

Because force is a vector it matters what direction it is acting in. Since all charges are positive the force is going to be repulsive. Draw a diagram if it helps.

So the left charge will produce a force that wants to push the centre charge to the right. The right charge will produce a force that wants to push the centre charge to the left. If we take right as the direction of the unit vector on the x-axis, then the force from the left charge will be a positive factor multiplied by the unit vector. The force from the right charge will be a negative factor of the unit vector. Now apply vector addition.