Triangle Electrostatics Problem

[saintv]

Homework Statement

Three positive particles of charges +7.0 uC are located at the corners of an equilateral triangle of side 20 cm. Calculate the magnitude and direction of the net force on each particle.

Answer: 19.0 N [AWAY FROM THE CENTRE]

Homework Equations

F = (kQ1Q2)/(r^2)

Where:

k is the Electrostatics Constant - 9 x 10^9 Nm^2/C^2
Q1 and Q2 are the charges in (C)
r is the separation in (m)
F is the Electrostatics Force in (N)

The Attempt at a Solution

Er, so I calculated the force on one of them from another, see below:

F = (kQ1Q2)/(r^2)
= [(9 x 10^9 Nm^2/C^2)(7.0 X 10^-6 C)(7.0 X 10^-6 C)]/(.20m)^2
= 11.0 N

And that's where I'm stuck: I have absolutely no idea how I would even obtain a number close to 19.0 N!

What I do understand is the direction: by [AWAY FROM THE CENTRE] they mean that because they are all positive charges, they will all repel away from one another- right? Do correct me if I'm wrong.

I really do appreciate it!

 

[denverdoc]

You need to consider both charges influencing the third. Hint: you need to resolve the forces into their horizontal and vertical components and the horizontals are oppositely directed so cancel out.

 

[saintv]

Technically, all of the forces acting on the charge are opposite another- so by that logic, wouldn't they all cancel out?

I truly do not understand. :-(

 

[denverdoc]

No, if the forces have the same sign and are operating in the same direction they add.

You have studied vectors a bit, right? Forces are vectors and the best way to add them is to break the forces down into horizontal and vertical components? See my picture, the red and blue arrows are the components. Equilateral triangles have what angles? Use sin, cos functions to compute the components then add, paying attn to sign! There are two balls each effecting the third, so my picture isn't complete.

That help?