Electric Force Calc: 4µC, -3µC, 4µC at 10cm, 4cm,-2cm

[redhot209]

Note: ^i means carrot on top of variable

Homework Statement

A particle with charge 4 µC is located on the x-axis at the point 10 cm, and a second particle with charge -3 µC is placed on the x-axis at 4 cm. What is the magnitude of the total electrostatic force on a third particle with a charge 4 µC placed on the x-axis at -2 cm? Answer in units of N.

The Attempt at a Solution

1) I converted all my variables to µC = C , and cm = m
q1 = 4 µC = 4e-6
q2 = -3 µC = -3e-6
q3 = 4 µC = 4e-6
x1 = 10 cm = 0.1 m
x2 = 4 cm = 0.04 m
x3 = -2 cm = -0.02 m

2 By using Coulom's law in vector form:
F₁₃= ke(q1q3/r²)^r₁₃
where ^r₁₃ is a unit vector directed from q1 to q3; i.e., ->r13 = ->r3 - ->r1

x13 = x3 - x1
(-0.02m)-(0.1m) = -0.12m
x23 = x3 - x2
(-0.02m)-(0.04) = -0.06 m

^x₁₃ = x₃-x₁/ √(x₃-x₁)^2 = -1 or -^i
^x₂₃ = x₃-x₂/ √(x₃-x₂)^2 = 1 or +^i

Since the forces are collinear, the force on the third particle is the algrebracis sum of the forces between the first and third and the second and third particles.

->F = ->F₁₃ + ->F₂₃
= k_e [q₁/ r²₁₃ X ^r₁₃ + q₂ / r²₂₃ x^r₁₃ ] q₃

After putting in the variables, I got 33.9444444445 N as my answer, but was wrong. What did i do wrong?

 

[rl.bhat]

First of all find the force between 4micro C at -2 cm and 4 micro C at 10 cm. It is repulsive force. Next find the force between 4micro C at -2 cm and -3 micro C at 4 cm. It is attractive force. Now find the net electrostatic force on the third charge.