Particle Physics: Net Force on Charge q1 Calculation

[Amelia]

Homework Statement

Three point charges are located on the positive x-axis of a coordinate system. Charge q1 = 2.0 nC is 2.0 cm from the origin, charge q2 = -3.5 nC is 4.0 cm from the origin and charge q3 = 5.0 nC located at the origin.

What is the net force ((a)magnitude and (b) direction) on charge q1 = 2.0 nC exerted by the other two charges?.[/B]

Homework Equations

F= k (q1q2/r^2)

The Attempt at a Solution

I know that the direction (b) is 0 degrees as all the points are all located on the x-axis, I just can't figure out part (a). I keep mixing up my diagram so that when the equation is applied, my answer comes out incorrectly. Any help would be great! [/B]

 

[Bystander]

Superposition of two forces. Do them one at a time.

 

[Amelia]

Superposition of two forces. Do them one at a time.

Okay, so would it be...
F of 2 on 1 = 9*10^-6((-3.5*10^-6)*(-2*10^-6)/(0.04)^2 = 39.375 N (+ve therefore forces repel)
F of 3 on 1 = 9*10^-6((2*10^-6)*(-5*10^-6)/(0.02)^2 = -225N (-ve therefore forces attract)
F total = 39.375 - (-225) = 264.375 N

 

[Bystander]

"n" means what? Like charges attract or repel?

 

[Amelia]

"n" means what? Like charges attract or repel?

'N' in the final result was for Newtons

 

[Bystander]

Lower case "n," as in "nC."

 

[Amelia]

Lower case "n," as in "nC."

nC = nanocoulomb

 

[Bystander]

"Nano" means what?

 

[Amelia]

"Nano" means what?

ohhh... 1*10^-9 not 1*10^-6 thank you I didn't even see that !
so the new solutions come out at
F of 2 on 1 = -0.000225 N
F of 3 on 1 =0.000039 N
So the final answer would be -0.000225 N - 0.000039 N = -2.46*10^-4 I just tried to enter that in and it still came out incorrect... I'm just not sure what I'm doing wrong

 

[Bystander]

F of 2 on 1

4-2=?

 

[Amelia]

4-2=?

4-2=2 but I'm not sure where that fits in..

 

[Bystander]

F of 2 on 1 = 9*10^-6((-3.5*10^-6)*(-2*10^-6)/(0.04)^2 = 39.375

 

[Amelia]

Okay so it should be
F of 2 on 1 = 9*10^9((-3.5*10^-9)*(-2*10^-9)/(0.02)^2 = 0.000158 N
F of 3 on 1 = 9*10^9((2*10^-9)*(-5*10^-9)/(0.02)^2 = -0.00025N
Therefore, Ft= 0.000158-(-0.00025) = 3.8*10^-4 N

Which I just put in and it's correct! Thank you so much for your help! :) :)

 

[Bystander]

You are welcome. Things to remember: one step at a time; and, pay attention to details.