Recent content by 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! :) :)

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

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...

nC = nanocoulomb

'N' in the final result was for Newtons

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

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...