QUESTION: What is the force F on the 1nC on the charge at the bottom of the figu
ID: 2002236 • Letter: Q
Question
QUESTION:
What is the force F on the 1nC on the charge at the bottom of the figure?
(Express answer as two vector components)
ATTEMPT:
F1 = Force of 2.0nC on 1.0nC (left side) r=0.05
F2 = Force of -6.0nC on 1.0nC (middle) r=0.0354 *2
F3 = Force of 2.0nC on 1.0nC (right side) r=0.05
F1:
x = 7.192*10^(-6)cos(45) = 5.08551*10^(-6)
y = 7.192*10^(-6)sin(45) = -5.08551*10^(-6)
F1 = <5.08551*10^-6, -5.08551*10^-6>
F3:
x = 7.192*10^(-6)cos(45) = -5.08551*10^(-6)
y = 7.192*10^(-6)sin(45) = -5.08551*10^(-6)
F3 = <-5.08551*10^-6, -5.08551*10^-6>
F2:
= kQq/r^2 = [ (8.99*10^9) (1.0*10^-9) (6.0*10^-9) ]/[ 0.0354 + 0.0354 ]^2 = 1.076*10^-5
x = 1.076*10^(-5)cos(90) = 0
y = 7.192*10^(-6)sin(45) = 1.076*10^(-5)
F3 = <0, 1.076*10^(-5)>
Fnet:
x=F1x + F2x + F3x = [ (-5.08551*10^-6) + (5.08551*10^-6) + (0) ] = 0
y=F1y + F2y + F3y = [ (-5.08551*10^-6) + (-5.08551*10^-6) + (1.076*10^(-5)) ] = 5.9*10^(-7)
Fnet = <0, 5.9*10^(-7)>
RESULTS:
My resulting net force(Fnet) is wrong and I have been attempting to find the issue for 2 days but can't find it :( if anyone can help me out I would really appreciate it. Also if there is a better way to do this I am open to suggestions.
Explanation / Answer
For the right and left side the charges and the distance seperated by each charge with the central charge,(1 nC) is same.Middle one has a charge of -6 nC. So to find the total force, just add them together.
So we get,
q1=q3=2 nC
q2=-6 nC
q4=1 nC
F=2(kq1*q4/r2)cos45-(kq2*q4/r2)
F=2*[{(9*109)*(2*1*10-18)}/0.052]*cos45 - [{(9*109)*(6*1*10-18)}/0.052]
F= (3.6*10-6)*[4*cos 45 - 6] = -11.4*10-6 N.
This force is towards - 6nC charge, that is 90 degrees above horizontal.