I’m not asking you to do my homework, only to help me find where I’m messing up.
Here is how one guy did it, but I couldn’t follow his notation:
This is how I attempted it:
If particle 3 is to remain stationary, then the force on particle 3 from particle 1 must equal the force from particle 2 on particle 1.
Using coulombs law to state this:
F=[K(q1)(q3)]/(r^2) = [K(q2)(q3)]/(r^2)
But, particle 3 has to be to the right of particle 1, if it was in the middle between particle 1 and particle 2, then it would be pulled to the left towards particle 2, because particle 1 would repel particle 3 to the right, and particle 2 would pull particle 3 left.
So to make up for this in our equation we make the distance from charge 2 to charge 3, “R+L” where r is the distance between charge 1 and charge 2:
F=[K(q1)(q3)]/(®^2) = [K(q2)(q3)]/((r+L)^2)
Now we can place some numbers in this equation, and solve for L:
F=[(8.99x10^9)(3.5x10^-6)(5.5x10^-6)]/((.0024)^2) = [(8.99x10^9)(-4.0x10^-6)(5.5x10^-6)]/((.0024+L)^2)
Coulombs constant and Q3 cancel from the equation, so I’m left with
F=[(3.5x10^-6)]/((.0024)^2) = [(-4.0x10^-6)]/((SQRT(.0026)+L)^2)
(I know that the two original particles are SQRT(.0026) m apart)
So now we have an equation with only one unknown, L, so I solved the equation for L and got that L was .03765662 meters.
So we now know that particle 3 is 3.765662 CM from particle 1.
We can add that to the original coordinates of particle 1, to find out where particle 3 is:
the angle that these three particles lie on is -11.3
x= 3.765662cos(-11.3) + 3 cm = 6.692663 CM
y= 3.765662sin(-11.3) + .5 cm = -.2378669183 CM
However the online grader says that those are both wrong. AND I ONLY HAVE ONE ATTEMPT LEFT TO GET CREDIT