Solving for Charges on Suspended Spheres

[twotaileddemon]

Hi ^^! I was doing my physics homework and had a little trouble doing this one, so I was wondering if anyone could help me out if they had time. It would mean a lot, but thanks anyway just for reading this ^_^! I appreciate it =)

Homework Statement

Two spheres, each having a mass of 50 mg, are suspended from a common point by massless threads 50 cm long. One of the spheres has been given a charge twice that of the other. The strings make an angle of 6 degrees.

What are the charges on each of the spheres?
What kind of charge is on each sphere (+ or -)? Explain

Homework Equations

F = kq1q2/r^2
Q = Ne
F = ma
q_1 = 2q_2

The Attempt at a Solution

Well, assuming that the charges are not coming close together, they must be repelling each other. As such, I thought there must be some kind of acceleration... F = 2ma = 2(50 x 10^-6 kg)(9.8 m/s^2) = 9.8 x 10^-4 N
In addition, if I use sin6 = x/.5m, I can tell the distance separating the charges is .0523 m.
Now that I found the F, I plugged it into the first equation
9.8 x 10^-4 N = (9x10^9)(q_1)(q_2)/(.0523 m^2)
(q_1)(q_2) = 2.978 x 10^-16 C
2q_2^2 = 2.978 x 10^-16 C
q_2 = 1.22 x 10^-8 C
q_1 = 2q_2
q_1 = 2.44 x 10^-8 C

As for the kind of charge, I said that it is impossible to tell except that the must both be either positive or negative, not one and the other, because otherwise there would not be a 6* angle separating them.

 

[twotaileddemon]

^^;; Does anyone agree/disagree? I'm just a bit worried, because I usually don't understand how to do these kinds of problems even though I understand the equations individually.. I guess I have trouble combining two or more topics? At any rate, if possible,any direction on where I went wrong would be very helpful :). Thank you (even just for reading this!)

 

[BobG]

Force of gravity pulls the spheres straight down. The force on each sphere is its mass times gravity (i.e. - look at each sphere separately). The force from the charge pushes the spheres apart horizontally. The vector sum of the forces is enough to deflect each string 3 degrees - i.e - This statement is not true: 9.8 x 10^-4 N = (9x10^9)(q_1)(q_2)/(.0523 m^2)

By the way, you got lucky on the small angle. You don't have a right triangle with a 6 degree angle; you have two right triangles with a 3 degree angle. With angles that small, the difference is negligle, especially since you're using a rounded off value for k.

 

[twotaileddemon]

Oh, I see, I can set up a diagram with mg pointing down and F as the horizontal component. As such,
tan 3* = x / (50 x 10^-6)(9.8)

I solve for x as the force, and then the other equation I had before was right.. I just used the wrong force.

Thanks! :D