# Finding the charge given distance apart + some more

1. Jan 10, 2005

### thursdaytbs

First, I said that the distance between the two spheres is 0.0523m since you can form a right triangle using 0.5m [50cm] as the hypotonose, and 3 degrees as an angle. and multiplying that answer by 2 giving me the distance between the two spheres.

From that I know one sphere = 2q, while the other sphere = q.

So... F = (K)(q)(2q) / (r^2)

From there i'm stuck, since you have two unknowns (charge 'q' and force 'F')

I'm thinking of using F=ma = 0 (a=0), then finding using a Free Body diagram to find the force on each sphere and adding them and setting them to zero?

So sphere 1 and 2's forces = mg, tension, Electric Force (F/q)
where the Forces are equal?

I'm not too sure, can anyone give me any suggestions? Thanks for any help.

2. Jan 10, 2005

### dextercioby

1.The 2 charges have to have opposite charge signs,else they wouldn't repel each other and would not make an angle of 6°.
2.The charges need to be in equilibrium,which means they have total acceleration zero,this meaning that the net force acting one each other is zero.
3.Write Newton's principle for each charge,knowing the 3 forces that act on each of them.
4.Use projections of forces on axes of coordinates conveninetly chosen and geomtry of the figure to find the unknowns.

Daniel.

3. Jan 10, 2005

### thursdaytbs

So finding exactly whether it's both positive or negative can not be done?

Also, would my equations look like this?

er... would the vectors added be: m1g, m2g, k(2q)(q) / r^2, k(2q)(q) / r^2, T1, and T2?

add them as vectors and set them equal to zero and solve for q?

Last edited: Jan 10, 2005
4. Jan 10, 2005

### thursdaytbs

I got it. Thank you.

I did F = ma = 0.

and one sphere = equilibrium, created a right triangle with mg, Tension, and Electrical force. Knowing [mg] and the angles, I solved for the Electrical Force.

Then set that = to (k)(2q)(q) / r^2, where r = 0.0523 and k = 9x10^9.

q = 1.96 x 10^-9
2q = 3.93 x 10^-9

Last edited: Jan 10, 2005
5. Jan 10, 2005

### Galileo

Me too. This is the charge on the lesser charged sphere.

6. Jan 10, 2005

### thursdaytbs

The problem with that was that I had [mg] wrong. For the mass, I used the gram value of the mass instead of the kilogram value.