# Homework Help: Finding Magnitude of a Charge on Sphere

1. Aug 27, 2011

### cheerspens

1. The problem statement, all variables and given/known data
Two small metallic spheres, each of mass 0.20 g, are suspended as pendulums by light strings from a common point as shown in the figure I attached. The spheres are given the same electric charge, and it is found that they come to equilibrium when each string is at an angle of 5.0 degrees with the vertical. If each string is 30.0 cm long, what is the magnitude of the charge on each sphere?

2. Relevant equations
Coulomb's Law

3. The attempt at a solution
I've drawn a force diagram for the first ball and I find that it's Fg=.00196, the tension, T, is .00197, and the Force of ball 2 on ball one, F21, is 1.71x10-4. When I plug into this equation: F=(ke|q|)/(r2), I get 5.1x10-9.

However, I know this is wrong because the correct answer is 7.2nC. How do I do this correctly?

#### Attached Files:

• ###### photo.jpg
File size:
30.8 KB
Views:
321
Last edited: Aug 27, 2011
2. Aug 27, 2011

### Tomer

Notice that F = K * q * q / r^2...
Your formula lacks one q... Maybe that's where it fell? I'm too tired to make the calculation :)

3. Aug 27, 2011

### cheerspens

Ok I've made sure that I account for both q's.
In order to solve this, do I have to break the vectors into components and set them equal to each other?

4. Aug 27, 2011

### Tomer

Yes. It seems to me you did everything properly.
You get the tension from comparing the y-axis forces: Gravity and T*sin5.
Then by comparing the x-axis forces: T*cos5 = Kq^2 / r^2 you can isolate q.
notice that r = 0.3 * sin5 * 2.
It should work. If you still get the same answer, then the book's wrong...

5. Aug 27, 2011

### cheerspens

I'm not sure if you took the time to actually solve it (I wouldn't blame you if you didn't) but for my final answer I got 6.88nC. If the book's answer is 7.2nC would it be safe to say this is accurate?

6. Aug 28, 2011

### Tomer

I'm getting the book's answer after trying.

You get the equation:

0.00197 * sin5 = 9 * 10^9 * q^2 / (0.3*(sin5)*2)^2)

solving it leads to q = 7.22 nano C.