Electricity and Gravity Question

[Alameen Damer]

Two pith balls, each with a mass of 5.00g, are attached to non-conducting threads and suspended from the same point on the ceiling. Each thread has a length of 1.00m. The balls are then given an identical charge, which causes them to separate. At the point that the electrical and gravitational forces balance, the threads are separated by an angle of 30.0 degrees. Calculate the charge on each pith ball."

My answer isn't matching the textbooks of 6.27 x 10^-7 C

Relevant Equations:

Fg=Gm1m2/r^2
Fe=kq1q1/r^2

Attempt:

(G)(m)^2/r^2=(k)(q)^2/r^2

r^2 cancels out

G(m)^2=k(q)^2
root [(G)(m)^2]/k
=4.3 x 10^-13

Am i missing something? Or is the book wrong?

 

[Simon Bridge]

Your equation is for the charge that balances the pith balls gravitational attraction for each other ... is there another source of gravity in the room?

 

[Alameen Damer]

Yes the gravity pulling the balls down. However, how would i write this gravity, it has me confused. Does the angle play a part?

 

[zabini]

You can ignore the gravitational attraction force between them as their masses are very small. Three forces acting on each ball. 1.mg vertically downward 2.electric force along the line joining them.
3.Tension acting towards the point of suspension.
Since the balls are in equilibrium, the net force along the string is zero. Resolve the forces along that direction and equate to 0.Also, the net force along the line joining them is 0. Resolve the forces in that direction and equate to 0. You now have 2 equations with 2. Variables that is Tension and Charge. Solve for charge.