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Two 0.09g pith balls are suspended from the same point by threads 40cm long. (Pith is a light insulating material once used to make helmets worn in tropical climates.) When the balls are given equal charges, they come to rest 19 cm apart
What is the magnitude of the charge on each ball? (Neglect the mass of the thread.)
q = C

I got this from drwls but I am still unable to figure it out.

I got the angle A as 12.27 and then
tan12.27 = F/Mg = F/.09g x9.8m/sec^2
.2174 x .09g x 9.8m/sec^2 = F = .1917468
I used F so F=E = kq/r^2
.1917468 = 9.0 x 10^9 x q/.19^2
.1917468 x 19^2/9.0 x 10^9 = 7.69117722 x 10^13 C = q
This the wrong answer. Please help.
This is the help from drwls.

physics - drwls, Sunday, January 24, 2010 at 2:50am
Compute the angle of the strings from vertical. It is sin^-1 (8.5/40)
Call this angle A

If the string tension is T

T sin A = F (the Coulomb force)
T cos A = M g

tan A = F/Mg

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