Finding the charge on pith balls

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hmparticle9
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Homework Statement
Two identical equally charged pith balls of mass ##m## and negligible radius are hanging from the same point on strings of length ##L##. The strings have an angle ##\theta## between them. What is the charge on each and the electric force on each ball?
Relevant Equations
$$F = \frac{q^2}{4\pi \epsilon_0 4 L^2 \sin^2 \frac{\theta}{2}}$$
The force is easily found to be:

$$F = \frac{q^2}{4\pi \epsilon_0 4 L^2 \sin^2 \frac{\theta}{2}}$$

I am trying to think how to get the charge ##q## from this. Since the balls are stationary the force felt on each of the balls must equal the force of gravity on the balls:

The force of gravity on the balls is ##mg \sin \frac{\theta}{2} ##.

However, if I set ##F## equal to this I do not get the answer in the book. The answer in the book is

$$\sqrt{(mg) \tan \frac{\theta}{2}4\pi \epsilon_0 4 L^2 \sin^2 \frac{\theta}{2}}$$

This,to me, suggests that the force on the ball is given by ##(mg) \tan \frac{\theta}{2}##. Could someone explain to me why this is the case?
 
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##F_e## is the repulsive force from the other charge, ##T## is the tension from the string and ##mg## is the force due to gravity. The above comments are obviously correct: ##F_e## and the other forces point in different directions. I am not sure where to go from here.
 
Last edited:
hmparticle9 said:
View attachment 362505

##F_e## is the repulsive force from the other charge, ##T## is the tension from the string and ##mg## is the force due to gravity. The above comments are obviously correct: ##F_e## and the other forces point in different directions. I am not sure where to go from here.
Nicely drawn.

With the drawing in mind and given a value for ##\theta##, can you find a formula for ##T##?

With the formula for ##T## in hand, can you then find a formula for ##f_e##?
 
So I get
$$T \sin \frac{\theta}{2} = F_e \text{ and } T \cos \frac{\theta}{2} = mg$$
Dividing and rearranging I get
$$mg \tan \frac{\theta}{2} = F_e$$
which is exactly what I want?
 
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hmparticle9 said:
$$mg \tan \frac{\theta}{2} = F_e$$
which is exactly what I want?
Presumably the question mark at the end is not needed?
 
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Yes, my bad. I was thinking that @jbriggs444 had a different solution method and I am all ears.