Two tiny spheres of mass = 4.80 carry charges of equal magnitude, 72.0 , but opposite sign. They are tied to the same ceiling hook by light strings of length 0.530 . When a horizontal uniform electric field that is directed to the left is turned on, the spheres hang at rest with the angle between the strings equal to 50 degree in the following figure.

Question

Two tiny spheres of mass  = 4.80  carry charges of equal magnitude, 72.0 , but opposite sign. They are tied to the same ceiling hook by light strings of length 0.530 . When a horizontal uniform electric field  that is directed to the left is turned on, the spheres hang at rest with the angle between the strings equal to 50 degree  in the following figure.

anonymous · Answer

Find magnitude of E field

anonymous · Answer

Not sure if this is right, but I used a work/energy approach.

The total work done in the system is given by

W =  We + Wm

Where We is the electrical work and Wm is the mechanical work.

The Mechanical work is then 
$$W_m = mgL(1-\cos(\theta/2))$$

And the electrical work is then
$$W_e = q^2 / (Lsin(\theta/2)\epsilon_0)$$

Also, the total energy in the electric field, W, is then also given by
W = $$(\epsilon_0 / 2)E^2Lsin(\theta/2)$$

If this is an intro level class, these equations may not be in the book but they are in Griffiths...

Putting it all together and solving for E, I got

$$E = \sqrt{(2/\epsilon_0) [mg(1-\cos(\theta/2))/\sin(\theta/2)] + q^2/(L^2 \sin^2(\theta/2) \epsilon_0)}$$

The units check out but I'm not 100% sure on this one.  You didn't specify units in the problem so I'm not gonna crunch the numbers.