Alarge electroscope is made with "leaves" that are 70 cm long wires with 19 g spheres at the ends. when charged, nearly all the charge resides on the spheres. if the wires each make a 30° angle with the vertical, what total charge q must have been applied to the electroscope? ignore the mass of the wires.

The only force for which we can make a direct calculation is Fg: Fg = mg. So we must ultimately find Fe in terms of Fg, then use Coulomb's Law to find the charge. 

The electromotive force (Fe) must be equal and opposite the horizontal compontent of the tension (Tx): Fe = Tx. Find Tx in terms of T: T sin 30°. Therefore: 

Fe = T sin 30°. 

The gravitational force (Fg) must be equal and opposite the vertical component of tension (Ty): Fg = Ty. Find Ty in terms of T: T cos 30°. So: 

Fg = T cos 30°. 

Solve both for T: 

Fe = T sin 30° 
T = Fe / sin 30° 

Fg = T cos 30° 
T = Fg / cos 30° 

Then set them equal and solve for Fe: 

Fe / sin 30° = Fg / cos 30° 
Fe = Fg sin 30° / cos 30° 
Fe = Fg tan 30° 
Fe = mg tan 30° 

By Coulomb's Law, Fe is also given by: Fe = KQq / r², where K is the electrostatic constant, Q and q are the two charges, and r is the distance between the charges. This gives: 

KQq / r² = mg tan 30° 

Both charges are equal, so: 

Kq² / r² = mg tan 30°. 

Solve for q: 

q² = mgr² tan 30° / K 
q = √(mgr² tan 30° / K) 

Find r. The mass is deflected by a horizontal distance of: (70 cm) sin 30° = 0.7 sin 30°, but since both masses are deflected an equal amount, the distance between them is given by: r = (2)(0.7 sin 30°) = 0.7 m. 

Finally, you can plug in your knowns (in SI units) and calculate: 

q = √[(0.033 kg)(9.8 m/s²)(0.7 m)²(0.577) / (8.988x10^9 Nּm²/C²) 
q ≈ ±3.19 x 10^-6 C = ±3.19 μC 

This is the charge on ONE sphere, so the total charge will be double this number. 

2q ≈ ±6.4 μC (rounded to two significant digits