Three capacitors (3.83, 5.89, and 12.9 μF) are connected in series across a 58.0 V battery. Calculate the voltage across the 3.83 μF capacitor.

Answer 1

C = q/V
so V = q/C
so
q1/C1 + q2/C2 + q3/C3 = 58
but q1=q2=q3
becaue where else are the electrons to come from or go to?
so
1/C1 + 1/C2 + 1/C3 = 58/ q
10^6 (1/3.83+1/5.89+1/12.9 = 58 /q
10^6 (.508) = 58 / q
q = (58/.508)10^-6

V = (58/.508)/3.83 = 29.8 volts