posted by Valerie .
A 11-V battery is connected to three capacitors in series. The capacitors have the following capacitances: 5.3 µF, 10.7 µF, and 31.8 µF. Find the voltage across the 31.8-µF capacitor. Please Help!
A charge Q comes out of the battery.
That charge shows up on the top of the first C
It draws -Q from top of second, leaving + Q on second
That draws -Q from third, leaving +Q on top of #3
each capacitor has charge Q
C = Q/V
V1 = Q/C1
V2 = Q/C2
V3 = Q/C3
11 = V1+V2+V3
11 = Q (1/C1 + 1/C2 + 1/C3)
solve for Q
then V3 = Q/(31.8*01^-6)