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
so
each capacitor has charge Q
C = Q/V
so
V1 = Q/C1
V2 = Q/C2
V3 = Q/C3
and
11 = V1+V2+V3
11 = Q (1/C1 + 1/C2 + 1/C3)
solve for Q
then V3 = Q/(31.8*01^-6)

To find the voltage across the 31.8 µF capacitor, you can use the concept of capacitors in series. In a series circuit, the total voltage across the capacitors is equal to the sum of the voltages across each individual capacitor.

Here's how you can calculate the voltage across the 31.8 µF capacitor:

1. Write down the capacitances of the capacitors:
C1 = 5.3 µF
C2 = 10.7 µF
C3 = 31.8 µF

2. Calculate the total capacitance (Ctotal) of the capacitors in series:
Ctotal = 1 / (1/C1 + 1/C2 + 1/C3)

Substituting the values:
Ctotal = 1 / (1/5.3 + 1/10.7 + 1/31.8)

3. Simplify the expression:
Ctotal = 1 / (0.1887 + 0.09346 + 0.03145)
= 1 / 0.31361
= 3.1882 µF (rounded to four decimal places)

4. Now, you can use the formula for the voltage across a capacitor in a series circuit:
Vcapacitor = Vtotal * (Ccapacitor / Ctotal)

Substituting the values:
V31.8 µF = 11 V * (31.8 µF / 3.1882 µF)
= 11 V * 9.9971

5. Calculate the voltage across the 31.8 µF capacitor:
V31.8 µF ≈ 109.9691 V (rounded to four decimal places)

So, the voltage across the 31.8 µF capacitor is approximately 109.9691 volts.