# Capacitors in Series

## Homework Statement

Two capacitors with capacitances of 1.0 μC and 0.50 μF, respectively, are connected in series. The system is connected to a 100 V battery. What charge accumulates on the 1.0 μF capacitor?

## Homework Equations

For capacitors in series: 1/C(equiv) = 1/C1 + 1/C2

V = Q/C

## The Attempt at a Solution

I used the second equation (V=Q/C) to find the charge with V = 100 and C = 1x10^(-6) but I did not get the right answer. However when I found the equivelant capacticance and used that in the second equation to find Q I got the right answer. Can anyone explain whey the second why to do this problem is the correct way?

## Answers and Replies

Because there are two capacitors in the circuit.
If you use C=C2 (1e-6), then you are just considering a circuit having *one* capacitor.

You basically need to find C to see how the capacitance is split between the two capacitors. That would also give you how the charge is split.

gneill
Mentor
I used the second equation (V=Q/C) to find the charge with V = 100 and C = 1x10^(-6) but I did not get the right answer. However when I found the equivelant capacticance and used that in the second equation to find Q I got the right answer. Can anyone explain whey the second why to do this problem is the correct way?

With two capacitors in series you will end up with separate voltages across each, summing to the total voltage (100V). FYI, the voltages across capacitors connected in this way go in inverse proportion to their fraction of the total capacitance values. So if C1 and C2 are initially discharged and placed in series, when 100V is applied you end up with

Vc1 = 100V * C2/(C1 + C2)

Vc2 = 100V * C1/(C1 + C2)

Another feature of capacitors in series is that if you try to push a charge in from "one end", it pushes the same amount of charge through to all the capacitors along the line. That's why you obtained the correct answer when you used the equivalent capacitance and the total voltage; each of the capacitors is holding the same charge as the equivalent capacitance.

Ceq = 1/(1/C1 + 1/C2) = C1*C2/(C1 + C2)

Qceq = 100V * Ceq = 100V * C1*C2/(C1 + C2)

Qc1 = 100V * C2/(C1 + C2) * C1

Qc2 = 100V * C1/(C1 + C2) * C2

All the same.