# Capacitors in series: Why is the equivalent charge the same as the individual charges

Consider two parallel-plate capacitors $C_1, C_2$ in series. For the "equivalent" circuit, clearly $$Q_{equiv} = C_{equiv}V$$ should hold, where V is the total voltage drop between input and output. It is also obvious that $$Q_1 = Q_2$$ where $Q_1$ and $Q_2$ are the charges (plus and minus on opposite sides) accumulated on each of the two sets of plates. Obviously in general, $$V_1 \neq V_2 \qquad \text{}$$

However, just because we have two capacitors with equal charge inside the circuit why is it true that for the equivalent circuit

$$Q_{equiv} = Q_1 = Q_2$$

I see no obvious reason why this should be true, and this is assumed, though not explained in the discusssion below: http://farside.ph.utexas.edu/teaching/302l/lectures/node46.html

Here's what I do (or rather, what I just did) to convince myself of your result.

We're trying to "reduce" the two capacitors in series to an equivalent capacitor. Equivalent in every way, including how much charge would flow when discharging the capacitor.

Imagine discharging the two capacitors that are series. How much total charge would flow? Since, as you've stated, the "input" and "output" sides store equal magnitude charges $|Q_1|=|Q_2|\equiv Q$, then we should be able to recover this much charge during discharging.

By discharging the equivalent capacitor, then, one should expect to have a total charge $Q$ flow during discharging. Thus, $Q_\mathrm{equiv}=Q$.

• I like this answer, I'm not 100% sure if should accept it though, i'm curious if there is anything slightly more airtight – Timtam Feb 10 '14 at 4:06

Here goes my answer :-) Being $(V_1,Q_1) \;\&\; (V_2,Q_2)$ the voltage and charge of each series cap $(C_1\;\&\; C_2)$, we know (KVL) that the voltage of the ensemble is: $$V_e=V_1+V_2. \tag{Eq.1}$$ and as you said (KCL), $$Q_1=Q_2 \tag{Eq. 2}$$ Where ${1\over2}\cdot Q_1\cdot V_1 \;\&\; {1\over2}\cdot Q_2\cdot V_2$ are the energy stored in $C_1 \;\&\; C_2$ respectively. Being the energy of the ensemble: ${1\over 2}\cdot Q_e\cdot V_e$.

It holds that (energy conservation) "Energy in $C_1$" plus "Energy in $C_2$" equals "Energy of the ensemble", then: $${1\over 2}\cdot Q_1\cdot V_1 + \frac{1}{2}\cdot Q_2\cdot V_2 = {1\over 2}\cdot Q_e\cdot V_e \tag{Eq. 3}$$

The only way that $\text{Eq.1, Eq.2 & Eq.3}$ holds is that $Q_1=Q_2=Q_e$.

and Problem solved! :-)

• Hi, welcome to PSE. Don't make multiple edits for minor changes. It is not a good etiquette. – user36790 Aug 21 '15 at 9:15
• wanted it to be perfect, because myself used to forget the answer :-)) – Dwight Aug 21 '15 at 9:17
• Isn't it more perfect now?? – user36790 Aug 21 '15 at 9:24

The same question comes to me today. And my answer and understanding are:

In the circuit, the left capacitor is C1, connected to positive pole; the right one is C2, connected to negative.

After fully charged, we have +Q1 and -Q1 on two sides of Capacitor C1, and +Q2 & -Q2 on C2. Same quantity of opposite charges ensure there's no net electric flux around C1 & C2, from Gauss Theorem. Actually, at this equivalence stage, there's no electric field within the wires, only electric field within the capacitors.

Now suppose Q1 not equal to Q2. Using Gauss theorem respectively for C1 and C2, there's still no net electric flux. Then look at the part with one plate of C1 & C2 connecting via the wire, i.e. |-------| . Since -Q1+Q2 not vanishes, the net electric flux not vanishes, leading us to different electric fields E1 and E2 within capacitors.

In formulas: (-Q1+Q2)/epsilon=-E1*A+E2*A =(-U1/d1 +U2/d2)*A= (-U1*C1/epsilon +U2*C2/epsilon)=(-Q1+Q2)/epsilon

where E1&E2 are electric fields, U1&U2 are voltages, d1&d2 are capacitors' separations, A is the plate area which is supposed to be equal for two capacitors.

Therefore that Q1 not equal to Q2 does not go against any physics rules. It's only the result of reaching the fully charged equivalence, together with E1=E2.

So why should the same amount of charges when equivalence? I understand as the usage of a battery in a circuit. Battery affects the free electrons to move, so there're extra electrons with -Q, also holes with +Q. Here C2 is connected to the negative pole of battery, so -Q2 is firstly accumulated on right plate of C2, along with Q1 on left plate of C1. There'are electric fields everywhere currently, and we got Q2, -Q1 accumulated under such electric fields. Yet in this process, |+Q2| not equal to |-Q2|; on the contrast, |+Q1| always equal to |-Q2|. The free electrons move under electric fields, until |+Q2| equaling to |-Q2|, the so-called equivalence condition. And electric fields within the wire also disappeared, they finished their job of "transporting electrons".