# How can I theoretically describe the potential between two capacitors in series?

Suppose to have two capacitors in series:

The voltage in the middle point will be:

$$V_X = V_1 \frac{C_1}{C_1+C_2}$$

How can this be explained? It's been asked in electronics, and explained in terms of impedence and charge equality, but none of the explanations is satisfying to me, as I think it should involve charge conservation (Gauss theorem?) and/or electric fields/potentials.

Could you enlighten me?

• The charge equalization version is a charge conservation argument. Note that there is an isolated conductor between the capacitors and that it starts off bulk neutral... Commented Jun 11, 2012 at 14:33
• @dmckee: it is, but it's not quite well explained there. Note that I mentioned it in a comment, and it's been reprised in the answer, but without further explanation Commented Jun 11, 2012 at 14:35

Suppose you imagine the battery to be a variable voltage, and start with the voltage at zero. Obviously everything is uncharged.

Now turn the battery up to 1V. As you do this positive charge leaves the positive terminal and an equal and opposite negative charge leaves the negative terminal. We know the charges leaving the positive and negative terminals must be the same because the battery is a conductor and can't develop a net charge like a capacitor. Let's call the charge that leaves the battery $Q$.

The only place the charge that leaves the battery can go is onto the capacitors, so both capacitors now have a charge of $Q$ on them. We know that for a capacitor of capacitance $C$, the voltage across the capacitor is given by:

$$V = \frac{Q}{C}$$

Call the voltage of the top (1$\mu$F) capacitor $V_1$, and the voltage of the bottom (2$\mu$F) capacitor $V_2$. Then:

$$V_1 = \frac{Q}{C_1}$$ $$V_2 = \frac{Q}{C_2}$$

Dividing the first equation by the second plus a bit of quick rearrangement gives:

$$V_1 = \frac{C_2}{C_1} V_2$$

The two voltages must add up to 1V because we have a 1V battery, therefore:

$$V_1 + V_2 = 1$$

If you substitute for $V_1$ you get:

$$\frac{C_2}{C_1} V_2 + V_2 = 1$$

and dividing through by $(1 + C_2/C_1)$ gives:

$$V_2 = \frac{1}{1 + C_2/C_1}$$

Tidy this up by multiplying to top and bottom of the right hand side by $C_1$ and you get the equation you're trying to prove:

$$V_2 = \frac{C_1}{C_1 + C_2}$$

Just to check, feed in $C_1 = 1$ and $C_2 = 2$ and $V_2$ does indeed come out as 1/3V.

• Detailed, but it doesn't explain the assumption that both capacitors have charge Q Commented Jun 11, 2012 at 18:10
• If charge +Q leaves the battery anode then charge -Q must leave the cathode because the battery can't have a net charge. That means the top plate of the top capacitor has a +Q charge and the bottom plate of the bottom capacitor has a -Q charge. But these charges are now attracting/repelling the electrons in the wire between the two capacitors. The +Q charge on the top capacitor will attract electrons from the wire until it's bottom plate builds up a charge of -Q. Likewise the -Q charge on the bottom capacitor will repel electrons until it's top plate has a +Q charge. So the charge ... Commented Jun 11, 2012 at 18:15
• ... on both capacitors is Q. The circuit doesn't start out with any charge, so if you add up the net charge it must sum to zero. That's how we know the charges on the top plate of the top capacitor and the bottom plate of the bottom capacitor must be equal and opposite. Commented Jun 11, 2012 at 18:15
• But why the potential in $A$ is equal to the lower voltage? i.e. why $V_A=V_2$? @JohnRennie Commented Feb 1, 2017 at 22:42

This is really just a restatement of John Rennie's answer, but it might be a bit easier to follow...

Assume both capacitors are initially uncharged (important, since otherwise the voltage of their common node is undefined) and the voltage source is 0.

Now ramp the voltage source up to 1V. During this ramp up, the same current i flows through both capacitors (since they're connected in series), so $$i= C_1\frac{dv_{c1}}{dt} = C_2\frac{dv_{c2}}{dt}$$

So the rates of change of the capacitor voltages are inversely proportional to their capacitances, and so will the final capacitor voltages after integrating (using the fact that the capacitor voltages were initially zero).

From that relation, it's straightforward to get to the expression in your question.

I think the quickest way to get an intuitive feel for this situation is to recognise the voltage at that part of the circuit is the potential difference between the two capacitors. And is in reference to the $0V$ terminal of the battery.

Kirchoff's voltage law tells us that we have to sum to $0V$ in a loop. As C2 = $2 *$C1 we have to lose twice as much voltage over C2. As we are dealing with $1V$ that means we lose $.66V$ over C2 and $.33V$ over C1. Now tracing around the circuit in either direction quickly leads to the answer that we have a voltage of $.33V$ between the capacitors...