# Why is charge the same on every capacitor in series?

Why is the amount of charge on every capacitor in series equal, regardless that capacitance values of capacitors are not the same? What really happens here so they are the same?

simulate this circuit – Schematic created using CircuitLab

• Charge is proportional to amount of current & length of time: the simple formula being Q = I x t. The capacitance plays no role in determining charge. Dec 16 '15 at 22:55
• NOTE this is only true if all the capacitors are discharged before you build the circuit. If one of them was charged to voltage V1, i.e. 1V, then no current would flow when you completed the circuit, so the others would remain discharged. Dec 17 '15 at 0:04
• Charge is the integral of current and current is the same on all of them. Jan 11 '16 at 10:35

Charge cannot be created or destroyed. Since you only have one possible current path through all the capacitors (and current is just flowing charge) the charge on all 3 capacitors has to be the same. The capacitance of the capacitor indicates how much voltage a particular amount of charge corresponds to Q/C = V. Put more charge into a cap, get a bigger voltage difference. Put the same charge in a smaller cap, get a bigger voltage difference. So what happens in your circuit is that the charge is distributed evenly, but the applied voltage is distributed according to the capacitor sizes, with the smallest cap ending up with the largest fraction of the applied voltage.

• Another observation would be that the number of electrons flowing into one plate must be very close to the number of electrons that flow out of the other. It's possible for a capacitor--like almost any other object--to have a net positive or negative charge relative to its environment, but the numbers of electrons involved are tiny compared with the number that must flow through a cap to build up a significant relative voltage. Feb 14 '17 at 20:31

There is no particular reason (except for "practicality") that the capacitors do have equal charge. There is an unstated assumption/convention in such examples that the circuit can be treated as if it started as a zero-volt source connected to capacitors which all have zero charge. Once you realize this, it's clear that this assumption can be violated and a number of capacitors with different charges can be assembled into the final circuit.

Part of the definition of an ideal capacitor is that its' resistance is infinite. As a result, once charge is placed on the two sides of an ideal capacitor there is no path which would allow for changes in the charge, except for the leads. In the normal case, this means that if charge flows out one lead it must flow into the lead of another capacitor (the voltage source obeys KCL) so all the capacitors must have equal charge.

In the non-ideal case, of course, this does not apply. Two capacitors in series can be considered as 3 plates. The two outer plates will have equal charge, but the inner plate will have charge equal to the sum of the two outer plates.

For various practical reasons, you would probably want resistors in parallel to help balance the DC charge on the capacitors.

But the theory of your question can be explained by imagining each of the capacitors as a rubber membrane. If you pull on one side, that creates or displaces volume on the other side that needs to go somewhere.

Physically, when charge is applied to one side of a capacitor, opposite charge is attracted to the opposing plate. This charge needs to come from somewhere, and in your diagram it comes from the next capacitor...which in turn charges that capacitor's opposing plate... which must come from somewhere... etc, until the loop is closed around the loop.

I see that you've drawn this with a DC voltage rather than an AC voltage though.. Note that this explanation only makes sense in a dynamic situation: either a turn on or turn off transient, or a constant AC input voltage.

In a pure DC environment the leakage current will act like resistors and balance the charge according to leakage current which is a completely different mechanism.

• adding parallel resistors doesn't balance the charge on the capacitors, it balances the voltage by creating an unbalanced charge in each capacitor. Dec 17 '15 at 1:02
• @helloworld922 yes, that is more technically correct. Thanks for pointing that out. The purpose is to avoid exceeding the breakdown voltage at those undefined nodes in between the caps. Dec 17 '15 at 7:29

Simple answer: same current flows through all capacitors for the same amount of time (once fully charged, no more current flow). Q = I x t is the same since each dependent variable is the same for each cap

The charge on a capacitor is the integral of the current: $$Q(t) = \int_0^t I(t) dt + Q(0)$$ A lot of the time, you're working with capacitors that were originally uncharged, so $$\Q(0) = 0\$$.

Then, your three capacitors are in series. Kirchoff says that they must all have the same current, so they must all have the same charge, too!

Note that the voltage across the capacitors is $$\V = Q/C\$$, so the larger capacitors will have smaller voltages across them and the smaller capacitors will have larger voltages. This intuitively makes sense - big capacitors can store a lot of energy without a large increase in voltage.

Allow a circuit with a supply voltage, E, applies across a series combination of $N$ capacitors. Further, as described above by WhatRoughBeasWhatRoughBeast, make no assumption that the charge across the the capacitors are identicle (i.e., in general, $Q_k \neq Q_l$, where $k$ and $l$ refer to the $k^{th}$ and $l^{th}$, respectively.

Allow by Kirchhoff's current law that the current in the series are identically given by I. Thus, according to the relation between the current and the flow of charge through time. \begin{align} I_k & = I_l \\ \dfrac{\Delta Q_{C,k}}{\Delta t} & = \dfrac{\Delta Q_{C,l}}{\Delta t} \\ \end{align} Thus $$\Delta Q_{C,k} = \Delta Q_{C,l} = \Delta Q \quad (*)$$

Considering that the voltage across each capacitor is given by $$V_k = \dfrac{Q_k}{C_k}$$ and that a change in voltage, $\Delta V_k$, induces a change in charge, $\Delta Q_k$, as $$\Delta V_k = \dfrac{\Delta Q_k}{C_k}, \quad (**)$$ we aply Kirchhoff's voltage law \begin{align} 0 & =- E + \sum_{k=1}^N {V_{c,k} } \\ E & = \sum_{k=1}^N {\dfrac{Q_{c,1}}{C_1} } \end{align} and \begin{align} \Delta E & = \sum_{k=1}^N {\dfrac{\Delta Q_{c,k}}{C_k} } \end{align} Further, from Eq. (*) \begin{align} \Delta E & = \Delta Q \, \sum_{k=1}^N {\dfrac{1}{C_k} } \\ \Delta E & = \dfrac{\Delta Q}{\dfrac{1}{\sum_{k=1}^N {\dfrac{1}{C_k} } }} \end{align} We can compare the above equation to the form of Eq (**) and write the effective capacitance, $C_{eff}$, of the series circuit as \begin{align} C_{eff} & = \dfrac{1}{ \sum_{k=1}^N {\dfrac{ 1}{C_k}} } \end{align}

In conclusion, for a network of capacitors in series, one can derive the well known equation for the effective capacitance without the need to state that the charge across each capacitance is equal.

Allow a circuit with a supply voltage, $E$, applied across a series combination of $N$ capacitors. Further, as described above by WhatRoughBeast, do not make any assumption that the charge on the plates of the $k^{th}$ capacitor is equal to the charge on the plates of the $l^{th}$ capacitor (i.e., in general, $Q_k \neq Q_l$).

Allow by Kirchhoff's current law that the current in the series circuit is equal everywhere in the series circuit. Thus, according to the relation between the current (i.e., $I_k, I_l$) and the change in charge (i.e., $\Delta{Q_k}, \Delta{Q_l}$) in the time frame $\Delta{t}$ \begin{align} I_k & = I_l \\ \dfrac{\Delta Q_{C,k}}{\Delta t} & = \dfrac{\Delta Q_{C,l}}{\Delta t} \\ \end{align}

Thus $$\Delta Q_{C,k} = \Delta Q_{C,l} = \Delta Q \quad \textrm{Eq. (*)}$$

Considering that the voltage across each capacitor is given by $$V_k = \dfrac{Q_k}{C_k}$$ and that a change in voltage, $\Delta V_k$, induces a change in charge, $\Delta Q_k$, as $$\Delta V_k = \dfrac{\Delta Q_k}{C_k}, \quad \textrm{Eq. (**)}$$ we apply Kirchhoff's voltage law \begin{align} 0 & =- E + \sum_{k=1}^N {V_{c,k} } \\ E & = \sum_{k=1}^N {\dfrac{Q_{c,1}}{C_1} } \end{align} and now for a change in the source voltage, $\Delta E$, we write \begin{align} \Delta E & = \sum_{k=1}^N {\dfrac{\Delta Q_{c,k}}{C_k} } \end{align} Further, from Eq. (*) \begin{align} \Delta E & = \Delta Q \, \sum_{k=1}^N {\dfrac{1}{C_k} } \\ \Delta E & = \dfrac{\Delta Q}{\dfrac{1}{\sum_{k=1}^N {\dfrac{1}{C_k} } }} \end{align} We can compare the above equation to the form of Eq (**) and write the effective capacitance, $C_{eff}$, of the series circuit as \begin{align} C_{eff} & = \dfrac{1}{ \sum_{k=1}^N {\dfrac{ 1}{C_k}} } \end{align}

In conclusion, for a network of capacitors in series, one can derive the well known equation for the effective capacitance without the need to state that the charge across each capacitance is equal.