When the capacitors arrange as below,

enter image description here

In the parallel arrangement, when we considering the net charge of the system,we take it as Q1+Q2. But in the series arrangement, The net charge is get as Q1.

I can't understand why is it not 2Q1. Why don't we consider the both charges when consider the net charge?

I searched this in google but I couldn't find a satisfied answer. I hope an answer without mathematical description. Thanks

  • $\begingroup$ You can assume that positive part of emf draws electron ($Q$) from right plate of $C_2$ which produces positive charge on that plate . Through induction , the left plate of $C_2$ becomes negatively charged .By same amount .... Similar happens for $C_1$ ... $\endgroup$ Commented May 4, 2018 at 7:58

2 Answers 2


As an example assume that the two capacitors are identical.

enter image description here

So in this case bring the capacitors very close together so that there is one central plate (middle diagram) and then remove that middle plate to get a capacitor with plate separation twice what it was before and with charge $Q$.


I'd like to add to the very nice and simple answer from @Farcher by giving a description and way of imagining the electrons moving step-by-step in deeper detail.

Think of the battery as giving a certain "push" (by means of its electromotive force / voltage).

  • For just one capacitor, the battery will "push" electrons to the plate. But as more and more electrons accumulate here, their total repulsion force grows. They repel new incoming electrons more and more, slowing down the current gradually. Soon the repulsion balances out the "push", the current becomes zero, and the maximum charge accumulation has been reached (stored).

We call that total charge $-Q$. This in turn induces the same but opposite charge on the opposite plate, $+Q$.

  • For two capacitors in parallel, the battery "push" is the same. The "pushed" electrons reach the T-split and must choose which plate to go towards. Electrons will always move to the plate from which they are repelled the least (the plate of lowest potential). So if one electrons picks the upper plate, then the next electron will pick the lower plate, roughly spoken. The repulsion from each plate now only corresponds to having one electron stored, although we in total have two stored.

In this way, parallel capacitors are sharing the charge amount. The same battery "push" will be able to push double as many electrons before the repulsion from each plate balances out the "push", because each plate only gets half. Double as much charge can be stored in this configuration for the same "push".

  • For two capacitors in series, we are back to something similar to the one-capacitor scenario. The battery is again "pushing" towards one plate only, with no other plates for the charge to spread over. The repulsion on the first plate will therefore accumulate in a similar manner as for just one capacitor.

The only difference in the series scenario is that some reconfiguration happens in between the two capacitors where each plate induced an equal but opposite charge on the next.

Now, the narrower the gap between the plates of a capacitor, the bigger is the attraction between the two oppositely charged plates. This attraction "helps" out the battery "push" and allows for a bit more charge to be stored. That is why you see the separation $d$ in the formula for capacitance:

$$C=\varepsilon_0 \varepsilon \frac Ad$$

The bigger the gap, the less "help" and the smaller is the charge that can be stored (the smaller is the capacitance $C$ that indicate the storing capability). With capacitors in series, some extra gaps are introduced along the same line. The overall "help" from the plate-to-plate attraction therefore decreases. The charge pushed to the first plate is there "helped" less by this effect and is thus a bit smaller.

Having two equal capacitors in series, therefore corresponds to having a capacitor with a doubled gap - as Farcher's answer clearly shows. In other words, two in series are worse (lower total capacitance) than one alone.

You also see the plate area $A$ in the formula; the larger the area, the more charge can be stored there and can spread out before the repulsion grows too big. Area thus increases the capacitance by allowing for more charge to be stored for the same push. The capacitors in parallel increases the total amount of charge that can be stored as well, and two parallel capacitors thus correspond to a capacitor of double the area.

The formulae for total capacitances in parallel and series are the following, with the total parallel capacitance being larger than either of the individual ones, while the total series capacitance is smaller than either of them:

$$\begin{align} \text{In parallel:}\qquad C_{total}&=C_1+C_2+C_3+\cdots \\ ~&~\\ ~&~\\ \text{In series:}\qquad \frac 1{C_{total}}&=\frac 1{C_1}+\frac 1{C_2}+\frac 1{C_3}+\cdots \quad\Leftrightarrow \\ C_{total}&=\frac 1{\frac 1{C_1}+\frac 1{C_2}+\frac 1{C_3}+\cdots} \end{align}$$

  • 1
    $\begingroup$ Thank you very much for spending your time for give such successful answer. This is not only an answer, this will be a good lesson for every one who are studying about capacitors. All theories of capacitors can be covered by this description. $\endgroup$ Commented May 5, 2018 at 0:01
  • 1
    $\begingroup$ Thank you, @OsalThuduwage, very kind words. I am glad it helped clear out the doubt. $\endgroup$
    – Steeven
    Commented May 5, 2018 at 7:58

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