Okay maybe a repeated one. I am asking this because none of the other explains how charges redistribute.

Okay here's the problem.

Suppose I have a $4mF$ capacitor and a $2mF$ capacitor. I charge the 4mF capacitor to a charge $Q$, then remove the battery. Now I connect the uncharged $2mF$ capacitor to the charged $4mF$ one. How does the charges on both the capacitor change?

My argument is this:

The charges on the 4mF capacitor will be equally divided between the 2mF and 4mF, that is Q/2 on each of them. Because if we take one plate of each capacitor and connect them, then the charge will be equally divided. That is if we take the +Q charged plate and connect it to one uncharged plate of the 2mF, then each one will have +Q/2 charge. Similarly taking the negatively charged plate we get -Q/2 on each of the plates. Now if we rejoin them we will have two capacitors each charged to Q/2.

But, obviously my argument is wrong and please explain why?


5 Answers 5


The potential across either capacitor must be equal. The potential difference of the original capacitor is $\frac{Q}{4*10^{-3}}$V Say the 4mF capacitor loses a charge $x$ which is gained by the 2mF capacitor. Then, $$\frac{Q-x}{4*10^{-3}}=\frac{x}{2*10^{-3}}$$

Solving for $x$ we get $x=\frac{Q}{3}$. So the final charge on the 4mF capacitor is $\frac{2Q}{3}$ and on the 2mF capacitor it's $\frac{Q}{3}$.

  • 2
    $\begingroup$ Can you please explain why the potential must be equal? $\endgroup$
    – Allen
    Mar 12, 2017 at 8:25
  • 4
    $\begingroup$ Because of Kirchoff's voltage law. In a circuit that comprises only of wires and two capacitors, the voltage across either must be the same. Otherwise there will be a net voltage and charge will continue to flow (ie-there will be a current) until the voltages become equal. $\endgroup$
    – GeeJay
    Mar 12, 2017 at 9:11
  • $\begingroup$ If the potentials were not equal, then charge would continue to flow until the potentials were equal. (Or, in theory, the charges could flow back in forth in an oscillation as there is not resistance to remove energy from the system.) $\endgroup$
    – Jiminion
    Jul 29, 2020 at 1:54
  • $\begingroup$ This maybe a trivial question, Is connecting a charged capacitor to an uncharged capacitor in series the same as connecting them in parallel? $\endgroup$
    – Toxin D
    Jul 3, 2021 at 18:49

The question you asked has a straightforward answer: the potential difference across the two capacitors must be equal, so the charges distribute in inverse proportion to the capacitances.

$$V_1 = V_2 \implies \frac{Q_1}{C_1} = \frac{Q_2}{C_2}$$ where $Q_1 + Q_2$ is your total charge.

I wanted to highlight a more subtle point though, concerning the energy of the system.

Consider two capacitors of the same size, with one initially charged to $V_i$, and the other initially uncharged. They are connected end to end via a switch (see diagram).

From what has been said, we expect the charge, $Q = CV_i$ on the first capacitor to distribute itself evenly between the two capacitors because they have the same capacitance.

Therefore the final voltage over each capacitor will be the same and correspond to holding a charge of $\frac{1}{2}Q$. This is $V = Q/C = \frac{1}{2}Q/C = \frac{1}{2}V_i$.

But what has happened to the energy? It was initially just $\frac{1}{2}V^2/C$ but it is now $2 \times \frac{1}{2}(V/2)^2/C = \frac{1}{4}V^2/C$.

So the energy has halved. What? Where has the other half gone? There is no resistance, so no energy can be dissipated, yet it is clearly halved.

This problem might trip you up and indeed there is a Wikipedia page titled "Two capacitor paradox", that explains this in detail.

Source: https://en.wikipedia.org/wiki/Two_capacitor_paradox

The solution is that, in reality, our circuit is not ideal. We have assumed the wires in the circuit has zero resistance and inductance. If this really were the case, the current would be infinite when the switch was closed due to a potential difference with zero resistance!

If we were to just introduce resistance, we would see that the current in the circuit when the switch is flipped would decay exponentially to reach the steady state solution we found above (equal charge distribution. This is an RC circuit. In the process half the initial energy would be lost.

And if we introduced inductance, then we would have an LC circuit and thus perpetual oscillation of charge between the two capacitors. Here no energy would be lost - when the charge is distributed evenly between the two capacitors, the remaining half of the energy will be stored in the magnetic field of the inductor.

In reality there would be both inductance and resistance in the wires, so we would have an RLC circuit. In this case the steady state would eventually be reached again, but instead of exponentially decaying, as in an RC circuit, the current would take the form of an exponentially decaying sinusoidal wave (well depending on the relative values of R and L, it might just decay without any oscillation anyway, in the so-called "over damped" case).


Your answer is wrong in the sense that the charge will not be equally divided between the plates of the two capacitors. And the reason is that the capacitors have different capacitances. In this case you need to apply Kirchoff's voltage rule to get the correct equation. If $Q$ is the total charge on the charged capacitor before discharging starts and $Q_0$ is the charge on the uncharged capacitor, then by Kirchoff's voltage rule, we have $$\frac{Q-Q_0}{C}+\frac{Q_0}{C_0}=0$$

Now solve this.

Hope this helps you.

  • $\begingroup$ But won't the charge be equally divided if if were just two metal plates and not capacitors connected together? Also please say what do you have against my argument. Can you also please remove the other answer you have accidentally posted $\endgroup$
    – Allen
    Mar 12, 2017 at 8:21
  • $\begingroup$ @Allen Yes it would if it were two metal plates only and not 2 capacitors. Read a little more about capacitors to know what actually makes them different from just 2 metal plates. I have mentioned in my first line what I have against your argument. Your arguments starts with the declaration that the charges will be equally divided without a logical reasoning and my answer clearly states that the charges will not be equally divided but that the charges will be divided as per KVL rule. Or in other words, the potential across each capacitor must be equal and the sum of the potentials must be 0. $\endgroup$ Mar 12, 2017 at 8:54
  • $\begingroup$ @Allen (Contd.) The other answer was mistakenly posted. And I am on the app right now. I will remove it as soon as I get to my computer. $\endgroup$ Mar 12, 2017 at 8:58

Great question!

  • If it were just 2 metal plates, charges would be equally distributed.
  • But in a capacitor arrangement, the NET electric field between the plates (due to both plates) contribute to the voltage across them.(Δ V = − ∫ E ⋅ dl).
  • When the two are connected, KVL tells us that the sum of voltages across any closed loop must be zero. Therefore charges must redistribute such that voltage is same.



Look my friend,first of all during charging of capacitor ,CHARGED AND UNCHARGED CAPACITOR ARE NOT IN CONTACT.SO EQUAL CHARGE IS NOT TRANSFERRED.Now process of charging of capacitor.From definition, how did you define a capacitor,two plates separated by an insulating material like air.Now I an going to tell you how actually charging of capacitor takes place.

Take a charged capacitor like positive charged and one uncharged capacitor and don't touch them.now induction happens.The end nearer to positive plate will have negative and other have positive polarity.Now you will know that negative charges will reduce the potential of charged plate and positive positive charges will increase its potential. So if we earth or ground the positive charge on inducted plate.Now there is only negative charges.So it will decrease the potential at charged plate all time when its potential becomes maximum due to positive charges.So all time it charges and discharges.

FOR BETTER UNDERSTANDING,Go to YouTube and see examfear capacitance part 6.Watch this and you will understand what I am saying.

Hope it helps.

  • 2
    $\begingroup$ @Shobit Swami Did you really read my question? $\endgroup$
    – Allen
    Mar 12, 2017 at 8:24
  • $\begingroup$ @Allen I explain the process $\endgroup$ Mar 12, 2017 at 14:35
  • $\begingroup$ @Shobit Swami Frankly that wasn't the question. $\endgroup$
    – Allen
    Mar 12, 2017 at 17:53
  • $\begingroup$ Sorry for that it was not my intention $\endgroup$ Mar 14, 2017 at 5:59

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