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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?

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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 $\frac{Q}{3}$.

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    $\begingroup$ Can you please explain why the potential must be equal? $\endgroup$ – Allen Mar 12 '17 at 8:25
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    $\begingroup$ 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 '17 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 '20 at 1:54
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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.

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  • $\begingroup$ 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 '17 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. $\endgroup$ – SchrodingersCat Mar 12 '17 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. $\endgroup$ – SchrodingersCat Mar 12 '17 at 8:58
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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.

Thanks!

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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.

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

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