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What happens if two capacitors having different charges is connected in series?

I see three possible scenarios

  1. The capacitors are connected in series using wires without resistance.
  2. The capacitors are connected in series using a resistor.
  3. The capacitors are connected in series with a cell

(there may be more scenarios but this is all I can think off) Here is what I think will happen for the three cases

  1. I learnt that when the terminals of a single capacitor are connected in series using wires without resistance, the charge oscillates between the two plates in a Simple Harmonic Motion. I believe the same is going to happen here too.

  2. I believe in this case the charge on each plate becomes zero because I learnt that in the case of 2 plates of a single capacitor being joined, the charge on each plate becomes $0$.

  3. This one I got no clue. I believe the two capacitors form a single mega-capacitor with capacitance $C$. Since the potential across two plates of a single capacitor attached to a cell is equal to the emf of the cell, I am going to apply the same logic here too. [$Q=\frac {C}V$] Also, Q is the same for each of the plate

Also, do the scenarios depend on which of the terminals we are connecting? Is connecting the positive terminals of two capacitors different from connecting the positive terminal to the negative terminal of two different capacitors? I am really hoping it isnt.

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  • $\begingroup$ Why charges oscillate in the first case? $\endgroup$ Oct 1, 2021 at 16:58
  • $\begingroup$ @SamyakMarathe Let's assume there's 5J of Potential energy due to excess charge on one of the plates. Usually, PE gets converted into KE or heat. But here the 5J goes completely to the other plate (assuming there's no resistance, else joules law comes in), but now there's an excess 5J on the other plate which continues to oscilate. $\endgroup$ Oct 1, 2021 at 17:36
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    $\begingroup$ It's not clear if your three scenarios involve complete circuits. $\endgroup$
    – Bob D
    Oct 1, 2021 at 18:29
  • $\begingroup$ @AdilMohammed but the charge will flow until the potential is same. After some time the potential will be equal and both plates will be neutral. $\endgroup$ Oct 2, 2021 at 13:40
  • $\begingroup$ @Bob D yes that's why I explicitly mentioned the three cases. They were scenarios in which you could make a complete circuit $\endgroup$ Oct 2, 2021 at 16:33

2 Answers 2

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Resistors dissipate energy but not charge. If two charged plates are connected together, the total charge on the two does not change. If the wires (and plates) have no resistance, then connecting two charged capacitors can result in LC oscillations (depending on voltages and polarities). If two charged capacitors are connected together (with resistance) they will come to the same voltage. If they are connected in series to a cell, charge can flow from the terminals of the cell until the sum of the two voltages is equal to that of the cell; but the sum of the charges on the two connected plates must remain the same.

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  • $\begingroup$ Oh I meant would the charge on the plates of the capacitor change? $\endgroup$ Oct 2, 2021 at 17:02
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I will deal with case 1 after dealing with cases 2 and 3.

Cases 2 and 3 are essentially the same with case 2 having a voltage source with no output.
I think that the easiest way to illustrate what might happen is to do a numerical example which is shown below.

enter image description here

The initial state was two capacitors, $4\,\rm F$ with charge $8\,\rm C$ and $2\,\rm F$ with charge $4\,\rm C$, connected in a circuit as shown below.

In the final state the charges of the capacitors are $Q$ and $q$.

Charge conservation dictates that $+8 +(-4) = 4 = -Q +q$ and Kirchhoff's voltage law gives $E = \dfrac Q 4 +\dfrac q2$.

Solving for $Q$ and $q$ for case 2, $E = 0\,\rm V$, the final state is as follows,

enter image description here

For case 3 with $E = 6\,\rm V$ the final state is as follows,

enter image description here

If you work out the energies stored in case 2 at the start and at the end you will find that there is a decrease in energy stored.
This is because to get from the initial to the final state some charge must have moved and if it moves through a resistor some of the stored energy is dissipated as heat.
There is therefore a problem if $R=0$ (case 1) as the circuit has no was of dissipating the energy.
A way forward is to look at what happens if $R\to 0$.

In all if the three cases it is in fact an capacitor, resistor and inductor (look at the loop in the diagram) series circuit but for most examples the inductance of the circuit is ignored and the circuit can be considered as heavily damped so the transition from initial state to final state follows an exponential function; remember what happens with a discharging $CR$ circuit.

If $R$ becomes low enough the circuit is now under-damped and the passage from the initial to the final state is via a decaying sinusoidal (ie oscillating) function with the rate of decay decreasing as the resistance in the circuit is decreased. So for no resistance in the circuit one gets non decaying oscillatory behaviour.

That is not the end of the story because unbound accelerating charges emit electromagnetic radiation and when the behaviour is oscillatory the system loses energy vis the emission of electromagnetic radiation.
Further information about this can be gained by reading Kirk T. McDonald's paper A Capacitor Paradox.

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