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Suppose I have two capacitors of different capacitance $C_1$, $C_2$and they have been charged prior connecting by voltage of $V_1$ and $V_2$ respectively the positive plate of one capacitor is connected to negative plate of the other.The other plates are not connected , we thus have an incomplete loop.

Had it been a complete loop , I would have easily redistributed charges . The two plates which have been connected are at different potentials , I'm interested to know what happens next , what changes will occur in the wire and all the plates , will a change be even there or not ? If there is no change then what happens to free electrons in wire why won't they move provided the wire's end are at different potentials ?

Please help , I'm unable to think around this. enter image description here

Here is what I think I going to happen , I've taken a case where capacitors are charged with 3 microcoulomb and 2 microcoulomb and then their plates are joined as shown , the system is now 3conductors - leftmost plate , the two inner plates connected taken as whole and rightmost plate and therefore charges in beginning and end on them will be same but there will be redistribution as shown in my diagram which will ensure a system of lowest energy , because obviously the two inner plates won't just stand when there are different charges on them and they are connected .

if it suites you then assume that both capacitors were identical .

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The two plates which have been connected are at different potentials , I'm interested to know what happens next , what changes will occur in the wire and all the plates , will a change be even there or not ?

Once you disconnect the individual capacitors from their respective voltage sources they no longer have a well-defined potential. They are "floating". The potential difference is fixed by the original voltage sources, but the absolute potential is not (absolute potential is arbitrary anyway).

So what happens is that when the two wires are connected then the potentials of the two connected plates becomes the same. There is no redistribution of charges (neglecting any small parasitic or self capacitance of the wire) because there is no need for it. If you ground any of the three points then the voltages are all well-defined, but until then the voltages are free to float as needed, with only the voltage difference across the plates being fixed. There is no need for them to redistribute charges to become equipotential.

EDIT (to address new content in the question):

The thing that you have labeled “potential difference” does not exist. There is a potential difference across each capacitor established by the sources, but when they are disconnected from the sources then the voltages are floating. When the wires touch they are already equipotential, having both “floated” to the ambient potential.

there will be redistribution as shown in my diagram which will ensure a system of lowest energy

Any redistribution will raise the energy of the system. With the charges on each capacitor as they originally are the E field is concentrated in the small volume between the plates. The field lines leaving the positive plate are very short and terminate on the negative plate. As you move charges away from the plate some of the field lines will become much longer. You will therefore increase the volume occupied by the field and thereby increase the energy.

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  • $\begingroup$ how can the potentials be arbitrary , I read this same reasoning elsewhere too but I'm not sure how far I can agree to it. Even after I have unplugged my capacitors from their original circuits , they have been fully charged and have a concrete , well defined charge on either of their plates , so how can the potential with respect to infinity be arbitrary ? $\endgroup$ – ADITYA PRAKASH May 24 at 13:39
  • $\begingroup$ I'm more or less a noobie so I'm seeing this from a electrostatic point of view which I just completed . since the plates have a well defined charge , therefore they must have a well defined potential with respect to infinity and a negative plate of whatever charge will always be lower in potential than a positive plate , so I'm seeing a potential difference here - well defined. $\endgroup$ – ADITYA PRAKASH May 24 at 13:44
  • $\begingroup$ given the potential difference I'm highly unconvinced why even the slightest of charge cannot redistribute (neglect stray capacitance) $\endgroup$ – ADITYA PRAKASH May 24 at 13:46
  • $\begingroup$ One plate has a well defined potential relative to the other plate. Neither plate has a well defined potential relative to infinity (ground). Moving the disconnected (but charged) capacitor can do miniscule amounts of work making the potential relative to ground vary in an undefined manner. Suppose you were to measure the voltage on one plate relative to ground, the voltmeter itself is a grounding path and you would get 0V. Then if you disconnect that plate and measure the other you would again get 0V. This is not because both plates have the same voltage, but because the voltage is floating. $\endgroup$ – Dale May 24 at 13:52
  • $\begingroup$ I don't get it at all , as I said I'm seeing from purely electrostatic point of view .. suppose I have a charge Q and I bring a test charge from infinity , now you're saying the potential w.r.t infinity of Q is arbitrary and floating , doesn't make sense to me what is it that I'm missing $\endgroup$ – ADITYA PRAKASH May 24 at 14:07
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When you remove the battery there will be nowhere for the charge at the lead ends to go, so it will stay the same. As there is no redistribution on the lead ends, by symmetry there will be no redistribution internally.

You can also thinking of it as the capacitors charging up while current is flowing. Once equilibrium is reached the current has stopped and the voltage across the capacitor matches that of the battery. With the removal of the battery, nothing has changed from this equilibrium state.

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  • $\begingroup$ this description comes when both capacitors are in series and one voltage source is connected , here I had charged my two different capacitors in two different circuits , disconnected them and then hooked them as described in my question $\endgroup$ – ADITYA PRAKASH May 24 at 11:36
  • $\begingroup$ If you charge them separately then they will likely have a different charge and thus there will be some redistribution in that case. $\endgroup$ – Paul Childs May 24 at 11:50
  • $\begingroup$ Exactly this is what is confusing me I read an answer where they said i wont redistribute. physics.stackexchange.com/questions/33699/… $\endgroup$ – ADITYA PRAKASH May 24 at 11:59
  • $\begingroup$ @Paul Child’s Not necessarily. Prior to connecting them together each has a charge of CV if the products voltage times capacitance is the same for each then they have the same charge on each. Assuming that were the case how would you answer the question? $\endgroup$ – Bob D May 24 at 12:04
  • $\begingroup$ Ah but that one is a special case where the capacitors have the same charge. In that case there will be no redistribution as equilibrium has been artifically created. $\endgroup$ – Paul Childs May 24 at 12:06
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The charges on the unconnected plates have no where to go. The key point is the charges on the other plate of each capacitor are bound to those plates by the electrostatic force of the unconnected plates.

Bottom line: In my opinion, nothing happens. You simply wind up with two capacitors in series with a voltage of $V_{1}$ on one and $V_{2}$ on the other. Total charge remains the same (conservation of charge). It doesn't matter if the voltages were originally equal or not, or if the original charge on each were equal or not. I am amenable to change my opinion given sound technical arguments to the contrary.

I would have to add, however, that if after connecting the plates together you then connect the open plates together, than charge can redistribute themselves causing the voltages to change (one capacitor charging another). Because for a complete circuit with capacitors in series, the charge on each capacitor will be the same.

Hope this helps.

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  • $\begingroup$ and what happens to the fact that the wire has been connected to plates having different potential plates , there is an electric field on the free electrons of wire as a result , would they just sit inside wire even if they are being forced by potential difference , I'm just shifting my attention to inner plates for a while , other things are nonexistent for a while and I see plates of different potential connected , wont there be redistribution of some sort ? $\endgroup$ – ADITYA PRAKASH May 24 at 12:33
  • $\begingroup$ @ADITYAPRAKASH See my edit that I made before seeing your comment. If the open plates are connected together then the charges on both the outer plates and inner plates are no longer constrained. But if they are not connected, the charges on the inner plates are constrained by the forces exerted by the outer plates. $\endgroup$ – Bob D May 24 at 12:37
  • $\begingroup$ I have no problem with a complete circuit , I can understand that .. so should I conclude that there will be no change even if the wire is connected to plates of different potential ? $\endgroup$ – ADITYA PRAKASH May 24 at 12:41
  • $\begingroup$ @ADITYAPRAKASH I think I know what you are thinking. But it depends on what you mean by "different potential". Consider two resistors in series carrying current. There is a voltage drop (difference in potential) across each. In the direction of the current flow the polarities of the voltages at the resistor terminals are + - + -. There is difference in potential between the connection between the resistors and the other sides of the resistors, but the connection point is at the same potential even though the polarities are different. $\endgroup$ – Bob D May 24 at 12:53

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