The discussion revolves around a problem involving two capacitors with different capacitances and initial voltages, specifically focusing on the effects of connecting their terminals in opposition. The subject area is primarily related to electrical circuits and capacitor behavior.
There is an ongoing exploration of the conditions for equilibrium and conservation of charge. Some participants have provided insights into the nature of charge flow and the relationship between the voltages across the capacitors, while others are seeking clarification on the concepts involved.
Participants note discrepancies between their calculations and expected outcomes, indicating potential misunderstandings about charge conservation and the behavior of capacitors in this configuration.
What can you say about the final voltages across the capacitors? What can you say about the polarities of the final voltages?carlyn medona said:No I mean initial charge, before connection, then I thought charge would flow between the plates till potentials become equal, but answer I got does not match with solution
Can you share more detail about what you actually did for this attempt? What were your thoughts about how the charge would flow? What calculations did you attempt and what were your results?carlyn medona said:No I mean initial charge, before connection, then I thought charge would flow between the plates till potentials become equal, but answer I got does not match with solution
carlyn medona said:No I mean initial charge, before connection, then I thought charge would flow between the plates till potentials become equal, but answer I got does not match with solution
That's correct.carlyn medona said:Okay guys I solved it, thanks for helping, I got charge on first capacitor = 21.8μC and second capacitor to be equal to 26.2μC
Ok I suppose its correct (seems to be an equilibrium state since the new voltages are equal), but someone explains me please why conservation of charge doesn't hold here. Charge before =120+72=192μC, charge after 21.8+26.2=48μC.cnh1995 said:That's correct.
Conservation of charge holds.Delta² said:Ok I suppose its correct, but someone explains me please why conservation of charge doesn't hold here. Charge before =120+72=192μC, charge after 21.8+26.2=48μC.
Sorry but you got me confused. Only definite conclusion I can make from your statements is that net (positive+negative) charge is zero before and is zero after in both capacitors. Ok in that sense, conservation of charge holds.gneill said:Conservation of charge holds.
When the plates are connected, one of the paired plates has a positive charge while the other a negative charge. This is true for both of the pairs that are connected. Unlike charges will meet and annihilate via the available conduction paths leaving the net sum the same, hence conservation of the net charge.
Note that (usually) on a charged capacitor the two plates carry equal but opposite charges. Hence the net charge on the device as a whole is zero. When we talk about the charge on a capacitor, we typically refer to the magnitude of the charge on one plate, since both plates have the same charge but of opposite sign. When charged capacitors are connected in parallel, the charges on the connected plates sum. If the charge signs are opposite, they will combine and cancel until no differently signed charges are left. Whatever un-combined charges are left represents the net remaining charge on that plate.
Right.Delta² said:Ok I see now conservation of charge holds for the net charge of the plates that are connected (don't know if you actually was trying to say that). Since we connect the positive to negative and vice versa, charge before=120-72=48 (or 72-120=-48), charge after=48 (or -48 if we sum the charges on the other pair of connected plates that carry negative charges) as well.