I thought my life is a lie when I solved this exactly like you. Cuz the charges on capacitors in parallel are not same at the end. Then I googled and saw u r video and realised one imp. Point that capacitirs in series only have same charge if their intital charge is same. Thanks da for not making my life a lie😘😘😘 Love you so much❤❤❤
+Ankit Giri Yeah, the potential difference between C2 and C1 threw me off at first too. However, we can always count on the Kirchoff Voltage Law (a.k.a. loop rule) - that the voltage differences around a loop add up to zero. With the capacitors in series, we can also count on the idea that the charges always increase or decrease by the same amount. It can help to think of this as one charge at a time, or that the current is the same everywhere in this loop - because the elements are in series. You can't have more electrons in part of the loop because there are no junctions where extra electrons could come from.
Another way to look at this problem perhaps? Applying Kirchoff's Voltage Law, { (emf-battery) - dV(2-initial) = dV(1-final) + dV(2-final) } Ultimately you end with the same algebraic equation for Q(1), except you have to fix Q(1) as equal to Q(2). This, I feel sits better, because in a series connection, flow of current always tries to keep the charges of the connected plates equal, lest electric field builds up to neutralize that anyway... In essence however, the initial dV(2) never disappears, so if one asks the charge on C2, you would have to add the charges due to both the potential differences. Is this logic flawed? Anyway, thanks for a concise, lucid video!
Thanks a lot! was looking for the solution to this problem for many days..really nice. just one doubt by the way, there is already a potential diff between c2 and c1 in the beginning, so that wont contribute to more electrons travelling to c1?
Hi RPT,Thanks to you, very interesting presentation indeed,would you help me please, I am investigating the probability to reach resonance between 2 capacitors in series,if treated in differential equations, or exponential solutions, my preliminary conclusion says that if the current of one CAP respect the other is opposed in phase, they can resonate in steady state,natural response for i and v,of course amp-op stabilized if suitable.?????johann wegmann, many thanks in advance.For overunity perhaps,one CAP opposed to a solids flywheel,may run for ever?I mean a Supercap graphene.
This is a question I'd typically reflect back to my student. ;-) The loop rule will still hold, and the fundamental way in which electrons move from one plate to another will also still be the same but the equation Q2 = Q1 + 300 microCoulombs will be different.
Yeah this threw me off at first too. It's tempting to think that capacitors in series have the same charge, but this example proves that we can't always assume that. Instead, capacitors in series build/lose the same amount of charge. So if they start out with different charges they can end up with different charges.
+Aman Sharma Neat! A complicated variation. If I understand your question, it would be pretty much the same except that C2 would be flipped over before being connected. Then C2 would initially have a charge of +300 micro Coulombs on its bottom plate. An electron taken from the left plate of C1 would then be deposited onto the (initially positive) bottom plate of C2, making the bottom plate of C2 less positive. After 1 micro Coulomb is transferred, Q1 would be 1 micro Coulomb and Q2 would be 299 micro Coulombs. This changes how Q1 and Q2 are related - instead of Q2 = Q1 + 300, it'll be something else (and you should give it a try :-). Electrons will continue to flow until Kirchoff's Voltage Law is satisfied, that is when the voltage across C2 and C1 is the same as the emf.
3:04 Interesting, but I disagree. Indeed, consider that there is no emf source. None the less, C2 will now see C1 and may be able to do a redistribution of the "charges" so that, clearly, Q2-Q1 may become something else than 300 uC. The assertion seems without strong foundation.
Yeah, if we don't have the emf then the charges would redistribute differently than in this case. Where would the electrons go though? Once the switch is closed, each electron that leaves one capacitor has to end up at the other capacitor (assuming the only change is to remove the emf). There's nowhere else for the electrons to accumulate. If you have $300 and I have $0, you could give me $100 and our combined total would still be $300. Or I could borrow $100 from my friend and give it to you, so you'd have $400 and I'd have -$100 (I'd owe my friend), and our total would still be $300. I may be wrong though. I've been away from this material for a long time now. Best of luck to you!
@@ScottMRedmondRedmond: Consider that the second cap is "lightly" charged. Then some Q from the charged cap can recombine with positive atoms of the second cap, and thus, the "conservation of Q_initial, just redistributed" is wrong, because of that POSSIBLE recombination, or possibility (why not) of creation of additional pairs "electrons-holes" (even if hole to describe a locally positive atom is far stretched for a conductor). If the total initial Q seems to be ok for the specific case of adding a totally uncharged cap, that does not look like a general principle for all the possible cases, and that is the research of the more global PRINCIPE which could explain all the conditions of charges (and orientation) of the second cap which would be better to use. I am not sure to have a correct model yet, though.
@@snnwstt Yes, I think you've got it. There are common exam problems in which we connect charged capacitors with like plates attached, or opposite plates attached, and we have to derive the conservation equation differently. It's just like you said: some of the positives and negatives cancel each other out when opposite plates are attached. Nice!
I thought my life is a lie when I solved this exactly like you. Cuz the charges on capacitors in parallel are not same at the end. Then I googled and saw u r video and realised one imp. Point that capacitirs in series only have same charge if their intital charge is same. Thanks da for not making my life a lie😘😘😘
Love you so much❤❤❤
+Ankit Giri Yeah, the potential difference between C2 and C1 threw me off at first too. However, we can always count on the Kirchoff Voltage Law (a.k.a. loop rule) - that the voltage differences around a loop add up to zero. With the capacitors in series, we can also count on the idea that the charges always increase or decrease by the same amount. It can help to think of this as one charge at a time, or that the current is the same everywhere in this loop - because the elements are in series. You can't have more electrons in part of the loop because there are no junctions where extra electrons could come from.
Another way to look at this problem perhaps? Applying Kirchoff's Voltage Law,
{ (emf-battery) - dV(2-initial) = dV(1-final) + dV(2-final) }
Ultimately you end with the same algebraic equation for Q(1), except you have to fix Q(1) as equal to Q(2). This, I feel sits better, because in a series connection, flow of current always tries to keep the charges of the connected plates equal, lest electric field builds up to neutralize that anyway...
In essence however, the initial dV(2) never disappears, so if one asks the charge on C2, you would have to add the charges due to both the potential differences.
Is this logic flawed? Anyway, thanks for a concise, lucid video!
Thanks a lot! was looking for the solution to this problem for many days..really nice. just one doubt by the way, there is already a potential diff between c2 and c1 in the beginning, so that wont contribute to more electrons travelling to c1?
Great work
Hi RPT,Thanks to you, very interesting presentation indeed,would you help me please,
I am investigating the probability to reach resonance between 2 capacitors in series,if treated in differential equations, or exponential solutions, my preliminary conclusion says that if the current of one CAP respect the other is opposed in phase, they can resonate in steady state,natural response for i and v,of course amp-op stabilized if suitable.?????johann wegmann, many thanks in advance.For overunity perhaps,one CAP opposed to a solids flywheel,may run for ever?I mean a Supercap graphene.
What about parallel charged capacitor initially
what if in above problem polarity of c2 is reversed.
This is a question I'd typically reflect back to my student. ;-) The loop rule will still hold, and the fundamental way in which electrons move from one plate to another will also still be the same but the equation Q2 = Q1 + 300 microCoulombs will be different.
@@RedmondPhysicsTutoringVideo is the equation Q1=Q2+300
I do not understand if the series capacitors should have the same charge because then why? the final charge of q1 and q2 are different?
Yeah this threw me off at first too. It's tempting to think that capacitors in series have the same charge, but this example proves that we can't always assume that. Instead, capacitors in series build/lose the same amount of charge. So if they start out with different charges they can end up with different charges.
@@dasgoood2811 Awesome! You did the hard part of understanding the details. Thanks for letting me know that it helped!
What if the polarity of c2 is reversed and then connected???
+Aman Sharma Neat! A complicated variation. If I understand your question, it would be pretty much the same except that C2 would be flipped over before being connected. Then C2 would initially have a charge of +300 micro Coulombs on its bottom plate. An electron taken from the left plate of C1 would then be deposited onto the (initially positive) bottom plate of C2, making the bottom plate of C2 less positive. After 1 micro Coulomb is transferred, Q1 would be 1 micro Coulomb and Q2 would be 299 micro Coulombs. This changes how Q1 and Q2 are related - instead of Q2 = Q1 + 300, it'll be something else (and you should give it a try :-). Electrons will continue to flow until Kirchoff's Voltage Law is satisfied, that is when the voltage across C2 and C1 is the same as the emf.
3:04 Interesting, but I disagree. Indeed, consider that there is no emf source. None the less, C2 will now see C1 and may be able to do a redistribution of the "charges" so that, clearly, Q2-Q1 may become something else than 300 uC. The assertion seems without strong foundation.
Yeah, if we don't have the emf then the charges would redistribute differently than in this case. Where would the electrons go though? Once the switch is closed, each electron that leaves one capacitor has to end up at the other capacitor (assuming the only change is to remove the emf). There's nowhere else for the electrons to accumulate.
If you have $300 and I have $0, you could give me $100 and our combined total would still be $300. Or I could borrow $100 from my friend and give it to you, so you'd have $400 and I'd have -$100 (I'd owe my friend), and our total would still be $300.
I may be wrong though. I've been away from this material for a long time now. Best of luck to you!
@@ScottMRedmondRedmond: Consider that the second cap is "lightly" charged. Then some Q from the charged cap can recombine with positive atoms of the second cap, and thus, the "conservation of Q_initial, just redistributed" is wrong, because of that POSSIBLE recombination, or possibility (why not) of creation of additional pairs "electrons-holes" (even if hole to describe a locally positive atom is far stretched for a conductor). If the total initial Q seems to be ok for the specific case of adding a totally uncharged cap, that does not look like a general principle for all the possible cases, and that is the research of the more global PRINCIPE which could explain all the conditions of charges (and orientation) of the second cap which would be better to use. I am not sure to have a correct model yet, though.
@@snnwstt Yes, I think you've got it. There are common exam problems in which we connect charged capacitors with like plates attached, or opposite plates attached, and we have to derive the conservation equation differently. It's just like you said: some of the positives and negatives cancel each other out when opposite plates are attached. Nice!
Thank you Sir.
thanks a lot
with Laplace transform you can solve this problem MUCH easier