The problem that I'm having is that the initial charges would have been 300 and 600 microcoulomb on the 3 and 6 microfarad capacitors respectively. But when the switch is closed there's some charge that flows through it(300 microcoulomb to be precise) but i can't get to this conclusion. Please help...
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$\begingroup$ We need more details to be able to help you with homework-like problems. So why don't you start us off by showing a little more work. Show us how you reasoned about the amounts of charges on the capacitors when the switch was open. Since they're disconnected, maybe just start with the left side, for example, and explain how what charge gets where. Is it just the same but backwards for the right side? $\endgroup$– MikeCommented Dec 4, 2017 at 15:21
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$\begingroup$ I have added the picture of my attempt to this question. I know I have done something horrendous and upset the physics god. $\endgroup$– Kunal KumarCommented Dec 4, 2017 at 17:07
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$\begingroup$ Please note that this is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems. $\endgroup$– Kyle KanosCommented Dec 5, 2017 at 11:08
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2 Answers
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"the initial charges would have been 300 and 600 microcoulomb on the 3 and 6 microfarad capacitors respectively". Although we talk about the charges on capacitors, this is a loose way of speaking and one that lets you down with this sort of question. The point is that capacitors have equal and opposite charges on their plates. The overall charge is zero.
If the circuit had been set up with the switch open, then, additionally, for both the right hand and left hand pairs, the charge on the bottom plate of the top capacitor is equal and opposite to the charge on the top plate of the bottom capacitor, since these plates and the link between them form an 'island' that can't have acquired any net charge. Treating the left hand pair and the right hand pair as capacitors in series, you'll find that the charges on the plates of each capacitor are ±400 $\mu$C. Ask if you don't understand this.
When the switch is closed, each capacitor has 100 V across it (again, ask if not clear) and the 3$\mu$F and 6$\mu$F have different charges. By considering the 'island plates' (now joined by a bridge to the other island!) you should easily work out how much charge flows through the switch.
answered Dec 4, 2017 at 15:41
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$\begingroup$ I understood how each capacitor has 400 microcoulomb charge but i didn't understand how the potential difference is 100V across each capacitor after the switch is closed. $\endgroup$ Commented Dec 4, 2017 at 16:57
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$\begingroup$ It's like this… When the switch is closed the two capacitors above the switch are in parallel and the combination has a capacitance of 9 $\mu$F. Exactly the same for the pair below the switch. So, by symmetry, there's half of the 200 V across each combination, and therefore across each individual capacitor. $\endgroup$ Commented Dec 4, 2017 at 18:16
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$\begingroup$ Ohh... Right. I feel so dumb now to not understand it. Thank you sir for ur kind help😊 $\endgroup$ Commented Dec 4, 2017 at 18:19
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$\begingroup$ Not dumb at all. There's more to understanding capacitors than one first thinks! $\endgroup$ Commented Dec 4, 2017 at 19:35
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$\begingroup$ I'm afraid that I can't follow the other question. Suggest you have another go at drawing the diagram, indicating clearly where the plates of the capacitor are. $\endgroup$ Commented Dec 5, 2017 at 11:22
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Your new drawing looks right to me. You started with two separate paths, and correctly added the capacitances in series, then got the amount of charge on each plate of each capacitor. It's the same, but backwards, along each path initially. You also know that those central segments are isolated initially, which means charges can't flow onto or off of them. You correctly showed this in your diagram, since each capacitor had 400μC, but opposite signs.
Now, the top two are connected in parallel, and basically act as one capacitor. Similarly, the bottom two are connected in parallel and act as one capacitor. You seem to have correctly added each pair's capacitance in parallel, and then added them in series to get the right individual charges on each capacitor.
The important part to remember is the sign of the charge on each capacitor. On the left side, the bottom plate of the top capacitor has -600μC, while the top plate of the bottom capacitor has +300μC, giving us a net -300μC on the left side. Similarly, the right side has net +300μC. But the part of the circuit with the switch in it is still isolated, so charge can't come from outside; it can only move from one side to the other. That's why current flowed through.
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$\begingroup$ Sure thing. If my answer solved your problem, click the big checkbox to accept it as the answer. And it's standard practice to upvote any answers that were helpful. $\endgroup$– MikeCommented Dec 4, 2017 at 17:50
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$\begingroup$ Ok so the charge flowing through the switch is 300 microcoulomb. But what is the potential difference across a and b? $\endgroup$ Commented Dec 4, 2017 at 18:35
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$\begingroup$ After the switch is closed (and things have settled down), there's no potential difference. So I assume you mean the potential difference before the switch is closed. You already have all you need to calculate this. Start with the bottom-left. It has charge 400μC, and its capacitance is 3μF. Just plug this into $V=Q/C$ to get the voltage between the bottom and top plates of that capacitor: 133V. You know that the bottom plate is at ground, so the top is at ~133V relative to ground. Same deal on the other side, and you get ~67V, so their potential relative to each other is ~67V. $\endgroup$– MikeCommented Dec 4, 2017 at 19:20
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1$\begingroup$ Got it. Sorry i'm new to this forum so don't know much about the system. $\endgroup$ Commented Dec 4, 2017 at 19:41