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Charge Flown Through Battery In Charging Capacitor 
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#1
Jul309, 11:29 AM

P: 582

1. The problem statement, all variables and given/known data
Image URL: Before the switch was closed, Capacitor 1 had a charge of magnitude CE/3 on its plates and Capacitor 2 had a charge of magnitude CE/6. The third capacitor was initially neutral. Find the charge flown through battery when switch 'S' is made to close where E is the emf of the battery. 3. The attempt at a solution Firstly, I solve this problem using the conventional method. I assume that a charge 3q flows out from the positive end of the battery. It divides itself in the ration 2:1 between capacitors 1 and 2, in order to keep the potential across them equal, since they are connected in parallel. Therefore the net final charges on the capacitors are: 1== (CE/3)+2q 2== (CE/6)+q 3== 3q From potential balancing in the circuit when steady state is achieved, I get: (((CE/3)+2q)/C)+(3q/C)=E q=2CE/15 therefore charge flown through battery=3q=2CE/5  Till now, I had no problem. But after this method, an idea struck me: I calculated the initial total +VE charge=(CE/3)+(CE/6)=CE/2 Now, I calculated the final total +VE charge. C(equivalent for the circuit)=3C/5 The net positive charge on the equivalent single capacitor would have been 3CE/5. This extra positive charge must have come from the charge flown through the battery. Charge flown through battery=3CE/5  CE/2= CE/10 (which is different from the one calculated previously) This method worked right until now and I thought it was another version of charge conservation. But this problem really shocked me literally. Please explain the anomaly with the second method. 


#2
Jul309, 11:58 AM

HW Helper
P: 4,435

Your second approach is correct.



#3
Jul309, 12:21 PM

P: 582

PS. Sir, please answer me only if you are sure about it. 


#4
Jul309, 01:14 PM

P: 211

Charge Flown Through Battery In Charging Capacitor
Your second approach is actually the typical, compact way to solve capacitor questions. Do you mean you don't understand why it's correct? 


#5
Jul309, 01:34 PM

P: 582

across 2 is also E/3 and across 3 is 0 initially. So why wont the charge divide itself in ratio 2:1 ??? Yes, please explain to me the 2nd method via the method of charge flow from the battery. I am simply confused. 


#6
Jul409, 06:02 PM

P: 582




#7
Jul409, 08:36 PM

P: 211

Well, think about the amount of charge you start with in the problem. Is the assertion in your first method still correct?
I would stay put and type a more detailed explanation, but unfortunately I'm busy too =( But if u can wait another 12 hours or so, I should be able to come up with something. 


#8
Jul409, 10:41 PM

HW Helper
P: 4,435

1== (CE/3)+2q
2== (CE/6)+q 3== 3q This assumption is wrong. If the capacity 3 has 3q charge, then the parallel combination of 1 and 2, which is in series with 3, must have 3q charge including the initial charges. 


#9
Jul509, 01:09 AM

P: 582

See, if I take this assumption: 1== (CE/3)+2q 2== (CE/6)+q 3== 3q to be true, 3q=2CE/5 1== 0.6CE 2== 0.3 CE 3== 0.4 CE 0.6CE+ 0.3 CE is not equal to 0.4 CE, yet there seems to be no problem in the circuit, if I analyze it. Potential drop across the parallel combination of 1 & 2=0.6 E Potential drop across capacitor 3= 0.4 E Net drop= E which is the emf of the battery. Which law of physics did my assumptions in method 1 violate? 


#10
Jul509, 01:14 AM

P: 582

Thanks. 


#11
Jul509, 03:29 AM

HW Helper
P: 4,435

If two capacitors are connected in series to a battery, plate one of the first capacitor is connected to the battery, plate 2 of the first capacitor is connected to the plate one of the second capacitor. Plate two of the capacitor is connected to the battery. According to the conservation of charges, charge on the plate one of the first capacitor must be equal and opposite to the charge on the plate 2 of the second capacitor. 


#12
Jul509, 04:20 AM

Sci Advisor
HW Helper
Thanks
P: 26,157

Hi Ritwik! thanks for your PM of last night.
imagine the switch is closed, but there is no emf … what happens to the charge then? 


#13
Jul509, 06:20 AM

P: 211

now coupled with the hints from other posters, hopefully you can now understand what's going on. 


#14
Jul509, 08:10 AM

P: 582

Lets shift our discussion to a 2 capacitor system in diagram given below. Capacitor 1 has a charge of 0.6 CE initially. Capacitor 2 has a charge of 0.4 CE initially. I connect their series combination to a battery of emf 'E'. I would say that no charge would flow from the battery. And according to you, some charge must flow from battery since this is not the steady state wherein plate 1 of capacitor 1 has an equal and opposite charge as plate 2 of capacitor 2, right? And my question is what would make the charges flow? There is no potential difference anywhere to support charge flow. I have marked potential regions in my diagram V1,V2 and V3(initially). V1V2=0.6E ..................... via capacitor 1 V2V3=0.4E ..................... via capacitor 2 V1V3=E ..................... via the battery adding equation one and two: V1V3=E ........................ which seems perfectly in agreement with the battery potential difference. So will the charges yet flow??? 


#15
Jul509, 08:47 AM

P: 211

Imagine both capacitors in your example were uncharged to start with. Connect the batteries, and positive charges flow to the left plate of C1. Now, for every positive charge entering the left plate, a negative charge leaves the right plate. This negative charge MUST go somewhere, and it will go to the left plate of C2. And this SAME amount of negative charge will leave the right plate of C2 and return to the battery, satisfying the conservation of charge.
Do you see it now? The charges displaced by the right plate of C1 can only end up on the C2 If not you'll have a net charge on the system which is not what your started with. Now do you understand why capacitors in series MUST have the same charge? So in your example, we know effective capacitance is 0.5C, so overall charge is 0.5CE. Now we have to be careful: what we mean here is that the left plate of C1 has 0.5CE of charge, as does the right plate of C2. After all we're now treating the system as one capacitor. Of course, correspondingly there is 0.5CE of charge on the right plate of C1 and left plate of C2, but it does NOT affect our analysis here. It does NOT contribute to the "overall charge" we just calculated. Thus what happens is that 0.1CE of charge flows from right plate of C1 to left plate of C2. 0.1CE of charge flows from left plate of C1 to battery, and 0.1 CE of charge flows from battery to right plate of C2. Now let's see what went wrong with your argument. You are quite right to say that the potential differences add up to EMF of battery, but the problem is that this is NOT a steady state! Cuz the right plate of C1 is at a higher potential than the left plate of C2, so there is an emf driving charges around. Now, are you able to answer tim's question, and subsequently your original question, correctly? 


#16
Jul509, 11:28 AM

HW Helper
P: 4,435

Here the battery does not supply the charges. It acts like a pump, which transfer the electrons from one plate to other plate. 


#17
Jul709, 07:46 PM

P: 582

I cannot think of any such way. Please just prove to me that there is a potential difference between the plate 2 of capacitor 1 and plate 1 of capacitor 2, I will be able to understand the whole of it then. 


#18
Jul709, 09:52 PM

P: 211

Let's try another angle of looking at things. You can always find the effective capacitance of a network of capacitors, which means it is equivalent with replacing all the capacitors with a single "effective capacitor". This single capacitor must have the same charge on each plate right? In our example, these two plates are the left plate of C1 and right plate of C2. This effectively means that C1 and C2 have the same charge.



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