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Charging a capacitor 
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#1
Feb1314, 09:56 PM

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I am quite puzzled about charging a capacitor.
Given a resistor, in series with a capacitor and a battery, when the switch is closed 1. Why there is a current immediately The capacitor is separated by insulator, hence this is not a closed circuit. The appearance of current seems to be a bit confusing. I thought through this, and conclude that this current is caused by the electronics flow which is for sure due to potential difference. However, when I further think of potential difference, I cannot form a logic explanation to this. The two plates have no potential difference since at the very start it is not charged. How can the electrons flow from a plate to another? I know something has to do with the battery. Could you help me explain more about this? If there is a circuit with battery, switch, resistor, and the wire split at one place, will something happen like what happens in RC circuit Wire has its cross area also, and when they are separated apart, can I imagine it as a capacitor? If so, will it also be the case of charging a capacitor? Thank you for answering all my doubts!! 


#2
Feb1314, 10:25 PM

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If you put an ammeter in the circuit you will be able to measure the current that flows. If there is a circuit with battery, switch, resistor, and the wire split at one place, will something happen like what happens in RC circuit[/quote]Yes  the ends of the wire become charged. The maths is a bit more complicated than the platecapacitor though. People who build compact electric circuits have to be careful to avoid accidentally adding too much unwanted capacitance when they put breaks in the wires or even when two wires run alongside each other. 


#3
Feb1314, 10:26 PM

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1. The battery introduces a potential difference to the circuit when you close the switch. Current starts to flow through the circuit, but when it gets to the capacitor it encounters an open since it can't get through. However, a capacitor is designed to have a significant amount of capacitance, aka the ability to store charge. Current, in this case electrons, flows into one plate, making it more and more negatively charged, and out of the other plate, making that plate positively charged. This process continues until the charges on each plate build up enough to counteract the applied voltage. Since the insulator keeps current from flowing from one plate to the other, you now have two electrically charged plates and no more current flow.
2. Yes, a cut wire is exactly like a very low capacitance capacitor. The principals are identical. 


#4
Feb1414, 12:10 AM

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Charging a capacitor



#5
Feb1414, 12:52 AM

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#6
Feb1414, 09:38 AM

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Thank you for your answer. now my doubt is this: If we apply V=Q/C here Initially Q=0 Then it means V is also equal to zero? 


#8
Feb1414, 10:08 AM

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#9
Feb1414, 12:13 PM

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Th voltage across the terminals of the capacitor is dependant upon how much charge is held within the capacitor. If the capacitor is uncharged, then Q=0, and then V=0. By closing the circuit, charge will begin to flow into the capacitor. As more charge flows into the capacitor, the voltage across the capacitor terminals will increase to satisfy the equation V=Q/C. Since there is a resistor in series with the capacitor, at the moment of closing the circuit, the voltage across the resistor, Vr, equals the battery voltage E, and the voltage across the capacitor, Vc= 0. When the capacitor is fully charged, Vr = 0 and Vc = E. At all times E = Vr + Vc. 


#10
Feb1414, 09:38 PM

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It means that the voltage due to the charge on the capacitor is zero when there is no charge separation  which is what it would be if there was no battery. But there is a battery. Check it against real life  actually get a capacitor and a resistor and a battery and a voltmeter and see. Try to get values for R and C so that RC is a decent amount of time  like 510seconds. Hook the voltmeter up so the negative terminal is on the battery's negative terminal and the positive terminal is between the resistor and the capacitor. Go do it. 


#11
Feb1514, 12:49 AM

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Which is positive terminal and negative terminal when we make our own capacitor as shown in this video. Thank you for the help.
http://www.youtube.com/watch?v=WC2sZ5iJ7gQ 


#12
Feb1514, 01:18 AM

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Parallelplate capacitors do not have positive and negative terminals.
The positive terminal on a voltmeter is the one marked with a "+" sign. On a multimeter, it is usually colorcoded red. 


#13
Feb1614, 03:51 AM

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#14
Feb1614, 05:09 PM

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E = V_{R} + V_{C} (Kirchoff II) Because I = V/R, this means that the current will also decrease exponentially. The time constant (time to 1/e of any given value) will be RC. 


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