kartikwat
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Why does the potential drops when charge pass through a capacitors connected in series.
Capacitors in Series: Potential Drops Explained
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kartikwat
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See, we have seen that capacitors are connected in series the same charge flows through each capacitor but when charge passes through each capacitor its potential drops.so why do this potential drops
Consider two capacitors in series. If they are connected to a 12 volt battery, then obviously the total voltage across both capacitors is only 12 volts, as that's all the battery can supply. Since the voltage drop must add up to 12 volts, then each capacitor must be charged to less than 12 volts individually. If they are of equal capacitance, then both would be 6 volts. Since capacitance is c=q/v, we can re-write this as q=cv to find the total charge given by this reduced voltage. Since the voltage is half what it would be if we had a single capacitor, that means that the total number of charges on each capacitor is also halved.
Remember that the number of charges moving onto or off of the two plates in a capacitor always remains equal. Since the middle two plates (one plate of each capacitor, connected by the conductor between the capcitors) are separated from the rest of the circuit by the dielectrics, no charge can leave them. In the end, the first plate of the first capacitor and the last plate of the second capacitor have equal and opposite charges, but at half of what a single capacitor would have. So this series of 2 capacitors acts like a single capacitor with half the capacitance.
Does that help?
Remember that the number of charges moving onto or off of the two plates in a capacitor always remains equal. Since the middle two plates (one plate of each capacitor, connected by the conductor between the capcitors) are separated from the rest of the circuit by the dielectrics, no charge can leave them. In the end, the first plate of the first capacitor and the last plate of the second capacitor have equal and opposite charges, but at half of what a single capacitor would have. So this series of 2 capacitors acts like a single capacitor with half the capacitance.
Does that help?
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If you are asking how come the middle plates are in same potential though they have equal and opposite charges, the answer is that the potential depends not only on the charge of the plates but on the charge of the adjacent plates. The left middle plate with charge -q has to its left another plate with charge +q, while the right middle plate with charge +q has to its right another plate with charge -q.
kartikwat
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@drakkith i am not getting what u r explaining
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kartikwat said:
Hard to see how it could be explained any more simply or lucidly than what Drakkith has presented. I'm thinking maybe we don't really understand your question but it does seem like you've given a very simple question, that has a very simple answer and that Drakkith has given that answer. I can't think of any way to explain it better. Maybe Drakkith will come up with something.
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kartikwat said:
WHAT paragraph?
kartikwat
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kartikwat said:
Yes, this says exactly what Drakkith said, except that it shows you the math and gives the generalized case rather than a simple example as Drakkith did. I would suggest that you look at Drakkith's explanation again, since it is easier to understand than the general case, then when you've figured that out, the general case should be easier to understand.
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There is potential drop between the plates of the first capacitor because there is electric field between them due to the equal and opposite charges in each of the plates. The potential drop V between two points A and B usually is defined as the integral [tex]V_{AB}=\int_A^B\vec{E}\vec{dl}[/tex].
Where the path of the integral can be any curve that connects the points A and B. It doesn't depend on the choosing of the curve as long as we consider the electrostatic or quasi static case. You can see that the above integral is not zero, assuming that the electric field between the plates is homogeneous and vertical to the plates , the integral is equal to E*d where d is the distance between the plates.
Similarly there would be potential drop between the plates of the 2nd capacitor.
Where the path of the integral can be any curve that connects the points A and B. It doesn't depend on the choosing of the curve as long as we consider the electrostatic or quasi static case. You can see that the above integral is not zero, assuming that the electric field between the plates is homogeneous and vertical to the plates , the integral is equal to E*d where d is the distance between the plates.
Similarly there would be potential drop between the plates of the 2nd capacitor.
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