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Well what I read that in the process of charging a Capacitor, charges are transferred from one plate to another. The work done to move a charge from one plate to another stores as electrical potential energy in it and the capacitor is charged up.

Before this I read that when a Capacitor is placed in a circuit with switch closed, positive charges pile up at one end/plate of the capacitor inducing same amount of negative charges on the other end of the capacitor. This continues till the voltage across the capacitor becomes equal to the voltage of battery. This way a capacitor is charged up.

In both of these what I found that there is only induction of charges at another plate due to charge present at the 1st plate. There is no transference of charges between the plates of capacitor while charging up the capacitor.

Why this is so?

I know I am wrong somewhere but where I don't know. Please tell me an appropriate answer for this doubt. Thanx

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  • $\begingroup$ It is not clear what you are asking. Your 1st sentence ( in the process of charging a Capacitor, charges are transferred from one plate to another) is contradicted by a later sentence (There is no transference of charges between the plates of capacitor while charging up the capacitor). $\endgroup$ – sammy gerbil Sep 10 '16 at 14:15
  • $\begingroup$ Yup the two statements are contradictory to each other. While reading how energy stored in a Capacitor, I got that work is done to move a charge from one plate to another which stored as electrical energy in it. But before this I read that same amount of charges are induced at one end of capacitor due to charges at another end of the capacitor . @sammygerbil I just want to know which statement is correct? $\endgroup$ – Perspicacious Sep 10 '16 at 14:34
  • $\begingroup$ The 1st statement is correct : in the process of charging a capacitor, charges are transferred from one plate to the other. $\endgroup$ – sammy gerbil Sep 10 '16 at 21:49
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There are no charges traversing between the plates because between the plates there is a strong insulating dielectric material. Charges on both plates are supplied by the battery. Through the electric field that crosses the dielectric they feel the presence of the charges on the other plate.

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  • $\begingroup$ If charging a Capacitor is due to induction between the charges on the plates then why we do work to transfer a charge from one to another plate while evaluating electric energy stored in the capacitor? Is there any reason @Jan Bos $\endgroup$ – Perspicacious Sep 10 '16 at 17:21
  • $\begingroup$ Not sure what you mean by induction but the battery effectively moves charges from one plate to other (through the circuit) building the electric field between the plates. This requires work and this energy is stored in the electric field. $\endgroup$ – Jan Bos Sep 11 '16 at 1:11
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I think you are asking whether the capacitor plates becomes charged because of (1) the transfer of charge caused by the battery, or (2) the mutual induction of charge due to the presence of charges on the other plate.

Both these 2 processes are occurring at the same time and they reinforce each other. The presence of the 2nd plate makes it easier for the battery to charge the capacitor. For the same battery voltage, more charge can be stored on the 2 plates - ie the capacitance is higher.

The plates of the capacitor could intially be separated by a large distance, so that each plate is isolated from the electric field of the other. Each is then a separate capacitor with the 2nd plate grounded at infinity. The capacitance of each in this configuration is $8\epsilon R$ where $R$ is the radius of the circular plates.

The total capacitance in this situation is $C_{\infty} = 16\epsilon R$, which can be much smaller than the capacitance if the 2 plates were placed very close together in a single parallel-plate capacitor.

If the 2 plates are brought close together at separation $d<<R$ the capacitance is now
$C_{||}=\epsilon A/d=\epsilon \pi R^2/d = 16\epsilon R (\pi R/16d)$
which is higher than $C_{\infty}$ by a factor $\pi R/16d$.

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  • $\begingroup$ I think that due to induction of charge at 2nd plate due to charge at 1st plate is the correct way to charge up a capacitor. I am considering 2nd statement to be correct because charge flow from one plate to another in a capacitor only when there is a dielectric breakdown which neutralize the charge in the capacitor.This was my opinion. I didn't get ur opinion on the 2 statements. $\endgroup$ – Perspicacious Sep 10 '16 at 17:19

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