Charging Up a Capacitor 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
 A: 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.
A: 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 $\pi R/16d$.
