The P.D of a charged capacitor should be more than the charging source .

[dE_logics]

The P.D of a charged capacitor should be more than the charging source...

That's cause the charge felt by the cell on its terminals might be less that the net charge (depends on what level of charging is the capacitor in), as a result charge accumulation continues until the plates are saturated with charge and it finally gets exposed to the terminals of the battery. After this level of saturation, we can be ensured that the P.D by virtue of the net charge stored in each plate will be more than that of the charging source's. That's cause, there much of the charge on the plates which are being mutually held by each other and cannot be felt by the terminals of the charging source.

If the plates are dismantled at this stage, the potential difference gained by the individual pates will be equal and having more potential than the charging source (cause now the net charge can be exposed and felt).

 

[dE_logics]

How am I wrong?

 

[dE_logics]

Are you all not getting it?

 

[TVP45]

Are you all not getting it?

No. Can you provide a diagram or some equations (or both?)?

 

[Dadface]

When you charge a capacitor the charge continues to flow until the pd across the capacitor terminals becomes equal and opposite to that of the supply terminals.Are you saying that during charging the capacitor can get an even higher voltage?.The voltage across the capacitor can be increased by disconnecting it from the battery and then reducing its capacitance for example by pulling the plates further apart.

 

[dE_logics]

You mean, in a fully charged capacitor, the P.D across the terminals will be higher than that of the charging source?

 

[Born2bwire]

You mean, in a fully charged capacitor, the P.D across the terminals will be higher than that of the charging source?

Not without help. Disconnect the voltage source without dissipating the charge across the capacitor and then you can manipulate the voltage across the capacitor. This is done by moving the plates or inserting a dielectric. But in each of these cases you are doing work that gets put into the system. So the increase in the potential does not come free. If you do these things with the voltage source still connected then the potential will go back to what it was before (after the transients work out the charge distributions again).

 

[dE_logics]

Yes of course...thanks for the confirmation.