assume an 80 mF capacitor is initially connected to a 9V battery. The battery is then removed and a wire is connected between the 2 terminals of the capacitor. What is the charge on the capacitor after the wire is connected?

 

since a wire is connected between 2 terminals of the capacitor the capacitor should be discharged so charge=0

 

If both were ideal (no resistance) and the connections made were ideal (no sprakes) then where did the potential energy stored in the charges go? So it can't be zero, can it?

 

If both were ideal (no resistance)

Hypothesis contrary to fact. Wires heat up. Thermodynamics are not something that a slick lawyer can bypass. (Although many would-be slicks have tried.)

Optionally, we could play the "ideal" game and say that the shorted capacitor forms an "ideal" tank with the stray inductance of the "ideal" wire. Perpetual oscillation, anyone?

 

the electron when it reaches the lower potential will have K.E which was the potential stored in capacitor since 0 resistance was offered to its flow no loss in energy.

 

for the moment after the battery is removed, many electrons are stored on the plate which was connected to the batt cathode, and many holes are on the anode plate. the cap is charged to a 9 volt potential. when the wire (near 0 ohm) is connected, the cap does a discharge at the stanard discharge rate of t=r x c. a cap is 99% discharged (or charged) after 7 time constants. since the resistance is extremely low, the cap will discharge at a very fast rate.

 

Electrolytic caps discharge at a slower rate, and even if momentarily shorted out, the can conserve measurable charge for a while until they are shorted again. That is due to the fact that the charging process also motivates a chemical reaction that thickens an aluminum oxide layer in one of the electrodes. The aluminum oxide layer acts as a dielectric. In a fast discharge like the one caused by a short, the dielectric is not depleted at the same rate of the discharge. Thus the capacitor appears to have no charge until time is given for the aluminum oxide layer to reform to its original state. This is true because a capacitor, once charged, will be always charged, and a residual charge will deliver a greater voltage if the dielectric thickens.