A capacitor is charged by an 8V battery. When 1 C passes through the capacitor, it goes through the first plate with 8 J  and comes out of the second plate with 0 J. Does this mean that if we connect a voltmeter to the capacitor it will read a potential difference of 8 V just because of 1 C of charg

I smell confusion. Let's clear a few things up and see if it helps. First note that it's a bit misleading to say a coulomb of charge goes "through" the capacitor. Rather, a coulomb goes in one side and a different coulomb comes out the other side. Second, the voltage on the capacitor will build up gradually as more charge goes "through". If it really does come to 8 volts after 1 coulomb, then it must have been a 1/8 farad capacitor. (You didn't mention that so I have to hope you realize it matters.) Third, if you imagine connecting an ideal constant voltage supply (8 V say) to an ideal capacitor, there's a paradox. Because the power supply gives 8 V, every coulomb that leaves the supply takes 8 joules with it. If it's charging a 1/8 F capacitor, 1 coulomb leaves, and with it 8 joules. The voltage on the capacitor only builds up gradually as it charges. The energy it ends up storing is less than 8 V * 1 C = 8 J. In fact it's half that. Thus the idealized situation can't be physically realizable. In reality the current missing 4 J of energy will go into either heating of the wires leading to the capacitor or radio waves from the sudden current pulse.

When the capacitor is fully charged (ideally) no more current flows. With no current flow, the voltmeter will read the full voltage of the source. This assumes an ideal voltmeter with infinite impedance. In real life, any voltmeter will ever so slightly drain some current, gradually reducing the voltage a bit below 8 volts.

When we connect the capacitor to the battery, charge gets stored in the capacitor, which is the form of energy stored in the capacitor. Since no resistor is connected to the capacitor, the capacitor gets instantly charged. This work is actually converted into energy which is stored.This energy gives rise to potential difference

Electronic technician for lo these many years. This explanation ignores all inconvient realities. If the capacitor alone (no resistance) is left connected to the battery after it has reached full charge, it will stop conducting current and act as an open circuit. An ammeter in the circuit will show first the full current capacity of the battery and then fall off to zero. A volt meter will measure 1st no voltage and then rise to the full voltage of the battery. If, at this time, the battery is disconnected from the capacitor the cap will read the full voltage of the battery as well. (Remember, no inconvient realities.) The coulombs in the circuit will be at maximum when the current is max and the voltage is 0. At full charge there will be no amps therefore no coulombs. The missing realities: There will be some resistance. There could be some leakage through the capacitor resulting in some current after it's charged. The battery will likely be at less than rated voltage. The capacitor probably won't charge to the full voltage supplied anyway. There are probably more that I'm not thinking of because of my brain, one of my inconvient realities.