Let us consider an ideal parallel plate capacitor in a circuit. The capacitor is connected to a battery which generates a potential difference of V across the terminals.
1) The work required to charge the capacitor is calculated by assuming that a small charge dq is taken from the lower potential plate to the higher potential plate. Let us say that the potential difference between the plates at this point is q/C (C is the capacitance). When the charge dq goes through the battery it acquires a potential of V (assuming that the negative terminal is at 0 potential). The work done to take this charge from the high potential to low potential plate is q*dq/C. Where did the rest of the energy go ?
2) In order to do deposit the charge on the high potential plate some work is required to move the charge against the electric field of the capacitor. This work is said to be done by the battery. My question is : If the battery gives some electric potential energy to the charge then moving it to the high potential plate of the capacitor should increase the potential energy of the charge . Why is it then said that there is a potential drop across a capacitor ?
(EDIT : NEW QUESTIONS)
3) While charging the capacitor in the above figure a charge dq is extracted from the negative plate. The charge gains a potential V after going through the battery. while approaching the high potential plate will the charge loose the Electric potential energy ?
4)The potential drop across a capacitor is Q/C. Let us assume that the emf of the battery is 5V. a charge goes through the battery and acquires some potential energy . Let us say that the potential drop across the capacitor is 2V. The potential of the charge after crossing the capacitor (displacing another charge on the low potential plate) will be 3V. What confuses me is the assumption that the high plate will be at 5V as well since only then the potential of the low potential plate can be said to be 3V. Where have I gone wrong?