As soon as the switch is put in position 2 a 'large' current starts to flow and the potential difference across the capacitor drops. (Figure 4). As charge flows from one plate to the other through the resistor the charge is neutralised and so the current falls and the rate of decrease of potential difference also falls.
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.
This process will be continued until the potential difference across the capacitor is equal to the potential difference across the battery. Because the current changes throughout charging, the rate of flow of charge will not be linear. At the start, the current will be at its highest but will gradually decrease to zero.
When a voltage is placed across the capacitor the potential cannot rise to the applied value instantaneously. As the charge on the terminals builds up to its final value it tends to repel the addition of further charge. (b) the resistance of the circuit through which it is being charged or is discharging.
The potential of the charge after crossing the capacitor (displacing another charge on the low potential plate) will be 3V 3 V. What confuses me is the assumption that the high plate will be at 5V 5 V as well since only then the potential of the low potential plate can be said to be 3V 3 V. Where have I gone wrong?
A higher capacitance means that more charge can be stored, it will take longer for all this charge to flow to the capacitor. The time constant is the time it takes for the charge on a capacitor to decrease to (about 37%). The two factors which affect the rate at which charge flows are resistance and capacitance.