OCR A Physics 2018 paper 2 capacitors

The capacitors question says:

A capacitor is charged through a resistor

The cell has e.m.f. 1.50 V and negligible internal resistance.
The capacitor is initially uncharged. The time constant of the circuit is 100 s.
The switch is closed at time t = 0.
What is the potential difference across the capacitor at time t = 200 s?

The answer is 1.30V, which uses the formula V = V0 e^(-t/CR).
However, as the capacitor is being charged, I thought the p.d. across it is given by V = V0 (1 - e^(-t/CR))

Can somebody explain why the discharging equation is used in this case?

Reply 1

alkjsto

heres the question with the circuit diagram

Reply 2

Joinedup

What the discharge equation really gives you is the difference between the voltage across the capacitor at time t and what the final voltage across the capacitor is going to be. when you're discharging a capacitor the final voltage is going to be 0 so the difference is just the same as the PD at time t.

in this question the final voltage across the capacitor is going to be 1.5 so the voltage at time t is 1.5 minus the value you get from the discharge equation.

the extra 'junk' in the charge equation is just there to subtract the value you'd get from the discharge equation from the final fully charged voltage.