Question

Capacitor X is charged to a potential difference 68 V by a battery. It is disconnected from the battery and then connected in parallel to an uncharged 12 µF capacitor. A voltage of 20 V is measured across the parallel combination. Calculate the capacitance of capacitor X. Express your answer in µF.
Answer is 5, but how do I get this?

Answer 1

C =q/V
so
x = q/68

that q is all the charge we have now for both capacitors

total C = 12*10^-6 + x

q = 20 (12*10^-6 + x)

68 x = = 240*10^-6 + 20 x

48 x = 240 * 10^-6

x = 5 *10^-6 Farads or 5 µF

Answer 2

Ah yes, just noticed you said 5, good