Using a battery, a 100 pF capacitor is charged to 60 V and then the battery is removed. After that, a second uncharged capacitor is connected to the first capacitor in parallel. If the final voltage across the second capacitor is 20 V, its capacitance is: (in pF)
- 600
- 100
- 400
- 200
Answer
(4) 200
Solution
Charge on the first capacitor: Q = C1V1 = 100 pF 60 V = 6000 pC
The battery is removed, this charge is shared between the two capacitors (charge is conserved).
In parallel both have the same final voltage V = 20 V:
Q = (C1 + C2)V
6000 = (100 + C2) 20
100 + C2 = 300
C2 = 200 pF
Answer: (4)