Question

Two capacitors C1 = 6.8 μF, C2 = 14.6 μF are charged individually to V1 = 19.6 V, V2 = 3.5 V. The two capacitors are then connected together in parallel with the positive plates together and the negative plates together.Calculate the amount of charge (absolute value) that flows from one capacitor to the other when the capacitors are connected together. By how much (absolute value) is the total stored energy reduced when the two capacitors are connected? The final potential difference across the plates of the capacitors once they are connected was calculated to be 8.616 V

Answer 1

a)
Charge stored in the first capacitor initially, Q1 = V1 x C1
= 19.6 x 6.8 F
= 133.28 C
For the second capacitor, initial charge, Q2 = V2 x C2
= 3.5 x 14.6 F
= 51.1 C

Final charge on the first capacitor, Q1f = Vf x C1
= 8.616 x 6.8 F
= 58.5888 C
Final charge on the second capacitor, Q2f = Vf x C2
Q2f = 8.616 x 14.6 F
= 125.7936 C
Q1 - Q1f = 74.69 C charge flow from the initial state to the final state for the first capacitor and since the total charge is conserved, the same amount of charge flow into the second capacitor.

b)
Initial energy, E1 = 1/2 C1 (V1)² + 1/2 C2 (V2)²
= 0.5 x (6.8 x 10^-6) x (19.6)² + 0.5 x (14.6 x 10^-6) x (3.5)²
= 1.396 mJ
Final energy, E2 = 1/2 C1 (Vf)² + 1/2 C2 (Vf)²
= 0.5 x (6.8 x 10^-6) x (8.616)² + 0.5 x (14.6 x 10^-6) x (8.616)²
= 0.794 mJ

E2 - E1 = 1.396 mJ - 0.794 mJ
= 0.6 mJ