Question

Two capacitors, one that has a capacitance of 7 µF and one that has a capacitance of 21 µF are first discharged and then are connected in series. The series combination is then connected across the terminals of a 13-V battery. Next, they are carefully disconnected so that they are not discharged, and they are then reconnected to each other--positive plate to positive plate and negative plate to negative plate.

1. Find the potential difference across each capacitor after they are reconnected.
2. Find the energy stored in the capacitor before they are disconnected from the battery, and find the energy stored after they are reconnected

Answer 1

Given,

C1 = 7 uF ; C2 = 21 uF

Ceq = 7 x 21/(7 + 21) = 5.25 uF

Q = 5.25 x 13 = 63 uC

Charge remains conserved.

Now the two are connected in parallel

Ceq = 7 + 21 = 28 uF

Q1 + Q2 = 63 uF

V1 = V2

Q1/C1 = Q2/C2

Q1/7 = Q2/21

Q2 = 3Q1

Q1 + 3Q1 = 63

Q1 = 15.75 uC

Q2 = 47.25 uC

V1 = 15.75/7 = 2.25 V

V2 = 47.25/21 = 2.25 V

Hence, V1 = V2 = 2.25 V

b)E = 1/2 Ceq V^2 = 1/2 Q^2/C

E1 = 1/2 x (63 x 10^-6)^2/(7 x 10^-6) = 2.84 x 10^-4 J

E2 = 1/2 x (63 x 10^-6)^2/( 21x 10^-6) = 9.45 x 10^-5 J

E1' = 1/2 x 7 x 10^-6 x 2.25^2 =1.77 x 10^-5 J

E2' = 1/2 x 21 x 10^-6 x 2.25^2 = 5.32 x 10^-5 J