Two capacitors A and B with capacities `3 muF` and `2 muF` are charged to a potential differece of `100 V` and `180 V`, respectively. The plates ot the capacitors are connected as shown in, with one wire free from each capacitor. The upper plate of A is positive and that of B is negative An uncharged `2muF` capacitor C with lead wires falls on the free ends to complete the circuit.
i. Calculate the final charge on the three capacitors,
ii. Find the amount of electrostatic energy stored in the system before and after the completion of the circuit.
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Explanation

(i) Method 1 : Initial charge on capacitor `A` is `q_(A) = 3 xx 100 = 300 mu C`. Initial charge on capacitor `B` is `q_(B) = 2 xx 180 = 360 muC`. After completing the circuit, let the charge on capacitor `A` be `q_(1)`, on `B` be `q_(2)`, and on `C` be `q_(3)` with polarities as shown in. Applying conservation of charge for plates `2` and `3`, we get
`q_(1) + q_(3) = -360 + 0 = 300` (i)
Applying conservation of charge for plates `4` and `5`, we get
`-q_(2) - q_(3) = -360 + 0` or `q_(2) + q_(3) = 360` (ii)
Applying Kirchhoff's law in the whole cirucuit, we get
`(q_(1))/(3)- (q_(3))/(2) + (q_(2))/(2) = 0` (iii)
On solving, we get `q_(1) = 90 muC, q_(2) = 150 mu C`, and `q_(3) = 210 mu C`

Method 2 : After completing the circuit, let the charge `q` flow in the circuit in the clockwise direction. Then the final charges on the capacitors will be as shown in.

Applying Kirchhoff's law, we get
`(300 - q)/(3) - (q)/(2) + (360 - q)/(2) = 0` or `q = 210 mu C`
Now charge on all three capacitors can be found.
(ii) Initially, energy stored is
`U_(i) = (1)/(2) xx 3 xx 10^(-6) xx (100)^(2)+(1)/(2)xx2xx10^(-6)(180)^(2)`
`= 4.74 xx 10^(-2) J`
Finally, the energy stored is
`U_(f) = ((90 xx 10^(-6))^(2))/(2 xx 3 xx 10^(-6))+((150 xx 10^(-6))^(2))/(2 xx 2 xx 10^(-6)) + ((210 xx 6^(-6))^(2))/(2 xx 2 xx 10^(-6))`
`= 1.8 xx 10^(-2) J`