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

Explanation

(i) `q_A=90 muC,q_B=10 muC,q_C=210 muC` (ii) 47.4,8
(i) Charge in capacitor A,before joining with uncharged capacitor
`q_A=CV=(100)(3)muC=300 muC`
Similarly,charge on capacitor B
`q_B=(180)(20)muC=360muC`
Let`q_1,q_2` and `q_3` be the charge on three capacitor after joining them as shown in figure .
(`q_1,q_2` and `q_3` are in microculombs)
From conservation of charge
Net Charge on plates 2 and 3 before joining =net charge after joining
`300=q_1+q_1` ...(i)
Similarly,net charge on plate 4 and 5 before joining
=net charge after joining `-360=-q_2-q_3`
or `360=q_2+q_3` ....(ii)
Applying Kinchoff's secon law en closed loop ABCDA
`q_1/3-q_2/2+q_3/2=0 or 2q_1-3q_2+3q_3=0`
Solving Eqs (i),(ii) and (iii),we get
`q_1=90muC,q_2=210 muC and q_3=150 muC`
(a) Electrostatic energy stored before, completing of circuit
`U_i=1/2(3xx10^(-6))(1000)^(2)+1/2(2xx10^(-6))(180)^(2)(Q U=1/2CV^(2))=4.74xx10^(-2)J orU_i=47.4m J`
(b) Electrostatic energy stored after, completing the circuit
`U _f=1/2((90xx10^(-6))^(2))/((3xx10^(-6)))+1/2((210xx10^(*6)))/((2xx10^(-6)))+1/2((150xx10^(-6))^(2))/((2xx10^(-6)))[U=1/2q^(2)/c=18 /c]=1.8xx10^(-2)J orU_f=18 mJ`