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.

Answer

A

Explanation

i. charge on capcitor before joining wilth an uncharged capacitor.

`q_A=CV=(100_(3)muC=300muC)`
Similarly, charge on capacitor B
`q_B=(180)(2)muC=360muC`
Let `q_1,q_2` and `q_3` be the charge onthe three capacitors after joining them as shown in figure.
`(q_1,q_2` and `q_3` are in microcoulombs)
From conservtion of charge
Net charge on plates 2 and 3 before joining =net charge after joining
`:. 300=q_1+q_2`...........i
Similarly, net chrge on plantes 4 and 5 before joiing =net chasrge after joining
`-360=-q_2-q_3`
`or 360=q_2+q_3`........ii
Applying Kirchoff's second law in closed loop
`q_1/3-q_2/2+q_3/2=0`
or `2q_1-3q_2+3q_3=0`........iii
Solving eqn i , ii, iii we get
`q_1=90muC`
`q_2=210muC`
`q_3=150muC`
ii. a. Electrostatic energy stored before completing the circuit.
`U_i=1/2(3xx10^-6)(100)^2+1/2(2xx10^-60(180))^2 (U=1/2CV^2)`
`=4.74xx10^-2J or U_i=47.4mJ `
b. electrostatic energy stored after completing the circuit,
`U_f=1/2((90xx0^-6)^2)/((3xx10^-6))1/2((210xx10^-6)^2)/((2xx10^-6))`
`+1/2((150xx10^-6)^2)/((2xx10^-6) [U=1/2q^2/C]`
`=1.8xx10^I-2J` or `U_f=18mJ`