Question

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

Answer 1

(a) Charge on capacitor A, before joining with an uncharged
capacitor
$q_A=CV=(100)(3)\mu C=300\mu C$
Similarly, charge on capacitor B
$q_B=(180)(2)\mu C=360\mu C$
Let $q_1,q_{2}and{q}_3$ be the charges on the three capacitors
after joining them as shown in figure.
$(q_1,q_{2}and{q}_3$ 3are in microcoulombs)
From conservation of charge
Net charge on plates 2 and 3 before joining
= net charge after joining
$\therefore 300=q_1+q_2...(i)$
Similarly, net charge on plates 4 and 5 before joining
= net charge after joining
$-360=-q_2-q_3$
or $360=q_2+q_3...(ii)$
Applying Kirchhoff s second law in closed loop ABCDA
$\frac{q_1}{3}-\frac{q_2}{2}+\frac{q_3}{2}=0$
or $2q_1-3q_2+3q_3=0...(iii)$
Solving Eqs. (i), (ii) and (iii), we get
$q_1=90\mu C,q_2=210\mu C$
and $q_3=150\mu C$
(b) (i) Electrostatic energy stored before, completing the
circuit
$U_i=\frac{1}{2}(3\times 10^{-6})(100{)}^2+\frac{1}{2}(2\times 10^{-6}(180{)}^2$
$(U=\frac{1}{2}C{V}^2)$
$=4.74\times 10^{-2}Jor{U}_i=47.4mJ$
(ii) Electrostatic energy stored after, completing the
circuit
$U_f=\frac{1(90\times 10^{-6}{)}^2}{2(3\times 10^{-6})}+\frac{1(210\times 10^{-6}{)}^2}{2(2\times 10^{-6})}$
$+\frac{1(150\times 10^{-6}{)}^2}{2(2\times 10^{-6})}[U=\frac{1q^2}{2C}]$
$=1.8\times 10^{-2}Jor{U}_f=18mJ$