Question

Two capacitor $C_{1}and{C}_2$ are connected across a 60 V battery as shown in figure 1. In this situation, the final charges appearing on $C_{1}and{C}_2$ are 120mC and 180mC respectively. In another arrangement, the same capacitors $C_{1}and{C}_2$ are completely discharged and then connected with the same batteries along with a switch ‘S’ as shown in figure 2.

Consider the charges which will flow after shorting the switch ‘S’ in figure 2 through sections 1, 2 and 3 in the directions indicated by arrows as $q_1,q_2,and{q}_3$ respectively. Then,

• A. $q_1=+24\mu C;q_2=-36\mu C$
• B. $q_1=+24\mu C;q_3=-60\mu C$
• C. $q_2=-36\mu C;q_3=-60\mu C$
• D. $q_1=+24\mu C;q_3=-60\mu C$

Answer 1

The correct option is B $q_1=+24\mu C;q_3=-60\mu C$


$C_1=\frac{Q_1}{V}=\frac{120}{60}=2\mu F{C}_2=\frac{Q_2}{V}=\frac{180}{60}=3\mu F$


Before switch is connected
$V_{01}=P.Dacross{C}_1$ before shorting the switch
$2E\times \frac{C_2}{C_1+C_2}=72V$
$V_{02}=$ P.d across $C_2$ before shorting the switch
$2E\times \frac{C_1}{C_1+C_2}=48V$
After switch connected:
$V_1=$ final p.d across $C_1=E$
$V_2=$ final p.d across $C_2=E$
The charge $q_1$ goes on to the upper plate of $C_1$
$\Rightarrow q_1=\Delta Q_1=C_1{V}_1-C_1{V}_{01}=C_1[E-\frac{\left(2E\right)C_2}{C_1+C_2}]=\frac{C_{1}E(C_1-C_2)}{C_1+C_2}=-24\mu C$
The charge $q_2$ goes to the lower negative plate of $C_2$.
$q_2=\Delta Q_2=(-C_2{V}_2)-(-C_2{V}_{02})=\frac{C_{2}E(C_1-C_2)}{C_1+C_2}=-36\mu C$
Also from conservation of charge,
$q_1+q_2+q_3=0\Rightarrow q_3=60\mu C$