### In the circuit shown, the switch S is connected to position P for a long time so that the charge on the capacitor becomes $q_1$ μC. Then S is switched to position Q. After a long time, the charge on the capacitor is $q_2$ μC.

Q.1 The magnitude of $q_1$ is ___ .
Q.2 The magnitude of $q_2$ is ___ .

Solution

The figure below shows the situation when switch S is connected to position P.

No current flows through the capacitor in steady state.

So, $I=\frac {2-1}{1+2}=\frac {1}{3} A$

$V_C =2-2I=2-2\times \frac {1}{3}=\frac {4}{3} V$

$q_1 =CV_C =1\times \frac {4}{3} = \frac {4}{3} \mu C$

Actually, the answer to Q.1 is $q_1 =\frac {4}{3}=1.33$ as the unit is outside.

The figure below shows the situation when switch S is connected to position Q.

Again, no current flows through the capacitor in steady state.

$I=\frac {2}{1+2}=\frac {2}{3} A$

$V_C = \frac {2}{3} \times 1 = \frac {2}{3} V$

$q_2 = 1 \times \frac {2}{3} = \frac {2}{3} \mu C$

Actually, the answer to Q.2 is $q_2 =\frac {2}{3}=0.67$ as the unit is outside.