In the circuit shown below the switch S is connected to position P for a long time:IIT JEE Adv 2021

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In the circuit shown below, the switch S is connected to position P for a long time so that the charge on the capacitor become q1 µC. The S is switched to position Q. After a long time, the charge on the capacitor is q2 µC. The magnitude of q1 and q2 is:

At steady state current passing through capacitor is 0
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How is this possible, when q1 was 1.33 and then C2 was charging sir where does that 1.3 charge gone, All online videos shows wrong answer of 2nd option according to u and others ans of second is 0.66 but correct anser is 0.67 which is q1 - 0.66 then correct answer will be 0.67 but the only reason of my question is that hy we did minus why not addition, for which I was looking but u all had done wrongly

kajalbhaskar