Answer :

Given

Capacitance of capacitor C_{1} = 6C = 6μF

Capacitance of capacitor C_{2} = 3C = 3μF

Capacitance of capacitor C_{3} = 3C = 3μF

Voltage of battery E = 9V

When the switch K_{1} is closed and K_{2} is open, then the capacitors C_{1} and C �_{2} get charged and voltage V_{1} and V_{2} develops across the capacitors respectively. Applying Kirchhoff’s voltage law in the circuit, we get

We know that the charge stored in a capacitor and the voltage applied across it are related as:

Then for the capacitors C_{1} and C_{2},

Then from equations (1) and (2) we have

and

Now the charges on the capacitors

and

After this, when K_{1} was opened and K_{2} is closed the capacitor C_{2} becomes parallel to C_{3}.

By conservation of charge we have

Where Q_{2}’ is the new charge on capacitor C_{2} and Q_{3} is the charge flown from C_{2} to C_{3}. Since they are connected in parallel, they have a common potential difference across them. Let the common potential be V’, then

Therefore, the charges on the capacitors are:

and

Therefore, the final charges on all the capacitors are:

,

and

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