Answer :

Given


Capacitance of capacitor C1 = 6C = 6μF


Capacitance of capacitor C2 = 3C = 3μF


Capacitance of capacitor C3 = 3C = 3μF


Voltage of battery E = 9V


When the switch K1 is closed and K2 is open, then the capacitors C1 and C �2 get charged and voltage V1 and V2 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 C1 and C2,




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



and



Now the charges on the capacitors


and



After this, when K1 was opened and K2 is closed the capacitor C2 becomes parallel to C3.


By conservation of charge we have



Where Q2’ is the new charge on capacitor C2 and Q3 is the charge flown from C2 to C3. 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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