When switch `S` is thrown to the left in figure, the plates of capacitor 1 acquire a potential difference `V_0`. Capacitors 2 and 3 are initially uncharged. The switch is now thrown to the right. What are the final charges `q_1, q_2` and `q_3` on the capacitors?

Initially capacitor 'A' is charged to a potential drop epsilon and capacitor B is uncharged At t=0, switch S is closed, then the maximum current through the inductor is

Charges q_(1) , q_(2) and q_(3) are placed on capacitors of capacitance C_(1) , C_(2) and C_(3) , respectively, arranged in series as shown. Switch S is there closed. What are the final charges q'_(1) , q'_(2) and q'_(3) on the capacitors? Given q_(1)=30muC , q_(2)=muC , q_(3)=10muC , C_(1)=10muF , C_(2)=20muF , C_(3)=30muF and epsilon=12"volt"

A capacitor of capacitance C is charged to a potential difference V from a cell and then disconnected from it. A charge +Q is now given to its positive plate. The potential difference across the capacitor is now

A capacitor of capacitance C is charged to a potential difference V from a cell and then disconnected from it. A charge +Q is now given to its positive plate. The potential difference across the capacitor is now.

The plates of a capacitor of capacitance C are given the charges Q_1 and Q_2 as shwon in figure. Now the switch is closed at t =0. Find the charges on plates after time t.

If A is the area of each plate, charge on it is q and potential difference is V then the distance between the parallel plate capacitor is

A battery is used to charge a parallel plate combination of two identical capacitors. If the potential difference across the battery is V and the total charge Q flows through the battery during the charging process then 1) The charge on each capacitor is Q/2 and the potential of each capacitor is V/2 2) The charge on each capacitor is Q and the potential of each capacitor is V 3) The charge on each capacitor is Q/2 and the potential of each capacitor is V 4) The charge on each capacitor is Q and the potential of each capacitor is V/2

The left plate of the capacitor shown in the figure above carries a charge +Q while the right plate is uncharged at t=0 . The total charge on the right plate after closing the switch will be: