Five identical capacitors, each of capacitance $20 \mu F$, are connected to a battery of $150 V$ in the combination shown in the diagram. What is the total amount of charge stored?

$A$ parallel plate capacitor of capacity $C_0$ is charged to a potential $V_0$.
$(i)$ The energy stored in the capacitor when the battery is disconnected and the separation is doubled is $E_1$.
$(ii)$ The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is $E_2$.
Then the value of $E_1/E_2$ is:

The numerical value of the charge on either plate of the capacitor $C$ shown in the figure is

Two capacitors $C_1 = 2\,\mu F$ and $C_2 = 6\,\mu F$ in series are connected in parallel to a third capacitor $C_3 = 4\,\mu F$. This arrangement is then connected to a battery of $e.m.f. = 2\,V$,as shown in the figure. How much energy is lost by the battery in charging the capacitors?

$A$ parallel plate capacitor is charged and then disconnected from the source of steady $E.M.F.$ The plates are then drawn apart farther. Again it is connected to the same source. Then:

The potential difference that must be applied across the parallel and series combination of three identical capacitors such that the energy stored in them becomes the same. The ratio of potential difference in parallel to series combination is