$A$ capacitor of capacitance $C$ is initially charged to a potential difference of $V$ $volt$. Now it is connected to a battery of $2V$ with opposite polarity. The ratio of heat generated to the final energy stored in the capacitor will be

$A$ 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 plate 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:

$A$ thin metallic partition of negligible thickness is inserted between two shaded metallic plates as shown. The remaining ends are then packed with insulating plates to form a container-like structure. Two taps shown are opened at $t = 0$ and finally closed at $t = 5 \ s$. Find the capacitance of the system between $A$ and $B$ after closing the taps. (Assume the liquid to be non-conducting). Volumetric flow rates and dielectric constants of the liquid are given.

In the circuit shown,each capacitor has a capacitance $C$. The emf of the cell is $E$. If the switch $S$ is closed:

As shown in the figure,what will be the charge on each capacitor and the potential difference across each,respectively?

When a potential difference of $10^{3} \, V$ is applied between $A$ and $B$, a charge of $0.75 \, mC$ is stored in the circuit. The value of $C$ is (in $\mu F$):