The given circuit shows two capacitors connected to a battery. After the capacitors are completely charged,the battery is removed and the capacitors are connected with plates of opposite polarity together. Then the charge on $2C$ and the energy lost in the process are respectively

Two thin conducting shells of radii $R$ and $3R$ are shown in the figure. The outer shell carries a charge $+Q$ and the inner shell is neutral. The inner shell is earthed with the help of a switch $S$.

Three point charges $1 \ C$,$-2 \ C$,and $-2 \ C$ are placed at the vertices of an equilateral triangle of side $1 \ m$. The work done by an external force to increase the separation of the charges to $2 \ m$ in joules is ($\varepsilon_0 =$ permittivity of air).

$A$ capacitor $C_1 = 1.0 \, \mu F$ is charged up to a voltage $V = 60 \, V$ by connecting it to a battery $B$ through switch $(1)$. Now,$C_1$ is disconnected from the battery and connected to a circuit consisting of two uncharged capacitors $C_2 = 3.0 \, \mu F$ and $C_3 = 6.0 \, \mu F$ in series through a switch $(2)$ as shown in the figure. The sum of the final charges on $C_2$ and $C_3$ is ...... $\mu C$.

$A$ 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 plates $S$ and $T$ of an uncharged parallel plate capacitor are connected across a battery. The battery is then disconnected and the charged plates are now connected in a system as shown in the figure. The system shown is in equilibrium. All the strings are insulating and massless. The magnitude of charge on one of the capacitor plates is: [Area of plates = $A$]