If the charge on a capacitor is increased by $2 \ C$,the energy stored in it increases by $21\%$. The original charge on the capacitor is....$C$

Two capacitors,one of capacitance $C$ and the other of capacitance $C/2$,are connected in parallel to a battery of potential difference $V$. Calculate the heat produced in the connecting wires.

Initially, $n$ identical capacitors are joined in parallel, and are charged to potential $V$. Now they are separated and joined in series. Then:

What are the charges stored in the $1 \,\mu F$ and $2 \,\mu F$ capacitors in the circuit below,once the currents become steady?

Two capacitors are shown in the diagram. Each is charged to potential $V_0$. If capacitor $C$ is filled with a dielectric of constant $2$ and capacitor $2C$ is filled with a dielectric of constant $3$,then the key is closed. The final voltage on the capacitor $C$ 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: