Two capacitors $C_1$ and $C_2 = 2C_1$ are connected with a switch $S$ as shown in the figure. Initially,the switch is open and the charge on capacitor $C_1$ is $Q$. Now,when the switch is closed,the final charges on the capacitors are:

The circuit shows two capacitors $A$ and $B$ of capacitances $C$ and $2C$ respectively. When they are fully charged,the cell is removed and the capacitors are connected with their plates of opposite polarities touching each other. Then:
$(a)$ Charge on $A$ is $\frac{4CE}{9}$
$(b)$ Charge on $B$ is $\frac{8CE}{9}$
$(c)$ Loss of energy in this process is $\frac{CE^2}{3}$
The correct statement$(s)$ is/are:

$A$ capacitor of capacitance $C$ and potential $V$ has energy $E$. It is connected to another capacitor of capacitance $2C$ and potential $2V$. Then the loss of energy is $\frac{x}{3} E$,where $x$ is . . . . . . .

$A$ capacitor of capacitance $4 \mu F$ is charged to a potential difference of $6 \ V$ with a battery. The battery is then removed and in its place another capacitor of capacitance $8 \mu F$ is introduced and the circuit is closed. The potential difference attained by each of the capacitors in $V$ is

$A$ capacitor of capacitance $C$ is charged to a potential difference $V_0$. The charging battery is removed and the capacitor is now connected to an uncharged capacitor of unknown capacity. The potential difference across the combination becomes $V$. The unknown capacitance is

Two charged spheres of radii $a$ and $b$ are connected by a wire. The ratio of the electric fields on their surfaces $E_a/E_b$ is: