$A$ capacitor of capacitance $10\,\mu F$ has been charged to $100\,V$. It is now connected to another uncharged capacitor in parallel. The common potential becomes $40\,V$. The capacitance of the other capacitor is......$\mu F$.

$A$ $20 \mu F$ capacitor is charged to $5 V$ and isolated. It is then connected in parallel with an uncharged $30 \mu F$ capacitor. The decrease in the energy of the system will be (in $\text{J}$)

$A$ parallel plate capacitor of capacitance '$C$' is connected to a battery and charged to a potential '$V$'. Another capacitor of capacitance '$3C$' is charged to a potential '$3V$'. The charging battery is then disconnected and both the capacitors are connected in parallel to each other such that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is (in $CV^2$)

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 $C$ 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

The energy stored in a capacitor is $U$. If an uncharged capacitor of the same capacity is connected in parallel with it,then the energy stored in each capacitor is: