$A$ $10\,\mu F$ capacitor is charged to a potential difference of $1000\, V$. The terminals of the charged capacitor are disconnected from the power supply and connected to the terminals of an uncharged $6\,\mu F$ capacitor. What is the final potential difference across each capacitor?......$V$

$A$ capacitor $P$ with capacitance $10 \times 10^{-6} \text{ F}$ is fully charged with a potential difference of $6.0 \text{ V}$ and disconnected from the battery. The charged capacitor $P$ is connected across another capacitor $Q$ with capacitance $20 \times 10^{-6} \text{ F}$. The charge on capacitor $Q$ when equilibrium is established will be $\alpha \times 10^{-5} \text{ C}$. (Assume capacitor $Q$ does not have any charge initially). The value of $\alpha$ is . . . . . . .

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

Consider two uncharged capacitors of equal capacitance $200 \text{ pF}$. One of them is charged by a $100 \text{ V}$ supply and disconnected. Now, this capacitor is connected to the uncharged capacitor. The amount of electrostatic energy lost in the process is:

$A$ condenser of capacity $C_1$ is charged to a potential $V_0$. The electrostatic energy stored in it is $U_0$. It is connected to another uncharged condenser of capacity $C_2$ in parallel. The energy dissipated in the process is

Two capacitors having capacitance $C_{1}$ and $C_{2}$ respectively are connected as shown in the figure. Initially,capacitor $C_{1}$ is charged to a potential difference $V$ volt by a battery. The battery is then removed and the charged capacitor $C_{1}$ is now connected to uncharged capacitor $C_{2}$ by closing the switch $S$. The amount of charge on the capacitor $C_{2}$ after equilibrium is