$A$ $20\,F$ 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.......$J$

Two capacitors $C_1$ and $C_2 = 2C_1$ are connected in a circuit with a switch between them as shown in the figure. Initially,the switch is open and $C_1$ holds charge $Q$. The switch is closed. At steady state,the charge on the capacitors will be:

Two capacitors of $2\ \mu F$ and $3\ \mu F$ are charged to $300\ V$ and $500\ V$ respectively. When they are connected in parallel, the energy lost is: (in $J$)

Two identical capacitors are joined in parallel,charged to a potential $V$ and then separated and then connected in series,i.e.,the positive plate of one is connected to the negative plate of the other. What happens to the system?

An uncharged parallel plate capacitor having a dielectric of constant $K$ is connected to a similar air-cored parallel capacitor charged to a potential $V$. The two share the charge and the common potential is $V'$. The dielectric constant $K$ is

Two capacitors of $100 \mu F$ and $50 \mu F$ are connected in parallel. If the potential difference across $100 \mu F$ is $20 \text{ V}$ and across $50 \mu F$ is $40 \text{ V}$,then the common potential of the parallel combination will be (assuming same polarities of the capacitors are connected together).