$A$ $4 \mu F$ capacitor is charged by a $200 \ V$ battery. It is then disconnected from the supply and connected to another uncharged $2 \mu F$ capacitor. During the process,the loss of energy (in $J$) is:

$A$ conducting sphere $S_1$ of radius $r_1 = 3 \text{ cm}$ is connected by a conducting wire to another conducting sphere $S_2$ of radius $r_2 = 2 \text{ cm}$. Before they are connected,$S_1$ carries a charge of $10 \text{ units}$. The electric potential at a point which is at a distance $4 \text{ cm}$ from the centre of $S_1$ and a distance $3 \text{ cm}$ from the centre of $S_2$ is:

$A$ $10\,\mu F$ capacitor is fully charged to a potential difference of $50\, V$. After removing the source voltage, it is connected to an uncharged capacitor in parallel. Now, the potential difference across them becomes $20\, V$. The capacitance of the second capacitor is $\dots \mu F$.

Two identical capacitors are connected in parallel across a potential difference $V$. After they are fully charged,the positive plate of the first capacitor is connected to the negative plate of the second,and the negative plate of the first is connected to the positive plate of the other. The loss of energy will be

Two metallic charged spheres whose radii are $20\,cm$ and $10\,cm$ respectively,have each $150\,\mu C$ positive charge. The common potential after they are connected by a conducting wire is

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?