$A$ capacitor of capacity $C$ is connected to a battery of potential $V$ in parallel. The distance between its plates is reduced to half at once,while the battery remains connected. Then,the energy supplied by the battery to charge the capacitor to potential $V$ again is:

Two capacitors of $6 \, pF$ are connected in series and a potential difference of $5000 \, V$ is applied across the combination. If the combination is disconnected and then reconnected in parallel, what will be the potential difference between the plates?

The distance between the plates of a parallel plate capacitor is $5d$. The positively charged plate is at $x=0$ and the negatively charged plate is at $x=5d$. Two slabs,one of a conductor and the other of a dielectric,both of equal thickness $d$,are inserted between the plates as shown in the figure. The potential versus distance graph will look like:

$A$ capacitor of $2\,\mu F$ capacitance is charged steadily from $0$ to $5\,C$. Which of the following graphs correctly represents the variation of potential difference $(V)$ across its plates with respect to the charge $(Q)$ on the capacitor?

$A$ parallel plate capacitor of capacity $100 \mu F$ is charged by a battery of $50 \text{ V}$. The battery remains connected and if the plates of the capacitor are separated so that the distance between them becomes double the original distance,the additional energy given by the battery to the capacitor in joules is

$A$ parallel plate capacitor of capacitance $2\; F$ is charged to a potential $V$. The energy stored in the capacitor is $E_1$. The capacitor is now connected to another uncharged identical capacitor in parallel combination. The energy stored in the combination is $E_2$. The ratio $E_2 / E_1$ is