$A$ parallel plate capacitor is charged by a battery until the potential difference between the plates equals the electromotive force of the battery. What is the ratio of the energy stored in the capacitor to the work done by the battery?

$A$ parallel plate capacitor having a plate separation of $2\, mm$ is charged by connecting it to a $300\, V$ supply. The energy density is.....$J/m^3$

The charge $Q$ on a capacitor varies with voltage $V$ as shown in the figure,where $Q$ is taken along the $X$-axis and $V$ along the $Y$-axis. The area of triangle $OAB$ represents

$A$ fully charged capacitor has a capacitance $C$. It is discharged through a small coil of resistance wire,embedded in a block of specific heat $s$ and mass $m$ under thermally isolated conditions. If the temperature of the block is raised by $\Delta T$,the potential difference $V$ across the capacitor initially is

An uncharged capacitor is connected to a battery. While charging the capacitor,how much is the energy lost,from the energy supplied by the battery (in $\%$)?

$A$ parallel plate capacitor has a uniform electric field $E \ (V/m)$ in the space between the plates. If the distance between the plates is $d \ (m)$ and the area of each plate is $A \ (m^2)$,then the energy (joules) stored in the capacitor is: