Two capacitors of equal capacitance $(C_1 = C_2)$ are shown in the figure. Initially, while the switch $S$ is open, one of the capacitors is uncharged and the other carries charge $Q_0$. The energy stored in the charged capacitor is $U_0$. Sometimes after the switch is closed, the capacitors $C_1$ and $C_2$ carry charges $Q_1$ and $Q_2$, respectively; the voltages across the capacitors are $ V_1$ and $V_2$; and the energies stored in the capacitors are $U_1$ and $U_2$. Which of the following statements is INCORRECT ?
$Q_0$ $=$ $\frac{1}{2}$ $(Q_1 + Q_2)$
$Q_1 = Q_2$
$V_1 = V_2$
$U_0 = U_1 + U_2$
If initial charge on all the capacitors were zero, work done by the battery in the circuit shown is ........... $mJ$
A parallel plate capacitor has a uniform electric field ' $\overrightarrow{\mathrm{E}}$ ' in the space between the plates. If the distance between the plates is ' $\mathrm{d}$ ' and the area of each plate is ' $A$ ', the energy stored in the capacitor is : $\left(\varepsilon_{0}=\right.$ permittivity of free space)
A parallel plate capacitor has circular plates of $10\, cm$ radius separated by an air-gap of $1\, mm$. It is charged by connecting the plates to a $100\, volt$ battery. Then the change in energy stored in the capacitor when the plates are moved to a distance of $1\, cm$ and the plates are maintained in connection with the battery, is
A parallel plate capacitor carries a charge $q$. The distance between the plates is doubled by application of a force. The work done by the force is
The energy stored in a condenser is in the form of