A parallel plate capacitor is connected to a battery. The plates are pulled apart with a uniform speed. If $x$ is the separation between the plates, the time rate of change of electrostatic energy of capacitor is proportional to
$x^{-2}$
$x$
$x^{-1}$
$x^{2}$
The plates of a parallel plate capacitor have an area of $90 \,cm ^{2}$ each and are separated by $2.5\; mm .$ The capacitor is charged by connecting it to a $400\; V$ supply.
$(a)$ How much electrostatic energy is stored by the capacitor?
$(b)$ View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume $u$. Hence arrive at a relation between $u$ and the magnitude of electric field $E$ between the plates.
A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system
A parallel plate capacitor is charged to a certain potential and the charging battery is then disconnected. Now, if the plates of the capacitor are moved apart then:
capacitor is used to store $24\, watt\, hour$ of energy at $1200\, volt$. What should be the capacitance of the capacitor
$(a)$ A $900 \;p\,F$ capacitor is charged by $100 \;V$ battery [Figure $(a)$]. How much electrostatic energy is stored by the capacitor?
$(b)$ The capacitor is disconnected from the battery and connected to another $90\; p\,F$ capacitor [Figure $(b)$]. What is the electrostatic energy stored by the system?