Work done by an external agent in separating the parallel plate capacitor is
$CV$
$\frac{1}{2}{C^2}V$
$\frac{1}{2}C{V^2}$
None of these
capacitor is used to store $24\, watt\, hour$ of energy at $1200\, volt$. What should be the capacitance of the capacitor
A parallel plate capacitor of capacitance $C$ is connected to a battery and is charged to a potential difference $V$. Another capacitor of capacitance $2C$ is connected to another battery and is charged to potential difference $2V$. The charging batteries are now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is
$A$ $2$ $\mu F$ capacitor is charged to a potential $=$ $10\,V$. Another $4$ $\mu F$ capacitor is charged to a potential $=$ $20\,V$. The two capacitors are then connected in a single loop, with the positive plate of one connected with negative plate of the other. What heat is evolved in the circuit?......$\mu J$
Three plates $A, B$ and $C$ each of area $0.1 \ m^2$ are separated by $0.885\ mm$ from each other as shown in the figure. $A$ $10$ $V$ battery is used to charge the system. The energy stored in the system is
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.