2. Electric Potential and Capacitance
medium

A parallel plate capacitor whose capacitance $C$ is $14\, pF$ is charged by a battery to a potential difference $V =12\, V$ between its plates. The charging battery is now disconnected and a porcelin plate with $k =7$ is inserted between the plates, then the plate would oscillate back and forth between the plates with a constant mechanical energy of $..........pJ$. (Assume no friction)

A

$872$

B

$972$

C

$784$

D

$864$

(JEE MAIN-2021)

Solution

$U _{ i }=\frac{1}{2} \times 14 \times 12 \times 12 pJ \quad\left(\because U =\frac{1}{2} CV ^{2}\right)$

$=1008 pJ$

$U _{ f }=\frac{1008}{7} pJ =144 pJ \quad\left(\because C _{ m }= k C _{0}\right)$

Mechanical energy $=\Delta U$

$=1008-144$

$=864 pJ$

Standard 12
Physics

Similar Questions

Consider a simple $RC$ circuit as shown in Figure $1$.

Process $1$: In the circuit the switch $S$ is closed at $t=0$ and the capacitor is fully charged to voltage $V_0$ (i.e. charging continues for time $T \gg R C$ ). In the process some dissipation ( $E_D$ ) occurs across the resistance $R$. The amount of energy finally stored in the fully charged capacitor is $EC$.

Process $2$: In a different process the voltage is first set to $\frac{V_0}{3}$ and maintained for a charging time $T \gg R C$. Then the voltage is raised to $\frac{2 \mathrm{~V}_0}{3}$ without discharging the capacitor and again maintained for time $\mathrm{T} \gg \mathrm{RC}$. The process is repeated one more time by raising the voltage to $V_0$ and the capacitor is charged to the same final

take $\mathrm{V}_0$ as voltage

These two processes are depicted in Figure $2$.

 ($1$) In Process $1$, the energy stored in the capacitor $E_C$ and heat dissipated across resistance $E_D$ are released by:

$[A]$ $E_C=E_D$ $[B]$ $E_C=E_D \ln 2$ $[C]$ $\mathrm{E}_{\mathrm{C}}=\frac{1}{2} \mathrm{E}_{\mathrm{D}}$ $[D]$ $E_C=2 E_D$

 ($2$) In Process $2$, total energy dissipated across the resistance $E_D$ is:

$[A]$ $\mathrm{E}_{\mathrm{D}}=\frac{1}{2} \mathrm{CV}_0^2$     $[B]$ $\mathrm{E}_{\mathrm{D}}=3\left(\frac{1}{2} \mathrm{CV}_0^2\right)$    $[C]$ $\mathrm{E}_{\mathrm{D}}=\frac{1}{3}\left(\frac{1}{2} \mathrm{CV}_0^2\right)$   $[D]$ $\mathrm{E}_{\mathrm{D}}=3 \mathrm{CV}_0^2$

Given the answer quetion  ($1$) and  ($2$)

normal
(IIT-2017)

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