Question

The capacitance of a parallel plate capacitor is $C_0$ when the plates have air between them. This region is now filled with a dielectric slab of dielectric constant $K$ and capacitor is connected with a battery of emf $E$ and zero internal resistance. Now slab is taken out, then during removal of slab:

$1.)$ Charge $E{C}_0(K-1)$ flows through the cell.

$2.)$ Energy $E^2{C}_0(K-1)$ is absorbed by cell.

$3.)$ The energy stored in the capacitor is reduced by $E^2{C}_0(K-1)$.

$4.)$ The external agent has to do $E^2{C}_0(K-1)$ amount of work to take out the slab.

Select the correct phenomenon happening during the process:

• A. $1\&3$
• B. $2\&3$
• C. $2,3\&4$
• D. $1\&2$

Answer 1

The correct option is D $1\&2$

Initial charge on capacitor when slab inside the plates,

$Q_i=K{C}_{0}E$

Charge after removing dielectric slab,

$Q_f=C_{0}E$

Amount of charge that flows through the cell is,

$\Delta Q=K{C}_{0}E-C_{0}E=C_{0}E(K-1)$

After removing dielectric, the charge on capacitor is decreased. Thus more charge will be supplied by the battery

Now, energy absorbed by cell (workdone on battery),

$\Delta E^{\prime }=\Delta Q\times E$

$\therefore \Delta E^{\prime }=C_0{E}^2(K-1)$

Initial energy stored in capacitor,

$U_i=\frac{1}{2}K{C}_0{E}^2$

Final energy stored in capacitor,

$U_f=\frac{1}{2}{C}_0{E}^2$

$\therefore$ loss of energy will be

$\Delta U_{lost}=\frac{1}{2}K{C}_0{E}^2-\frac{1}{2}{C}_0{E}^2$

$\Rightarrow \Delta U_{lost}=\frac{1}{2}{C}_0{E}^2(K-1)$

Workdone by external agent is equal to the energy dissipated in circuit.

$\therefore W_{ext}=\frac{1}{2}{C}_0{E}^2(K-1)$

Correct phenomenon $\to \left(1\right)\&\left(2\right)$ only.

Hence, option $\left(d\right)$ is correct.

Why this question? Caution: If a positive charge flows through a cell such that it leaves the +ve terminal of battery and enters through its -ve terminal, then work is done by battery i.e. $W_{battery}>0$ and for the reverse case $W_{battery}<0$.