Question

A parallel-plate air condenser of plate area A and separation d is charged to potential V and then the battery is removed. Now a slab of dielectric constant k is introduced between the plates. If Q, E and W denote respectively the magnitude of charge on each plate, the electric field between the plates (after introduction of dielectric slab) and work done on the system in the process of introducing the slab, then

• A. $W=\frac{{\epsilon }_{0}AV{h}^2}{2d}(1-1/k)$
• B. $Q=\frac{{\epsilon }_{0}KAV}{d}$
• C. $Q=\frac{{\epsilon }_{0}AV}{d}$
• D. $E=\frac{V}{kd}$

Answer 1

Answer: (A), (C), (D)

$W=\frac{{\epsilon }_{0}AV{h}^2}{2d}(1-1/k)$
$Q=\frac{{\epsilon }_{0}AV}{d}$
$E=\frac{V}{kd}$

As battery is removed so charge remains constant. $Q_0=C_{0}V$
here $C_0=\frac{A{ϵ}_0}{d}$, $V=\frac{Q_{0}d}{A{ϵ}_0}$
after insert dielectric, the filed, $E=\frac{Q_0}{Ak{ϵ}_0}=\frac{V}{kd}$
and charge $Q=Q_0=C_{0}V=\frac{A{ϵ}_{0}V}{d}$ and $C=k{C}_0$
The work done , $W=$ energy loss in capacitor $=\frac{Q_0^2}{2C_0}-\frac{Q_0^2}{2k{C}_0}$
$W=\frac{Q_0^2}{2C_0}(1-\frac{1}{k})=1/2C_0{V}^2(1-\frac{1}{k})=\frac{A{ϵ}_0{V}^2}{2d}(1-\frac{1}{k})$
Ans :(C),(D)