Question

A parallel plate capacitor of plate area $A$ and plate separation $d$ is charged to potential difference $V$ and then the battery is disconnected. A slab of dielectric constant $k$ is then inserted between the plates of the capacitor so as to fill the space between the plates. If $Q,E$ and $W$ denote respectively, the magnitude of charge on each plate, the electric field between the plates (after the slab is inserted) and the work done on the system, in question, in the process of inserting the slab, then:

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

Answer 1

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

The correct options are
A $Q=\frac{{ϵ}_{0}AV}{d}$
C $E=\frac{V}{kd}$
D $W=-\frac{{ϵ}_{0}A{V}^2}{2d}(1-\frac{1}{k})$

As battery is removed, 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 inserting dielectric, the field, $E=\frac{Q_0}{Ak{ϵ}_0}=\frac{V}{kd}$
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 $=\frac{Q_0^2}{2C_0}-\frac{Q_0^2}{2k{C}_0}$

$W=\frac{Q_0^2}{2C_0}(1-\frac{1}{k})=1/2C_0{V}^2(1-\frac{1}{k})=\frac{A{ϵ}_0{V}^2}{2d}(1-\frac{1}{k})$

Note: sign convention , work done by the system is positive sign and work done on the system is negative sign.