Question

A parallel plate capacitor of plate area A and plate separation d is charged to p.d. 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 insertion of dielectric and work done on the system 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)

$Q=\frac{{ϵ}_{0}AV}{d}$
$E=\frac{V}{Kd}$
$W=\frac{{ϵ}_{0}A{V}^2}{2d}(1-\frac{1}{K})$

Before insert dielectric, capacitance, $C_0=\frac{A{ϵ}_0}{d}$
Magnitude of charge on each plate, $Q=C_{0}V=\frac{A{ϵ}_0}{d}V$
Before disconnect the battery, filed between the plates, $E_0=\frac{V}{d}$
Afe insert dielectric, electric filed, $E=\frac{E_0}{K}=\frac{V}{Kd}$
Before insert dielectric, work done, $=W_i=\frac{1}{2}{C}_0{V}^2$
After insert work done, $W_f=\frac{1}{2}\left(K{C}_0\right)(\frac{V}{K}{)}^2$. $(C_f=\frac{A{ϵ}_{0}K}{d},V_f=\frac{E}{d})$
So net work done,$W=W_i-W_f=\frac{1}{2}{C}_0{V}^2-\frac{1}{2}\left(K{C}_0\right)(\frac{V}{K}{)}^2=\frac{1}{2}{C}_0{V}^2(1-\frac{1}{K})=\frac{A{ϵ}_0{V}^2}{d}(1-\frac{1}{K})$