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 capacitor so as to fill the space between the plates. If $Q,E$ and $W$ denote respectively, the magnitude of charge on each plate electric field between the plates (after the slab is inserted), and work done on the system, in question, in the process of inserting the slab, then which is wrong?

• 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

The correct option is B $Q=\frac{{ϵ}_{0}KAV}{d}$

Initial capacitance, $C_0=\frac{A{ϵ}_0}{d}$
So initial charge on the capacitor is: $Q_0=C_{0}V=\frac{AV{ϵ}_0}{d}$
As the battery is disconnected so charge remains unchanged.

So, $Q=Q_0=\frac{AV{ϵ}_0}{d}$
After insert the dielectric the capacitance becomes, $C^{\prime }=K{C}_0$ where K dielectric constant.
Now potential becomes, $V^{\prime }=Q/C^{\prime }=Q_0/K{C}_0=V/K$
Field, $E=V^{\prime }/d=\frac{V}{Kd}$
Initial energy, $U_0=\frac{1}{2}{C}_0{V}^2=\frac{A{V}^2{ϵ}_0}{2d}$ and
final energy $U_f=\frac{1}{2}{C}^{\prime }{V}^{\prime 2}=\left(1/2\right)\left(K{C}_0\right)(V/K{)}^2=\frac{A{V}^2{ϵ}_0}{2Kd}$

Since, work done $=$Decrease in Potential Energy
Work done, $W=-\Delta U=-(U_f-U_0)=U_0-U_f=\frac{1}{2}\frac{A{V}^2{ϵ}_0}{d}(1-1/K)$