Question

The capacitance of a parallel plate capacitor is $C$ when the region between the plate has air. This region is now filled with a dielectric slab of dielectric constant $k$. The capacitor is connected to a cell of emf $E$, and the slab is taken out

• A. charge $CE(k-1)$ flows through the cell
• B. energy $E^{2}C(k-1)$ is absorbed by the cell
• C. the energy stored in the capacitor is reduces by $E^{2}C(k-1)$
• D. the external agent has to do $\frac{1}{2}{E}^{2}C(k-1)$ amount of work to take the slab out

Answer 1

Answer: (A), (D)

the external agent has to do $\frac{1}{2}{E}^{2}C(k-1)$ amount of work to take the slab out
charge $CE(k-1)$ flows through the cell

After inserting the dielectric the capacitance will be $C^{\prime }=kC$
The charge flow through the cell when the slab insert is $Q^{\prime }=C^{\prime }E=kCE$
The charge flow through the cell when the slab is taken out $=Q=CE$
Net charge flow through the cell $Q_n=Q^{\prime }-Q=kCE-CE=CE(k-1)$
The amount of work for taking out slab is $W=\frac{1}{2}(Ck{E}^2-C{E}^2)=\frac{1}{2}C{E}^2(k-1)$