Question

The capacitance of a parallel-plate capacitor is $C_0$ when the region between the plates has air. This region is now filled with a dielectric slab of dielectric constant K. The capacitor is connected to a cell emf ${\epsilon }_1$ and the slab is taken out.

• A. Charge $\epsilon C_0(K-1)$ flows through the cell.
• B. Energy ${\epsilon }^2{C}_0(K-1)$ is absorbed by the cell.
• C. The energy stored in the capacitor is reduced by ${\epsilon }^2{C}_0(K-1)$.
• D. The external agent has to do $\frac{1}{2}{\epsilon }^2{C}_0(K-1)$ amount of work to take the slab out.

Answer 1

Answer: (A), (B), (D)

The capacitance of capacitor with dielectric is $C=K{C}_0$
Initial charge on capacitor is $q=C\epsilon =K{C}_0\epsilon$
As cell is still connect so potential remain constant.
after slab is taken out the charge on capacitor is $q^{\prime }=C_0\epsilon$
The charge flows through the cell is $\Delta q=q-q^{\prime }=K{C}_0\epsilon -C_0\epsilon =C_0\epsilon (K-1)$
Energy absorbed by cell $E=\Delta q\epsilon =C_0{\epsilon }^2(K-1)$
Energy stored in capacitor $U=\frac{1}{2}q\epsilon -\frac{1}{2}{q}^{\prime }\epsilon =\frac{1}{2}{C}_0{\epsilon }^2(K-1)$
Work done $W=E-U=C_0{\epsilon }^2(K-1)-\frac{1}{2}{C}_0{\epsilon }^2(K-1)=\frac{1}{2}{C}_0{\epsilon }^2(K-1)$