Question

A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect?

• A. The change in energy stored is $\frac{1}{2}C{V}^2(\frac{1}{K}-1)$
• B. The potential difference between the plates decreases K times
• C. The charge on the capacitor is not conserved
• D. The energy stored in the capacitor decreases K times

Answer 1

Answer: (C)

$C_f=K{C}_i$
$Q_i=C_{i}V$
$U_i=\frac{1}{2}{C}_i{V}^2$
$U_f=\frac{1}{2}\frac{Q_f^2}{C_f}$
After the capacitor is disconnected from the battery, charge on the capacitor cannot change.
$\therefore Q_f=Q_i$
$\therefore U_f=\frac{1}{2}\frac{Q_f^2}{C_f}=\frac{1}{2}\frac{C_i^2{V}^2}{K{C}_i}$
$\Delta U=U_f-U_i=\frac{1}{2}\frac{C_i^2{V}^2}{K{C}_i}-\frac{1}{2}{C}_i{V}^2=\frac{1}{2}{C}_i{V}^2(\frac{1}{K}-1)=\frac{1}{2}C{V}^2(\frac{1}{K}-1)$ since $C_i=C$ as given