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

Answer 1

Answer: (B)

The charge on the capacitor is not conserved

Answer 2

$q = CV \Rightarrow V = q/c$
Due to dielectric insertion, new capacitance $C_2 = CK$
Initial energy stored in capacitor, $U_1 = \frac{q^2}{2C}$
Final energy stored in capacitor, $U_2 = \frac{q^2}{2KC}$
Change in energy stored, $\Delta U = U_2 - U_1$
$\Delta U = \frac{q^2}{2C} \bigg(\frac{1}{K} - 1 \bigg) = \frac{1}{2} CV^2 \bigg(\frac{1}{K} - 1 \bigg)$
New potential difference between plates
$V' = \frac{q}{CK} = \frac{V}{K}$