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 potential difference between the plates decreases $K$ times.
• B

  The energy stored in the capacitor decreases $K$ times.
• C

  The change in energy stored is $\frac{1}{2}C{V}^2\left(\frac{1}{K}-1\right)$.
• Correct answer
  D

  The charge on the capacitor is not conserved.

Solution

For air $K=1$,
$C_1=C$
$q=C_{1}V \dots \left(i\right)$.
The charge remains constant.
$U_1=\frac{q^2}{2C}$
$U_2=\frac{q^2}{2CK}$
Change in energy stored in the capacitor,
$U=U_2-U_1$
$=\frac{q^2}{2C} \left(\frac{1}{K}-1\right) =\frac{1}{2}C{V}^2 \left(\frac{1}{K}-1\right)$
$V\text{'}=\frac{q}{CK}\Rightarrow V\text{'}=\frac{V}{K}$
Here,1- Potential difference between the plates decreases by $K$ times.
2- The change in energy stored $∆U=\frac{1}{2} C{V}^2 \left(\frac{1}{K}-1\right)$.
3- Charge is conserved.
4- The energy stored in capacitor decrease by $K, U_2=\frac{1}{2}\frac{q^2}{CK}$.

Hence, the correct answer is option: 'D'.