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 charge on the capacitor is not conserved.
• B. The potential difference between the plates decreases K times.
• C. The energy stored in the capacitor decreases K times.
• D. the change in energy stored is (1/2)CV²[(1/K)-1]

Answer 1

The correct option is A

The charge on the capacitor is not conserved.

A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it.

The charge on the capacitor is given by

Q = CV

The energy stored in the capacitor is

$E=\frac{1}{2}C{V}^2$

When a dielectric slab of dielectric constant K is in it, the charge Q is conserved. The capacitance becomes K times the original capacitance. (C’ = KC)

The voltage becomes $\frac{1}{K}$ time the original voltage.

$V‘=\frac{1}{K}$

The change in energy stored is

$\frac{Q^2}{2C^{\prime }}-\frac{Q^2}{2C}=\frac{Q^2}{2KC}-\frac{Q^2}{2C}=\frac{Q^2}{2C}[\frac{1}{K}-1]$
$=\frac{1}{2}C{V}^2[\frac{1}{K}-1]$