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

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Solution

Q.
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 [2015]

1 The change in energy stored is $\frac{1}{2} C V^{2} (\frac{1}{K} - 1)$

2 The charge on the capacitor is not conserved.

3 The potential difference between the plates decreases $K$ times.

4 The energy stored in the capacitor decreases $K$ times.

Ans.
(2)

$q = C V \Rightarrow V = \frac{q}{C}$

Due to dielectric insertion, new capacitance

$C_{2} = C K$

Initial energy stored in capacitor, $U_{1} = \frac{q^{2}}{2 C}$

Final energy stored in capacitor, $U_{2} = \frac{q^{2}}{2 K C}$

Change in energy stored, $Δ U = U_{2} - U_{1}$

$Δ U = \frac{q^{2}}{2 C} (\frac{1}{K} - 1) = \frac{1}{2} C V^{2} (\frac{1}{K} - 1)$

New potential difference between plates $V' = \frac{q}{C K} = \frac{V}{K}$

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