First slide

Capacitance

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?

Moderate

A

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

Solution

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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