Question

Assertion A parallel plate capacitor is connected across battery through a key. A dielectric slab of dielectric constant $K$ is introduced between the plates. The energy which is stored becomes $K$ times.
Reason The surface density of charge on the plate remains constant or unchanged.

• A. Both assertion and reason are true and reason is the correct explanation of assertion
• B. Both assertion and reason are true but reason is not the correct explanation of assertion
• C. Assertion is true but reason is false
• D. Both Assertion and reason are false

Answer 1

Answer: (C)

If a dielectric slab of dielectric constant $K$ is filled in between
the plates of a condenser while charging it, the potential
difference between the plates does not change but the
capacity becomes $K$ times, therefore $V^{\prime }=V,C^{\prime }=KC$
Energy stored in the capacitor
$U^{\prime }=\frac{1}{2}{C}^{\prime }{V}^{\prime 2}$
$=\frac{1}{2}(KC)(V^2)$
$=[\frac{1}{2}C{V}^2]K$
$=KU$
Thus, energy stored becomes $K$ times. Surface charge density
$\sigma =\frac{q^{\prime }}{A}=\frac{C^{\prime }{V}^{\prime }}{A}=\frac{KCV}{A}$
$=K\frac{q}{A}=K\sigma$