Question

A parallel plate capacitor (without dielectric) is charged by a battery and kept connected to the battery. A dielectric slab of dielectric constant $‘k^{\prime }$ is inserted between the plates fully occupying the space between the plates. The energy density of electric field between the plates will:

• A. Decrease $k$ times
• B. Increase $k^2$ times
• C. Decrease $k^2$ times
• D. Increase $k$ times

Answer 1

The correct option is D Increase $k$ times

We know that,
electric field energy density in vacuum
$=\frac{1}{2}ϵE^2$
in dielectric, electric field energy density
$=\frac{1}{2}kϵE^2$
Where, $k$= dielectric constant
$E$= Electric field between plates $=\frac{v}{d}$
$V$= potential difference,
$d$= distance between plates
Since the capacitor is connected to the battery, $V$ will remain the same. Thus, $E$will also remain same within the capacitor.

Energy density will become $k$ times the initial value $i.e$ increases $k$ times
Hence, correct option is (c).