Question

The energy and capacity of a charged parallel plate capacitor are $E$ and $C$ respectively. Now a dielectric slab of ${ϵ}_r=6$ is inserted in it, then energy and capacity becomes
(Assuming charge on plates remains constant)

• A. $6E,6C$
• B. $E,C$
• C. $\frac{E}{6},6C$
• D. $E,6C$

Answer 1

The correct option is C $\frac{E}{6},6C$

The capacitance of parallel plate capacitor is given by, $C=\frac{{\epsilon }_{0}A}{d}$
In the presence of medium,
$C^{{}^{\prime }}=\frac{{ϵ}_r{\epsilon }_{0}A}{d}$
$C^{{}^{\prime }}=\frac{6{\epsilon }_{0}A}{d}\{\because {ϵ}_r=6\}$
$C^{{}^{\prime }}=6C$

And energy stored in capacitor is–
$E=\frac{q^2}{2C}$
In the presence of medium $E^{{}^{\prime }}=\frac{q^2}{2C^{{}^{\prime }}}$
$E^{{}^{\prime }}=\frac{q^2}{\left(2C\right)6}=\frac{E}{6}$