The energy and capacity of a charged parallel plate capacitor are U and C respectively. Now a dielectric slab of εr=6 is inserted in it then energy and capacity becomes (Assuming charge on plates remains constant)

Question

The energy and capacity of a charged parallel plate capacitor are U and C respectively. Now a dielectric slab of εr=6 is inserted in it then energy and capacity becomes (Assuming charge on plates remains constant) (A) 6 U , 6 C (B) U, C (C) (U/6), 6 C (D) U, 6C

Tardigrade Admin · Accepted Answer

C '=(ϵ0 ϵr A /d) ⇒ C '=6 C U =(q2/2 C) ⇒ U '=(q2/2 × C')=(q2/2 × 6 C) =(U/6)

Q. The energy and capacity of a charged parallel plate capacitor are $U$ 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)

$6 U, 6 C$

$U, C$

$\frac{U}{6}, 6 C$

$U, 6 C$

$C^{'} = \frac{ϵ _{0} ϵ _{r} A}{d}$
$\Rightarrow C^{'} = 6 C$
$U = \frac{q ^{2}}{2 C}$
$\Rightarrow U^{'} = \frac{q ^{2}}{2 \times C ^{'}} = \frac{q ^{2}}{2 \times 6 C}$
$= \frac{U}{6}$

Q. The energy and capacity of a charged parallel plate capacitor are U and C respectively. Now a dielectric slab of εr​=6 is inserted in it then energy and capacity becomes (Assuming charge on plates remains constant)

6U,6C

U,C

6U​,6C

U,6C