A parallel plate capacitor filled with a medium of dielectric constant 10 , is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant 15. Then the energy of capacitor will :

Question

A parallel plate capacitor filled with a medium of dielectric constant 10 , is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant 15. Then the energy of capacitor will : (A) increase by 50 % (B) decrease by 15 % (C) increase by 25 % (D) increase by 33 %

Tardigrade Admin · Accepted Answer

E ⇒ (1/2)( KC ) v 2 ∴ % change ⇒ ((1/2) K 2 CV 2-(1/2) K 1 CV 2/(1)2 K 1 CV 2) =( K 2- K 1/ K 1) × 100 ⇒ (15-10/10) × 100=50 %

Q. A parallel plate capacitor filled with a medium of dielectric constant $10$ , is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant $15$ . Then the energy of capacitor will :

increase by $50%$

decrease by $15%$

increase by $25%$

increase by $33%$

$E \Rightarrow \frac{1}{2} (K C) v^{2}$
$∴ %$ change
$\Rightarrow \frac{\frac{1}{2} K _{2} C V ^{2} - \frac{1}{2} K _{1} C V ^{2}}{\frac{1}{2} K _{1} C V ^{2}}$
$= \frac{K _{2} - K _{1}}{K _{1}} \times 100$
$\Rightarrow \frac{15 - 10}{10} \times 100 = 50%$

Q. A parallel plate capacitor filled with a medium of dielectric constant 10 , is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant 15. Then the energy of capacitor will :

increase by 50%

decrease by 15%

increase by 25%

increase by 33%