(B) Initial energy stored in the capacitor is $U_i = \frac{1}{2} C V^2$.Since the capacitor is disconnected from the source,the charge $Q$ on the plat

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(B) Initial energy stored in the capacitor is $U_i = \frac{1}{2} C V^2$.Since the capacitor is disconnected from the source,the charge $Q$ on the plates remains constant.When a dielectric of constant $K$ is inserted,the new capacitance becomes $C' = KC$.The new energy stored is $U_f = \frac{Q^2}{2C'} = \frac{Q^2}{2KC} = \frac{U_i}{K}$.Given that the energy decreases by $25 \%$,the final energy is $U_f = U_i - 0.25 U_i = 0.75 U_i = \frac{3}{4} U_i$.Equating the two expressions for $U_f$: $\frac{U_i}{K} = \frac{3}{4} U_i$.Solving for $K$,we get $K = 4/3$.

Class 12 Physics Electric Potential and Capacitance Effect of Dielectric Inside Capacitor

Medium

$A$ capacitor of capacitance $2 \mu F$ is charged to $50 V$ and then disconnected from the source. Later,the gap between the plates of the capacitor is filled with a dielectric material. If the energy stored in the capacitor is decreased by $25 \%$ of its initial value,then the dielectric constant of the dielectric material is

AP EAMCET 2025 All AP EAMCET

A
$2/3$
B
$4/3$
C
$3/4$
D
$3/2$

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Class 12

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Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

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MediumMHT CET 2024

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$A$ parallel plate capacitor having capacitance $12 \, pF$ is charged by a battery to a potential difference of $10 \, V$ between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant $6.5$ is slipped between the plates. The work done by the capacitor on the slab is.......$pJ$

DifficultJEE MAIN 2019

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The energy and capacity of a charged parallel plate capacitor are $U$ and $C$ respectively. Now a dielectric slab of dielectric constant $K = 6$ is inserted in it. Then,the new energy and capacity become (Assume the charge on the plates remains constant).

Medium

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Two capacitors,each having capacitance $40\,\mu F$,are connected in series. The space between the plates of one of the capacitors is filled with a dielectric material of dielectric constant $K$ such that the equivalent capacitance of the system becomes $24\,\mu F$. The value of $K$ is:

The energy and capacity of a charged parallel plate capacitor are $U$ and $C$ respectively. Now a dielectric slab of dielectric constant $K = 6$ is inserted in it. Then,the new energy and capacity become (Assume the charge on the plates remains constant).

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