(C) The initial capacity is $C$ and the initial energy is $U = \frac{Q^2}{2C}$,where $Q$ is the constant charge on the plates.When a dielectric slab o

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(C) The initial capacity is $C$ and the initial energy is $U = \frac{Q^2}{2C}$,where $Q$ is the constant charge on the plates.When a dielectric slab of dielectric constant $K = 6$ is inserted,the new capacity becomes $C' = KC = 6C$.Since the charge $Q$ remains constant,the new energy $U'$ is given by $U' = \frac{Q^2}{2C'}$.Substituting $C' = 6C$,we get $U' = \frac{Q^2}{2(6C)} = \frac{1}{6} \left( \frac{Q^2}{2C} \right) = \frac{U}{6}$.Therefore,the new energy is $\frac{U}{6}$ and the new capacity is $6C$.

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

Medium

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).

A
$6U, 6C$
B
$U, C$
C
$\frac{U}{6}, 6C$
D
$U, 6C$

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Class 12

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Electrostatic Potential and Capacitance

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Effect of Dielectric Inside Capacitor

Two identical parallel plate air capacitors are connected in series to a battery of emf $V$. If one of the capacitors is completely filled with a dielectric material of constant $K$,then the potential difference across the other capacitor will become:

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In the given circuit,two identical capacitors $C_1$ and $C_2$ are connected to a battery. The space between the plates of $C_1$ is filled with air,while the space between the plates of $C_2$ is filled with a dielectric material of dielectric constant $K$. Then,

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Two parallel plate capacitors are connected in series and then connected to a $100 \ V$ battery. $A$ dielectric slab of dielectric constant $K = 4.0$ is inserted between the plates of the second capacitor. What will be the potential difference across each capacitor respectively?

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$A$ capacitor stores $60 \mu C$ charge when connected across a battery. When the gap between the plates is filled with a dielectric,a charge of $120 \mu C$ flows through the battery. The dielectric constant of the material inserted is:

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$A$ capacitor has a capacitance of $15\ \mu F$ and a plate separation of $2\ mm$. If a dielectric slab of dielectric constant $K = 2$ and thickness $t = 1\ mm$ is inserted between the plates,what will be the new capacitance (in $\mu F$)?

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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).

Similar Questions

Two identical parallel plate air capacitors are connected in series to a battery of emf $V$. If one of the capacitors is completely filled with a dielectric material of constant $K$,then the potential difference across the other capacitor will become:

In the given circuit,two identical capacitors $C_1$ and $C_2$ are connected to a battery. The space between the plates of $C_1$ is filled with air,while the space between the plates of $C_2$ is filled with a dielectric material of dielectric constant $K$. Then,

Two parallel plate capacitors are connected in series and then connected to a $100 \ V$ battery. $A$ dielectric slab of dielectric constant $K = 4.0$ is inserted between the plates of the second capacitor. What will be the potential difference across each capacitor respectively?

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