(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 Electric 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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Chapter

Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

The capacitance of a spherical capacitor in air is $50 \,\mu F$. When it is immersed in oil,its capacitance becomes $110 \,\mu F$. Calculate the dielectric constant of the oil.

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In the figure,a capacitor is filled with dielectrics of constants $K_1$,$K_2$,and $K_3$. The resultant capacitance is

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$A$ parallel plate capacitor of area $A$ and plate separation $d$ is filled with two dielectrics as shown. What is the equivalent capacitance of the arrangement?

Medium

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$A$ capacitor of capacitance $9 \, nF$ having a dielectric slab of $\varepsilon_{r} = 2.4$,dielectric strength $20 \, MV/m$,and potential difference $V = 20 \, V$. The area of the plates is ....... $\times 10^{-4} \, m^{2}$.

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$A$ voltmeter reads $4 \ V$ when connected to a parallel plate capacitor with air as a dielectric. When a dielectric slab is introduced between the plates for the same configuration,the voltmeter reads $2 \ V$. What is the dielectric constant of the material?

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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

The capacitance of a spherical capacitor in air is $50 \,\mu F$. When it is immersed in oil,its capacitance becomes $110 \,\mu F$. Calculate the dielectric constant of the oil.

In the figure,a capacitor is filled with dielectrics of constants $K_1$,$K_2$,and $K_3$. The resultant capacitance is

$A$ parallel plate capacitor of area $A$ and plate separation $d$ is filled with two dielectrics as shown. What is the equivalent capacitance of the arrangement?

$A$ capacitor of capacitance $9 \, nF$ having a dielectric slab of $\varepsilon_{r} = 2.4$,dielectric strength $20 \, MV/m$,and potential difference $V = 20 \, V$. The area of the plates is ....... $\times 10^{-4} \, m^{2}$.

$A$ voltmeter reads $4 \ V$ when connected to a parallel plate capacitor with air as a dielectric. When a dielectric slab is introduced between the plates for the same configuration,the voltmeter reads $2 \ V$. What is the dielectric constant of the material?

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