(C) The dielectric constant $K$ is defined as the ratio of the electric field in a vacuum $(E_0)$ to the electric field in the dielectric medium $(E)$

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(C) The dielectric constant $K$ is defined as the ratio of the electric field in a vacuum $(E_0)$ to the electric field in the dielectric medium $(E)$.Given:Electric field in vacuum,$E_0 = 2 \times 10^5 \ V/m$Electric field in dielectric,$E = 1 \times 10^5 \ V/m$Formula:$K = \frac{E_0}{E}$Calculation:$K = \frac{2 \times 10^5}{1 \times 10^5} = 2$Therefore,the dielectric constant of the material is $2$.

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

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Two parallel plates have equal and opposite charges. When the space between them is evacuated,the electric field between the plates is $2 \times 10^5 \ V/m$. When the space is filled with a dielectric,the electric field becomes $1 \times 10^5 \ V/m$. Find the dielectric constant of the dielectric material.

A
$1/2$
B
$1$
C
$2$
D
$3$

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Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

$A$ combination of parallel plate capacitors is maintained at a certain potential difference. When a $3 \, mm$ thick slab is introduced between all the plates,in order to maintain the same potential difference,the distance between the plates is increased by $2.4 \, mm$. Find the dielectric constant of the slab.

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The capacitance of a parallel plate capacitor with air as medium is $6\, \mu F$. With the introduction of a dielectric medium,the capacitance becomes $30\, \mu F$. The permittivity of the medium is..........$C^{2} N^{-1} m^{-2}$.
$(\varepsilon_{0} = 8.85 \times 10^{-12} C^{2} N^{-1} m^{-2})$

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$A$ capacitor is charged by using a battery which is then disconnected. $A$ dielectric slab is introduced between the plates which results in

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If the equivalent dielectric constant of the given system is $K$,then...

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The area of the plates of a parallel plate capacitor is $A$ and the distance between the plates is $10\,mm$. There are two dielectric sheets in it,one of dielectric constant $10$ and thickness $6\,mm$ and the other of dielectric constant $5$ and thickness $4\,mm$. The capacity of the capacitor is

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The capacitance of a parallel plate capacitor with air as medium is $6\, \mu F$. With the introduction of a dielectric medium,the capacitance becomes $30\, \mu F$. The permittivity of the medium is..........$C^{2} N^{-1} m^{-2}$.
$(\varepsilon_{0} = 8.85 \times 10^{-12} C^{2} N^{-1} m^{-2})$