A $1 m m$ thick paper of dielectric constant 4 lies between the plates of a parallel-plate capacitor. It is charged to 100 volt the intensity of electric field between the plates of the condenser will be :

100

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100000

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400000

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25000

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Solution

The correct option is C 25000
The electric field in between the plates in the presence of dielectric is
$E = \frac{V}{k d} = \frac{100}{4 \times 10^{- 3}} = 25000$ V/m

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[Pb PMT 2004]

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A 1 mm thick paper of dielectric constant 4 lies between the plates of a parallel-plate capacitor. It is charged to 100 volt the intensity of electric field between the plates of the condenser will be :

The correct option is C 25000The electric field in between the plates in the presence of dielectric is E=Vkd=1004×10−3=25000V/m

A $1 m m$ thick paper of dielectric constant 4 lies between the plates of a parallel-plate capacitor. It is charged to 100 volt the intensity of electric field between the plates of the condenser will be :

The correct option is C 25000
The electric field in between the plates in the presence of dielectric is
$E = \frac{V}{k d} = \frac{100}{4 \times 10^{- 3}} = 25000$ V/m