A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has thickness $3 d / 4$ , where d is the distance between plates. How is the capacitors changed when the slab is inserted between the plates?

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Answer 1

$C = \frac{A ϵ _{0}}{d}$

$C^{'} = \frac{A ϵ _{0}}{d - t + \frac{t}{k}}$

Put $t = \frac{3 d}{4}$ ;

$C^{'} = \frac{4 k}{3 + k} . \frac{A ϵ _{0}}{d}$

$C^{'} = \frac{4 k}{3 + k} . C$

Answer 2

C = \frac{A ϵ _{0}}{d}

C^{'} = \frac{A ϵ _{0}}{d - t + \frac{t}{k}}

Put

t = \frac{3 d}{4}

;

C^{'} = \frac{4 k}{3 + k} . \frac{A ϵ _{0}}{d}

C^{'} = \frac{4 k}{3 + k} . C

A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has thickness $3 d / 4$ , where d is the distance between plates. How is the capacitors changed when the slab is inserted between the plates?

Solution

$C = \frac{A ϵ _{0}}{d}$

$C^{'} = \frac{A ϵ _{0}}{d - t + \frac{t}{k}}$

Put $t = \frac{3 d}{4}$ ;

$C^{'} = \frac{4 k}{3 + k} . \frac{A ϵ _{0}}{d}$

$C^{'} = \frac{4 k}{3 + k} . C$

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A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has thickness 3d/4, where d is the distance between plates. How is the capacitors changed when the slab is inserted between the plates?

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Electrostatic Potential and Capacitance

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A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has thickness $3 d / 4$ , where d is the distance between plates. How is the capacitors changed when the slab is inserted between the plates?

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Put $t = \frac{3 d}{4}$ ;

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$C^{'} = \frac{4 k}{3 + k} . C$

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