Two thin dielectric slabs of dielectric constants and are inserted between plates of a parallel plate capacitor, as shown in the figure. The variation of electric field between the plates with distance as measured from plate is correctly shown by

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Solution

Q.
Two thin dielectric slabs of dielectric constants $K_{1}$ and $K_{2}$ $(K_{1} < K_{2})$ are inserted between plates of a parallel plate capacitor, as shown in the figure. The variation of electric field $E$ between the plates with distance $d$ as measured from plate $P$ is correctly shown by [2014]

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Ans.
(3)

$E_{medium} = \frac{E_{vacuum}}{K}$

The electric field inside the dielectrics will be less than the electric field in vacuum. The electric field inside the dielectric could not be zero. As $K_{2} > K_{1}$ , the drop in electric field for $K_{2}$ dielectric must be more than $K_{1}$ .

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