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Two thin dielectric slabs of dielectric constants K₁ and K₂ (K₁ < K ₂) are inserted between the plates of a parallel plate capacitor, as shown in the figure alongside. The variation of electric filed E between the plates with distance d as measured from plate P is correctly shown by

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answer is C.

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Detailed Solution

Electric field inside a dielectric slab is given by $\vec{E} = \frac{{\vec{E}}_{0}}{K}$

Since K₁ < K₂ , the electric field is less in second slab.

Will be the most appropriate diagram.

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