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Two thin dielectric slabs of dielectric constants K₁ and K₂ (K₁<K₂) 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

Asked in CBSE AIPMT 2014 · Compound and non-uniform dielectrics

• (1) Graph (1): E steps up inside the first slab, back down in the gap, up by a larger step inside the second slab, then falls to zero beyond Q
• (2) Graph (2): E steps down inside the first slab, back up in the gap, rises above the air value inside the second slab, then falls to zero beyond Q
• (3) Graph (3): E steps down inside the first slab, back up in the gap, down by a larger step inside the second slab, then falls to zero beyond Q
• (4) Graph (4): E drops to zero inside both slabs, has the same value in the three air gaps, and is zero beyond Q

Step-by-step solution

Why the other options are wrong

• (1) Takes the field inside a dielectric to be KE₀; the bound charges oppose the field, so it is (E₀)/K, smaller.
• (2) No single rule gives a dip in one slab and a rise in the other: both K₁ and K₂ exceed 1, so the field must fall inside both.
• (4) Treats the slabs as conductors, where the field is zero; a dielectric only reduces the field, to (E₀)/K.

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