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For a capacitor the distance between the two plates is 4x and the electric field between them is E₀. Now a dielectric slab having dielectric constant 3 and thickness x is placed between them in contact with one plate. In this condition, what is the p.d. between its two plates?

Asked in GSEB Board March 2019 · Dielectric slab

• (1) (11E₀x)/3
• (2) (13E₀x)/3
• (3) (10E₀x)/3
• (4) (9E₀x)/3

Step-by-step solution

Why the other options are wrong

• (1) This uses E₀ - (E₀)/K = (2E₀)/3 inside the slab, which is the amount by which the field drops, not the field itself: 3E₀x + (2E₀x)/3 = (11E₀x)/3. The field inside the slab is (E₀)/3.
• (2) This keeps the whole 4x as air and then adds the slab's term: 4E₀x + (E₀x)/3 = (13E₀x)/3. The slab takes up x of the gap, leaving only 3x of air.
• (4) This drops the slab's contribution entirely, as if the field inside it were zero: E₀(3x) = 3E₀x = (9E₀x)/3. That would be a metal slab; a dielectric only reduces the field, to (E₀)/3.

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