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A slab of material of dielectric constant 3 has the same area as the plates of a parallel plate capacitor but has a thickness (3/4)d, where d is the separation of the plates. What is the electrical potential difference between the plates when the slab is inserted between them? The initial electrical potential difference is V₀.

Asked in GUJCET 2022 · Dielectric slab

Answer: (3) (V₀)/2

Step-by-step solution

Given: K = 3, slab thickness t = 3/4d, initial p.d. V₀ = E₀d. The capacitor is isolated, so its charge — and the field E₀ in the air — stays the same.

Idea: add the p.d. across the air and across the slab; inside the slab the field is (E₀)/K.

Air: E₀×d/4 = (V₀)/4.

Slab: (E₀)/3×(3d)/4 = (E₀d)/4 = (V₀)/4.

V = (V₀)/4 + (V₀)/4 = (V₀)/2.

(Equivalently, the capacitance doubles: C = (4K)/(K+3)C₀ = 2C₀.)

So the p.d. becomes (V₀)/2.

Why the other options are wrong

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