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A capacitor has capacitance 5 μ F when its parallel plates are separated by an air medium of thickness d. A slab of material of dielectric constant 1.5, having area equal to that of the plates but thickness d/2, is inserted between the plates. The capacitance of the capacitor in the presence of the slab will be ______ μ F.

Asked in JEE Main 25th Jan 2nd Shift 2023 · Dielectric slab

Answer: 6

Step-by-step solution

With a slab of thickness t=d/2 and constant K filling part of the gap, C=(ε₀ A)/((d-t)+t/K).

Denominator: d/2+(d/2)/(1.5)=d/2+d/3=(5d)/6.

C=(ε₀ A)/(5d/6)=6/5·(ε₀ A)/d=6/5C₀.

C=6/5×5=6 μ F.

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