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A parallel-plate capacitor has capacitance 2 pF in vacuum with plate…

Question

A parallel-plate capacitor has capacitance 2 pF in vacuum with plate separation d. A dielectric slab of dielectric constant K = 3 and thickness d/2, with the same area as the plates, is inserted so that it fills exactly half the gap (the remaining half being vacuum). Find the new capacitance, in picofarad.

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Step-by-step solution

The filled gap acts as two capacitors in series: a vacuum layer of thickness d/2 and a dielectric layer of thickness d/2 with constant K.

New capacitance C = eps0 A / (d/2 + (d/2)/K).
Dividing by C0 = eps0 A/d gives C/C0 = d / (d/2 + d/(2K)) = 1/(1/2 + 1/(2K)) = 2K/(K+1).
With K = 3: C/C0 = 6/4 = 1.5, so C = 1.5 x 2 = 3 pF.

Final answer3

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