(17%) Problem 6: The gap between the plates of a parallel-plate capacitor is filled with three equal-thickness layers of mica, paper, and a material of unknown dielectric constant. The area of each plate is 110 cm$^2$ and the capacitor's gap width is 3.25 mm. The values of the known dielectric constants are $K_{mica} = 5.5$ and $K_{paper} = 3.75$. The capacitance is measured and found to be 95 pF. Find the value of the dielectric constant of the unknown material. $K_{unknown} = $
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- Plate area A = 110 cm^2 = 110 × 10^-4 m^2 = 0.011 m^2. - Total gap d = 3.25 mm = 3.25 × 10^-3 m. - Each layer thickness = d/3 = 3.25×10^-3 / 3 = 1.0833×10^-3 m. - Dielectric constants: K_mica = 5.5, K_paper = 3.75, K_unknown = ?. - Measured capacitance C_eq = 95 Show more…
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