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calculate the area of a paper required to construct a parallel plate capacitor of 0.004 micro Farad if the dielectric constant of paper be 2.5 and its thickness 0.025 mm.​

Answer» N:CAPACITANCE of parallel plate capacitor, C = 0.004 μF = 0.004 * 10⁻⁶ FDielectric constant of paper, K = 2.5Thickness of paper, d = 0.025 MM = 0.025 * 10⁻³ mPermittivity of free SPACE, ε₀ = 8.854 * 10⁻¹² F/mThe AREA of paper is 4.52 * 10⁻³ m²


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