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InterviewSolution
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A parallel-plate capacitor consists of two square metal plates 500 mm on a side separated by 10 mm. A slab of Teflon (εr = 2.0) 6 mm thick is placed on the lower plate leaving an air gap 4 mm thick between it and the upper plate. If 100 V is applied across the capacitor, find the electric field (E0) in the air, electric field Et in Teflon, flux density Da in air, flux density Dt in Teflon and potential difference Vt across Teflon slab. |
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Answer» C = ε0A/(d1/εe1 + d2/εr2) = (8.854 x 10-12 x (0.5)2)/((6 x 10-3/2) + (4 x 10-3/1)) = 3.16 x 10-10F = CV = 3.16 × 10−10 × 100 = 31.6 × 10−9 C D = Q/A = 31.6 × 10−9/(0.5)2 = 1.265 × 10−7 C/m2 The charge or flux density will be the same in both media i.e. Da = Dt = D In air, E0 = D/ε0 = 1.265 × 10−7/8.854 × 10−12 = 14,280 V/m In Teflon, Et = D/ε0εr = 14,280/2 = 7,140V/m Vt = Et × dt = 7,140 × 6 × 10−3 = 42.8 V |
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