Figure 25-49 shows a parallel-plate capacitor with a platearea A = 5.77 cm 2 and
ID: 1764443 • Letter: F
Question
Figure 25-49 shows a parallel-plate capacitor with a platearea A = 5.77 cm2 and plateseparation d = 2.51 mm. The top half of the gapis filled with material of dielectric constant1 = 13.0; the bottom half is filled withmaterial of dielectric constant 2 = 14.9.What is the capacitance?The figure shows twoparallel plates(arranged hamburger style..one on top of the other). The section between the plates is divided into 2, the upperhalf with dialectic constant K2 and the lower half with dialectricconstant K1. The sections are even (therefore, d/2 has constant k1and d/2 has constant k2).
The figure shows twoparallel plates(arranged hamburger style..one on top of the other). The section between the plates is divided into 2, the upperhalf with dialectic constant K2 and the lower half with dialectricconstant K1. The sections are even (therefore, d/2 has constant k1and d/2 has constant k2). The figure shows twoparallel plates(arranged hamburger style..one on top of the other). The section between the plates is divided into 2, the upperhalf with dialectic constant K2 and the lower half with dialectricconstant K1. The sections are even (therefore, d/2 has constant k1and d/2 has constant k2).
Explanation / Answer
in the given problem for the simplicity letus assume that there is charge q on one plate and charge - q on theother the electric field in the lower half of theregion between the plates is E1 = q /1 o A where A is the plate area the electric field in the upper halfis E2 = q /2 o A let (d / 2) be the thickness of eachdielectric as the field is uniform in each region the potentialdifference between the plates is V = (E1 d / 2) + (E2 d/ 2) = (q d / 2o A) [(1 / 1) + (1 /2)] = (q d / 2o A) [(1 +2) (1 2)] so the capacitance will be C = q / V = (2o A / d)[(1 2)(1 + 2)] if 1 =2 C = 1 o A/ d C = 1 o A/ dRelated Questions
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