Engineering Electromagnetics
Engineering Electromagnetics
9th Edition
ISBN: 9780078028151
Author: Hayt, William H. (william Hart), Jr, BUCK, John A.
Publisher: Mcgraw-hill Education,
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Chapter 6, Problem 6.5P
To determine

The factor by which capacitance is increased.

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The plates of a parallel-plate capacitor are square and have the length of one side. 20 cm and the distance between the plates is 0.5 cm. The material between the plates is air.What is the capacitance of a parallel plate capacitor in farads?
(6) A parallel-plates capacitor with plate area of 100 cm? has a separation of 3mm between the pates. The separation is divided into three slabs, where each slab has 1 mm distance. The first slab is filled by air and has a constant of 1. The second slab is filled by polystyrene and has a constant of 2.5, while the third slab is filled by silicon dioxide and has a constant of 4. Compute the capacitance of the system. Note: for parallel slabs; Conductor 1 E = T 1 Air E1 83 Material A Material 3 Conductor Ans. C = 54 pF
5. Cylindrical capacitors of length L is filled up with an inhomogeneous dielectric as shown in the figure. Find the capacitance of each capacitor. $! 8x2 (a) Erl (b) &

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Engineering Electromagnetics

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