The two conducting planes illustrated inFigure 6.14 are defined by 0.001 < p <0.120 m, 0 < z < 0.1 m, o = 0.179 and 0.188rad. The medium surrounding the planes isair. For Region 1, 0.179 < o < 0.188; neglectfringing and find (a) V(); (b) E (p); (c) D (p);(d) ps on the upper surface of the lowerplane; (e) Q on the upper surface of thelower plane. (f) Repeat parts a through cfor Region 2 by letting the location of theupper plane be o = .188 – 27T, and thenfind ps and Q on the lower surface of thelower plane. (g) Find the total charge onthe lower plane and the capacitancebetween the planes.Figure 6.14 See Problem 6.39.%3|p= 1 mmRegion 2$=0.188, V 20 V$ = 0.179, V= 200 VRiGap

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The two conducting planes illustrated in Figure 6.14 are defined by 0.001

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

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Question

The two conducting planes illustrated in
Figure 6.14 are defined by 0.001 < p <
0.120 m, 0 < z < 0.1 m, o = 0.179 and 0.188
rad. The medium surrounding the planes is
air. For Region 1, 0.179 < o < 0.188; neglect
fringing and find (a) V(); (b) E (p); (c) D (p);
(d) ps on the upper surface of the lower
plane; (e) Q on the upper surface of the
lower plane. (f) Repeat parts a through c
for Region 2 by letting the location of the
upper plane be o = .188 – 27T, and then
find ps and Q on the lower surface of the
lower plane. (g) Find the total charge on
the lower plane and the capacitance
between the planes.
Figure 6.14 See Problem 6.39.
%3|
p= 1 mm
Region 2
$=0.188, V 20 V
$ = 0.179, V= 200 V
Ri
Gap

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