This document provides solutions to drill problems from Chapter 6 of the textbook Engineering Electromagnetics by William Hyatt, 7th Edition. The solutions include calculations for capacitance, electric field strength, electric displacement, and electric charge density based on given parameters such as dielectric constants, voltages, distances, and geometries. Calculations make use of Maxwell's equations and relationships between electromagnetic quantities.

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Chapter 6

2. 1
D6.1
Given data:
r = 3.8
D = 8 nC
m2
(a) E = D
or
= 237.7V
m
(b) We first find χe:
χe = r −1 = 2.8
Now,
P = χeoE = 5.9nC
m2
(c) We know that,
No. of Dipoles
V olume = P
AverageDipole Moment = 5.9×10−9
10−29 = 5.9 ×1020
m−3
Solved by Zaeem Ahmad Varaich
Com piled by www.EngineersTeam.tk
CHAPTER 6 DRILLS

3. D6.2
Given Data:
r1 (z < 0) = 3.2
r2 (z > 0) = 2
D1 = −30ax + 50ay + 70az
nC
m2
(a)DN1 = D1 .z = 70 nC
m2
(b) The tangential vector components of D1are perpedicular to z −axis, so:
Dt1 = −30ax + 50ay
nC
m2 (∵ x and y axes are ⊥)
(c) Magnitude of Dt1is:
Dt1 =
√
302
+ 502
= 58.3nC
m2
(d) Magnitude of D1is:
D1 =
√
302
+ 502
+ 702
= 91.1nC
m
2
(e) θ1 = cos−1
(DN1
D1
) = 39.8◦
(f) P1 = χeoE1 = (r1 −1)D1
r1
= −20.6ax+34.4ay+48.1az
nC
m2
Compiled by www.EngineersTeam.tk
2
D6.3
Given Data:
From previous drill problem!
(a)DN2 = DN1 = 70 nC
m2 , so, DN2 = 70az
nC
m2
(b) As, Dt1
Dt2
= r1
r2
⇒ Dt2 = r2×Dt1
r1
= −18.75ax+31.25ay
nC
m2
(c) D2 = DN2 + Dt2 = −18.75ax+31.25ay+70az
nC
m2 :
(d) P2 = χeoE2 = (r2 −1)D2
r2
= −9.38ax+15.63ay+35az
nC
m2
(e) θ2 = cos−1
(DN2
D2
) = 27.5◦
D6.4
(a) We first find C:
C = 2WE
V 2
o
= 13.9µF
Now,
r = Cd
So
= 1.04
(b) Here, WE = 100 ×d ×S = 100 ×45µ ×0.12 = 5.4 ×10−4
J
So, C =
2WE
V 2o
= 27µF .Now,
r = Cd
So
= 1.14
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3 of 4
(c) As, E = ρs

⇒ r = ρs
Eo
= 11.3
D6.5
(a) C = 2πL
ln(b/a) = 2πo(2.26)(1×30×10−2
)
ln(0.680/0.1045) = 20.05pF
(b) C = 4π
1
a
−
1
b
=
4πo(2.26)
1
2.5m
−
1
4.5m
= 1.41pF
(c) For each of three plates, S = 1 cm × 4 cm = 4 ×10−4
m2
and d = 0.1 mm,so
C = 1
d
1S
+ d
2S
+ d
3S
= 28.71pF
D6.6
Given Data:
b = 1cm
Vo = 20V
h = 5cm
r = 4.5
(a) C = 2π
cosh−1
(h/b)
= 109.205pF
(b) ρs,max = ρL
2π
 h−b+a
(h−b+a)2 − h−b−a
(h−b−a)2

, so we first find ρL and a:
Compiled by www.EngineersTeam.tk
3
ρL = 4πVo
ln(K1) , where
√
K1 = h+
√
h2
−b2
b = 9.89, so K1 = 97.98
Now, ρL = 2.18nC
m , and
a =
√
h2 −b2 = 0.049
Now,
ρs,max = 3.46 ×10−10
 0.089
(0.089)2 − −9×10−3
(−9×10−3
)2

= 42.33nC
m2
D6.7
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(a) C = o
NQ
NV
= o
2×12
3 = 70.8 pF
m
1
(b) E = ∆V
∆LN
= 20V
1mm
3
= 60 kV
m
2
(c) ρS = E = o(60k) = 531nC
m2
1
Two regions, each containing about 12 boxes
2
Radius divided by 3 gives the length of space between the 60V and 0V sphere on left side
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