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Solution to the Drill problems
of chapter 02
(Engineering
Electromagnetics,Hayt,A.Buck
7th ed)
BEE 4A,4B & 4C
Following Exercise questions are
IMPORTANT!
2.4, 2.5, 2.13, 2.14, 2.16 ,2.17,
2.18, 2.19, 2.22, 2.23, 2.27,
2.28,2.29,2.30,2.31
D2.1 (a). QA = −20µC located at
A(-6,4,7) ,QB = 50µC located at
B(5,8,-2)
Find ~
RAB
~
RAB = (5 − (−6))ˆax + (8 − 4)ˆay +
(−2 − 7)ˆaz = 11ˆax + 4ˆay − 9ˆaz
(b). | ~
p
RAB |= (112 ) + 4 2 + (−9)2 =
14.76m
(c). ~
FAB = QA QB ~
RAB/4 o | ~
RAB |3 = (−20 × 10−6 × 50
× 10−6 (11ˆax +
4ˆay − 9ˆaz ))/(4π × (10−9 /36π) |
14.76 |3 )
⇒ ~
FAB = 30.76ˆax +
11.184ˆay − 25.16ˆaz mN
(d). ~
FAB = QA QB ~
RAB/4 o | ~
RAB |3 = (−20 × 10−6 × 50
× 10−6 (11ˆax +
4ˆay − 9ˆaz ))/(4π × 8.85 × 10−12
| 14.76 |3 )
⇒ ~
FAB = 30.72ˆax +
11.169ˆay − 25.13ˆaz mN
D2.2(a). QA = −0.3µC located at
A(25,-30,15) in cm ,QB = 0.5µC
located at B(-10,8,12)
Find ~
E at the origin O(0,0,0).
Let ~
E at the origin is denoted by ~
Eo and it will be the sum of ~
EA ( ~
E due to QA located at point A)
and ~
EB ( ~
E due to QB located at point B)
~
E A = QA ~
ROA/4 o | ~
ROA |3
~
ROA = (0 − 25))ˆax + (0
− (−30))ˆay + (0 − 15)ˆaz =
(−25ˆax + 30ˆay − 15ˆaz )cm
| ~
p
ROA |= (−25)2 + (30)2 + (−15)2 =
41.83cm
~
EA =
(−0.3×10−6 )×(−25ˆax +30ˆay −15ˆ
az )×10−2 /4π×8.85×10−12× |
41.83×10−2 |3 = −368.55(−25ˆax +
30ˆay −15ˆaz )
~
E B = QB ~
ROB /4 o | ~
ROB |3
~
ROB = (0 − (−10)))ˆax + (0
− 8)ˆay + (0 − 12)ˆaz =
(10ˆax − 8ˆay − 12ˆaz )cm
| ~
p
ROB |= (10)2 + (−8)2 + (−12)2 =
17.55cm
~
EB =
(0.5×10−6 )×(−25ˆax +30ˆay −15ˆa
−2 /4π ×8.85×10−12× |
)×10
z
17.55×10−2 |3 =
8317.36(−25ˆax +30ˆay −15ˆaz )
~
Eo = ~
EA + ~
EB = (−368.55(−25ˆax +
30ˆay − 15ˆaz )) +
8317.36(10ˆax − 8ˆay − 12ˆaz ) =
(92.3ˆax − 77.6ˆay − 94.2ˆaz )KV
/m
(b). Find ~
E at the point P(15,20,50).
It is the same as part(a) but this
time we have to calculate ~
RP A and ~
RP B and the rest of the problem is
similar to
part(a)
D2.3 (a).
Σ5
m )/(m2 + 1)) = (1 +
((1
+
(−1)
0
(−1)0 )/(02 + 1) + (1 +
(−1)1 )/(12 + 1) + (1 +
(−1)2 )/(22 + 1) + (1 +
(−1)3 )/(32 +
1) + (1 + (−1)4 )/(42 + 1) + (1 +
(−1)5 )/(52 + 1) = 2 + 0 +
2/5 + 0 + 2 /17 + 0 = 2.52
(b). Similar to the part(a)
D2.4 (a).
0.1 ≤ (| x |, | y |, | z |) ≤ 0.2 ,
given ranges of x,y and z coordinates doesnot constitute a
cubical
R
volume so dv = 0 ⇒ Q = vol ρv dv
=0
(b). Differential volume in
cylindrical co-ordinates is given
by dv = ρdρdφdz , we have
Q=
R
vol ρv dv
⇒Q=
0
2
0
0
2
2 z 2 sin(0.6)φ)ρdρdφ
(ρ
vol
R 0.1
dz =
0
R
Rπ
R4
2 2
(ρ z sin(0.6)φ)ρdρdφdz =
3
ρ dρ
Rπ
(sin(0.6)φ)dφ
dz
R4
⇒ Q =| ρ4 /4 |0.1
0 × | (−cos(0.6φ))/0.6 |π
4
×
|
z
|
0
R 0.1
4 /4 | × |
=|
(0.1)
2
(−cos(1080 ) − (−cos(0)))/0.6 | ×
| (64 − 8)/3 |
⇒ Q =| (0.1)4 /4 | × | (1.31)/0.6 |
×
| 56/3 |= 1.018mC
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This document is prepared in L
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(C). Assuming this universe to be
a perfect sphere we have limits
as 0 ≤ r ≤ ∞, 0 ≤ φ ≤ 2π,
0≤θ≤π