Jason Smith
Instructor: Dr. Sudhanshu Choudhary
Homework Assignment 5
EE 316: Circuit Analysis
20 April 2024
1
1.
a.
´
(−106 z 1.5 az )pdφdza
´ 20
´ 2π
−106 z 1.5 0 pdp 0 dφ
-106 ∗ .11.5 ∗ .5(400 − 0) ∗ 2πE − 6 = −39.8µA
b.
J = ρv xV
ρv = VJ
−106 z 1.5 az
= −.0158c/m3
2E6
c.
J = ρv xV
V = ρJv
106 (.15)1.5 az
= 29m/s
−2000
2
2.
a.
J = µσe
J = (6.17E−7)(1.5E−6)
= 16520A/m2
.0056
b.
J = σE = (6.17E7)(1E − 3) = 61700A/m2
c.
(6.17E7)(.0004)
J = σV
= 9.872E6A/m2
L =
2500
d.
.5
2
J = SI = 2500
2 = 80000A/m
3
3.
a.
Convert to Metric:
.6”=.015m, 1200’=365.68m
L
365.8
R= σS
= (5.8E7)(π(7.5E−3)
2 ) = .035Ω
b.
50
J = SI = 1.7E−4
= .284E6A/m2
c.
V = I ∗ R = 50 ∗ .035 = 1.75V
d.
P = V ∗ I = 1.75 ∗ 50 = 87.5W
4
4.
a.
V = 100sinh(5x) ∗ sin(5y) = 100sinh(5 ∗ .1) ∗ sin(5 ∗ .2) = 100sinh(.5) ∗
sin(1) = 43.81V
b.
δV
δV
E = −[ δV
δx ax + δy ay + δz az ]
δ
δ
δ
E = −[ δx
100sinh(5x)∗sin(5y)ax + δy
100sinh(5x)∗sin(5y)ay + δz
100sinh(5x)∗
sin(5y)az ]
E = −[500cosh(5x) ∗ sin(5y)ax + 500sinh(5x) ∗ sin(5y)ay + 0az ]
E = −[500cosh(.5) ∗ sin(1)ax + 500sinh(.5) ∗ sin(1)ay ]
E = −474.43ax − 140.77ay
c.
|E| =
√
−474.432 − 140.772 = 494.87V /m
d.
ρs = 0 ∗ |E| = (8.854E − 12)(494.87) = 4.38E − 9C/m
5
0
5.
a.
V = k ∗ ( rλ1 − rλ2 ) = (9E9)( 40E−9
− 40E−9
) = 316V
1
3
b.
E = k ∗ ( rλ2 − rλ2 ) = (9E9)( 40E−9
− 40E−9
12
32 ) = −45.4N/C
1
2
6
6.
a.
8E−9
E = 0Dr = (8.854E−12)(3.8)
= 237.8V /m
b.
P = (r − 1) ∗ o E = (3.8 − 1)(8.854E − 12)(237.8) = 5.89E − 9C/m
c.
P
Pi
P
P= n
Pi = n10−29
n
∆V
= 5.89E20m−3
= 5.89E−9
10−29
7
7.
a.
Dn1 = D1 az
Dn1 = (−30ax + 50ay + 70az ).az
Dn1 = 70az nC/m2
b.
Dt1 = D1 − Dn1
Dt1 = (−30ax + 50ay + 70az ) − 70az
Dt1 = (−30ax + 50ay )nC/m2
c.
Dt1
)
θ1 = tan−1 ( D
n1
√
2
Dt1 = −30 + 502
θ1 = tan−1 (
√
−302 +502
) = 39.8deg
70
d.
P1 = D1 (1 − 11 )
1
P1 = (−30ax +50ay +70az )∗(1− 3.2
) = (−20.6ax + 34.4ay + 48.1az )n/m2
e.
Dn1 = Dn2 = 70nC/m2
f.
Dt2 = 21 ∗ Dt1
2
Dt2 = ( 3.2
) ∗ (−30ax + 50ay ) = (−18.75ax + 31.25ay )nC/m2
8