Jason Smith
Instructor: Dr. Sudhanshu Choudhary
Homework Assignment 4
EE 316: Circuit Analysis
14 April 2024
1
1.
a.
dW = −qE • dL
dL = 6E − 6aρ
dW = −(6E − 6)(100)(20E − 6) = −12E − 9J
b.
dL = 6E − 6aφ
dW = −(6E − 6)(−200)(20E − 6) = 24E − 9J
c.
dL = 6E − 6aφ
dW = −(6E − 6)(300)(20E − 6) = −3.6E − 9J
d.
dL = 6E − 6aφ
100a −200aφ +300az
aE = √ ρ 2
= .267aρ − .535aφ + .802az
2
2
(100 −200 +300 )
dW = −qE • dL
dW = (100aρ − 200aφ + 300az ) • (.267aρ − .535aφ + .802az )(6E − 6)(−20E −
6) = −44.9E − 9J
e.
2a −3ay +4az
aG = √x
= .371ax − .557ay + .743az
(4+9+16)
dW = (−20E −6)[100aρ−200aϕ+300az]·[0.371ax−0.557ay+0.743az](6E −
6) = (−20E − 6)(37.1(aρ · ax) − 55.7(aρ · ay) − 74.2(aϕ · ax) + 111.4(aϕ · ay) +
222.9)(6E − 6)
dW = (−20E −6)[28.4−35.8+47.7+85.3+222.9)(6E −6) = −41.8E − 9J
2
2.
a.
´
´
2
ρs a sinθ
1
dθdφ
4πε0 r dq =
0r
´4πε
2 ´π
2π
ρs a
ρs a2
V (r) = 4πεo r 0 sinθdθ 0 dφ = 4πε
or
2
V (1E − 2) = ρεs0(6E−3)
=
8.13V
(1E−2)
V (r) =
b.
2
2
2
1
1
( .02
− .03
) = 1.36V
VAB = Va − VB = ρ0sraA − ρ0sraB = (20E−9)(6E−3)
0
3
3.
a.
E = − dV
dx a
ρ0 d
E = − a
(1 − e−ax )a
0 dx
ρ0 −ax
E = − a0 e
a
b.
ρ0
ρ0
(1 − e−ad ) − a
(1 − e−a0 )
V (d) − V (0) = a
0
0
ρ0
ρ0
(1 − e−ad ) − 0 = a
V (d) − V (0) = a
(1 − e−ad )
0
0
c.
D = E
D = 0 eax (− ρ00 e−ax )a
D = −ρa
ρv = ∇ • D
ρv = ∇ • (−ρa)
d
ρv = dx
−ρ=0
d.
U=
´ d 1 ax ρ0 −ax 2
´
´
( e ( 0 e
) )dx dy dz
0 2 0
−ρ2
U = 2∗0 0∗a (−e−ad + 1)
4
4.
a.
2
W = k Qa (2 + √12 )
2
(.8E−9)
)(2 + √12 ) = .779E − 9J
Wtotal = 2W = 2( 4∗π∗(8.854E−12)∗(.04)
b.
2
2
2
2
Wtotal = 12 (8k Qa + 4k aQ√2 + 4k a√Q2/2 + 4k a√Q2/2 )
2
2
(.8E−9)
10
√4
√8
√8
√
Wtotal = kQ
2a (8 + 2 + 2 + 2 )= 4(.04)π(8.854E−12) (4 + 2 ) = 1.59E − 6J
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