ECE 329 Homework 1 — Solutions Due: Tue, Sep 21, 2021
1. Review of vectors : Following vectors are given in the problem :
A = x̂ + ŷ − ẑ,
B = (1, −2, −3),
C = 2x̂ − ŷ + 3ẑ.
a) D = A + B = 2x̂ − ŷ − 4ẑ.
b) A + B − 4C = 2x̂ − ŷ − 4ẑ − 4(2x̂ − ŷ + 3ẑ) = −6x̂ + 3ŷ − 16ẑ.
p √
c) |A + B − 4C| = (−6)2 + (3)2 + (−16)2 = 301 ≈ 17.35.
d) A + 2B − C = x̂ − 2y − 10ẑ. p √
The magnitude of the vector is |A + 2B − C| = 12 + (−2)2 + (−10)2 = 105. Then, the unit
vector is written as
A + 2B − C x̂ − 2ŷ − 10ẑ
û = = √ ≈ 0.098x̂ − 0.195ŷ − 0.976ẑ.
|A + 2B − C| 105
e) A · B = (x̂ + ŷ − ẑ) · (x̂ − 2ŷ − 3ẑ) = (1 × 1) + (1 × (−2)) + ((−1) × (−3)) = 2.
x̂ ŷ ẑ
f) B×C = (x̂−2ŷ −3ẑ)×(2x̂− ŷ +3ẑ) = 1 −2 −3 = x̂(−6−3)+ ŷ(−6−3)+ ẑ(−1−(−4)) =
2 −1 3
−9x̂ − 9ŷ + 3ẑ.
2. Electrostatic charges and fields :
a) The electric fields at point P3 and P4 are the superposition of those induced by the charges Q1
and Q2 . Referring to page 2 of Lecture 2, the field at point P3 is given by
2
X Qn r3 − rn ẑ (−ẑ) V
E3 = · = 9 3 − 18 3 = 27ẑ ,
4πϵo |r3 − rn |2 |r3 − rn | |ẑ| |ẑ| m
n=1
and the field at point P4 is calculated as
2
X Qn r4 − rn x̂ + ẑ (x̂ − ẑ) 9 V
E4 = 2
· =9 3
− 18 3
= √ (−x̂ + 3ẑ) .
4πϵo |r4 − rn | |r4 − rn | |x̂ + ẑ| |x̂ − ẑ| 2 2 m
n=1
b) Bonus problem : Plot of these fields in part (a) we obtained using Mathematica is shown in
the following figure.
1
1. Review of vectors : Following vectors are given in the problem :
A = x̂ + ŷ − ẑ,
B = (1, −2, −3),
C = 2x̂ − ŷ + 3ẑ.
a) D = A + B = 2x̂ − ŷ − 4ẑ.
b) A + B − 4C = 2x̂ − ŷ − 4ẑ − 4(2x̂ − ŷ + 3ẑ) = −6x̂ + 3ŷ − 16ẑ.
p √
c) |A + B − 4C| = (−6)2 + (3)2 + (−16)2 = 301 ≈ 17.35.
d) A + 2B − C = x̂ − 2y − 10ẑ. p √
The magnitude of the vector is |A + 2B − C| = 12 + (−2)2 + (−10)2 = 105. Then, the unit
vector is written as
A + 2B − C x̂ − 2ŷ − 10ẑ
û = = √ ≈ 0.098x̂ − 0.195ŷ − 0.976ẑ.
|A + 2B − C| 105
e) A · B = (x̂ + ŷ − ẑ) · (x̂ − 2ŷ − 3ẑ) = (1 × 1) + (1 × (−2)) + ((−1) × (−3)) = 2.
x̂ ŷ ẑ
f) B×C = (x̂−2ŷ −3ẑ)×(2x̂− ŷ +3ẑ) = 1 −2 −3 = x̂(−6−3)+ ŷ(−6−3)+ ẑ(−1−(−4)) =
2 −1 3
−9x̂ − 9ŷ + 3ẑ.
2. Electrostatic charges and fields :
a) The electric fields at point P3 and P4 are the superposition of those induced by the charges Q1
and Q2 . Referring to page 2 of Lecture 2, the field at point P3 is given by
2
X Qn r3 − rn ẑ (−ẑ) V
E3 = · = 9 3 − 18 3 = 27ẑ ,
4πϵo |r3 − rn |2 |r3 − rn | |ẑ| |ẑ| m
n=1
and the field at point P4 is calculated as
2
X Qn r4 − rn x̂ + ẑ (x̂ − ẑ) 9 V
E4 = 2
· =9 3
− 18 3
= √ (−x̂ + 3ẑ) .
4πϵo |r4 − rn | |r4 − rn | |x̂ + ẑ| |x̂ − ẑ| 2 2 m
n=1
b) Bonus problem : Plot of these fields in part (a) we obtained using Mathematica is shown in
the following figure.
1
