EEE 2030-03
Electricity and Magnetism I
1st Exam.
Sep. 16, 2020 AM 10:00~12:00
▪ Name:
▪ Student ID number:
1. [20] Answer the following questions.
(a) Write the integral form of the divergence theorem and Stokes’s theorem (4)
(b) Write the meaning of the ‘solenoidal(divergence-less)’ and ‘irrotational(curl-free)’ field. (4)
(c) What is the Helmholtz’s theorem? (4)
1
𝑅𝑅
(d) Derive the answer of ∇′ � �? (4)
1
𝑅𝑅
(e) Derive the answer of ∇2 � �? (4)
2. [30] Solve problems satisfying each condition.
(a) Given that a vector field formed as 𝐹𝐹⃑ = 𝑥𝑥 2 𝑦𝑦𝑥𝑥� + 𝑦𝑦 2 𝑦𝑦�, evaluate ∭𝑉𝑉 (∇ × 𝐹𝐹⃑ )dV where V is the
volume under the plane z = x+y+2, and above z = 0 within -1 ≤ x ≤ 1, -1 ≤ y ≤ 1. [10]
(b) Given that a vector field formed as 𝐹𝐹⃑ = 𝑟𝑟 2 𝑟𝑟̂ − 𝑟𝑟 sin 𝜙𝜙𝜙𝜙�, evaluate ∮ 𝐹𝐹� ∙ 𝑑𝑑𝑙𝑙 ̅ where closed
𝐶𝐶
contour C is denoted in Fig. 1. [10]
(c) Given that a vector field formed as 𝐹𝐹⃑ = 𝑅𝑅 4 𝑅𝑅�, evaluate ∭𝑉𝑉 (∇ ∙ 𝐹𝐹� )𝑑𝑑𝑑𝑑 where V is the region
defined by the inequality 𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 ≤ 1. [10]
Figure 1
3. [15] The electric field intensity (𝐄𝐄) has a correlation with scalar electric potential V within its
negative gradient (𝐄𝐄 = −𝛁𝛁𝑽𝑽). Find the electric field intensity 𝐄𝐄 for given electric potentials.
(a) V(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = 𝐸𝐸0 𝑒𝑒 −𝑥𝑥 𝑐𝑐𝑐𝑐𝑐𝑐
𝜋𝜋𝜋𝜋
4
(b) V(𝑟𝑟, 𝜙𝜙, 𝑧𝑧) = E0 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 (5)
(5)
cos𝜃𝜃
1
+ ) (5)
𝑅𝑅 2
𝑅𝑅
(c) V(𝑅𝑅, 𝜃𝜃, 𝜙𝜙) = 𝐸𝐸0 (
4. [35] Prove the following equations.
(a) ∇ ∙ (𝑓𝑓∇𝑔𝑔) = 𝑓𝑓∇2 𝑔𝑔 + ∇𝑓𝑓 ∙ ∇𝑔𝑔
Using the above identity, show that
∮𝑠𝑠 (𝑓𝑓∇𝑔𝑔 − 𝑔𝑔∇𝑓𝑓) ∙ 𝑑𝑑𝑠𝑠̅ = ∫𝑣𝑣 (𝑓𝑓∇2 𝑔𝑔 − 𝑔𝑔∇2 𝑓𝑓) 𝑑𝑑𝑑𝑑 (10)
(b) ∇ ∙ 𝑟𝑟̅ = 3 (𝑟𝑟̅ ∶ 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣) (10)
(c) ∇2 𝑅𝑅𝑛𝑛 = 𝑛𝑛(𝑛𝑛 + 1)𝑅𝑅 𝑛𝑛−2 (15)
Appendix
Cylindrical coordinates (r, 𝜙𝜙, z )
Cartesian coordinates (x, y, z)
∇𝑉𝑉 = 𝑥𝑥�
𝜕𝜕𝑉𝑉
𝜕𝜕𝑉𝑉
𝜕𝜕𝑉𝑉
+ 𝑦𝑦� + 𝑧𝑧̂
𝜕𝜕𝑥𝑥
𝜕𝜕𝑦𝑦
𝜕𝜕𝑧𝑧
∇𝑉𝑉 = 𝑟𝑟̂
𝜕𝜕𝐹𝐹
𝜕𝜕𝐹𝐹
𝜕𝜕𝐹𝐹
∇ ∙ 𝐹𝐹� = 𝑥𝑥 + 𝑦𝑦 + 𝑧𝑧
𝜕𝜕𝑥𝑥
𝑥𝑥�
𝜕𝜕
∇ × 𝐹𝐹� = �𝜕𝜕𝑥𝑥
𝐹𝐹𝑥𝑥
𝜕𝜕2 𝑉𝑉
𝜕𝜕𝑦𝑦
1 𝜕𝜕(𝑟𝑟𝐹𝐹𝑟𝑟 )
1 𝜕𝜕𝐹𝐹𝜙𝜙
𝜕𝜕𝐹𝐹
∇ ∙ 𝐹𝐹� =
+
+ 𝑧𝑧
𝜕𝜕𝑧𝑧
𝑦𝑦�
𝑧𝑧̂
𝜕𝜕
𝜕𝜕𝑦𝑦
𝜕𝜕
�
𝜕𝜕𝑧𝑧
𝐹𝐹𝑦𝑦
𝜕𝜕2 𝑉𝑉
Spherical
𝜃𝜃, 𝜙𝜙)
∇2 𝑉𝑉 = 2coordinates
+ 2 + (R,
2
𝜕𝜕𝑥𝑥
𝜕𝜕𝑦𝑦
𝜕𝜕𝑧𝑧
𝜕𝜕𝑉𝑉
1 𝜕𝜕𝑉𝑉
1
𝜕𝜕𝑉𝑉
�
�
∇𝑉𝑉 = 𝑅𝑅 + 𝜃𝜃
+ 𝜙𝜙�
𝜕𝜕𝑅𝑅
𝑅𝑅 𝜕𝜕𝜕𝜕
𝑅𝑅 sin 𝜃𝜃 𝜕𝜕𝜕𝜕
2
1 𝜕𝜕(𝑅𝑅 𝐹𝐹𝑅𝑅 )
∇ ∙ 𝐹𝐹� = 2
+
𝑅𝑅
∇ × 𝐹𝐹� =
∇2 𝑉𝑉 =
𝜕𝜕𝑅𝑅
𝑅𝑅�
𝜕𝜕
1
�
𝑅𝑅2 sin 𝜃𝜃 𝜕𝜕𝑅𝑅
𝐹𝐹𝑅𝑅
𝑟𝑟
𝜕𝜕𝑟𝑟
𝑟𝑟̂
1 𝜕𝜕
∇ × 𝐹𝐹� = �𝜕𝜕𝑟𝑟
𝑟𝑟
𝐹𝐹𝑧𝑧
𝜕𝜕2 𝑉𝑉
𝜕𝜕𝑉𝑉
1 𝜕𝜕𝑉𝑉
𝜕𝜕𝑉𝑉
+ 𝜙𝜙�
+ 𝑧𝑧̂
𝜕𝜕𝑟𝑟
𝑟𝑟 𝜕𝜕𝜕𝜕
𝜕𝜕𝑧𝑧
∇2 𝑉𝑉 =
𝐹𝐹𝑟𝑟
𝑟𝑟 𝜕𝜕𝜕𝜕
𝑟𝑟𝜙𝜙�
𝜕𝜕
𝜕𝜕𝜕𝜕
𝑟𝑟𝐹𝐹𝜙𝜙
𝜕𝜕
𝜕𝜕𝜕𝜕
𝑅𝑅𝐹𝐹𝜃𝜃
𝑅𝑅 sin 𝜃𝜃 𝜙𝜙�
𝜕𝜕
𝜕𝜕𝜕𝜕
𝑅𝑅 sin 𝜃𝜃 𝐹𝐹𝜙𝜙
𝜕𝜕
�
𝜕𝜕𝑧𝑧
𝐹𝐹𝑧𝑧
1 𝜕𝜕
𝜕𝜕𝑉𝑉
1 𝜕𝜕2 𝑉𝑉
𝜕𝜕2 𝑉𝑉
�𝑟𝑟 � + 2 2 + 2
𝑟𝑟 𝜕𝜕𝑟𝑟
𝜕𝜕𝑟𝑟
𝑟𝑟 𝜕𝜕𝜙𝜙
𝜕𝜕𝑧𝑧
1
𝜕𝜕(sin 𝜃𝜃𝐹𝐹𝜃𝜃 )
1 𝜕𝜕𝐹𝐹𝜙𝜙
+
𝑅𝑅 sin 𝜃𝜃
𝜕𝜕𝜕𝜕
𝑅𝑅 sin 𝜃𝜃 𝜕𝜕𝜕𝜕
𝑅𝑅𝜃𝜃�
𝑧𝑧̂
𝜕𝜕𝑧𝑧
�
1 𝜕𝜕
1
𝜕𝜕
𝜕𝜕𝑉𝑉
1
𝜕𝜕2 𝑉𝑉
𝜕𝜕𝑉𝑉
�𝑅𝑅 2 � + 2 sin 𝜃𝜃 �sin 𝜃𝜃 � + 2 sin2 𝜃𝜃 2
2
𝑅𝑅 𝜕𝜕𝑅𝑅
𝑅𝑅
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
𝑅𝑅
𝜕𝜕𝜙𝜙
𝜕𝜕𝑅𝑅
0
