EE
20
4B
| F
all 2
01
7 |
Jan
g, M
in S
eok
Electrostatics in Free Space
EE204B Fall 2017
Lecture 4
Jang, Min Seok
EE
20
4B
| F
all 2
01
7 |
Jan
g, M
in S
eok
SI Units
2
• Length in meters (m)
• Mass in kilograms (kg)
• Time in seconds (s)
• Force in newtons (N = kg × m / s2)
• Energy in joules (J = N × m = kg × m2 / s2)
• Charge in coulombs (C): elementary charge 𝑒 = 1.6019 × 10-19 C
• Electric potential in volts (V = J / C )
• Electric capacitance in farads (F = C / V)
EE
20
4B
| F
all 2
01
7 |
Jan
g, M
in S
eok
Coulomb’s Law
3
In 1784, Charles-Augustin de Coulomb found that the force between two point charges 𝑄1 and 𝑄2 is:
• Along the line joining them: 𝐅 ∥ 𝐑12 = 𝐫2 − 𝐫1• Directly proportional to 𝑄1𝑄2 of the charges: 𝐅 ∝ 𝑄1𝑄2• Inversely proportional to the square of the distance: 𝐅 ∝ 𝐑12
−2
EE
20
4B
| F
all 2
01
7 |
Jan
g, M
in S
eok
Coulomb’s Law
4
𝐅12 =𝑄1𝑄24𝜋𝜖0𝑅
2𝐚𝑅12 =
𝑄1𝑄24𝜋𝜖0 𝐑12
2
𝐑12
𝐑12=𝑄1𝑄2 (𝐫2 − 𝐫1)
4𝜋𝜖0 𝐫2 − 𝐫13
Vacuum permittivity 𝜖0 = 8.854×10-12 F/m
The force exerted on 𝑄2 due to 𝑄1:
EE
20
4B
| F
all 2
01
7 |
Jan
g, M
in S
eok
Many Charges: Superposition Principle
5
𝑂
𝑄1
𝑄𝑘
𝑄2
𝑄
𝐅
𝐅1
𝐅𝑘
𝐅2
𝐫
𝐫1
𝐫2
𝐫𝑘
𝐅 =
𝑘=1
𝑁
𝐅𝑘 =𝑄
4𝜋𝜖0
𝑘=1
𝑁𝑄𝑘(𝐫 − 𝐫𝑘)
𝐫 − 𝐫𝑘3= 𝑄𝐄
𝐄 =1
4𝜋𝜖0
𝑘=1
𝑁𝑄𝑘(𝐫 − 𝐫𝑘)
𝐫 − 𝐫𝑘3
Force exerted on 𝑄:
Electric field at position of 𝑄:
EE
20
4B
| F
all 2
01
7 |
Jan
g, M
in S
eok
E-Fields by Continuous Charge Distributions
6
𝐄 =1
4𝜋𝜖0
𝑄
𝑅2𝐚𝑅 𝐄 =
1
4𝜋𝜖0න𝐿
𝜌𝐿𝑑𝑙
𝑅2𝐚𝑅
Point charge (𝑄):
𝐄 =1
4𝜋𝜖0න𝑆
𝜌𝑆𝑑𝑆
𝑅2𝐚𝑅 𝐄 =
1
4𝜋𝜖0න𝑣
𝜌𝑣𝑑𝑣
𝑅2𝐚𝑅
Line charge (𝑑𝑄 = 𝜌𝐿𝑑𝑙):
Surface charge (𝑑𝑄 = 𝜌𝑆𝑑𝑆): Volume charge (𝑑𝑄 = 𝜌𝑣𝑑𝑣):
EE
20
4B
| F
all 2
01
7 |
Jan
g, M
in S
eok
Continuous Charge Distributions
7
• Exercise: Find the electric field at 𝑃due to an infinitely long straight line charge by using Coulomb’s law
𝑂
𝜌𝐿
𝜌
𝐑
𝑑𝐸𝜌
𝑑𝐸𝑧 𝑑𝐄
𝐚𝑅
𝐄 =1
4𝜋𝜖0න𝐿
𝜌𝐿𝑑𝑙
𝑅2𝐚𝑅
𝛼
Answer:𝑑𝑙 = 𝑑𝑧′ (primed coordinate variables for the source point)
𝑅 = 𝑧′2 + 𝜌2, 𝐚𝑅 = (−𝑧′𝐚𝑧 + 𝜌𝐚𝜌)/𝑅
𝐄 =1
4𝜋𝜖0න−∞
∞ 𝜌𝐿 −𝑧′𝐚𝑧 + 𝜌𝐚𝜌
𝑧′2 + 𝜌23 𝑑𝑧′ =
𝜌𝐿4𝜋𝜖0
න−∞
∞ 𝜌𝐚𝜌
𝑧′2 + 𝜌23 𝑑𝑧′
=𝜌𝐿𝐚𝜌
4𝜋𝜖0𝜌න−∞
∞ 1
1 + (𝑧′/𝜌)23
𝑑𝑧′
𝜌=
𝜌𝐿𝐚𝜌
4𝜋𝜖0𝜌න−𝜋/2
𝜋/2 sec2 𝛼 𝑑𝛼
sec3 𝛼
𝑧 component is odd function
𝑧′ = 𝜌 tan𝛼𝑑𝑧′ = 𝜌 sec2 𝛼 𝑑𝛼
=𝜌𝐿𝐚𝜌
4𝜋𝜖0𝜌න−𝜋/2
𝜋/2
cos 𝛼 𝑑𝛼 =𝜌𝐿
2𝜋𝜖0𝜌𝐚𝜌
EE
20
4B
| F
all 2
01
7 |
Jan
g, M
in S
eok
Continuous Charge Distributions
8
• Exercise: Find the electric field at 𝑃(0,0, ℎ)due to an infinite sheet of charge by using Coulomb’s law
𝐄 =1
4𝜋𝜖0න𝑆
𝜌𝑆𝑑𝑆
𝑅2𝐚𝑅
Answer:𝑑𝑆 = 𝜌′𝑑𝜌′𝑑𝜙′ (primed coordinate variables for the source point)
𝑅 = ℎ2 + 𝜌′2, 𝐚𝑅 = (ℎ𝐚𝑧 − 𝜌′𝐚𝜌)/𝑅
𝐄 =1
4𝜋𝜖0න0
∞
න0
2𝜋 𝜌𝑆 ℎ𝐚𝑧 − 𝜌𝐚𝜌
ℎ2 + 𝜌′23 𝜌′𝑑𝜙′𝑑𝜌′ =
𝜌𝑆ℎ
4𝜋𝜖0න0
∞
න0
2𝜋 𝐚𝑧𝜌′𝑑𝜙′𝑑𝜌′
ℎ2 + 𝜌′23
=(2𝜋)𝜌𝑆ℎ𝐚𝑧
4𝜋𝜖0න0
∞ 𝜌′𝑑𝜌′
ℎ2 + 𝜌′23 =
𝜌𝑆ℎ𝐚𝑧2𝜖0
−1
ℎ2 + 𝜌′20
∞
=𝜌𝑆2𝜖
𝐚𝑧
By symmetry,𝐄 ∥ 𝐚𝑧
EE
20
4B
| F
all 2
01
7 |
Jan
g, M
in S
eok
Continuous Charge Distributions
9
• Exercise: Find the electric field due to a uniformly charged sphere with charge density 𝜌𝑣 and radius 𝑎
𝐄 =1
4𝜋𝜖0න𝑣
𝜌𝑣𝑑𝑣
𝑅2𝐚𝑅
Answer:
(See Sadiku 6th ed. pp.121-123 for details)
𝐄 =𝐚𝑟
4𝜋𝜖0𝑟2
4𝜋𝑎3𝜌𝑣3
=𝑄total4𝜋𝜖0𝑟
2𝐚𝑟
Outside the sphere:
EE
20
4B
| F
all 2
01
7 |
Jan
g, M
in S
eok
Gauss’s Law
10
HyperPhysics
In 1813, Carl Friedrich Gauss found that
ර𝑆
𝐄 ⋅ 𝑑𝐒 =1
𝜖0න𝑣
𝜌𝑣𝑑𝑣 =𝑄
𝜖0𝛻 ⋅ 𝐄 =
𝜌
𝜖0↔
EE
20
4B
| F
all 2
01
7 |
Jan
g, M
in S
eok
Gauss’s Law: Point Charge
11
𝐄
ර𝑆
𝐄 ⋅ 𝑑𝐒 = න0
2𝜋
න0
𝜋
𝐄 ⋅ 𝐚𝑟𝑟2 sin 𝜃 𝑑𝜃𝑑𝜙 𝑑𝐒 = 𝐚𝑟𝑟
2 sin 𝜃 𝑑𝜃𝑑𝜙
= න0
2𝜋
න0
𝜋
𝐸𝑟 𝑟2 sin 𝜃 𝑑𝜃𝑑𝜙
By spherical symmetry,𝐄 = 𝐸𝑟𝐚𝑟
= 𝐸𝑟 4𝜋𝑟2 =𝑄
𝜖0↔ 𝐸 =
𝑄𝐚𝑟4𝜋𝜖0𝑟
2
EE
20
4B
| F
all 2
01
7 |
Jan
g, M
in S
eok
Gauss’s Law: Infinite Line Charge
12
• Exercise: Find the electric field at 𝑃due to an infinitely long straight line charge by using Gauss’s Law
Answer:
ර𝑆
𝐄 ⋅ 𝑑𝐒 = නtop
+නbottom
+නside
𝐄 ⋅ 𝑑𝐒
By azimuthal symmetry, 𝐄 = 𝐸𝜌𝐚𝜌
ර𝑆
𝐄 ⋅ 𝑑𝐒 = නside
𝐸𝜌𝐚𝜌 ⋅ 𝑑𝐒
= න0
2𝜋
𝐸𝜌𝐚𝜌 ⋅ 𝐚𝜌𝜌𝑙𝑑𝜙 = 2𝜋𝜌𝑙𝐸𝜌 =𝑄
𝜖0=𝜌𝐿𝑙
𝜖0
∴ 𝐄 =𝜌𝐿
2𝜋𝜖0𝜌𝐚𝜌
EE
20
4B
| F
all 2
01
7 |
Jan
g, M
in S
eok
Gauss’s Law: Infinite Sheet of Charge
13
• Exercise: Find the electric field at 𝑃 due to an infinite sheet of charge by using Gauss’s Law
Answer:
ර𝑆
𝐄 ⋅ 𝑑𝐒 = නtop
+නbottom
+නsides
𝐄 ⋅ 𝑑𝐒
𝐄 ∥ 𝐚𝑧 (by azimuthal symmetry),𝐸𝑧 is an odd function of 𝑧 (by mirror symmetry with respect to 𝑧 = 0).
∴ 𝐄 = ቊ𝜌𝑆𝐚𝑧/2𝜖0 (𝑧 > 0)−𝜌𝑆𝐚𝑧/2𝜖0 (𝑧 < 0)
𝐄 = ቊ𝐸𝑧𝐚𝑧 (𝑧 > 0)−𝐸𝑧𝐚𝑧 (𝑧 < 0)
ර𝑆
𝐄 ⋅ 𝑑𝐒 = නtop
+නbottom
𝐄 ⋅ 𝑑𝐒 = 2𝐸𝑧𝐴 =𝑄
𝜖0=𝜌𝑆𝐴
𝜖0
EE
20
4B
| F
all 2
01
7 |
Jan
g, M
in S
eok
Gauss’s Law: Uniformly Charged Sphere
14
• Exercise: Find the electric field due to a uniformly charged sphere with charge density 𝜌𝑣 and radius 𝑎
Answer:
ර𝑆
𝐄 ⋅ 𝑑𝐒 = ර𝑆
𝐸𝑟𝐚𝑟 ⋅ 𝐚𝑟𝑟2 sin 𝜃 𝑑𝜃𝑑𝜙 = 4𝜋𝑟2𝐸𝑟
∴ 𝐄 = ቊ𝑟𝜌𝑣/3𝜖0(𝑟 < 𝑎)
𝑎3𝜌𝑣/3𝜖0𝑟2(𝑟 ≥ 𝑎)
By spherical symmetry,𝐄 = 𝐸𝑟𝐚𝑟
=𝑄
𝜖0=
1
𝜖0න𝑣
𝜌𝑣𝑟2 sin 𝜃 𝑑𝑟𝑑𝜃𝑑𝜙 = ൝
4𝜋𝑟3𝜌𝑣/3𝜖0(𝑟 < 𝑎)
4𝜋𝑎3𝜌𝑣/3𝜖0(𝑟 ≥ 𝑎)
|𝐄|
𝑎𝜌𝑣3𝜖0
𝑟𝜌𝑣3𝜖0
𝑎3𝜌𝑣3𝜖0𝑟
2
EE
20
4B
| F
all 2
01
7 |
Jan
g, M
in S
eok
Electric Potential
15
Path 2
Path 1
E
B
E
A
Work done in moving 𝑄 from 𝐴 to 𝐵:
𝑊 = −න𝐴
𝐵
𝐅 ⋅ 𝑑𝐥 = −𝑄න𝐴
𝐵
𝐄 ⋅ 𝑑𝐥
Potential difference between 𝑃1 and 𝑃2:
𝑉𝐴𝐵 =𝑊
𝑄= −න
𝐴
𝐵
𝐄 ⋅ 𝑑𝐥
E-field is a conservative field (𝛻 × 𝐄 = 0) → 𝑉𝐴𝐵 is path independent
ර𝐄 ⋅ 𝑑𝐥 = න𝐴 path 1
𝐵
𝐄 ⋅ 𝑑𝐥 + න𝐵 path 2
𝐴
𝐄 ⋅ 𝑑𝐥 = 0
−න𝐴 path 1
𝐵
𝐄 ⋅ 𝑑𝐥 = න𝐵 path 2
𝐴
𝐄 ⋅ 𝑑𝐥 = −න𝐴 path 2
𝐵
𝐄 ⋅ 𝑑𝐥 = 𝑉𝐵 − 𝑉𝐴 = 𝑉𝐴𝐵
Electric potential at 𝑃 with reference to 𝑂:
𝐄 = −𝛻𝑉𝑉 = −න𝑂
𝑃
𝐄 ⋅ 𝑑𝐥
EE
20
4B
| F
all 2
01
7 |
Jan
g, M
in S
eok
Electric Potential: Point Charge
16
Electric potential due to a point charge 𝑄 at the origin:
𝑉 = −න𝑂
𝐫
𝐄 ⋅ 𝑑𝐥 = −න𝑂
𝑟 𝑄
4𝜋𝜖0𝑟2𝐚𝑟 ⋅ 𝐚𝑟𝑑𝑟 = −න
𝑂
𝑟 𝑄
4𝜋𝜖0𝑟2𝑑𝑟
Electric potential due to 𝑛 point charges 𝑄1, 𝑄2, … , 𝑄𝑛 located at points with position vectors 𝐫1, 𝐫2, … , 𝐫𝑛 (superposition principle)
= −න∞
𝑟 𝑄
4𝜋𝜖0𝑟2𝑑𝑟 =
𝑄
4𝜋𝜖0𝑟reference point 𝑂 𝑟𝑜 → ∞
Electric potential due to a point charge 𝑄 at 𝐫′:
𝑉 =𝑄
4𝜋𝜖0|𝐫 − 𝐫′|
𝑉 =1
4𝜋𝜖0
𝑘=1
𝑛𝑄𝑘
𝐫 − 𝐫𝑘
EE
20
4B
| F
all 2
01
7 |
Jan
g, M
in S
eok
Electric Potential: Dipole
17
Electric dipole: two point charges of equal magnitude but opposite singe that are separated by a small distance
𝑉 =𝑄
4𝜋𝜖0
1
𝑟1−1
𝑟2=
𝑄
4𝜋𝜖0
𝑟2 − 𝑟1𝑟1𝑟2
If 𝑟 ≫ 𝑑, 𝑟2 − 𝑟1 = 𝑑 cos 𝜃 and 𝑟2𝑟1 ≈ 𝑟2
𝑉 =𝑄
4𝜋𝜖0
𝑑 cos 𝜃
𝑟2=𝑝 cos 𝜃
4𝜋𝜖0𝑟2=
𝐩 ⋅ 𝐚𝑟4𝜋𝜖0𝑟
2
where 𝐩 = 𝑄𝐝 = 𝑄𝑑𝐚𝑧 is the dipole moment.
Electric field due to the dipole:
𝐄 = −𝛻𝑉 = −𝜕𝑉
𝜕𝑟𝐚𝑟 +
1
𝑟
𝜕𝑉
𝜕𝜃𝐚𝜃
=𝑝
4𝜋𝜖0𝑟32 cos 𝜃 𝐚𝑟 + sin 𝜃 𝐚𝜃
EE
20
4B
| F
all 2
01
7 |
Jan
g, M
in S
eok
Electric Potential: Charge Distributions
18
Surface charge:Line charge: Volume charge:
𝑉 =1
4𝜋𝜖0න𝐿
𝜌𝐿(𝐫′)
|𝐫 − 𝐫′|𝑑𝑙′
1
4𝜋𝜖0න𝑆
𝜌𝑆(𝐫′)
|𝐫 − 𝐫′|𝑑𝑆′
1
4𝜋𝜖0න𝑣
𝜌𝑣(𝐫′)
|𝐫 − 𝐫′|𝑑𝑣′
• Exercise: Find the electric potential and field at 𝑃(0,0, 𝑧 > 0) due to an uniformly charged disk with surface charge density 𝜌𝑆 and radius 𝑏.
𝜌′ 𝜌′
Answer:
𝑉 =1
4𝜋𝜖0න𝑆
𝜌𝑆(𝐫′)
|𝐫 − 𝐫′|𝑑𝑆′ =
1
4𝜋𝜖0න0
𝑏 𝜌𝑆
𝜌′2 + 𝑧22𝜋𝜌′𝑑𝜌′
=𝜌𝑆2𝜖0
𝜌′2 + 𝑧2
0
𝑏
=𝜌𝑆2𝜖0
𝑏2 + 𝑧2 − 𝑧
𝐄 = 𝐸𝑧𝐚𝑧 = −𝜕𝑉
𝜕𝑧𝐚𝑧 =
𝜌𝑆2𝜖0
1 −𝑧
𝑏2 + 𝑧2𝐚𝑧 By azimuthal symmetry, 𝐄 = 𝐸𝑧𝐚𝑧
EE
20
4B
| F
all 2
01
7 |
Jan
g, M
in S
eok
Electric Potential: Charge Distributions
19
Limiting behavior:
• 𝑧 ≪ 𝑏 → infinite sheet of charge
𝐄 = lim𝑧→∞
𝜌𝑆2𝜖0
1 −𝑧
𝑏2 + 𝑧2𝐚𝑧 =
𝜌𝑆2𝜖0
𝐚𝑧
• 𝑧 ≫ 𝑏 → point charge
𝐄 =𝜌𝑆2𝜖0
1 −1
1 + 𝑏/𝑧 2𝐚𝑧
=𝜌𝑆𝑏
2
4𝜖0𝑧2𝐚𝑧 =
1
4𝜋𝜖0
𝜌𝑆(𝜋𝑏2)
𝑧2𝐚𝑧
≈𝜌𝑆2𝜖0
1 − 1 −1
2
𝑏
𝑧
2
𝐚𝑧
=1
4𝜋𝜖0
𝑄
𝑧2𝐚𝑧
Taylor expansion:1
1 + 𝑥= 1 −
𝑥
2+⋯
EE
20
4B
| F
all 2
01
7 |
Jan
g, M
in S
eok
Electrostatic Energy
20
Electrostatic energy of a group of 3 point charges (𝑄1,𝑄2, 𝑄3):
• First charge:
• Second charge:
• Third charge:
• Total:
𝑊1 = 0
𝑊2 = 𝑄2𝑉21 =𝑄2𝑄1
4𝜋𝜖0𝑅21
𝑊3 = 𝑄3𝑉32 + 𝑄3𝑉31 =𝑄3𝑄2
4𝜋𝜖0𝑅32+
𝑄3𝑄14𝜋𝜖0𝑅31
𝑅12
𝑅23
𝑊𝐸 =
𝑗=1
3
𝑘=1
𝑗−1𝑄𝑗𝑄𝑘
4𝜋𝜖0𝑅𝑗𝑘=1
2
𝑗=1
3
𝑘=1𝑘≠𝑗
3𝑄𝑗𝑄𝑘
4𝜋𝜖0𝑅𝑗𝑘
=1
2
𝑗=1
3
𝑄𝑗 𝑘=1𝑘≠𝑗
3𝑄𝑘
4𝜋𝜖0𝑅𝑗𝑘=1
2
𝑗=1
3
𝑄𝑗𝑉𝑗
Electrostatic energy of a group of 𝑛 charges: 𝑊𝐸 =1
2
𝑗=1
𝑛
𝑄𝑗𝑉𝑗
EE
20
4B
| F
all 2
01
7 |
Jan
g, M
in S
eok
Electrostatic Energy
21
Surface charge:Line charge: Volume charge:
𝑊𝐸 =1
2න𝐿
𝜌𝐿𝑉 𝑑𝑙1
2න𝑆
𝜌𝑆𝑉 𝑑𝑆1
2න𝑣
𝜌𝑣𝑉 𝑑𝑣
Gauss’s law: 𝜌𝑣 = 𝜖0𝛻 ⋅ 𝐄𝑊𝐸 =1
2න𝑣
𝜌𝑣𝑉 𝑑𝑣 =1
2න𝑣
𝜖0(𝛻 ⋅ 𝐄)𝑉 𝑑𝑣
=1
2න𝑣
𝜖0(𝛻 ⋅ 𝑉𝐄) 𝑑𝑣 −1
2න𝑣
𝜖0(𝐄 ⋅ 𝛻𝑉) 𝑑𝑣𝛻 ⋅ 𝑉𝐄 =
𝐄 ⋅ 𝛻𝑉 + 𝑉(𝛻 ⋅ 𝐄)
=1
2ර𝑆
𝜖0𝑉𝐄 ⋅ 𝑑𝐒 −1
2න𝑣
𝜖0(𝐄 ⋅ 𝛻𝑉) 𝑑𝑣 Divergence theorem
= −1
2න𝑣
𝜖0(𝐄 ⋅ 𝛻𝑉) 𝑑𝑣 As 𝑟 → ∞, 𝐄 ∝ 1/𝑟2, 𝑉 ∝ 1/𝑟, 𝑆 ∝ 𝑟2
=1
2න𝑣
𝜖0(𝐄 ⋅ 𝐄) 𝑑𝑣 =1
2න𝑣
𝜖0𝐸2 𝑑𝑣 𝐄 = −𝛻𝑉
Electrostatic energy density: 𝑤𝐸 =1
2𝜖0𝐸
2
EE
20
4B
| F
all 2
01
7 |
Jan
g, M
in S
eok
Electrostatic Energy
22
• Exercise: Find the electrostatic energy of a uniformly charged sphere with charge density 𝜌𝑣 and radius 𝑎.
𝜌𝑣
𝑟
𝑑𝑟
𝑎
Answer:
𝑊 =1
2න𝑣
𝜌𝑣𝑉 𝑑𝑣 =1
24𝜋 න
0
𝑎
𝜌𝑣𝑉𝑟2𝑑𝑟 = 𝜋
𝜌𝑣2
3𝜖0න0
𝑎
3𝑎2 − 𝑟2 𝑟2𝑑𝑟
𝑉 = −න∞
𝑟
𝐄 ⋅ 𝐚𝑟𝑑𝑟′ = −න
∞
𝑎
𝐄 ⋅ 𝐚𝑟𝑑𝑟′ −න
𝑎
𝑟
𝐄 ⋅ 𝐚𝑟𝑑𝑟′
= −න∞
𝑎 𝑎3𝜌𝑣
3𝜖0𝑟′2𝑑𝑟′ −න
𝑎
𝑟 𝑟′𝜌𝑣3𝜖0
𝑑𝑟′ =𝜌𝑣𝑎
2
3𝜖0+𝜌𝑣𝑎
2
6𝜖0−𝜌𝑣𝑟
2
6𝜖0=
𝜌𝑣6𝜖0
3𝑎2 − 𝑟2
= 𝜋𝜌𝑣2
3𝜖0𝑎5 −
𝑎5
5=4𝜋𝜌𝑣
2𝑎5
15𝜖0=
3
20𝜋𝜖0
𝑄2
𝑎