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Electrostatics in Free Space

EE204B Fall 2017

Lecture 4

Jang, Min Seok

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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)

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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

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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:

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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:

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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):

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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

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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

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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:

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Gauss’s Law

10

HyperPhysics

In 1813, Carl Friedrich Gauss found that

ර𝑆

𝐄 ⋅ 𝑑𝐒 =1

𝜖0න𝑣

𝜌𝑣𝑑𝑣 =𝑄

𝜖0𝛻 ⋅ 𝐄 =

𝜌

𝜖0↔

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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

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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

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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

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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

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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

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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

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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

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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

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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+⋯

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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