ECE 329 – Fall 2021

Prof. Ravaioli – Office: 2062 ECEB

Section E – 1:00pm

Lecture 2

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Lecture 2 – Outline

• Maxwell’s Equations

• Electrostatics

– Coulomb’s Law

– Gauss’ Law

• Examples

Reading assignment

Prof. Kudeki’s ECE 329 Lecture Notes on Fields and Waves:

2) Static electric fields – Coulomb’s and Gauss’ laws

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FROM THE BUSINESS WORLD

CASE STUDY

Wave propagation speed is very relevant for our globally interconnected world.

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https://www.scientificamerican.com/article/an-internet-cable-will-soon-cross-the-arctic-circle/

Scientific American, June 2016

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The underwater fiber-optic cable would trace the shortest path between Europe and Asia

London-Tokyo via Northwest passage 𝐿1 ≈ 9,600 miles = 9,600 × 1,609 = 15,446,400 m

Assume dielectric constant of fiber optic glass core

𝜀𝑟 = 2.25

Propagation along the axis of the fiber (exact 𝒄)

𝑣𝑔 =𝑐

𝜀𝑟=299,792,458

2.25= 199,861,638.7

m

s

𝑡𝐿𝑇1 =𝐿

𝑣𝑔=

15,446,400

199,861,638.7= 0.0772855s

With approximate 𝒄

𝑡𝐿𝑇1 =𝐿

𝑣𝑔≈15,446,400

2 × 108= 0.077232s

time is underestimated by 53 s

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London-Tokyo via New York – San Francisco

𝐿2 ≈ 11,500 miles = 11,500 × 1,609 = 18,503,500 m

Assume again

𝑣𝑔 = 199,861,638.7m

s

Propagation along the axis of the fiber

𝑡𝐿𝑇1 =𝐿

𝑣𝑔=

15,446,400

199,861,638.7= 0.0772855 s

𝑡𝐿𝑇2 =𝐿

𝑣𝑔≈

18,503,500

199,861,638.7= 0.0925816 s

∆𝑡 ≈ 15.3 ms 6

London-Tokyo via satellite?

Signals move faster in air/vacuum than in fiber optic glass, but a geosynchronous satellite is at an altitude

𝐻 = 22,236 miles = 35,786 km

The speed of EM waves in air is 𝑣𝑔 ≈ 3 × 108 instead of

𝑣𝑔 ≈ 2 × 108 in glass but the link distance to be covered

would be at least more than five times the shortest one between London and Tokyo via the Northwest passage.

Nonetheless, for purposes other than trading, the additional delay may be quite reasonable and business entities are indeed exploring an expansion of satellite links:

https://www.businesswire.com/news/home/20210513005383/en/SoftBank-Corp.-and-OneWeb-Agree-on-Collaboration-Towards-Satellite-Communication-Service-Business-Deployment-in-Japan-and-Global-Markets

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4 times the height of Mt. Everest and 89% length of the equator (40,075km)

Static Regimes – Fields do not vary in time

ELECTROSTATICS

• The electric charges do not change position in time.

• Therefore, ρ, E and D are constant and there is no magnetic field H, since there is no current density J.

MAGNETOSTATICS

• The charge crossing a given cross-section (current) does not vary in time so that J, H and B are constant.

• Although charges are moving, the steady current maintains a constant charge density ρ in space and the electric field E is static.

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Maxwell’s Equations

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Static Electric Fields (Electrostatics)

Maxwell’s equations reduce to

The main goal of electrostatics is to find the electric field 𝐄 and the associated electric flux density 𝐃 induced by a specified distribution of charge.

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Coulomb’s Law – An isolated charge 𝑸 induces an electric field at points of distance 𝑅 given by

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+Q

Radial Symmetry - Field does not depend on direction but only on distance

If the charge is at the origin and in vacuum

where r is the position vector

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The field exerts a force on a test charge at distance

Charge away from the origin

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Difference between vectors

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Multiple charges (superposition of effects)

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Example

Two point charges with 𝑄1 = 2 × 10−5C and 𝑄2 = −4 × 10−5C are locate in free space at points with Cartesian coordinates (1, 3, -1) and (-3, 1, -2), respectively.

Find:

(a) the electric field at (3, 1, -2) and

(b) the force on a charge 𝑞 = 8 × 10−5C, located at that point.

All distances are in meters.

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𝑃

𝑄1

𝑄2

(1, 3, -1)

(-3, 1, -2)

(3, 1, -2)

(a)

𝐄 =1

4𝜋𝜖0𝑄1

𝐫 − 𝐫1𝐫 − 𝐫1

3+ 𝑄2

𝐫 − 𝐫2𝐫 − 𝐫2

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𝐫1 = 𝐱 + 3𝐲 − 𝐳 𝐫2 = −3𝐱 + 𝐲 − 2𝐳 𝐫 = 3𝐱 + 𝐲 − 2𝐳

𝐫 − 𝐫13 = 4 + 4 + 1

3= 27

𝐫 − 𝐫23 = 6 3 = 216

𝐄 =1

4𝜋𝜖022𝐱 − 2𝐲 − 𝐳

27− 4

6𝐱

216× 10−5 =

=1

4𝜋𝜖0

4𝐱 − 4𝐲 − 2𝐳

27−3𝐱

27× 10−5

=𝐱 − 4𝐲 − 2𝐳

108𝜋𝜖0× 10−5 V m

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Field

Location vectors

Distances from probe charge

(b)

𝐄 =𝐱 − 4𝐲 − 2𝐳

108𝜋𝜖0× 10−5 V m

𝑞 = 8 × 10−5C

𝐅 = 𝑞𝐄 = 8 × 10−5 ×𝐱 − 4𝐲 − 2𝐳

108𝜋𝜖0× 10−5 V m

𝐅 =2𝐱 − 8𝐲 − 4𝐳

27𝜋𝜖0× 10−10 (N)

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Field

Force on q

Probe charge

Example Two identical charges are located on the 𝑥 axis at

𝑥1 = 3 m 𝑥2 = 7 m

At what point in space is the net electric field zero?

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0 1 2 3 4 6 7 8 5

Example Two identical charges are located on the 𝑥 axis at

𝑥1 = 3 m 𝑥2 = 7 m

At what point in space is the net electric field zero? Not considering the zero field solution at infinity, the only possibility for the field to be zero is on a 𝑦, 𝑧 plane where all points are equidistant from the two charges. By placing a probe charge 𝑞 on this plane, one can see that the only location with zero field is 𝑥 = 5.

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0 1 2 3 4 6 7 8 5

𝑄1 𝑄2

Assume identical positive charges

𝐸1 𝐸2 𝑞

Enclose a charge in a spherical surface of radius 𝒓

with surface area 𝑆 = 4𝜋𝑟2

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+Q 𝑟

𝐸𝑟 =𝑄

4𝜋𝜖0𝑟2

𝜖0𝐸𝑟𝑆 = 𝑄 → 𝜖0 𝐄 ∙ 𝑑𝐒

𝑆

= 𝑄

flux of 𝐄 out of surface 𝐒

NOTE:

For a spherical surface surrounding a charge at its center, the actual field at the surface is radial and therefore normal.

In the general case, the flux considers the normal components of the field at any points of the surface, that is the projection of the surface field onto the direction of the unit vector normal to the surface element.

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𝒏

𝐄

𝐄𝐧

𝑑𝐒

For an arbitrary volume 𝑽 enclosed by surface 𝑺, which encloses an arbitrary net charge distribution 𝑸𝑽 we have Gauss’ Law

where 𝐃 = 𝝐𝟎 𝐄 is the displacement field.

Gauss’ Law provides an alternative to Coulomb’s Law for computing static fields.

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𝐃 ∙ 𝑑𝐒

𝑆

= 𝑄𝑉

NOTE: Coulomb’s Law is only valid in static conditions (fixed charges).

“Information” on charge position (i.e., the field caused by the charge) can only be transmitted at the speed of light.

Also, charge movements create currents generating magnetic fields, which must be considered to solve the electromagnetic field problem.

On the other hand, Gauss’s law is valid in all conditions, since the surface integral will only change when charges get to cross it, at whatever speed they might be travelling.

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Example

Use Gauss’s law to obtain an expression for 𝐄 due to an

infinitely long line with uniform charge density 𝜌 = C

m

that resides along the z axis in free space.

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𝜑 𝜑

Solution using Gauss’ Law

Since the charge density along the line is uniform, infinite in extent and residing along the 𝑧 axis, for symmetry 𝐃 is in the radial direction and is independent of 𝑧 or rotation angle 𝜑.

The total charge contained in the cylinder is

Top and bottom surfaces do not contribute to the integral for the flux, only the lateral cylindrical surface:

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Solution using Gauss’ Law

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𝐄 =𝐃

𝜖0= 𝒓

𝐷𝑟𝜖0

= 𝒓

2𝜋𝜖0𝑟

Solution using Coulomb’s Law

Distribute discrete charges 𝑄 along the 𝑧-axis, spaced by a distance ∆𝑧 = 𝑄 , at locations 𝑧 = 𝑛∆𝑧, where 𝑛 is an integer.

For 𝑟 ≫ ∆𝑧 (macroscopic limit)

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microscopic field

macroscopic field

Example

Consider a sphere of radius 𝑟0 filled with a space-charge cloud of positive uniform density 𝜌. Find the electric field inside and outside the sphere. Assume free space throughout.

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𝑟0

𝑟

Solution

Total charge contained by the sphere 𝑄 = 𝜌4

3𝜋𝑟0

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Outside the sphere (𝑟 > 𝑟0), the electric field is the same as the field of a point charge 𝑄

and on the surface of the sphere

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𝑟0

𝑟

𝐸 𝑟 ≥ 𝑟0 =𝑄

4𝜋𝜖0𝑟2=

𝜌

3𝜖0 𝑟03

𝑟2

𝐸 𝑟 = 𝑟0 =𝜌

3𝜖0𝑟0

(decay like 𝟏 𝒓𝟐 )

Inside the sphere (𝑟 < 𝑟0), the enclosed charge grows with increasing 𝑟

The field is obtained from

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𝑟0

𝑟

𝑄 𝑟 ≤ 𝑟0 =4𝜋𝜌

3𝑟3

𝐸 𝑟 = 𝑟0 =𝜌

3𝜖0𝑟0

𝜖0𝐸 𝑟 ≤ 𝑟0 4𝜋𝑟2 =4𝜋𝜌

3𝑟3

𝐸 𝑟 ≤ 𝑟0 =𝜌

3𝜖0𝑟

As before, at the surface we have

(linear growth)

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𝑟0 𝑟

𝜌𝑟03𝜖0

𝐸𝑟

0

0