ECE 329 – Fall 2021

Prof. Ravaioli – Office: 2062 ECEB

Section E – 1:00pm

Lecture 3

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Lecture 3 – Outline

• More examples on field calculations

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

3) Gauss’ laws and static charge densities

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Example (from previous lecture)

Use Gauss’ 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

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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

𝐷𝑟

𝜖0= 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

(decays 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

linear growth

decay 𝟏 𝒓𝟐

From the symmetry of the problem we deduce that the field is of the form

We can choose a convenient surface to calculate the flux, e.g., a cylinder of Area 𝐴 parallel to the 𝑥 axis. This time we will get non-zero flux only through the ends and not through the sides of the cylindrical surface.

Example – Charged Plane

Charges are distributed over the plane 𝑥 = 0 with average surface charge density 𝜌𝑠 C m2 . Determine the macroscopic electric field 𝐄 of this charge distribution using Gauss’ Law.

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The Field is an odd function of 𝒙. Why?

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𝑥

𝑧

−𝑥

𝐴 𝐴

Assume a positive charge density on the sheet

+ + 𝑞 positive probe charge

Charge is repelled by the sheet toward positive 𝑥 coordinates. Electric Field is positive

𝑞 positive probe charge

Charge is repelled by the sheet toward negative 𝑥 coordinates. Electric Field is negative

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in vector form

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To represent mathematically charge distributions in confined spaces, it is convenient to use the impulse or “delta” function. This allows us to define isolated charges as a density function that covers the whole space. Examples: • Point charge at point 𝐫 = 𝑥0, 𝑦0, 𝑧0 • Charged layer of infinitesimal thickness on plane 𝒙 = 𝑥0

𝜌 𝑥, 𝑦, 𝑧 = 𝜌𝑠(𝑦, 𝑧) 𝛿(𝑥 − 𝑥0)

𝜌 𝑥, 𝑦, 𝑧 = 𝑄 𝛿(𝑥 − 𝑥0) 𝛿(𝑦 − 𝑦0)𝛿(𝑧 − 𝑧0)

Example – Charged Slab

Now stretch the plane into an infinite slab of width 𝑊 with charge density 𝜌 C m3 . Determine the macroscopic electric field 𝐄 inside and outside the slab.

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Again, from symmetry the field is of the form

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𝑥

𝑧

−𝑥

𝐴 𝐴

𝜌 𝜌

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Inside the slab

𝒙 is positive Field is positive

𝒙 is negative Field is negative

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Outside the slab

𝒙 is positive Field is positive

𝒙 is negative Field is negative

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The charge sheet is a limit case for 𝑊 → infinitesimal (surface density)

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These elementary results can be used as building blocks for more complicated situations

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Example – Two planar surface charges

simple model of a parallel plate capacitor −

𝑊

2

𝑊

2 0

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Example – Two planar surface charges

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Ideal Model of Finite Capacitor

Field lines exist only between plates

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Realistic Model of Finite Capacitor

Fringing field lines at the edges

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Example – Two Slabs with opposite volumetric charge

The system is charge neutral when

You will see in ECE 340 that this is a simple model for the active region of a p-n junction diode

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Graphically

=

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Mathematically

Fields cancel out outside the region

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Example – Electric Dipole

An electric dipole consists of two point charges of equal magnitude but opposite polarity, separated by a distance d.

+

−

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Electric field pattern of the Electric Dipole

+

−

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Electric Field of the Electric Dipole from Coulomb’s Law

+

−

𝑥

𝑧

𝑦

𝑑

2

−𝑑

2

𝒓

𝑃 𝟏

𝟐

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Question: Find the electric flux across the {𝒙-𝒚} plane

𝑑𝐒

The total flux is the sum of the fluxes for the individual particles

Flux =

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Gauss’ Law gives the flux for a closed surface enclosing the charge

For our problem we can infer that half the flux crosses the {𝒙-𝒚} plane and write

𝑑𝐒

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Similarly, for the negative charge below the {𝒙-𝒚} plane

Since we are looking for the total flux crossing the plane upward (but the field points downward so it is negative), also for the negative charge we get the flux

upward 𝑑𝐒

𝑑𝐒

downward 𝑑𝐒

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In fact, if you place a positive probe charge in the middle of the dipole, the positive charge repels it downward and the negative charge attracts it downward

+

−

𝑑

2

−𝑑

2

+𝑞

𝟏

𝟐

The total flux is obtained by superposition

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Without taking shortcuts, one could simply solve the integral directly, which requires a change of variables (a lot more work)

with

change variables and integrate over 𝜑 which just gives a factor of 2𝜋 and eliminates 2𝜋 from the denominator

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Example – Electric field of a Ring of Charge

The field 𝑑𝐄1due to infinitesimal segment 1

segment length 𝑑𝑙 = 𝑏𝑑𝜑

𝑏 = radius

charge in segment 𝑑𝑞 = 𝜌𝑑𝑙 = 𝜌𝑏𝑑𝜑

We look for the Electric Field on points along the axis of the ring

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Example – Electric field of a Ring of Charge

The fields 𝑑𝐄1and 𝑑𝐄2 due to infinitesimal segments at diametrically opposite locations

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Example – Electric field of a Ring of Charge

The radial components are opposite and cancel out. The axial components add up.

Coulomb’s Law in differential form

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Example – Electric field of a Ring of Charge

The total field is obtained by integration in the range

total charge on the ring

×2𝜋

2𝜋

then multiply