ECE 536 – Integrated Optics and Optoelectronics Lecture 3 – … · 2021. 2. 2. ·...
Transcript of ECE 536 – Integrated Optics and Optoelectronics Lecture 3 – … · 2021. 2. 2. ·...
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