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Unformatted text preview: ECE 2317: Applied Electricity and Magnetism
Fall 2008
Prof. Jeffery T. Williams Dept. of Electrical & Computer Engineering University of Houston Introduction to course Introduction to this Course
All important information, announcements and assignments will be posted on the webpage for this class. • Check it regularly. • You are responsible for everything on it! http://www.egr.uh.edu/courses/ece/ece2317 ECE 2317 Section 18562 10:0011:30 AM, TTh W205D3
Instructor Jeffery T. Williams W432AD3 phone: 7137434455 fax: 7137434444 email: [email protected] Prerequisites
Prerequisites:
Math 2433: (Calculus III) Physics 1322 (University Physics II) Credit for or Registration in (CFORI):
Math 3321: (Engineering Math) ECE 2331 (Numerical Methods) Prerequisite Letter – What is it?
If you got a letter, the ECE Department computer system thinks you are missing the course requirements. If you got a letter you must respond to it and return it to the Department by the deadline or you will be dropped. Any questions about the form…please see Dr. Barr in room N311D. Individual Responsibilities
All students must be thoroughly familiar with all the requirements, regulations, and responsibilities described in the UH Student Handbook. Unless otherwise specified, these provisions will be followed as described in the handbook. http://www.uh.edu/dos/hdbk/ Academic Honesty Statement At the end of the syllabus is an academic honesty form. You must fill this out, sign it, and return it to the instructor by Monday, Sept 8th. Otherwise, you may be dropped from the course. Grading
Homework Project Exam 1 Exam 2 Final Exam: 10% 10% 25% 25% 30% Homework
Exercises
• not collected • not graded • detailed solutions provided Problems
• collected • graded • no solutions distributed General Notes
• No late assignments will be accepted. • No makeup assignments or exams. • Important Dates
August 25 September 1 September 8 November 4 November 2629 December 6 December 11 First day of classes Labor Day Holiday Last day to drop without receiving a grade Last day to drop Thanksgiving Holiday Last day of classes Final exam (11AM  2PM) What’s this course about ?
Introduction to electromagnetism (electric and magnetic fields)
• In this course we study STATICS (zero or low frequency) • In the next EM course, ECE 3317, you will study DYNAMICS Electromagnetics
• The basic sources of electromagnetic (electrical) forces are stationary and moving electric charges, which exert forces on other stationary and moving electric charges. To study these forces, the concept of force fields has been devised.
– When charges are stationary, the force field is an electric (electrostatic) field. – When charges are moving with a constant velocity, the force field is a magnetic (magnetostatic) field. – Accelerating charges produce dynamic electromagnetic fields. Statics
Definitions: Statics  f = 0 [Hz] Quasistatics – slow time variations The Static electromagnetic fields splits into two independent sets: Electrostatics: (q, E) Magnetostatics: (I, B) Static charge Constant current The (quasi)static approximation is generally valid if a signal's "travel time" t = d /c across a circuit of dimension d is much smaller than ∆t, the time for a significant change in the signal (e.g. "rise time" of a pulse) to occur. E.g., d ⎧c∆t , a pulse ⎪ c ⎨ cT = = λ , a sinusoid ⎪ f ⎩ f  frequency T  period = 1/ f λ  wavelength = c / f Frequency/Wavelength
Example: f = 60 [Hz]
λ0 = c / f c = 2.99792458 × 108 [m/s] f = 60 [Hz] This gives: λ0 = 4.9965×106 [m] = 4,996.5 [km] = 3,097.8 [miles] Radius of earth = 6.378 km (3.961 miles) Clearly, most circuits fall into the staticapproximation category at 60 [Hz]! How does this stuff fit in?
The following areas rely upon static/quasistatic EM field theory:
• circuit theory (e.g, ECE 2300) • electronics • power engineering • magnetics ECE 2317 The following areas rely upon dynamic EM field theory:
• antennas • transmission lines • microwaves • optics ECE 3317 ECE 2317: Applied Electricity and Magnetism
Fall 2008
Prof. Jeffery T. Williams Dept. of Electrical & Computer Engineering University of Houston Charge
Notes prepared by the EM group, University of Houston. Charge
proton: q = e = 1.602 x 1019 [C] electron: q = e = 1.602 x 1019 [C] 1 [C] = 1 / 1.602 x1019 protons = 6.242 x 1018 protons e p atom Charge Density
1) Volume charge density ρv [C/m3] ∆V ∆Q ρv uniform cloud of charge ∆Q ρv = [C/m3 ] ∆V Charge Density (cont.)
nonuniform (inhomogeneous) volume charge density ρv (x,y,z) dV
nonuniform cloud of charge dQ ∆Q dQ ρv ( x, y, z ) = lim = ∆V → 0 ∆V dV Charge Density (cont.)
ρv (x,y,z) dV ∆Q ρv ( x, y, z ) = lim ∆V → 0 ∆V ∆Q ρ v ( x, y , z ) ≈ ∆V
so dQ ∆Q ≈ ρ v ( x, y, z ) ∆V dQ = ρ v ( x, y, z ) dV
Q = ∫ ρ v ( x, y, z ) dV
V or Charge Density (cont.)
2) Surface charge density ρs [C/m2] ρs (x,y,z) ∆S
nonuniform sheet of charge ∆Q ∆Q ρ s = lim [C/m2 ] ∆S → 0 ∆S
nonuniform ρs = ∆Q [C/m2 ] ∆S
uniform Charge Density (cont.)
ρs (x,y,z) dS dQ ρs = [C/m2 ] dS dQ = ρ s ( x, y, z ) dS dQ Q = ∫ ρ s ( x, y, z ) dS
S Charge Density (cont.)
3) Line charge density ρ l [C/m]
nonuniform line of charge ∆l + + +++ + + + + + + ++ ∆Q ρ l (x,y,z) ∆Q ρl = lim [C/m] ∆l → 0 ∆l
nonuniform ∆Q ρl = [C/m] ∆l
uniform Charge Density (cont.)
dl + + +++ + + + + + + ++ dQ ρ l (x,y,z) dQ = ρl ( x, y, z ) dl Q = ∫ ρl ( x, y, z ) dl
C Example – Find the total charge in sphere.
z ρv = 10 [C/m3] a
y x Find Q Q = ∫ ρ v ( x, y, z ) dV
V Q = 10 ⋅ V ⎛4 3⎞ =10 ⋅ ⎜ π a ⎟ ⎝3 ⎠
Q= 40 3 π a [C] 3 = ∫ 10 dV
V = 10 ∫ dV
V Example – Find the total charge in sphere.
z ρv = 2r [C/m3] a
y x Find Q 2π Q = ∫ ρ v ( x, y, z ) dV
V Q= ∫ dφ
0 π ∫ sin θ dθ
0 ( 2r ) r 2 dr ∫
0 a = ∫ 2r dV
V 2π π a ⎛2 4⎞ = ( 2π )( 2 ) ⎜ a ⎟ ⎝4 ⎠
Q = 2π a 4 [C] = ∫ ∫ ∫ ( 2r ) r
0 00 2 sin θ dr dθ dφ Example – Find the Equivalent Surface Charge Example Density for a Thin Slab of Charge
y ρ v = 2 xyz ⎡C/m3 ⎤ , ⎣ ⎦
0 ≤ z ≤ ∆z
∆z ρV ( x, y, z ) ∆V x
∆S ∆Q = ∆V ∫ ρ dV = ∫ ∫
v ∆S ∆z z 0 ρv dz dS ≡ ∆S ∫ ρ S dS ⇒ ρ S ( x, y ) = ∫ ∆z 0 ρ v ( x, y, z ) dz = ∫ ∆z 0 2 xyzdz = xy∆z 2 ⎡C/m2 ⎤ ⎣ ⎦ Example – Find the Equivalent Line Charge Density for a Thin Cylinder of Charge
∆V
y z ρ v = 2 x 2 yz ⎡C/m3 ⎤ , ⎣ ⎦
0 ≤ x ≤ ∆x , 0 ≤ y ≤ ∆ y
∆Q =
dxdy x
z +∆ / 2 ∆ ∆V 3 ∫ ρv dV = z +∆ / 2 ∆y ∆x z −∆ / 2 0 0 2 ∫∫ 2 x 2 yz dxdydz ∫ ≡ z −∆ / 2 ∫ ρ dz ⎡ ( z + ∆ / 2 )2 − ( z − ∆ / 2 )2 ⎤ ⎛ 2∆x ⎞⎛ ∆y ⎞ =⎜ ⎥ ⎟⎜ ⎟⎢ 2 ⎥ ⎝ 3 ⎠⎝ 2 ⎠ ⎢ ⎣ ⎦ 2∆x 3 ∆y 2 2 z ∆ = 3 2 2 ∆Q ∆x 3∆y 2 ⇒ ρ ( z ) = lim = z [C/m] ∆ →0 ∆ 3 ECE 2317: Applied Electricity and Magnetism
Fall 2008
Prof. Jeffery T. Williams Dept. of Electrical & Computer Engineering University of Houston Current
Notes prepared by the EM group, University of Houston. e c t r i c c u r r e n t What is Electric Current?
Electric current is a flow of charged particles (usually electrons) through conductors (usually wires and electronic components). It can be likened to the flow of water through pipes. Just as water is pushed through pipes by a pump, electric current is pushed through wires by a battery or voltage source. Electrons will flow from the negative terminal of a battery, through the lamp, to the positive terminal. Direction of electron flow Electric current Current
Current is the flow of charge. The unit for current is the Ampere [Amp, A]. 1 Amp = 1 [C/s] Convention (Ben Franklin) : positive current is in the direction of positive charge flow. Example: +
velocity vector + + + +
current flows from left to right. Reference Direction Reference
Sign convention #1: a positive current flowing one
way is equivalent to a negative current flowing the other way. flow rate is 1 C per second + + + +
1 [A] + or 1 [A] Note: the arrow establishes the reference direction for the current. Reference Direction Reference
Sign convention #2: positive charges moving one way
is equivalent to negative charges moving the other way. flow rate is 1 C per second equivalent to:     flow rate is 1 C per second 1 [A]
or 1 [A] + + + + + Definition of Current
Mathematical definition of current
reference direction arrow i
∆Q ∆Q = amount of charge (positive or negative) that crosses the reference plane in the direction of the reference arrow in time ∆t. ∆Q i= ∆t more generally, dQ i (t ) = dt Types of Current
⎧ = constant  Direct (DC) current dQ ⎪ i (t ) = ⎨ ≠ constant  Alternating (AC) current dt ⎪ (time varying current) ⎩ Example Example
∆Q i(t) i ( t ) = 10 2 cos ( 2π 60t ) [A]
Find: charge Q that crosses dashed line going from left to right in the time interval (0, t) [S]. dQ i (t ) = dt Q ( t ) = ∫ i (t ) dt + C1 = ∫ i (t ) dt + C2
0 t t = ∫ i (t ) dt
0 (Q ( 0) = 0) Example (cont.) Example
Q ( t ) = ∫ i (t ) dt
0 t = ∫ 10 2 cos ( 2π 60t ) dt
0 t ⎡1 ⎤ sin ( 2π 60t ) ⎥ = 10 2 ⎢ ⎣ 2π 60 ⎦0
2 Q (t ) = sin ( 2π 60t ) 12π t [C] ∆S
+ Current Density Current
+ + + + I
Define current density magnitude as, v=v ˆ I J≡ ∆S [A/m 2 ] Define current density vector as, ˆ= I ˆ J≡J ∆S [A/m 2 ]
The current density vector points in the direction of current flow (positive charge motion) Current Density Vector (cont.)
∆S
∆Q + +
∆L I
+ + + ˆ ∆S v=v ˆ ∆L = distance traveled by charges in time ∆t. ∆Q / ∆t ( ∆Q / ∆t ) ∆L ⎛ ∆Q ⎞⎛ ∆L ⎞ I J= = = =⎜ ⎟⎜ ⎟ ∆S ∆S ∆S ∆ L ∆V ⎠⎝ ∆t ⎠ ⎝
or J = ρv v
or J = ρv v ˆ J = ρv v Current Crossing Surface Current
ˆ n ∆θ J
∆S ˆ ˆ J = n J n + tJ t ˆ n Jn J
ˆ n ˆ n Jn
ˆ t Jt ˆ t Jt ∆S ˆ ∆I = J n ∆S = ( J ⋅ n ) ∆S For a small differential surface dS ˆ dI = J n dS = ( J ⋅ n ) dS Current Crossing Surface
ˆ ∆I = ( ⋅ n ) the IntegratingJover ∆S entire surface,
ˆ I = ∫ J ⋅ n dS
S ˆ n J
S Note: The direction of unit normal vector determines whether current is going in or out. Example Example
cloud of electrons Ne = 1020 electrons / m3 (a) Find: current density vector ˆ v = 2.5 z [m/s] J = ρv v ρv = N e qe = (1020 [1/ m3 ]) ( −1.602 ×10−19 [C ])
ˆ J = ( −40.05) z [A/m 2 ] Example Example
(b) Find: current in wire with reference direction shown z
radius a = 1 [mm] I
ˆ J = ( −40.05) z [A/m 2 ] z ˆ I = ∫ J ⋅ n dS
S ˆ ˆ = ∫ ( − z 40.05 ) ⋅ z dS
S I = − 40.05 (π a 2 ) = −40.05 π ( 0.001) = ∫ − 40.05 dS
S ( 2 ) = − 40.05( Area ) I = −1.2582 × 10−4 [A] Example Example
ˆ ˆ ˆ J = x ( 3x 2 y ) + y ( 3 z 3 y 2 ) + z ( 3xy ) [A/m 2 ]
z
(0,1,0) (1,0,0) y x S Find: current I crossing surface in the upward direction ˆ ˆ I = ∫ J ⋅ n dS = ∫ J ⋅ z dS = ∫ 3 xy dS
S S S Example Example
I = ∫ 3 xy dS
S y ⎛ y( x) = ∫ ⎜ ∫ 3 xy dy ⎜ 0⎝ 0
1 ⎞ ⎟ dx ⎟ ⎠ dx y=1x
1 =∫
0 1 ⎡3 2⎤ ⎢ 2 xy ⎥ ⎣ ⎦0
1 y( x ) x
1 3⎡ 2 = ∫ x (1 − x ) ⎤ dx ⎦ 20 ⎣ 3 ⎡1 4 2 3 1 2 ⎤ = ⎢ x− x+ x⎥ 2 ⎣4 3 2 ⎦0
1 I = −0.125 [A] ECE 2317: Applied Electricity and Magnetism
Fall 2008 Prof. Jeffery T. Williams Dept. of Electrical & Computer Engineering University of Houston Coordinate Systems
Far Side by Gary Larson Review of Coordinate Systems Review
z P (x,y,z) ˆ r
y x An understanding of coordinate systems is critically important for solving multidimensional engineering problems. Rectangular Coordinates Rectangular
z P (x,y,z) Position vector ˆ ˆ ˆ r =xx+y y+zz
r = ( x, y , z ) r
y Short hand: Text uses: dS = dxdy
x r = ax x + ay y + az z dz dS = dydz dy dS = dxdz dV = dx dy dz dx Rectangular Coordinates Rectangular
z Path or Contour Integral (integrating from A to B along c)
B
d c A r r+d
y ˆ ˆ ˆ r = xx+y y+zz x ˆ ˆ ˆ d = x dx + y dy + z dz Cylindrical Coordinates
z
P (ρ, φ, z) y
ρ φ φ
x . ρ z y x x = ρ cos φ y = ρ sin φ z=z ρ = x2 + y 2 φ = tan −1 ( y / x )
z=z Cylindrical Coordinates
z ˆ z .
r
x y ˆˆˆ Unit Vectors (ρ, φ, z )
ˆ φ ˆ ρ ˆ ˆ Note: ρ and φ depend on ( ρ , φ )
y ˆ φ
φ ˆ ρ
x Position vector ˆ ˆ r =ρρ +zz Cylindrical Coordinates
Expressions for unit vectors
y ˆ φ
φ ˆ ρ
x ˆ ˆ ˆ Assume ρ = A1x + A2 y
Then we have: Similarly, ˆˆ ˆˆ ˆˆ ρ ⋅ x = A1x ⋅ x + A2 y ⋅ x ˆˆ A1 = ρ ⋅ x ˆˆ ρ ⋅ x = cos φ A1 = cos φ
Hence ˆˆ A2 = ρ ⋅ y ⎛π ⎞ = cos ⎜ − φ ⎟ ⎝2 ⎠ = sin φ ˆˆ ˆ ρ = x cos φ + y sin φ Cylindrical Coordinates
Summary of Results ˆˆ ˆ ρ = x cos φ + y sin φ ˆˆ ˆ φ = x ( − sin φ ) + y cos φ ˆ ˆˆ x = ρ cos φ + φ ( − sin φ ) ˆ ˆˆ y = ρ sin φ + φ cos φ Cylindrical Coordinates
Differential elements
z dS = ρ dρ dφ
Note: dS may be in three different forms ρ dz dρ dS = dρ dz ρ dφ dS = ρ dφ dz
y x dV = ρ d ρ dφ dz Cylindrical Coordinates
Path Integrals
First consider differential changes along any of the three coordinate directions: dφ
y y dρ
φ
x ρ dφ
x x z ρ dz
y ˆ d = ρdρ ˆ d = φ ( ρ dφ ) ˆ d = z dz Cylindrical Coordinates
z Path or Contour Integral (integrating from A to B along c)
B
d c A r y ˆ ˆ r =ρρ +zz x ˆ ˆ ˆ d = ρ d ρ + φ ρ dφ + z dz Spherical Coordinates Spherical
z z θ
r P (r, θ, φ) ρ θ
y P (r, θ, φ) φ
x . rz φ
x . y 0 ≤θ ≤π 0 ≤ φ ≤ 2π ρ = r sin θ Spherical Coordinates
z ρ θ
rz P (r, θ, φ) φ
x . x = r sin θ cos φ y = r sin θ sin φ z = r cos θ
y r = x2 + y2 + z 2 = ρ 2 + z 2 θ = cos −1 ( z / r ) ρ = r sin θ φ = tan −1 ( y / x ) Spherical Coordinates
z ˆˆˆ Unit Vectors (r, θ, φ) .
ˆ θ
x ˆ r ˆ φ Note:
y ˆ r depends on (θ , φ ) Position vector ˆ r =rr Spherical Coordinates
z Transformation of Unit Vectors .
ˆ θ
x ˆ r ˆ φ
y ˆˆ ˆ ˆ r = x sin θ cos φ + y sin θ sin φ + z cos θ ˆˆ ˆ ˆ θ = x cos θ cos φ + y cos θ sin φ + z ( − sin θ ) ˆˆ ˆ φ = x ( − sin φ ) + y cos φ ˆ ˆ ˆˆ x = r sin θ cos φ + θ cos θ cos φ + φ ( − sin φ ) ˆ ˆ ˆˆ y = r sin θ sin φ + θ cos θ sin φ + φ cos φ ˆ ˆˆ z = r cos θ + θ ( − sin θ ) Spherical Coordinates
dS = r sinθ dr dφ Differential elements
z r sinθ dφ
dφ ρ r dθ dS = r2 sinθ dθ dφ
dr
dθ y dS = r dr dθ
x Note: dS may be in three different forms. dV = r sin θ dr dθ dφ
2 Spherical Coordinates
Path Integrals
ρ dφ = r sin θ dφ z dr z r dθ y y x r dθ dφ z r y x x ˆ d = r dr ˆ d = θ r dθ ˆ d = φ r sin θ dφ Spherical Coordinates
z Path or Contour Integral (integrating from A to B along c)
B
d c A r y ˆ r =rr x ˆ ˆ ˆ d = r d ρ + θ rdθ + φ r sin θ dφ Example Example
Given: P1 (4, 60°, 1) [m] P2 (3, 180°, 1) [m] (cylindrical coordinates) Find d = distance between points d=
x = ρ cos φ y = ρ sin φ z=z ( x1 − x2 ) + ( y1 − y2 ) + ( z1 − z2 )
2 2 2 x1 = 4 cos ( 60° ) = 2 y1 = 4sin ( 60° ) = 3.4641 z1 = 1 x2 = 3cos (180° ) = −3 y2 = 3sin (180° ) = 0 z2 = −1 d = 6.403 [m] Example Example
Derive Let ˆˆ ˆ ˆ r = x sin θ cos φ + y sin θ sin φ + z cos θ
ˆˆ ˆ ˆ r = x A1 + y A2 + z A3
ˆˆ A1 = r ⋅ x ˆˆ A2 = r ⋅ y
z ˆ r
θ
y ˆˆ A3 = r ⋅ z ˆ ( x component of r ) ˆ ( y component of r ) ˆ ( z component of r ) φ
x
ˆ ρ Example (cont.) Example
z ˆ r
θ
(π /2)θ y ∆x
x L ⎛π ⎞ L = cos ⎜ − θ ⎟ = sin θ ⎝2 ⎠ ˆˆ r ⋅ x = L cos φ = sin θ cos φ ˆˆ r ⋅ y = L sin φ = sin θ sin φ ˆˆ r ⋅ z = cos θ φ ˆ ρ Hence ˆˆ ˆ ˆ r = x sin θ cos φ + y sin θ sin φ + z cos θ Example Example
Derive ˆ ˆˆ z = r cos θ + θ ( − sin θ )
Let ˆ ˆ ˆˆ z = r A1 + θ A2 + φ A3 ˆˆ A1 = z ⋅ r ˆˆ A2 = z ⋅ θ
ˆˆ A3 = z ⋅ φ Example (cont.) Example
z ˆ z
θ θ
y ˆ z
r
z z ˆ φ
θ ˆ z θ ˆ θ θ
y y x x x ˆˆ z ⋅ r = cos θ A1 = cos θ ˆ = cos ⎛ π + θ ⎞ ˆ z ⋅θ ⎜ ⎟ 2 ⎝ ⎠ A2 = − sin θ ˆˆ z ⋅φ = 0 A3 = 0 ˆ ˆˆ Result: z = r cos θ + θ ( − sin θ ) Example Example
Given: Spherical shell of charge ρv = 3x108 (cos2φ / r4) [C/m3] , 2 < r <5 [m] Find Q z Solution: a y Q = ∫ ρv dv
V 2π π b x a = 2 [m], b = 5 [m] = ∫∫∫
0 0a b ρv ( r , φ ) r 2 sin θ dr dθ dφ
2π π = −3 ⋅10−8 = −3 ⋅10−8 ∫∫∫
0 0a 0 b cos 2 φ 1 sin θ dr dθ dφ 2 r 2π cos 2 φ dφ ∫ ∫
a b 1 dr 2 r π ∫
0 sin θ dθ Example (cont.) Example
2π ∫
0 cos φ dφ =
2 2π 1 + cos 2φ ∫ 2 dφ 0
2π sin 2φ ⎤ ⎡1 =⎢ φ+ ⎥ =π 4 ⎦0 ⎣2 1 ⎡ 1⎤ dr = ⎢ − ⎥ ∫ r2 ⎣ r ⎦a a
b b =
π 11 − = 3 /10 ab
π ∫
0 sin θ dθ = [ − cos θ ]0 = 2 Q = 5.655x108 [C] ˆ ˆ ˆ E = x ( 3 xy ) + y ( x ) + z ( 3 xyz ) [V/m],
2 2 Example Example
∫ E⋅d
A B x, y and z are in [ m ] . = ∫ Ex dx + E y dy
A B Find VAB
E(x,y,z) z = ∫ ( 3xy ) dx + ( x 2 ) dy
A B y = 1− x
0 dy = − dx (0,1,0) (1,0,0) A x y .
1 .
B VAB = ∫ ⎡( 3x )(1 − x ) + x 2 ( −1) ⎤ dx ⎣ ⎦ y 1 c = ∫ ⎡ −4 x 2 + 3 x ⎤ dx ⎣ ⎦
1 0 = 4 ∫ x dx − 3∫ x dx
2 0 0 1 1 y = 1− x
1 x 1 ⎛1⎞ ⎛1⎞ = 4 ⎜ ⎟ − 3⎜ ⎟ = − 6 ⎝3⎠ ⎝ 2⎠ VAB = 1/6 [V] Example Example ˆ ˆ E = x ( x ) + y ( 2 y ) [V/m], x and y in [ m ]
y B VAB = ∫ E ⋅ d ˆ ˆ ˆ d = ρ d ρ + φ ( ρ dφ ) + zdz ˆ = φ ( 3 dφ ) x = ρ cos φ = 3cos φ ˆ ˆˆ x = ρ cos φ + φ ( − sin φ ) y = ρ sin φ = 3sin φ ˆ ˆˆ y = ρ sin φ + φ cos φ
A B Find VAB
A x 3 [m] c Example (cont.) Example
ˆ ˆ E = 3cos φ [ρ cos φ − φ sin φ ] ˆ ˆ +2 ( 3sin φ ) [ρ sin φ + φ cos φ ] E ⋅ d = 9sin φ cos φ dφ
π /2 ˆ d = φ ( 3 dφ ) VAB = = − ∫π 9sin φ cos φ dφ ∫π
9 sin ( 2φ ) dφ 2
π /2 π /2
− 9 ⎡ cos ( 2φ ) ⎤ = ⎢− ⎥ 2⎣ 2 ⎦ −π = 9 ⎡ 1 ⎛ 1 ⎞⎤ 9 ⎢ 2 − ⎜ − 2 ⎟⎥ = 2 2⎣ ⎝ ⎠⎦ VAB = 9/2 [V] ECE 2317: Applied Electricity and Magnetism
Fall 2008
Prof. Jeffery T. Williams Dept. of Electrical & Computer Engineering University of Houston Coulomb’s Law
Notes prepared by the EM group,
Charles Coulomb University of Houston. Coulomb’s Force Law
Charles A. Coulomb, a French military engineer, measured the forces between electric charges, reporting his results in 1785. He found that the electric force between two charged particles was
• proportional to the product of the charges, • inversely proportional to the square of the distance between them, • directed along the line joining them, and • repulsive for like charges and attractive for unlike charges.
z ˆ r
q2 F21 θ
r
q1 y φ
x Coulomb’s Law Coulomb
z Experimental law: ˆ r
r
q2
y q1q2 ˆ F2 = r 2 4πε 0 r [ N] [ F/m] ε 0 = 8.854187818 × 10−12 q1
x Note: c = speed of light = 2.99792458×108 [m/s] (exactly) c= 1 µ 0ε 0 µ 0 = 4π ×10−7 [ H/m ] (exactly) Electric Field Electric
Electric Field – Force per unit charge z F2 = ˆ r
r
q2
y q1q2 ˆ r 2 4πε 0 r But F2 = q2 E1 ( r2 ) ( E1 : electric field due to q1 ) Hence: q1
x Note: no selfforce on charge 2 due to its own electric field E1 = 4πε 0 r q1 2 ˆ r Generalizing Coulomb’s Law
Generalization (q1 not at the origin):
z r1= (x1, y1, z1) q1 (x1,y1,z1) R ˆ R
q2 (x2,y2,z2) r2= (x2, y2, z2)
R = r2 − r1
R = R = r2 − r1 r1 r2
y = ⎡( x2 − x1 ) + ( y2 − y1 ) + ( z2 − z1 ) ⎤ ⎣ ⎦
2 2 2 1 2 x E1 = 4πε 0 R 2 q1 ˆ R R= R R Example Example
q1=0.7 [mC] located at (3,5,7) [m] q2=4.9 [µC] located at (1,2,1) [m] q1 R q2 Find: F1, F2
F1 = force on charge q1 F2 = force on charge q2 For F2: F2 = q2 E1 E1 = electric field due to charge q1, evaluated at point r2 Example (cont.) Example
q1=0.7 [mC] located at (3,5,7) [m] q2=4.9 [µC] located at (1,2,1) [m] q1
E1 = 4πε 0 R q1
2 R
F2 = q2 E1 q2 R ˆ ˆ ˆ R = r2 − r1 = x (1 − 3) + y ( 2 − 5 ) + z (1 − 7 ) ˆ ˆ ˆ R = x ( −2 ) + y ( − 3 ) + z ( −6 ) R= R = R= ( −2 ) + ( −3) + ( −6 )
2 2 2 =7 [ m] R ⎛ −2 ⎞ ⎛ −3 ⎞ ⎛ − 6 ⎞ ˆ ˆ ˆ = x⎜ ⎟ + y⎜ ⎟ + z⎜ ⎟ R ⎝7⎠ ⎝7⎠ ⎝7⎠ Example (cont.) Example
E1 = 4π ( 8.854 ×10−12 ) ( 7 ) 0.7 × 10−3
2 ⎡ ⎛ −2 ⎞ ⎛ −3 ⎞ ⎛ − 6 ⎞ ⎤ ˆ ⎜ ⎟ + y ⎜ ⎟ + z ⎜ ⎟⎥ ˆ ˆ ⎢x 7 ⎝ 7 ⎠ ⎝ 7 ⎠⎦ ⎣⎝ ⎠ ˆ ˆ E1 = 1.834 ×104 ⎡ x ( −2 ) + y ( −3) + z ( −6 ) ⎤ ⎣ˆ ⎦ F2 = q2 E1 = 4.9 × 10−6 E1 ˆ ˆ F2 = 0.08988 ⎡ x ( −2 ) + y ( −3) + z ( −6 ) ⎤ ⎣ˆ ⎦ ˆ ˆ ˆ F2 = x ( −0.180 ) + y ( −0.270 ) + z ( −0.539 ) ˆ ˆ ˆ F1 = x ( +0.180 ) + y ( +0.270 ) + z ( +0.539 ) [N ] [N ] [N ] (Newton’s Law) General Case: Multiple Charges
z r
q1 q2 q1 : r1 = (x1,y1,z1) q2 : r2 = (x2,y2,z2) ... qN : rN = (xN,yN,zN) R1 R2 R3
q3 RN
qN
y R1 = r − r1 R 2 = r − r2 … R N = r − rN x E = E1 + E 2 + … + E N (superposition) qN q2 E= R1 + R 2 + ... + RN 2 2 2 4πε 0 R1 4πε 0 R2 4πε 0 RN q1 Field from Volume Charge
z r= (x,y,z) r ' ( x ', y ', z ') R
ρV (r´) = ρV (x´,y´,z´)
y x ˆ ˆ ˆ R = r − r′ = x ( x − x ') + y ( y − y ') + z ( z − z ') R = R = ⎡( x − x ' ) + ( y − y ' ) + ( z − z ' ) ⎤ ⎣ ⎦ R R= R
2 2 2 1/ 2 Field from Volume Charge (cont.)
z r = (x,y,z) R r ' ( x ', y ', z ') dV´ dQ dE = R 2 4πε 0 R
y dQ = ρV (r´) dV´ ρV ( x ', y ', z ') dV ' = R 2 4πε 0 R x ρV ( r ') E(r ) = ∫ R dV ' 2 4πε 0 R V Field from Surface Charge
z r = (x,y,z) R dQ = ρS (r´) dS´
dS´ dS y r ' ( x ', y ', z ') x ρ S ( r ') E=∫ R dS ' 2 4πε 0 R S Field from Line Charge
z r = (x,y,z) R dQ = ρl (r´) dl´
dl´ r ' ( x ', y ', z ') y x ρl ( r ' ) E=∫ R dl ' 2 4πε 0 R C Example Example
z q1 = +20 [nC] located at (1,0,0) q2 = 20 [nC] located at (0,1,0) r = (0,0,1) Find E (0,0,1) R1 R2
y Solution: R1 = (0,0,1)  (1,0,0) q2 = 20 [nC] R2 = (0,0,1)  (0,1,0)
R 2 = ( 0, −1,1) R2 = 2 R2 = 1 ( 0, −1,1) 2 x q1 = +20 [nC] R1 = ( −1, 0,1) R1 = 2 1 ˆ R1 = ( −1, 0,1) 2 Example (cont.) Example
E= 4πε 0 R12 q1 R1 + q2 R2 2 4πε 0 R2 −20 × 10−9 1 ( −1, 0,1) + 2 4π ( 8.854 ×10−12 ) 1 ( 0, −1,1) 2 = 20 × 10−9 4π ( 8.854 × 10
−12 )( 2 ) 2 ( 2) 2 = 63.55 ⎡( −1, 0,1) − ( 0, −1,1) ⎤ ⎣ ⎦ = 63.55 ⎡( −1, +1, 0 ) ⎤ ⎣ ⎦ ˆˆ E = 63.55 ( − x + y ) [ V/m] Example
z Find E(0,0,h) r = (0,0,h) R ρl ( r ') E=∫ R dl ' 2 4πε 0 R C
R = r − r′ ˆ = z ( h − z ')
y h>0 R=
x ( h − z ') 2 = h− z' = h− z' r´= (0, 0, z´ ) ρl = ρlo [C/m]
semiinfinite uniform line charge ˆ R=z Example (cont.) Example
ρlo E= 4πε 0
ˆ z ρlo = 4πε 0 ˆ z ρlo = 4πε 0
−∞ 0 ∫ ( h − z ') ∫ ( h − z ')
1 0 ˆ z 2 dz ' dz ' 2 0 −∞ ⎡1⎤ ⎢ ⎥ ⎢ ( h − z ') ⎥ −∞ ⎣ ⎦ ⎛ ρlo ⎞ ˆ E=z ⎜ ⎟ 4πε 0 h ⎠ ⎝ [V/m] Example Example
z r = (0,0,z)
R Find E(0,0,z) ρl = ρlo [C/m]
a y (uniform) x y r´= (a, φ´, 0 )
dφ´ ρl ( r ' ) E=∫ Rd ' 2 4πε 0 R C
dl´= a dφ´ φ´
a x dφ´ > 0 so C ∫( )d 2π '→ ∫ ( ) a dφ '
0 Example (cont.) Example
z r = (0,0,z)
R R = r − r′
ˆ ˆ ˆ R = x ( 0 − a cos φ ') + y ( 0 − a sin φ ') + z ( z − 0 ) ˆ ˆ ˆ = −a ( x cos φ '+ y sin φ ') + z z
y a x y r´= (a, φ´, 0 ) Note: ˆ ρ'
φ´
x ˆˆ ˆ ρ ' = x cos φ '+ y sin φ ' Example (cont.) Example
ˆ ˆ R = − a ρ '+ z z
R z x a R = a2 + z 2 ˆ ˆ − a ρ '+ z z R= a2 + z 2 Example (cont.) Example
ρl ( r ' ) E=∫ R dl ' 2 4πε 0 R C
⎛ ⎞ ⎛ ⎞ ⎜ − a ρ '+ z z ⎟ ⎛ ρlo ⎞ ˆ ˆ 1 ⎟⎜ a dφ ' = ∫⎜ ⎟⎜ 2 1⎟ 2 4πε 0 ⎠ ⎜ ( a + z ) ⎟ ⎜ 2 0⎝ ⎝ ⎠ (a + z2 )2 ⎟ ⎝ ⎠ ⎛ ⎞ 2π 2π ⎤ a ⎜ ρlo ⎟⎡ ˆ ˆ =⎜ −a ∫ ρ ' dφ '+ z z ∫ dφ '⎥ 3⎟⎢ 0 ⎦ ⎜ 4πε 0 ( a 2 + z 2 ) 2 ⎟ 0 ⎝ ⎠
2π Example (cont.) Example
ˆ ρ′
2π ˆ ∫ ρ ' dφ ' = 0
0 2π ˆˆ ˆ ( ρ ' = x cos φ '+ y sin φ ') ∫ dφ ' = 2π
0 ⎛ ρlo a ⎜ E=⎜ 3 ⎜ 4πε 0 ( a 2 + z 2 ) 2 ⎝ ⎞ ⎟ ⎟ ⎡ z z ( 2π ) ⎤ ⎣ˆ ⎦ ⎟ ⎠ ⎛ ⎛ ρlo a ⎞ ⎜ z ˆ E=z ⎜ ⎟ 3 2ε0 ⎠ ⎜ 2 ⎝ ⎜ (a + z2 )2 ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ [V/m] Example (cont.) Example
Limiting case: a → 0 (while the total charge remains constant) ⎛ ρlo a ⎞ ⎛ z ⎞ ˆ E = z⎜ ⎟⎜ 3 ⎟ ⎝ 2ε0 ⎠ ⎜ z ⎟ ⎝ ⎠ ρl 0 ( 2π a ) 1 ˆ =z ±1) 2( 4π ε0 z ˆ = ±z Q 4π ε0 z2 ( + when z > 0, − when z < 0 ) (pointcharge result) ECE 2317: Applied Electricity and Magnetism
Fall 2008 Prof. Jeffery T. Williams Dept. of Electrical & Computer Engineering University of Houston Examples of Coulomb’s Law
Notes prepared by the EM group, University of Houston.
Charles Coulomb Example Example
z r= (0,0,z) For the uniform annular ring of charge, find E (0,0,z) [C/m2]
y r
a R ρs = ρs0 dS ' = ρ ' d ρ ' dφ ' R = r − r′ ˆ ˆ = z z − ρ′ ρ ' r′
b x y dφ´ dS´ x R = ρ '2 + z 2 ˆˆ − ρ ' ρ′ + zz R= ρ '2 + z 2 Example (cont.) Example
ρ s ( r ') E=∫ R ds ' 2 4πε 0 R S
⎛ ⎛ ⎞ ⎜ − ρ ' ρ′ + zz ⎛ ρs0 ⎞ ˆˆ 1 ⎟⎜ E = ∫ ∫⎜ ⎟⎜ 1 2 2 4πε 0 ⎠ ⎜ ( ρ ' + z ) ⎟ ⎜ 2 22 0 a⎝ ⎝ ⎠ (ρ ' + z ) ⎝ ⎛ ⎞ 2π b ⎛ ρ s 0 ⎞ ⎜ − ρ ' ρ′ + zz ⎟ ˆˆ = ∫ ∫⎜ ρ ' d ρ ' dφ ' ⎟⎜ 3⎟ 4πε 0 ⎠ ⎜ 0 a⎝ ( ρ '2 + z 2 ) 2 ⎟ ⎝ ⎠
2π b ⎞ ⎟ ⎟ ρ ' d ρ ' dφ ' ⎟ ⎠ Example (cont.) Example
2π b E= ∫
0 ⎛ ⎛ ρ s 0 ⎞ ⎜ − ρ ' ρ′ + zz ˆˆ ∫ ⎜ 4πε 0 ⎟ ⎜ 2 2 3 ⎠ ⎜ ( ρ ' + z )2 a⎝ ⎝ ⎞ ⎟ ⎟ ρ ' d ρ ' dφ ' ⎟ ⎠ Switch integration order and do φ´ integral first: ⎛ ⎛ ρ ⎞ ⎜ − ρ ' ρ′ + zz ˆˆ E = ∫ ∫ ⎜ s0 ⎟ ⎜ 3 4πε 0 ⎠ ⎜ a 0⎝ ( ρ '2 + z 2 ) 2 ⎝
b 2π ⎞ ⎟ ⎟ ρ ' dφ ' d ρ ' ⎟ ⎠ 2π Note that ˆ ˆ ˆ ∫ ρ′ dφ ' = ∫ ( x cos φ ′ + y sin φ ′) dφ ' = 0
0 0 2π Example (cont.) Example
E=∫
a b 2π ⎛ ⎛ ρs0 ⎞ ⎜ ˆ zz ∫ ⎜ 4πε 0 ⎟ ⎜ 2 2 3 ⎠ ⎜ ( ρ ' + z )2 0⎝ ⎝
b ⎞ ⎟ ⎟ ρ ' dφ ' d ρ ' ⎟ ⎠ ⎛ ⎛ ρs0 ⎞ 1 ⎜ ˆ E=⎜ zz∫ ⎜ ⎟ 3 4πε 0 ⎠ a ⎜ ⎝ ( ρ '2 + z 2 ) 2 ⎝ ⎞ 2π ⎟ ⎟ ρ ' d ρ ' ∫ dφ ' 0 ⎟ ⎠ ⎛ ρs0 ⎞ b ρ' ˆ E = 2π ⎜ zz∫ dρ ' ⎟ 3 ⎝ 4πε 0 ⎠ a ( ρ '2 + z 2 ) 2 Example (cont.) Example
b ⎛ ρs0 ⎞ ρ' ˆ E = 2π ⎜ dρ ' ⎟ zz∫ 3 ⎝ 4πε 0 ⎠ a ( ρ '2 + z 2 ) 2 ⎛ ρs0 ⎞ ⎡ −1 ˆ E = 2π ⎜ zz⎢ ⎟ 4πε 0 ⎠ ⎢ ρ '2 + z 2 ⎝ ⎣ ⎤ ⎥ ⎥a ⎦ b ⎤ ⎛ ρs0 ⎞ ⎡ 1 1 ˆ z⎢ E = 2π ⎜ − ⎟z ⎥ 2 2 2 2 4πε 0 ⎠ ⎣ a + z b +z ⎦ ⎝ [ V/m] Example (cont.) Example
Annular ring of charge ⎤ ⎛ ρs0 ⎞ ⎡ 1 1 ˆ − E = 2π ⎜ ⎟z z ⎢ 2 ⎥ 2 2 2 4πε 0 ⎠ ⎣ a + z b +z ⎦ ⎝ [ V/m] Sheet of charge is the limiting case: a → 0, b → ∞
z ⎛ ρS 0 ⎞ z ˆ E = z⎜ ⎟ 2ε0 ⎠ z ⎝
E ρs0
x y ⎛ ρS 0 ⎞ ˆ E = ±z⎜ ⎟ 2ε0 ⎠ ⎝ [ V/m ] E + for z > 0  for z < 0 Example Example
For a uniform line of charge, find E (ρ, φ, 0)
z ρl = ρl0 [C/m] (uniform line charge)
+h r´ = (0, 0, z´ )
y R
x r = (ρ, φ, 0) h R Example (cont.) Example
R = r − r′ ˆˆ = ρρ−z z'
z ˆ ˆ ˆ = x ( ρ cos φ ) + y ( ρ sin φ ) − z z ' ˆ ˆ ˆ = ρ ( x cos φ + y sin φ ) − z z ' z´ R ρ
x ˆ z y R = R = ρ 2 + z ′2 R= ˆ ˆ ρ ρ − z z′ ˆ φ ˆ ρ ρ 2 + z ′2 Example (cont.) Example
ρl ( r′ ) E=∫ R dl ' 2 4πε 0 R C
⎛ ⎛ ⎞ ⎜ ρ ρ − z z′ ⎛ ρl 0 ⎞ ˆˆ 1 ⎜2 ⎟⎜ E= ∫⎜ ⎟ 1 4πε 0 ⎠ ⎜ ( ρ + z ′2 ) ⎟ ⎜ 2 −h ⎝ ⎝ ⎠ ( ρ + z ′2 ) 2 ⎝ ⎛ ⎞ +h ⎛ ρ ⎞ ⎜ ρ ρ − z z′ ⎟ ˆˆ dz ' E = ∫ ⎜ l0 ⎟ ⎜ 3⎟ 4πε 0 ⎠ ⎜ 2 −h ⎝ ( ρ + z ′2 ) 2 ⎟ ⎝ ⎠
+h ⎞ ⎟ ⎟ dz ' ⎟ ⎠ Example (cont.) Example
⎛ ⎛ ρl 0 ⎞ ⎜ ρ ρ − z z ˆˆ E= ∫⎜ ⎟ 3 4πε 0 ⎠ ⎜ 2 22 −h ⎝ ⎜ ( ρ + z′ ) ⎝
+h ⎞ ⎟ ⎟ dz ′ ⎟ ⎠ Note that symmetry,
+h ⎛ z′ ⎜ ∫h ⎜ 2 2 3 −⎜ ( ρ + z′ ) 2 ⎝ ⎞ ⎟ ⎟ dz ′ = 0 ⎟ ⎠ (integrand is an odd function of z’) ⎛ ⎛ ρl 0 ⎞ 1 ⎜ ˆ E=⎜ ρρ ∫ ⎜ ⎟ 3 4πε 0 ⎠ 2 22 ⎝ −h ⎜ ′ ⎝ (ρ + z )
+h ⎞ ⎟ ⎟ dz ′ ⎟ ⎠ Example (cont.) Example
⎛ ⎛ ρl 0 ⎞ 1 ⎜ ˆρ ∫⎜ E=⎜ ⎟ρ 3 4πε 0 ⎠ ⎝ −h ⎜ ( ρ 2 + z ′2 ) 2 ⎝
+h ⎞ ⎟ ⎟ dz ′ ⎟ ⎠ ⎞ ⎟ ⎟ dz ′ ⎟ ⎠ ⎤ ⎥ ⎥0 ⎦
h ⎛ ⎛ ρl 0 ⎞ 1 ⎜ ˆ E = 2⎜ ρρ ∫ ⎜ ⎟ 3 4πε 0 ⎠ 2 22 ⎝ 0⎜ ′ ⎝ (ρ + z )
+h ⎛ ρl 0 ⎞ ˆ = 2⎜ ⎟ρ ρ ⎝ 4πε 0 ⎠ ⎛ρ ⎞ ˆ = 2 ⎜ l0 ⎟ ρ ρ ⎝ 4πε 0 ⎠ ⎡ z′ ⎢ ⎢ ρ 2 ρ 2 + z ′2 ⎣ ⎡ h ⎢ ⎢ ρ 2 ρ 2 + h2 ⎣ ⎤ ⎥ ⎥ ⎦ ⎛ ρl 0 ˆ E = ρ⎜ ⎝ 2πε 0 ρ ⎞⎡ h ⎢ ⎟ 2 2 ⎠⎢ ρ +h ⎣ ⎤ ⎥ ⎥ ⎦ [ V/m] Example (cont.) Example
Finite length line of charge
z ⎛ ρl 0 ˆ E = ρ⎜ ⎝ 2πε 0 ρ ⎞⎡ h ⎢ ⎟ 2 2 ⎠⎢ ρ +h ⎣ ⎤ ⎥ ⎥ ⎦ [ V/m] Infinite line of charge is the limiting case: let h → ∞
y r = (ρ, φ, 0)
x ρ ⎛ ρl 0 ⎞ ˆ E = ρ⎜ ⎟ 2πε 0 ρ ⎠ ⎝ [ V/m] infinite uniform line charge Example Example
For a pair of oppositely charged infinite line charges, find E( x,y) y r = (x,y)
Single line charge ρl0 on the z axis: R1 R2 ⎛ ρl 0 ⎞ ˆ E=ρ ⎜ ⎟ 2πε 0 ρ ⎠ ⎝
x
From superposition: ρl0 h h ρl0 two infinite uniform line charges ⎛ ρl 0 ⎞ ⎛ − ρl 0 ⎞ E = R1 ⎜ ⎟ + R2 ⎜ ⎟ 2πε 0 R1 ⎠ 2πε 0 R2 ⎠ ⎝ ⎝ Example (cont.) Example
y r = (x,y) ⎛ ρl 0 ⎞ ⎛ − ρl 0 ⎞ E = R1 ⎜ ⎟ + R2 ⎜ ⎟ 2πε 0 R1 ⎠ 2πε 0 R2 ⎠ ⎝ ⎝ ˆ ˆ R1 = x ( x − ( − h ) ) + y ( y − 0 ) ˆ ˆ R2 = x ( x − h) + y ( y − 0)
x R1 R2 ρl0 h h ρl0 ˆ ˆ R1 = x ( x + h ) + y ( y ) ˆ ˆ R2 = x ( x − h) + y ( y ) R1 = R2 = ( x + h) + y2
2 ( x − h) R1 = ˆ ˆ x ( x + h) + y y 2 +y 2 ( x + h) 2 + y2 R2 = ˆ ˆ x ( x − h) + y y ( x − h) 2 + y2 Example (cont.) Example
⎛ ρl 0 ⎞ ⎛ − ρl 0 ⎞ E = R1 ⎜ ⎟ + R2 ⎜ ⎟ 2πε 0 R1 ⎠ 2πε 0 R2 ⎠ ⎝ ⎝ R1 = R2 = ( x + h) ( x − h) 2 +y +y 2 R1 = ˆ ˆ x ( x + h) + y y ( x + h) ( x − h) 2 + y2 2 2 R2 = ˆ ˆ x ( x − h) + y y
2 + y2 ˆ ˆ ˆ ˆ ⎛ ρ l 0 ⎞ ⎡⎛ x ( x + h ) + y y ⎞ ⎛ x ( x − h ) + y y ⎞ ⎤ ⎜ ⎟−⎜ ⎟⎥ E=⎜ ⎟ ⎢⎜ 2 2 2⎟ ⎜ 2⎟ ⎝ 2πε 0 ⎠ ⎢⎝ ( x + h ) + y ⎠ ⎝ ( x − h ) + y ⎠ ⎥ ⎣ ⎦ Example (cont.) Example
y E2 R2 E1 R1 ˆ ˆ ˆ ˆ ⎛ ρ l 0 ⎞ ⎡⎛ x ( x + h ) + y y ⎞ ⎛ x ( x − h ) + y y ⎞ ⎤ ⎟−⎜ ⎟⎥ E=⎜ ⎟ ⎢⎜ 2 2 2⎟ ⎜ 2⎟ ⎜ ⎝ 2πε 0 ⎠ ⎢⎝ ( x + h ) + y ⎠ ⎝ ( x − h ) + y ⎠ ⎥ ⎣ ⎦
x ρl0 ρl0 ⎤ ⎛ ρl 0 ⎞ ⎡ x+h x−h − Ex = ⎜ ⎥ ⎟⎢ 2 2 2 2 ⎝ 2πε 0 ⎠ ⎢ ( x + h ) + y ( x − h ) + y ⎥ ⎣ ⎦ ⎤ ⎛ ρl 0 ⎞ ⎡ y y − Ey = ⎜ ⎥ ⎟⎢ 2 2 2 2 ⎢ ⎥ ⎝ 2πε 0 ⎠ ⎣ ( x + h ) + y ( x − h ) + y ⎦ ECE 2317: Applied Electricity and Magnetism
Fall 2008 Prof. Jeffery T. Williams Dept. of Electrical & Computer Engineering University of Houston. Potential Difference (Voltage)
Notes prepared by the EM group,
Alessandro Volta (17451827) University of Houston. External Work External
B q Force on q F = qE
External Force required to oppose E A Fext = −qE Work required to move q from B to A Wext = − ∫ qE ⋅ d = PE ( A ) − PE ( B )
B A Potential Calculation Potential
_ VAB
+ B A VAB Wext 1 = lim = ⎡ PE ( A ) − PE ( B ) ⎤ ⎦ q →0 q q⎣ = −∫ E ⋅ d
B A = ∫ E⋅d = Φ ( A) − Φ (B)
VAB = Φ ( A ) − Φ ( B ) = ∫ E ⋅ d
A B B A Example Example
A
++ V0 [V] + ++++ ++++ ++ x=0 E
 x=h B Find: E
Assume: ˆ E ( x, y, z ) = x E0
B VAB = ∫ E ( x, y, z ) ⋅ d = V0 [V]
A Example (cont.) Example
VAB = ∫ E ( x, y, z ) ⋅ d = V0
A B Evaluate in rectangular coordinates: ˆ ˆ ˆ E = x Ex + y E y + z Ez
ˆ ˆ ˆ d = x dx + y dy + zdz
xB VAB = xA ∫E x dx = V0 Example (cont.) Example
VAB =
xB xA
h ∫E
0 x dx = ∫ E0 dx = E0 h = V0 E0 = V0 / h Recall that ˆ E ( x, y, z ) = x E0
⎛ V0 ⎞ ˆ E ( x, y, z ) = x ⎜ ⎟ [V/m] ⎝h⎠ Hence: Example
A
++ V0 [V] + ++++ proton ++++ (x,y,z) ++ x=0 E
 x=h B
V0 = 9 [V] h = 0.1 [m] A proton is released at point A (zero velocity). Find the velocity v(x) of the proton at distance x from the top plate. KE ( x ) − KE ( 0 ) = PE ( 0 ) − PE ( x )
1 m v 2( x ) = PE ( 0 ) − PE ( x ) = q ⎡Φ ( 0 ) − Φ ( x ) ⎤ ⎣ ⎦ 2 Example (cont.) Example
1 m v 2( x ) = q ⎡ Φ ( 0 ) − Φ ( x ) ⎤ ⎣ ⎦ 2 ⎛ V0 ⎞ ⎛ V0 ⎞ Φ ( 0 ) − Φ ( x ) = ∫ E ⋅ d = ∫ Ex dx = ∫ ⎜ ⎟ dx = ⎜ ⎟ x h⎠ ⎝h⎠ A 0 0⎝
x x B Hence: 1 ⎛ V0 ⎞ 2 m v ( x) = q ⎜ ⎟ x 2 ⎝h⎠
⎛ 2qV0 ⎞ v ( x) = ⎜ ⎟x ⎝ mh ⎠ Example (cont.) Example
⎛ 2qV0 ⎞ v ( x) = ⎜ ⎟x ⎝ mh ⎠
V0 = 9 [V] h = 0.1 [m] q = 1.602 x 1019 [C] m = 1.673 x 1027 [kg]
(=1836 x electron mass; see Appendix C of book) Hence: v ( x ) = 5.627 ×106 x [m/s] Potential Calculation
_ VAB
+ R =B r= A
R VAB = Φ ( A ) − Φ ( B ) = ∫ E ⋅ d
A B Hence: Φ (r ) − Φ ( R ) = ∫ E ⋅ d
r Φ (r ) = Φ ( R ) + ∫ E ⋅ d
r R = Φ (R) − ∫ E⋅ d
R r Example Example
z R C
r q
y Given: R = ∞ , Φ ( R) = 0 Find the potential function Choose a radial path for convenience ⎛q⎞ ˆ⎜ E(r ) = r ⎟ 4π ε 0 r 2 ⎠ ⎝ Φ (r ) = 0 + ∫ E ⋅ d
r R Example (cont.) Example
ˆ ˆ ˆ ˆ Φ (r ) = ∫ ( r Er ) ⋅ r dr + θ r dθ + φ r sin θ dφ
r R ( ) = ∫ Er dr
r R =∫
r ∞ q 4π ε 0 r −q
2 ∞ dr = Φ (r ) = 4π ε 0 r q 4π ε 0 r r [V] Example Example
z Given: R is at ρ = b, Φ (R) = 0 ρ l0 [C/m] Find the potential function ⎛ ρ0 ⎞ ˆ⎜ E=ρ ⎟ 2π ε 0 ρ ⎠ ⎝
ρ
x r y C
R (ρ = b) Φ(ρ) = 0 + ∫ E⋅d
r R ˆ ˆ ˆ ˆ = ∫ ( ρ Eρ ) ⋅ ( ρ d ρ + z dz + φρ dφ )
r R = ∫ ( Eρ d ρ )
Choose a radial path
ρ b Example (cont.) Example
Φ(ρ) = ∫ ρ0 dρ ρ 2π ε 0 ρ
b ρ0 b ln ρ ρ = 2π ε 0
⎛b⎞ ρ0 = ln ⎜ ⎟ 2π ε 0 ⎝ ρ ⎠ ⎛b⎞ ρ0 Φ(ρ) = ln ⎜ ⎟ 2π ε 0 ⎝ ρ ⎠ ⎡V ⎤ ⎣⎦ Note: b ≠ ∞ ( infinite voltage drop between ρ = ρ and ρ = ∞ ) Also, b ≠ 0. Example Example
Given: Φ (∞) = 0
Find the potential function on the z axis z r (0,0,z) ρ l0 [C/m]
y a x Note: We have only calculated E on the z axis Example Example
Given: Φ (∞) = 0
Find the potential function on the z axis z R (∞) C
r
y x Choose a vertical path Example (cont.) Example
Φ (r ) = 0 + ∫ E ⋅ d
r R R ˆ ˆ ˆ ˆ Φ ( r ) = ∫ ( z Ez ) ⋅ ( x dx + y dy + z dz )
r = ∫ Ez dz
z ∞ From previous calculations, we know that on the z axis ⎛ ⎛ ρ 0 a ⎞⎜ z Ez ( 0, 0, z ) = ⎜ ⎟ 2ε 0 ⎠ ⎜ z 2 + a 2 3 2 ⎜( ⎝ ) ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Example (cont.) Example
Hence: ⎛ρ 0a⎞ z Φ ( 0, 0, z ) = ∫ ⎜ dz ⎟ 3 2ε 0 ⎠ z 2 + a 2 2 z⎝ ( )
∞ −1 ⎤ ⎛ ρ 0 a ⎞⎡ =⎜ − ( z2 + a2 ) 2 ⎥ ⎟ 2ε 0 ⎠ ⎢ ⎣ ⎦z ⎝ ∞ ρ 0a 1 Φ ( 0, 0, z ) = 2ε 0 z 2 + a 2 [V] Adding constant to Φ Adding
r C
R Φ ( r ) = Φ ( R ) + ∫ E ⋅ dr
r R Changing the potential at the reference point is equivalent to adding a constant to the solution Φ 2 ( x, y, z ) = Φ1 ( x, y, z ) + C Example Example
Given: Φ (z = 1 [m]) = 10 [V]
Find the potential function on the z axis z r (0,0,z) R (0,0,1)
y a x ρ l0 [C/m] ρ 0a 1 Φ ( 0, 0, z ) = +C 2 2 2ε 0 z + a Example Example
Given: Φ (z = 1 [m]) = 10 [V]
Find the potential function on the z axis z r (0,0,z) R (0,0,1)
y a x ρ 0a 1 Φ ( 0, 0, z ) = +C 2 2 2ε 0 z + a
Φ ( 0, 0, 1) = 10 [V] Set: ρ l0 [C/m] ρ 0a 1 + C = 10 2 2 2ε 0 1 + a Example Example
z r (0,0,z) R (0,0,1)
y a x Hence: ρ 0a 1 + C = 10 2 2 2ε 0 1 + a ρ 0a 1 C = 10 − 2ε 0 12 + a 2 [V] ρ l0 [C/m] ⎞ ρ 0a ⎛ 1 1 Φ ( 0, 0, z ) = 10 + − ⎜2 ⎟ 2 2 2 2ε 0 ⎝ z + a 1 +a ⎠ [V] ECE 2317: Applied Electricity and Magnetism
Fall 2008
Prof. Jeffery T. Williams Dept. of Electrical & Computer Engineering University of Houston. Potential Integral
Notes prepared by the EM group, University of Houston. Potential for Point Charge Potential
Assuming z Φ (∞) = 0
R r
y q r′ Φ (r ) = Q 4π ε 0 R x where, R = R = r − r′ Potential Integral Formula Potential
Assuming z r R
ρv (r´)
y Φ (∞) = 0 r′ ρv ( r′ ) dV ' dΦ = = 4π ε 0 R 4π ε 0 R
dQ x ρv ( r′ ) dV ' Φ (r ) = ∫ , 4π ε 0 R V Φ (∞) = 0
R = R = r − r′ Potential Integral Formula: Potential
Similarly ρ s ( r′ ) dS ' Φ (r ) = ∫ 4π ε 0 R S ρ ( r′ ) dl ' Φ (r ) = ∫ 4π ε 0 R C
R = R = r − r′ Example Example
z r
R Find Φ (0,0, z ), assuming Φ (∞) = 0
ρ 0d ′ Φ (r ) = ∫ 4π ε 0 R C
y a ˆ ˆ R = r − r′ = z z − a ρ′ = z 2 + a 2 Φ (r ) = =
2π x ρl0 ∫ 4π ε
0 ρ 0 a dφ ′
0 z2 + a2
2 2 ρ 0a
4π ε 0 z + a ( 2π ) ⎛ρ 0a⎞ 1 Φ (r ) = ⎜ ⎟ 2ε 0 ⎠ z 2 + a 2 ⎝ Example (cont.) Example
⎛ρ 0a⎞ 1 Φ (r ) = ⎜ ⎟2 2ε 0 ⎠ z + a 2 ⎝
For [V] z→∞
ρ 0 a ρ 0 ( 2π a ) Φ(r ) ∼ = 2ε 0 z 4π ε 0 z
= Q 4π ε 0 z
(agrees with the point charge formula) z Example Example
r For a uniform spherical sheet of charge, find the potential at a point r>a. Φ (∞ ) = 0
a y ρs0
x ˆ ˆ R = r − r′ = r r − a r′ ρ S 0 dS ′ Φ (r ) = ∫ 4π ε 0 R S ˆ ˆ x(r sin θ cos φ − a sin θ ′ cos φ ′) + y (r sin θ sin φ − a sin θ ′ sin φ ′) = ˆ + z (r cos θ − a cos θ ′)
Without loss of generality, assume θ = 0, φ = 0 ˆ ˆ ˆ R = x(− a sin θ ′ cos φ ′) + y (− a sin θ ′ sin φ ′) + z (r − a cos θ ′) = r 2 + a 2 − 2ar cos θ ′ z Example (cont.) Example
r
2π π a y ρs0
x Details on following pages. Φ (r ) = ∫ ∫ 4π ε
00 ρ S 0 a 2 sin θ ′ dθ ′dφ ′
0 r 2 + a 2 − 2ar cos θ ′ ⎧ ρS 0 ⎛2⎞ 2π ) a 2 ⎜ ⎟ ; r ≥ a ⎪ 4π ε ( ⎝r⎠ ⎪ 0 =⎨ ⎪ ρ S 0 ( 2π ) a 2 ⎛ 2 ⎞ ; r ≤ a ⎜⎟ ⎪ 4π ε 0 ⎝a⎠ ⎩ ⎧ ⎛ ρ S 0 4π a 2 ⎞ 1 Q = net ; r ≥ a ⎪⎜ ⎟ 4π ε 0 ⎠ r 4π ε 0 r ⎪⎝ Φ (r ) = ⎨ ⎛ ρ S 0 4π a 2 ⎞ 1 Q ⎪ = net ; r ≤ a ⎪⎜ 4π ε ⎟ a 4π ε a 0 0 ⎠ ⎩⎝ Example (cont.) Example
Φ (r ) =
2π π ∫ ∫ 4π ε
00 2 ρ S 0 a 2 sin θ ′ dθ ′dφ ′
0 2 2 π ρS 0 a2 sin θ ′ dθ ′ (2π ) ∫ = 4π ε 0 r 2 + a 2 − 2ar cos θ ′ r 2 + a 2 − 2ar cos θ ′ 0 π ⎡ 2 (r + a)2 2 (r − a) 2 ⎤ ⎛ 2 r + a − 2ar cos θ ′ ⎞ ρS 0 a ρS 0 a2 (2π ) ⎜ (2π ) ⎢ = − ⎟= ⎥ ⎜ ⎟ 4π ε 0 2ar 4π ε 0 2ar 2ar ⎢ ⎥ ⎝ ⎠ ⎣ ⎦
0 ⎧ ρS 0 ⎛ 2(r + a ) − 2(r − a ) ⎞ 2π ) a 2 ⎜ ⎟; r≥a ⎪ 4π ε ( 2ar ⎝ ⎠ ⎪ 0 =⎨ ⎪ ρ S 0 ( 2π ) a 2 ⎛ 2(r + a ) − 2(a − r ) ⎞ ; r ≤ a ⎜ ⎟ ⎪ 4π ε 0 2ar ⎝ ⎠ ⎩ ⎧ ρS 0 2 ( 2π ) a 2 ⎛ ⎞ ; r ≥ a ⎜⎟ ⎪ 4π ε ⎝r⎠ ⎪ 0 =⎨ ⎪ ρ S 0 ( 2π ) a 2 ⎛ 2 ⎞ ; r ≤ a ⎜⎟ ⎪ 4π ε 0 ⎝a⎠ ⎩ z r
a y ρs0
x Example Example
A sharp point produces a strong electric field Q = qa + qb
b a qb qa The system is charged with Q [C]. Determine how much charge goes to each sphere, and the electric field at the surface of each sphere. Example Example
b a Φ a = Φb 4π ε 0 a qa = 4π ε 0b qb qb qa qa a = qb b
Also, qa + qb = Q ⎛a⎞ qa = Q ⎜ ⎟ a+b⎠ ⎝ ⎛b⎞ qb = Q ⎜ ⎟ a+b⎠ ⎝ Example (cont.) Example
Era = Erb = 4π ε 0 a 2 4π ε 0b 2
2 qa Erb Era qb so Era ⎛ b ⎞ ⎛ qa ⎞ =⎜ ⎟ ⎜ ⎟ Erb ⎝ a ⎠ ⎝ qb ⎠ Hence, Era ⎛ b ⎞ =⎜ ⎟ Erb ⎝ a ⎠ (A strong electric field exists on the smaller sphere) Example Example
A high electric field is created at a sharp point. sharp point + V  Lightning Rod Lightning
 Lightning is attracted to the rod, not the house. Important to have it grounded ! ECE 2317: Applied Electricity and Magnetism
Fall 2008 Prof. Jeffery T. Williams Dept. of Electrical & Computer Engineering University of Houston Examples of Gauss’ Law
Notes prepared by the EM group, University of Houston. Example Example
z Find E S ˆ ∫ D ⋅ n dS = Q
S encl h y Assume
x ˆ D = ρ Dρ ( ρ ) ρl = ρl0 [C/m] infinite uniform line charge Example (cont.) Example
ρ
St Qencl = ρl 0 h ˆ ˆˆ ∫ D ⋅ n dS = ∫ ( Dρ ρ ) ⋅ ρ dS
S h Sc Sb 0 ˆˆ + ∫ ( Dρ ρ ) ⋅ z dS St 0 ˆ ˆ + ∫ ( Dρ ρ ) ⋅ ( − z ) dS
Sb Sc Example (cont.) Example
ˆ ∫ D ⋅ n dS = ∫ Dρ dS = Dρ ∫ dS
S Sc Sc = Dρ ( 2πρ h )
Hence Dρ ( 2πρ h ) = ρl 0 h Dρ = ρl 0 2πρ So ⎛ ρl 0 ⎞ ˆ E = ρ⎜ ⎟ 2πε 0 ρ ⎠ ⎝ [ V/m] Example Example
z Find E a ˆ ∫ D ⋅ n dS = Q
S encl ρ
Assume h
x y ˆ D = ρ Dρ ( ρ ) S ρv = 3ρ 2 [C/m3] , ρ<a nonuniform infinite cylinder of volume charge density Example (cont.) Example
(a) ρ <a
Qencl = ∫ ρv dV
V ρ =∫ h 2π ρ 000 ∫ ∫ ρ ( ρ ) ρ d ρ dφ dz
v = h ( 2π ) ∫ ρ v ( ρ ) ρ d ρ ρ h S 0 = 2π h ∫ 3ρ 2 ρ d ρ
0 ρ = 2π h 3ρ 4 4ρ 0 3 Qencl = π h ρ 4 2 Example (cont.) Example
ˆ ˆ ˆ ∫ D ⋅ n dS = ∫ ( ρ Dρ ) ⋅ n dS = ∫ Dρ dS
S S Sc = Dρ ( 2πρ h ) 3 Dρ ( 2πρ h ) = π hρ 4 2 33 Dρ = ρ 4 ⎛ 3ρ 3 ⎞ ˆ E = ρ⎜ ⎟ ⎝ 4ε 0 ⎠ Hence So [ V/m] ρ≤a Example (cont.) Example
(b) ρ >a ρ Qencl 3 = π h a4 2 h 3 Dρ ( 2πρ h ) = π ha 4 2 S ⎛ 3a 4 ⎞ ˆ E = ρ⎜ ⎟ 4ε 0 ρ ⎠ ⎝ [ V/m] ρ≥a Example Example
z When Gauss’s Law is not useful h ˆ ∫ D ⋅ n dS = Q
S encl y x ˆ D ≠ ρ Dρ h ρl0 Not enough symmetry. E has more than one component z Example Example
ρs = ρs0
[C/m2] y ˆ ∫ D ⋅ n dS = Q
S encl x Assume ˆ D = z Dz ( z ) A S Example (cont.) Example
ˆˆ ∫ ( D z ) ⋅ n dS = Q
z S encl A S Stop ˆˆ ∫ ( D z ) ⋅ z dS
z + Sbottom ˆ ˆ ∫ ( D z ) ⋅ ( −z ) dS = Q
z encl Dz + A − Dz − A = Qencl
Assume Dz − = − Dz + 2 ADz + = Qencl Example (cont.)
Qencl = ρ s 0 A 2 ADz + = Qencl Dz =
+ r A S ρs0 A
2A = ρs0
2 so ⎛ ρs0 ⎞ ˆ E = z⎜± ⎟ 2ε 0 ⎠ ⎝ [V/m], + z>0 − z<0 Example Example
slab of uniform charge
x d ρv 0 [C/m3 ]
ˆ E = x Ex ( x )
Assume y Ex ( − x ) = − Ex ( x ) ⇒ Ex ( 0 ) = 0
(since Ex (x) is a continuous function) Example (cont.) Example
(a) x > d / 2 S A
x x ˆˆ ∫ ( D x ) ⋅ n dS = Q
x S encl St ˆˆ ˆ ˆ ∫ ( D x ) ⋅ x dS + ∫ ( D x ) ⋅ ( −x ) dS = Q
x x Sb encl 0 Dx ( x ) A − Dx ( 0 ) A = Qencl = ρv 0 A(d / 2)
Dx ( x ) = ρv 0 d / 2 ⎛ ρv 0 d ⎞ ˆ E = x⎜ ⎟ ⎝ 2ε 0 ⎠ [ V/m] , x > (d / 2) Example (cont.) Example
(b) 0 < x < d / 2
x r
S x=x x=0 y Dx A = Qencl = ρ v 0 A x Dx = ρ v 0 x ⎛ ρv 0 x ⎞ ˆ E = x⎜ ⎟ ε0 ⎠ ⎝ [ V/m] , 0< x< d /2 Example (cont.) Example
x y ⎧ ⎛ρ d⎞ ˆ x ⎜ v 0 ⎟ [ V/m ] , ⎪ ⎝ 2ε 0 ⎠ ⎪ ⎪⎛ ⎪ x ⎜ ρ v 0 x ⎞ [ V/m ] , ˆ ⎟ ⎪ ⎝ ε0 ⎠ ⎪ E=⎨ ⎪−x ⎛ ρv 0 x ⎞ V/m , ˆ ] ⎪ ⎜ε ⎟ [ ⎪ ⎝ 0⎠ ⎪ ⎛ ρv 0 d ⎞ ˆ ⎪ −x ⎜ ⎟ [ V/m ] , ⎪ ⎝ 2ε 0 ⎠ ⎩ x > (d / 2)
Ex 0< x< d /2 0 > x > −d / 2 x < (− d / 2) ρv0 d / (2ε0 ) d/2 d/2 x ECE 2317: Applied Electricity and Magnetism
Fall 2008
Prof. Jeffery T. Williams Dept. of Electrical & Computer Engineering University of Houston Gauss’ Law
Notes prepared by the EM group, University of Houston. Electric Flux Density Electric
E E=
q q 4πε 0 r
2 ˆ r Define electric flux density (D): D ≡ ε 0E D= q 4π r [C/m 2 ] ˆ r 2 z Gauss’ Law Gauss
S
(closed surface) ˆ n
q
y The charge q is inside the surface x D Outward pointing normal to S ˆ ψ e = ∫ D ⋅ n dS = q
S
Net electric flux out of S [C] Gauss’ Law Gauss
The charge q is now outside the surface ˆ ∫ D ⋅ n dS = 0
S q S Electric flux out of S is equal to the flux into S. Net electric flux is equal to zero. Gauss’ Law Gauss
By superposition, this law must be true for arbitrary charges.
Net charge enclosed by S [C] ˆ ∫ D ⋅ n dS = Q
S encl Net electric flux out of S [C] ˆ ∫ E ⋅ n dS =
S Qencl ε0 S is any arbitrary closed surface Example Example
2q q q S1 S2 S1 ˆ ∫ D ⋅ n dS = q ˆ ∫ D ⋅ n dS = 0
Note: E ≠ 0 on S2 ! S2 Note: charge 2q produces E on S1 and S2 even though it does not appear on RHS of Gauss’s law. Direct application of Gauss’ law Direct
Gauss’ law can be applied directly to find the electric field for problems where a high degree of symmetry exists. Must be able to • determine direction of the electric field. • find “Gaussian surface” over which electric field is constant. Choice of Gaussian Surface Choice
Rule 1: S must be closed surface Rule 2: Pick S to be ⊥ to E, as much as possible ˆ ψ e = ∫ D ⋅ n dS
S E S ⊥E ˆ ⇒ nD ˆ n S Example Example
Point Charge: Find D, E ˆ r
r ˆ ∫ D ⋅ n dS = Q
S encl =q ˆ Assume, D = r Dr (r )
q S ˆˆ ∫ ( D r ) ⋅ r dS = q
r S ˆˆ n=r Dr ∫ dS = q
S Example (cont.) Example
So 4π r 2 Dr = q q Dr = 4π r 2 ⎛q⎞ ˆ D = r⎜ ⎟ 4π r 2 ⎠ ⎝ ⎛q⎞ ˆ E = r⎜ ⎟ 4πε 0 r 2 ⎠ ⎝ ⎡C/m 2 ⎤ ⎣ ⎦ [ V/m ] Example Example
z ρs= ρs0
y Find D, E everywhere Assume ˆ D = r Dr ( r )
< a)
encl a
x Solution (r r<a
r ˆ ∫ D ⋅ n dS = Q
S r 0 ˆˆ ∫ ( D r ) ⋅ r dS = 0
S Dr 4π r 2 = 0 Dr = 0 Example (cont.) Example
r>a r a Dr 4π r 2 = Qencl = ρ s 0 4π a 2 4π a 2 ρ s 0 Dr = 4π r 2 Q Dr = 4π r 2 ⎧ˆ Q , ⎪r 2 E = ⎨ 4πε 0 r ⎪0 ⎩ r>a r<a z Example Example
ρv = ρv0
y Find E(r) everywhere a
x Solution (r > a)
encl r>a
r a ˆ ∫ D ⋅ n dS = Q
S Qencl = ∫ ρv ( r ') dv '
V Example (cont.) Example
Qencl = ∫ ρ v 0 dv
V ˆ ˆˆ ∫ D ⋅ n dS = ∫ ( D r ) ⋅ r dS
r S S = ρ v 0 ∫ dv
V = ∫D
S r dS ⎛4 3⎞ = ρv 0 ⎜ π a ⎟ ⎝3 ⎠ r>a
r a = Dr ∫ dS
S = Dr ( 4π r 2 ) Example (cont.) Example
Hence, for r > a ⎛4 3⎞ Dr ( 4π r ) = ρ v 0 ⎜ π a ⎟ ⎝3 ⎠
2 ⎛ a3 ⎞ Dr = ρ v 0 ⎜ 2 ⎟ ⎝ 3r ⎠ ρv 0 a3 ˆ E=r 3ε 0 r 2
or
Note: [ V/m] , r>a Q ˆ E=r 4π r 2 ⎛4 ⎞ Q = ρv 0 ⎜ π a3 ⎟ ⎝3 ⎠ (The electric field outside the uniform sphere of charge is the same as from a point charge at origin.) Example (cont.) Example
r<a
ˆ ∫ D ⋅ n dS = Q
S encl r D ⋅ n dS = Dr ( 4π r 2 ) ∫ˆ
S a ⎛4 ⎞ Qencl = ρ v 0 ⎜ π r 3 ⎟ ⎝3 ⎠ So, ⎛4 ⎞ Dr ( 4π r 2 ) = ρv 0 ⎜ π r 3 ⎟ ⎝3 ⎠ ⎛r⎞ Dr = ρv 0 ⎜ ⎟ ⎝3⎠ ⎛r⎞ ˆ E = rρv 0 ⎜ ⎟ [V/m] , r < a ⎝ 3ε 0 ⎠ Example (cont.) Example
r a ⎧ ⎛ ρV 0 r ⎞ ˆ ⎪r⎜ ⎟ 3ε 0 ⎠ ⎪⎝ E=⎨ 3 ⎪r ⎛ ρV 0 a ⎞ ˆ ⎪ ⎜ 3ε r 2 ⎟ ⎩⎝ 0 ⎠
Er [V/m], r < a [ V/m] , r>a ρv0 a/ 3ε0 a r ECE 2317: Applied Electricity and Magnetism
Fall 2008 Prof. Jeffery T. Williams Dept. of Electrical & Computer Engineering University of Houston. Conductors and Grounding
Notes prepared by the EM group, University of Houston. Perfect Electric Conductors (PEC) Perfect
σ [S/m]
PEC: J =σE (Ohm’s law) σ →∞ E=0
Also,
A B
B Φ = constant 0 since VAB = ∫ E ⋅ d = 0
A Perfect Electric Conductors (PEC) Perfect
Inside PEC: ρv = 0 Only ρs on the surface is allowed Proof: assume point where ρv > 0 S + + ++ ++ + ˆ ∫ D ⋅ n dS = Q
S encl 0 = Qencl > 0
ρv > 0 inside small volume Contradiction ! Perfect Electric Conductors (PEC) Perfect
Inside a hollow PEC shell there is no electric field Only ρs on the outer surface is allowed
+ + + + + + + “Faradaycage effect”: E=0 + + + + + + PEC shell Perfect Electric Conductors (PEC) Perfect
Proof: Assume an electric field exists inside the shell at some point
+ + + A + B + + + + + + The flux line must end on the shell (otherwise it contradicts Gauss’s law) + E≠0
+ + A flux line must exist through this point. VAB ≠ 0 Contradiction ! Hence, there is no electric field (and no flux lines) inside the cavity. There is also no charge density on the inner surface (otherwise, there would be a flux line coming from the inner surface). PEC shell Shielding and Grounding
a Spherical PEC shell b Neutral (no charge) Drill hole and insert point charge, then solder hole. q a Find E: q b Neutral shell (no charge) Shielding and Grounding
(a) r<a ˆ ∫ D ⋅ n dS = Q
S encl Dr ( 4π r 2 ) = q ⎛q⎞ ˆ⎜ E=r ⎟ 4πε 0 r 2 ⎠ ⎝
(b) a<r<b a q b E=0
Dr ( 4π r
2 (PEC) (c) r>b )=q Neutral shell ⎛q⎞ ˆ E = r⎜ 2⎟ ⎝ 4πε 0 r ⎠ Shielding and Grounding Shielding q ⎛q⎞ ˆ E = r⎜ ⎟ 4πε 0 r 2 ⎠ ⎝ (outside metal) The neutral metal shell does not block the static electric field Shielding and Grounding Shielding
Find QA, QB:
q QA QB + + + + + + +  + + +
QB qQA  + Neutral shell + + + + + Shielding and Grounding Shielding
+ + + + + + + + + +
QB qQA  ˆ ∫ D ⋅ n dS = Q
S encl + + + Dr ( 4π r 2 ) = Qencl
so Hence + + + S
A Gaussian surface is chosen inside the metal shell. Qencl = 0 QA = − q QB = + q
(neutral shell) and Next, “ground” the shell:
r>b: E=0
qq  E=0 proof: A VAB = ∫ E ⋅ d = 0
A B This must hold for every path. PEC wire
flux line If the electric field were not zero at a point, there would be a flux line through the point that would end on the conductors, which would correspond to a voltage drop between them B Earth The earth is modeled as a “big fat conductor.” charge on outer surface: r > b: E = 0 QBG
qq  ˆ ∫ D ⋅ n dS = Q
S encl Qencl = 0
PEC wire Qencl = q + ( −q ) + QB G Earth Hence QB G = 0 charge on outer surface (cont.):
 QBG = 0
q q q q q PEC wire + + + + q Earth before grounding after grounding The charge q on the outer surface flows down to ground. Example Example
x r h PEC y Given: ρ s 0 [C/m 2 ] (This is the total surface charge density per square meter on the metal slab, which is the sum of the surface charge densities on both the top and bottom surfaces.) Find E everywhere Example Example
x r h PEC y ρ s 0 [C/m 2 ]
Charge Model:
x x = h/2 + + + + + + ρs0 / 2
y x =  h/2 + + + + + + + + + + ρs0 / 2 Example Example
x x = h/2 + + + + + + ρs0 / 2
y x =  h/2 + + + + + ⎧ ⎛ ρs0 / 2 ⎞ ⎛ρ ⎞ ˆ ˆ x 2⎜ = x ⎜ s0 ⎟ ⎪ ⎟ 2ε 0 ⎠ ⎝ 2ε 0 ⎠ ⎪⎝ ⎪ ⎪ ⎪ E = ⎨ 0, − h / 2 < x < h / 2 ⎪ ⎪ ⎪ ⎛ − ρs0 ⎞ ˆ ⎪x ⎜ ⎟ 2ε 0 ⎠ ⎪⎝ ⎩ [V/m], [V/m], + + + + + ρs0 / 2 x > h/2 x < −h / 2 Example (cont.) Example
Alternative solution (Gauss’s law):
+ + + + x S
+ + + + y h A PEC ˆ ∫ D ⋅ n dS = Q
S encl Dx A = ρ s+ A ⎛ρ ⎞ = ⎜ s0 ⎟ A ⎝2⎠ Dx = ⎛ ρs0 ⎞ ˆ E = x⎜ ⎟ 2ε 0 ⎠ ⎝ [V/m], x > h/2 ρs0
2 (The electric field outside the slab is the same as from a single sheet of surface charge.) Example Example
x ρ s+0 = ρ s 0 / 2
h PEC
y ρ s−0 = ρ s 0 / 2
The surfaces are initially charged as shown above. The bottom of the plate is then grounded. Find the charge and field everywhere. Example (cont.) Example
(a) charge on lower surface A
+ + + + + + h PEC ρ s−0G + + E=0
Earth S D ⋅ n dS = Qencl = ρ s−0G A ∫ˆ
S Hence ρ s−0G = 0 Example (cont.) Example
(b) charge on earth PEC h ρ searth
Earth E=0 S ++++ ++++++++ ++++ A D ⋅ n dS = Qencl = ρ s earth A ∫ˆ
S Hence ρ s earth = 0 Example (cont.) Example
(c) charge on top surface
x + ++ ++ ++ + + ++ ++ ++ + + ++ + ρs 0+G ρ s−0G = 0 ρ searth = 0 0 ⎡ ρ s+0G ρ s−0G ρ searth ⎤ ˆ E = x ⎢− − − 0 ⎥=0 2ε 0 ⎦ ⎣ 2ε 0 2ε 0
So E = 0 (inside earth) ρ s+0G = 0 Hence ρ sG0 = ρ s+0G + ρ s−0G = 0
(After grounding) Example (cont.) Example
ρs 0+ = ρs0 / 2
PEC (before grounding) ρs0− = ρs0 / 2 zero PEC zero (after grounding) Earth ECE 2317: Applied Electricity and Magnetism
Fall 2008 Prof. Jeffery T. Williams Dept. of Electrical & Computer Engineering University of Houston. Circuit Grounding
Notes prepared by the EM group, University of Houston. Grounding Grounding
Grounding: connecting a conductor to the physical earth
• Safety: minimize chance of electrocution • Reduce circuit noise equipment water pipe Grounding (cont.) Grounding
• Ensures that there is no voltage drop between equipment and ground • Ensures that there is no electric field outside of equipment • Ensures that there is no charge build up on equipment + VAB = 0  equipment water pipe Grounding (cont.) Grounding
Ground symbol:
equipment ground Note: grounding is NOT is the same thing as a reference point. (However, when a circuit is grounded, the ground is OFTEN chosen as a reference point of zero volts.) Example Example
A
reference point (choose 0 [V]) 12 [V] + 100 Ω B
Find: voltage at the ground VG ground V ( A ) − V ( B ) = 12 0 − V ( B ) = 12
VG = −12[V] Grounding (cont.) Grounding hot neutral (grounded back at junction box) ground Grounding (cont.) Grounding
wall hot + I
junction box V neutral ground The neutral wire is grounded back at the junction box. Note: The voltage of the neutral wire at the device is not the same as ground, because the neutral wire has a resistance. Ground Protection Ground
The metal casing is normally connected to ground. hot neutral ground GFCI (Ground Fault Circuit Interrupter) GFCI
Ih
hot Ig = Ih  In neutral In
GFCI Circuit trips when Ig > 5 [mA] Double Insulated Tool Double
Ih
hot plastic case neutral In The device is surrounded by an insulating plastic case. It does not require a ground. ECE 2317: Applied Electricity and Magnetism
Fall 2008 Prof. Jeffery T. Williams Dept. of Electrical & Computer Engineering University of Houston. Divergence and differential form of Gauss’ Law
Notes prepared by the EM group, University of Houston. Divergence (physical concept) Divergence
Start by considering a sphere of uniform volume charge density z ρv = ρv0
S r < a: ˆ ∫ D ⋅ n dS = Q
y encl r
x a ⎛4 3⎞ 4π r Dr = ρ v 0 ⎜ π r ⎟ ⎝3 ⎠ ρv 0 r ⎡C/m 2 ⎤ Dr = ⎣ ⎦ 3
2 The electric field is calculated using Gauss's law. Divergence (physical concept)
r > a:
⎛4 3⎞ 4π r Dr = ρ v 0 ⎜ π a ⎟ ⎝3 ⎠ ρv 0 a3 ⎡ C/m 2 ⎤ Dr = ⎣ ⎦ 3r 2
2 z ρv = ρv0 y r
x a Divergence (physical concept)
Flux through a spherical surface: ˆ ψ = ∫ D ⋅ n dS = 4π r 2 Dr
S Dr =
Dr = ρv 0 r
3
ρv 0 a3
3r
2 ⎡C/m 2 ⎤ ⎣ ⎦
⎡C/m 2 ⎤ ⎣ ⎦ (r < a) (r > a) 43 ψ = π r ρv 0 3 (r < a) 43 ψ = π a ρv 0 3 (r > a) Divergence Divergence
D ∆S ∆V ∆ψ = ∆S ˆ ∫ D ⋅ n dS > 0 1 div D ≡ lim ∆V → 0 ∆V ∆S ˆ ∫ D ⋅ n dS Note: the limit exists independent of the shape of the volume Divergence Divergence
Apply divergence definition to small volume inside a region of charge 1 div D ≡ lim ∆V → 0 ∆V
∆V
∆S ∆S ˆ ∫ D ⋅ n dS
≈ ρ v ( r ) ∆V ˆ ∫ D ⋅ n dS = Q encl ρ v (r ) 1 div D ( r ) = lim ( ρv ( r ) ∆V ) ∆V → 0 ∆V = ρv ( r ) Gauss’s Law (Differential Form) Gauss div D ( r ) = ρ v ( r ) z Example Example
ρv = ρv0 y 1 div D ≡ lim ∆V → 0 ∆V ∆S ˆ ∫ D ⋅ n dS r
x a Choose ∆V to be small sphere of radius r: 4 ∆V = π r 3 3 ∆S D ⋅ n dS = Dr ( 4π r 2 ) ∫ˆ ⎛ 4π r 3 ⎞ ⎛ ρv 0 r ⎞ 2 =⎜ ⎟ ⎟ ( 4π r ) = ρ v 0 ⎜ 3⎠ 3⎠ ⎝ ⎝ ⎛ ρ v 0 4π r 3 ⎞ div D ≡ lim ⎜ ⎟ = ∆lim0ρ v 0 = ρ v 0 ∆V → 0 4 3 ⎠ V→ π r3 ⎝ 3 1 Calculation of Divergence Calculation
z div D ≡ lim 1 ∆V → 0 ∆x ∆y ∆z ˆ ∫ D ⋅ n dS
S ∆z
x (0,0,0) y ⎛ ∆x ⎞ ˆ D ⋅ n dS ≈ Dx ⎜ , 0, 0 ⎟ ∆y ∆z ∫ ⎝2 ⎠ S ⎛ ∆x ⎞ − Dx ⎜ − , 0, 0 ⎟ ∆y ∆z ⎝2 ⎠ ⎛ ∆y ⎞ + Dy ⎜ 0, , 0 ⎟ ∆x ∆z ⎝2⎠ ∆y ⎞ ⎛ − Dy ⎜ 0, − , 0 ⎟ ∆x ∆ z 2⎠ ⎝ ∆z ⎞ ⎛ + Dz ⎜ 0, 0, ⎟ ∆x ∆y 2⎠ ⎝ ∆z ⎞ ⎛ − Dz ⎜ 0, 0, − ⎟ ∆x ∆y 2⎠ ⎝ ∆x ∆y For simplicity, assume point of interest is at the origin. The integrals over the 6 faces are approximated by “sampling” at the centers of the faces. Calculation of Divergence (cont.) Calculation
div D ≡ lim
⎛ ∆x ⎞ ˆ D ⋅ n dS ≈ Dx ⎜ , 0, 0 ⎟ ∆y ∆z ∫ ⎝2 ⎠ S ⎛ ∆x ⎞ − Dx ⎜ − , 0, 0 ⎟ ∆y ∆z ⎝2 ⎠ ⎛ ∆y ⎞ + Dy ⎜ 0, , 0 ⎟ ∆x ∆z ⎝2⎠ ∆y ⎞ ⎛ − Dy ⎜ 0, − , 0 ⎟ ∆x ∆ z 2⎠ ⎝ ∆z ⎞ ⎛ + Dz ⎜ 0, 0, ⎟ ∆x ∆y 2⎠ ⎝ ∆z ⎞ ⎛ − Dz ⎜ 0, 0, − ⎟ ∆x ∆y 2⎠ ⎝ 1 ∆V → 0 ∆x ∆y ∆z ˆ ∫ D ⋅ n dS
S 1 ∆x ∆y ∆z ˆ ∫ D ⋅ n dS =
S ⎛ ∆x ⎞ ⎛ ∆x ⎞ Dx ⎜ , 0, 0 ⎟ − Dx ⎜ − , 0, 0 ⎟ ⎝2 ⎠ ⎝2 ⎠ ∆x ∆y ⎞ ⎛ ∆y ⎞ ⎛ Dy ⎜ 0, , 0 ⎟ − Dy ⎜ 0, − ,0⎟ 2⎠ 2⎠ ⎝ +⎝ ∆y ∆z ⎞ ∆z ⎞ ⎛ ⎛ Dz ⎜ 0, 0, ⎟ − Dz ⎜ 0, 0, − ⎟ 2⎠ 2⎠ ⎝ +⎝ ∆z Calculation of Divergence (cont.) Calculation
1 div D ≡ lim ∆V → 0 ∆x ∆y ∆z ˆ ∫ D ⋅ n dS
S ⎛ ∆x ⎞ ⎛ ∆x ⎞ Dx ⎜ , 0, 0 ⎟ − Dx ⎜ − , 0, 0 ⎟ 1 ⎝2 ⎠ ⎝2 ⎠ ˆ D ⋅ n dS = ∆x ∆y ∆z ∫ ∆x S ∆y ⎞ ⎛ ∆y ⎞ ⎛ ,0⎟ Dy ⎜ 0, , 0 ⎟ − Dy ⎜ 0, − 2⎠ 2⎠ ⎝ +⎝ ∆y ∆z ⎞ ∆z ⎞ ⎛ ⎛ Dz ⎜ 0, 0, ⎟ − Dz ⎜ 0, 0, − ⎟ 2⎠ 2⎠ ⎝ +⎝ ∆z Hence 1 div D = lim ∆x → 0 ∆x ∆y ∆z ∆y → 0
∆z → 0 ∂Dx ∂Dy ∂Dz ˆ ∫ D ⋅ n dS = ∂x + ∂y + ∂z S Calculation of Divergence (cont.) Calculation ∂Dx ∂Dy ∂Dz div D = + + ∂x ∂y ∂z “del operator”
∂ ∂ ∂ ˆ ˆ ˆ ∇≡x +y +z ∂x ∂y ∂z
Examples of derivative operators: d : dx d ˆ x: dx d ( sin x ) = cos x dx d ˆ ˆ x ( sin x ) = x cos x dx d ⎛ d⎞ ˆ ⎟ ⋅ ( x sin x ) = x ⋅ x ( sin x ) = cos x ˆ ˆˆ ⎜x dx ⎝ dx ⎠ “del operator”
Now consider: ⎛∂ ∂ ∂⎞ ˆ ˆ ˆ ˆ ˆ ˆ ∇ ⋅ D = ⎜ x + y + z ⎟ ⋅ ( x Dx + y Dy + z Dz ) ∂y ∂z ⎠ ⎝ ∂x ∂Dy ∂Dx ∂Dz ˆˆ ˆˆ ˆˆ = x⋅x + y⋅y + z⋅z ∂x ∂y ∂z ∂Dx ∂Dy ∂Dz ∇⋅D = + + ∂x ∂y ∂z ∇ ⋅ D = div D “del operator”
Rectangular: ∂Dx ∂Dy ∂Dz ∇⋅D = + + ∂x ∂y ∂z
Cylindrical: 1∂ 1 ∂Dφ ∂Dz ∇⋅D = ( ρ Dρ ) + ρ ∂φ + ∂z ρ ∂ρ
Spherical: 1∂ 2 1 ∂ 1 ∂Dφ ∇ ⋅ D = 2 ( r Dr ) + ( Dθ sin θ ) + r ∂r r sin θ ∂θ r sin θ ∂φ Example Example
r < a:
z ρv = ρv0
y x a 1 ∇⋅D = 2 r 1 =2 r 1 =2 r ∂2 ( r Dr ) ∂r ∂ ⎛ 2 ρv 0 r ⎞ ⎜r ⎟ ∂r ⎝ 3⎠ ρv 0 r 2 ∇ ⋅ D = ρv 0
Note: This agrees with our previous example for the sphere of charge. Example Example
r > a:
z ρv = ρv0
y ⎛ ρv 0 a3 ⎞ ˆ D = r⎜ 2⎟ ⎝ 3r ⎠ 1 ∂ ⎛ 2 ρv 0 a3 ⎞ ∇⋅D = 2 ⎜r 2⎟ 3r ⎠ r ∂r ⎝ ∇⋅D = 0 x a Divergence Theorem Divergence
S V V ˆ ∫ ∇ ⋅ A dV = ∫ A ⋅ n dS
S In words: The volume integral of the divergence (flux per volume) equals the total flux! Divergence Theorem Divergence
Proof: N cubes ∆V … V ∫ ∇ ⋅ A dV = lim ∑ ( ∇ ⋅ A )
∆V → 0 n =1 N rn ∆V rn is the center of cube n Divergence Theorem Divergence
From the definition of divergence: ( ∇ ⋅ A )r = lim
n 1 ∆V → 0 ∆V ∆Sn ˆ ∫ A ⋅ n dS 1 ≈ ∆V ∆Sn ˆ ∫ A ⋅ n dS Hence: V ∫ ∇ ⋅ A dV = lim ∑ ∆V ( ∇ ⋅ A )
∆V → 0 n =1 N rn = lim ∆V → 0 ˆ ∑ ∫ A ⋅ n dS
n =1 ∆Sn N Divergence Theorem
N cubes ∆V … V ˆ ∫ ∇ ⋅ A dV = lim ∑ ∫ A ⋅ n dS
∆V → 0 n =1 ∆Sn N Consider two adjacent cubes: 1 ˆ A ⋅n
1 ˆ n2 ˆ n1 is opposite on the two faces 2 Hence: the surface integral cancels on all INTERIOR faces. Divergence Theorem
N cubes ∆V … V ˆ ∫ ∇ ⋅ A dr = lim ∑ ∫ A ⋅ n dS
∆V → 0 n =1 ∆Sn N = lim ∆V → 0 outside ∆Sn faces ˆ ∑ ∫ A ⋅ n dS Hence: 1 V ˆ ˆ ∫ ∇ ⋅ A dV = lim ∑ ∫ A ⋅ n dS ≈ ∫ A ⋅ n dS
∆V → 0 outside ∆Sn faces S Therefore: V ˆ ∫ ∇ ⋅ A dV = ∫ A ⋅ n dS
S (proof complete) Example Example
z ˆ A = x 3x
Verify Divergence Theorem ˆ ˆˆ ˆˆ ∫ A ⋅ ndS = x ⋅ ( x 3 ( 3) ) ⋅ ( 2 ) + − x ⋅ ( x 3 ( 0 ) ) ⋅ ( 2 )
S 1 3 2
x y = 18
∂Ax ∂Ay ∂Az ∇⋅A = + + ∂x ∂y ∂z ∂ = ( 3x ) = 3 ∂x V ∫ ∇ ⋅ A dV = ∫ 3 dV = 3V = 3 ⋅ (1⋅ 2 ⋅ 3) = 18
V Gauss’s Law (Differential Form)
Integral from of Gauss’ Law: Using the Divergence theorem: ˆ ∫ D ⋅ n dS = Q
S V encl Qencl
= ∫ ρv dV
V V ∫ ( ∇ ⋅ D ) dV = Q encl ρv S Valid for any volume, so let V ∆V ∆V ∫ ( ∇ ⋅ D ) dV = ∫ ρ
∆V v dV ρv ( ∇ ⋅ D ) ∆ V ≈ ρ v ∆V
∇ ⋅ D = ρv ∆V→ 0 Hence: Gauss’s Law (Differential Form) Gauss
ˆ ∫ D ⋅ n dS = Q
S encl Integral (volume) form of Gauss’s law Divergence theorem ∇ ⋅ D = ρv Differential (point) form of Gauss’s law ECE 2317: Applied Electricity and Magnetism
Fall 2008 Prof. Jeffery T. Williams Dept. of Electrical & Computer Engineering University of Houston. Faraday’s Law and Curl
Notes prepared by the EM group,
Michael Faraday, 17911867 University of Houston. Path independence of ∫ E ⋅ d Path
Consider a point charge E = a q 4πε 0 R
2 R = ERR b a C2 C6 C5 C1 bC 3 C4 ∫E⋅d
b a = ∫ E⋅d + ∫ E⋅d = 0 +
C1 C2 Ra Rb ∫ E dR
R Rb Ra ∫E⋅d
b a = ∫ E⋅d + ∫ E⋅d + ∫ E⋅d + ∫ E⋅d
C3 C4 C5 C6 = ∫ ER dR + 0 + ∫ ER dR + 0
Rb R4 R4 Ra =
a Ra Rb ∫E R dR Path independent for point charge. By superposition, ∫ E⋅d
b is path independent for any charge distribution. Static Faraday’s Law Static
(Integral Form)
Path independence a ∫E⋅d +∫E⋅d
b a a b =0 Thus, b ∫ C E⋅d = 0 Circulation of a vector field Circulation
V ( x, y, z ) = vector function ˆ n ˆ circulation of V around n axis ≡ ∫ C V⋅d C V Note: path is defined according to the “righthand rule” Curl of a vector Curl
z ∆S Cz ∆S ∆S x y Cy Cx 1 ˆ x ⋅ curl V ≡ lim ∆s → 0 ∆S 1 ˆ ⋅ curl V ≡ lim y ∆s → 0 ∆S 1 ˆ z ⋅ curl V ≡ lim ∆s → 0 ∆S ∫ ∫ ∫ Cx V⋅d V⋅d V⋅d Cy Cz Note: paths are defined according to the “righthand rule” Curl of a vector Curl
“curl meter” ˆ n
Assume that V represents the velocity of a fluid. V ˆ n ⋅ curl V = velocity of rotation (in the sense indicated) Curl Calculation Curl
Path Cx :
z 4 Cx
1 y ∆z 2
3 ∆y
(1) ∫ Cx V⋅d = ⎛ ∆y ⎞ , 0 ⎟ ∆z Vx dx + Vy dy + Vz dz ≈ Vz ⎜ 0, ∫ Cx ⎝2 ⎠ ∆y ⎞ ⎛ − Vz ⎜ 0, − , 0 ⎟ ∆z 2 ⎝ ⎠ ∆z ⎞ ⎛ + Vy ⎜ 0, 0, − ⎟ ∆y 2⎠ ⎝ ∆z ⎞ ⎛ − Vy ⎜ 0, 0, ⎟ ∆y 2⎠ ⎝ (2) (3) (4) Curl of a vector Curl
⎡ ⎛ ∆y ⎞ ∆y ⎞ ⎤ ⎛ , 0 ⎟ − Vz ⎜ 0, − , 0⎟ ⎥ Vz ⎜ 0, ⎢ 2 2 ⎠ ⎝ ⎠⎥ V ⋅ d ≈ ( ∆y ∆z ) ⎢ ⎝ ∫ Cx ∆y ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎡⎛ ∆z ⎞ ∆z ⎞ ⎤ ⎛ − Vy ⎜ 0, 0, − ⎟ ⎥ Vy ⎜ 0, 0, ⎟ ⎢ 2⎠ 2⎠ ⎝ ⎥ − ( ∆z ∆y ) ⎢ ⎝ ∆z ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ∂Vy ∂Vz = ∆S − ∆S ∂y ∂z ⎛ ∂Vz ∂Vy ⎞ ∫ V ⋅ d ≈ ∆S ⎜ ∂y − ∂z ⎟ ⎝ ⎠ Cx Curl of a vector Curl
Similarly, ⎛ ∂Vx ∂Vz ⎞ ⎜ ⎟ ∫ V ⋅ d ≈ ∆S ⎝ ∂z − ∂x ⎠ Cy ⎛ ∂Vy ∂Vx ⎞ ⎜ ⎟ ∫ V ⋅ d ≈ ∆S ⎝ ∂x − ∂y ⎠ Cz Hence, ⎛ ∂Vz ∂Vy ⎞ ⎛ ∂Vx ∂Vz ⎞ ⎛ ∂Vy ∂Vx ⎞ ˆ ˆ ˆ curl V = x ⎜ − − − ⎟+ y⎜ ⎟ ⎟+ z⎜ ∂z ⎠ ∂x ⎠ ⎝ ∂x ∂y ⎠ ⎝ ∂z ⎝ ∂y Del operator Del
⎛∂ ∂ ∂⎞ ˆ +y +z ⎟ ˆ ˆ ∇ ≡ ⎜x ∂y ∂z ⎠ ⎝ ∂x
⎛∂ ∂ ∂⎞ ˆ ˆ ˆ ˆ ˆ ˆ ∇ × V = ⎜ x + y + z ⎟ × ( x Vx + y Vy + z Vz ) ∂y ∂z ⎠ ⎝ ∂x ˆ x = ∂ ∂x Vx ˆ y ∂ ∂y Vy ˆ z ∂ ∂z Vz ⎛ ∂Vz ∂Vy ⎞ ⎛ ∂Vx ∂Vz ⎞ ⎛ ∂Vy ∂Vx ⎞ ˆ ˆ ˆ = x⎜ − − − ⎟+ y⎜ ⎟ ⎟+ z⎜ ∂y ∂z ⎠ ∂z ∂x ⎠ ⎝ ∂x ∂y ⎠ ⎝ ⎝ Del operator Del
Hence, curl V = ∇ × V Example Example
ˆ ˆ ˆ V = x ( 3 xy 2 z ) + y ( 2 x 2 − z 3 ) + z ( 2 xz )
⎛ ∂Vz ∂Vy ⎞ ⎛ ∂Vx ∂Vz ˆ ˆ ∇×V = x⎜ − − ⎟+ y⎜ ∂z ⎠ ∂x ⎝ ∂z ⎝ ∂y ⎞ ⎛ ∂Vy ∂Vx ⎞ ˆ − ⎟ ⎟+ z⎜ ∂y ⎠ ⎠ ⎝ ∂x ˆ ˆ ˆ ∇ × V = x ( 0 + 3 z 2 ) + y ( 3xy 2 − 2 z ) + z ( 4 x − 6 xyz ) Example Example
ˆ V=xy ⎛ ∂Vz ∂Vy ⎞ ⎛ ∂Vz ∂Vx ⎞ ⎛ ∂Vy ∂Vx ⎞ ˆ ˆ ˆ ∇×V = x⎜ − − − ⎟−y⎜ ⎟ ⎟+ z⎜ ∂y ∂z ⎠ ∂x ∂z ⎠ ⎝ ∂x ∂y ⎠ ⎝ ⎝ ˆ ∇ × V = −z
y x Example Example
ˆ ∇ × V = −z ˆ (∇ × V ) ⋅ z < 0 y x Summary of Curl Formulas Summary
⎛ ∂Vz ∂Vy ⎞ ⎛ ∂Vx ∂Vz ˆ ˆ ∇×V = x⎜ − − ⎟+ y⎜ ∂z ⎠ ∂x ⎝ ∂z ⎝ ∂y ⎞ ⎛ ∂Vy ∂Vx ⎞ ˆ − ⎟ ⎟+ z⎜ ∂y ⎠ ⎠ ⎝ ∂x
⎞ ⎟ ⎟ ⎠ ⎛ 1 ∂Vz ∂Vφ ⎞ ⎛ ∂Vρ ∂Vz ⎞ 1 ⎛ ∂ ( ρVφ ) ∂Vρ ˆ ˆ ˆ⎜ ∇× V = ρ⎜ − − − ⎟ + φ⎜ ⎟+z ⎜ ρ ⎝ ∂ρ ∂z ⎠ ∂ρ ⎠ ∂φ ⎝ ρ ∂φ ⎝ ∂z 1 ∇×V = r sin θ ⎡ ∂ (Vφ sin θ ) ∂V −θ ⎢ ∂θ ∂φ ⎢ ⎣ ⎤ 1 ⎡ 1 ∂Vr ∂ ( rVφ ) ⎤ ˆ 1 ⎡ ∂ ( rVθ ) ∂Vr ⎤ ˆ ˆ − − ⎥r + ⎢ ⎥θ+ ⎢ ⎥φ ∂r ⎥ ∂θ ⎦ r ⎢ sin θ ∂φ r ⎣ ∂r ⎥ ⎦ ⎣ ⎦ ECE 2317: Applied Electricity and Magnetism
Fall 2008
Prof. Jeffery T. Williams Dept. of Electrical & Computer Engineering University of Houston. Stoke’s Theorem and Faraday’s Law
Notes prepared by the EM group,
Michael Faraday, 17911867 University of Houston. Stokes’ Theorem Stokes
ˆ n
S (open) C ˆ n : chosen from “righthand rule” applied to the surface ˆ ∫ ( ∇ × V ) ⋅ n dS = ∫ V ⋅ d
S C Proof Proof
Divide S into rectangular patches that are normal to x, y, or z axes. ˆ n
∆S S ˆ ni
C ˆ ˆˆ ˆ ( ni = x, y, or z ) ˆ ∫ ( ∇ × V ) ⋅ n dS ≈ ∑ ( ∇ × V )
S n ri ˆ ⋅ n i ∆S Proof (cont.) Proof
S C ˆ ∫ ( ∇ × V ) ⋅ n dS ≈ ∑ ( ∇ × V )
S n ri ˆ ⋅ n i ∆S ˆ ni ⋅ ( ∇ × V ) = lim ˆ n i ⋅ ( ∇ × V )r 1 ∆s → 0 ∆S
i ∫ ∫ Ci V⋅d V⋅d 1 ≈ ∆S ˆ n i ⋅ ( ∇ × V ) r ∆S ≈
i ∫ Ci V⋅d Ci ˆ ˆˆˆ n i = x, y , z Proof (cont.) Proof
S C Hence, ˆ ∫ ( ∇ × V ) ⋅ n ds ≈ ∑ ∫ V ⋅ d
S n Ci → C ∫ V⋅d
C (Interior edges cancel) Example Example
y Verify Stokes’ Theorem ρ=a
C C A B
x C ∫ V⋅d = ∫ Vx dx + Vy dy
C = C ∫ x dy = I A + I B + IC ˆ V = xy
IA = 0 IC = 0
(dy = 0) (x = 0) Example (cont.) Example
IB =
y B
CB ∫ x dy
ρ=a
CB
x A y a −y a −1 ⎛ y ⎞ IB = + sin ⎜ ⎟ 2 2 ⎝a⎠
2 2 2 a 0 a2 = sin −1 (1) 2 a2 ⎛ π ⎞ = ⎜⎟ 2 ⎝2⎠ I= π a2
4 I B = ∫ a 2 − y 2 dy
0 a Example (cont.) Example
Alternative evaluation (use cylindrical coordinates):
y IB = ∫ V ⋅ d
Α Β Β ρ=a
C C A Now use: B
x ˆ ˆ ˆ = ∫ V ⋅ ( ρ d ρ + φ a dφ + z dz )
A π = ∫ Vφ a dφ
0 2 ˆ ˆˆ ˆˆ Vφ = V ⋅ φ = ( x y ) ⋅ φ = x y ⋅ φ = x cosφ or Vφ = ( a cosφ ) cosφ = a cos 2φ Example (cont.) Example
π Hence IB = ∫
0 2 a 2 cos 2φ dφ
π = a2 ⎛ 1 + cos2φ ⎞ ∫ ⎜ 2 ⎟ dφ ⎠ 0⎝
2 ⎡ φ sin ( 2φ ) ⎤ 2 = a2 ⎢ + ⎥ 2 4 ⎦0 ⎣ ⎛π ⎞ =a ⎜ ⎟ ⎝4⎠
2 π ⎛π ⎞ I =a ⎜ ⎟ ⎝4⎠
2 Example (cont.) Example
Now Use Stokes’ Theorem: C ∫ V⋅d ˆ = ∫ ( ∇ × V ) ⋅ z ds
S ˆˆ (n = z ) ˆ V = xy ⎛ ∂Vz ∂Vy ⎞ ⎛ ∂Vz ∂Vx ˆ ˆ ∇×V = x⎜ − − ⎟−y⎜ ∂z ⎠ ∂z ⎝ ∂x ⎝ ∂y ⎞ ⎛ ∂Vy ∂Vx ⎞ ˆ − ⎟ ⎟+ z⎜ ∂y ⎠ ⎠ ⎝ ∂x ˆ ∇×V = z
ˆˆ I = ∫ z ⋅ z dS = ∫ dS = A =
S S 1 π a2 ) ( 4 I= πa
4 2 Vector Identity Vector
∇ ⋅ (∇ × V ) = 0
Proof: ⎛ ∂Vz ∂Vy ⎞ ⎛ ∂Vz ∂Vx ⎞ ⎛ ∂Vy ∂Vx ⎞ ˆ ˆ ˆ ∇×V = x⎜ − − − ⎟−y⎜ ⎟ ⎟+ z⎜ ∂y ∂z ⎠ ∂x ∂z ⎠ ⎝ ∂x ∂y ⎠ ⎝ ⎝ ∂Ax ∂Ay ∂Az ∇ ⋅ (∇ × V ) = + + ∂x ∂y ∂z
A ⎛ ∂ 2Vz ∂ 2Vy ⎞ ⎛ ∂ 2Vz ∂ 2Vx ⎞ ⎛ ∂ 2Vy ∂ 2Vx ⎞ =⎜ − − − + − ⎜ ∂x ∂y ∂x ∂z ⎟ ⎜ ∂y ∂x ∂y ∂z ⎟ ⎜ ∂z ∂x ∂z ∂y ⎟ ⎟ ⎜ ⎟ ⎠⎝ ⎝ ⎠⎝ ⎠ =0 Example Example
Find curl of E: 1 2 3 q ρs0
sheet of charge (side view) ρl0
line charge point charge Example Example
1 x ⎛ ρs0 ⎞ ˆ E = x⎜ ⎟ ⎝ 2ε 0 ⎠ ρs0 ⎛ ∂Ez ∂E y ⎞ ⎛ ∂Ez ∂Ex ⎞ ⎛ ∂E y ∂Ex ⎞ ˆ ˆ ˆ ∇×E = x⎜ − − − ⎟−y⎜ ⎟ ⎟+ z⎜ ∂y ∂z ⎠ ∂x ∂z ⎠ ⎝ ∂x ∂y ⎠ ⎝ ⎝
ˆ ˆ ˆ ∇ × E = x ( 0 − 0) + y ( 0 − 0) + z ( 0 − 0) =0 Example Example
2 ⎛ ρ0 ⎞ ˆ E = ρ⎜ ⎟ ⎝ 2π ε 0 ρ ⎠
ρl0 ⎛ 1 ∂Ez ∂Eφ ⎞ ⎛ ∂Eρ ∂Ez ⎞ 1 ⎛ ∂ ( ρ Eφ ) ∂Eρ ˆ ˆ ˆ ∇×E = ρ⎜ − − − ⎟ + φ⎜ ⎟+z ⎜ ∂z ⎠ ∂ρ ⎠ ∂φ ρ ⎜ ∂ρ ⎝ ρ ∂φ ⎝ ∂z ⎝ =0 ⎞ ⎟ ⎟ ⎠ Example
3 q ⎛q⎞ ˆ E = r⎜ ⎟ 4π ε 0 r 2 ⎠ ⎝ 1 ∇×E = r sin θ =0 ⎡ ∂ ( Eφ sin θ ) ∂E −θ ⎢ ∂θ ∂φ ⎢ ⎣ ⎤ 1 ⎡ 1 ∂Er ∂ ( rEφ ) ⎤ ˆ 1 ⎡ ∂ ( rEθ ) ∂Er ⎤ ˆ ˆ − − ⎥r + ⎢ ⎥θ+ ⎢ ⎥φ r ⎢ sin θ ∂φ r ⎣ ∂r ∂r ⎥ ∂θ ⎦ ⎥ ⎦ ⎣ ⎦ Faraday’s Law (Differential Form) Faraday
Stokes’ Th.: ˆ ∫ ( ∇ × E ) ⋅ n dS = ∫ E ⋅ d
S C ˆ n
Let S → ∆S
∆S small planar surface Hence ∆S ˆ ∫ ( ∇ × E ) ⋅ n dS = 0 Let ∆S 0: ˆ n ⋅ ( ∇ × E ) ∆S = 0 ˆ n ⋅ (∇ × E) = 0 Faraday’s Law (Differential Form) Faraday
ˆ n ⋅ (∇ × E) = 0
∆S ˆ n ˆ ˆˆˆ Let n = x, y, z : ˆ x ⋅ (∇ × E) = 0 ˆ y ⋅ (∇ × E) = 0 ˆ z ⋅ (∇ × E) = 0 Hence, ∇×E = 0 Faraday’s Law (Summary) Faraday C ∫ E⋅d = 0 Integral form of Faraday’s law Stokes’ theorem ∇×E = 0 Differential (point) form of Faraday’s law Path Independence Path
Assume ∇×V = 0
C1 B A C2 I1 = ∫ V ⋅ d
C1 I2 =
I1 = I 2 C2 ∫ V⋅d Path Independence (cont.) Path
Proof
A B C C = C2  C1 C ∫ V⋅d ˆ = ∫ ( ∇ × V ) ⋅ n dS = 0
S S is any surface that is attached to C. I 2 − I1 = 0 (proof complete) Path Independence (cont.) Path
∇×V = 0 Stokes’ theorem Definition of curl path independence Summary of Electrostatics Summary ∇ ⋅ D = ρv ∇×E = 0 D = ε 0E
Differential (point) form ECE 2317: Applied Electricity and Magnetism
Fall 2008
Prof. Jeffery T. Williams Dept. of Electrical & Computer Engineering University of Houston. Gradient and Poisson’s Equation
Notes prepared by the EM group, University of Houston. Gradient Gradient
Φ ( x, y, z ) = scalar function ∂Φ ∂Φ ∂Φ ˆ ˆ ˆ +y +z grad Φ ≡ x ∂x ∂y ∂z ⎛∂ ∂ ∂⎞ ˆ ˆ ˆ grad Φ = ⎜ x + y + z ⎟ Φ ∂y ∂z ⎠ ⎝ ∂x grad Φ = ∇Φ Directional Derivative Property Directional
r + dr = (x+dx ,y+dy, z+dz) d Φ = Φ ( r + dr ) − Φ ( r ) ∂Φ ∂Φ ∂Φ = dx + dy + dz ∂x ∂y ∂z dr r =(x,y,z) Hence Let Then d Φ = ∇Φ ⋅ dr dr = d d Φ = ∇Φ ⋅ ( )d “directional derivative” dΦ = ∇Φ ⋅ ˆ d Physical Interpretation Physical
∇Φ
dΦ = ∇Φ ⋅ ˆ d dΦ = ∇Φ cos θ d θ
ˆ • The gradient is a vector which points in the direction of maximum increase of the function. • The magnitude of the gradient vector is the rate of change of the function in this direction. Gradient Gradient
ˆ ∇Φ = x ∂Φ ∂Φ ∂Φ ˆ ˆ +y +z ∂x ∂y ∂z ∂Φ 1 ∂Φ ∂Φ ˆ ˆ ˆ ∇Φ = ρ +φ +z ∂ρ ρ ∂φ ∂z
∂Φ ˆ 1 ∂Φ 1 ∂Φ ˆ ˆ ∇Φ = r +θ +φ ∂r r ∂θ r sin θ ∂φ Relation Between E and Φ Relation
VAB = Φ ( A ) − Φ ( B ) = ∫ E ⋅ d
Α Β Also, Φ ( A) − Φ ( B ) = ∫ d Φ
Β Α = ∫ ∇Φ ⋅ d
Β Α Hence Α ∫E⋅d Β = ∫ −∇Φ ⋅ d
Α Β Relation Between E and Φ Relation
Α ∫E⋅d Β = ∫ −∇Φ ⋅ d
Α Β ∆x
Assume a small path in the x direction: A B x Α ∫E⋅d Β ˆ = ∫ E ⋅ ( x dx ) =
Α Β xB xA ∫ E dx ≈ E ∫ dx = E ∆x
x x x xA xB Similarly for the second integral: Α ∫ −∇Φ ⋅ d Β ≈ ( −∇Φ ) x ∆x Relation Between E and Φ Relation
Hence: Ex = − ( ∇Φ ) x Similarly, E y = − ( ∇Φ ) y Ez = − ( ∇Φ ) z Hence, E = −∇Φ Note: the choice of the reference point does not affect E (the derivative of a constant is zero). Vector Identity Vector
∇ × ( ∇ψ ) = 0
⎛ ⎜ˆ ⎜x ⎜∂ ∇ × ( ∇ψ ) = ⎜ ⎜ ∂x ⎜ ∂ψ ⎜ ⎝ ∂x ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Proof: ˆ y ∂ ∂y ∂ψ ∂y ˆ z ∂ ∂z ∂ψ ∂z ⎛ ∂ 2ψ ∂ 2ψ ⎞ ⎛ ∂ 2ψ ∂ 2ψ ⎞ ⎛ ∂ 2ψ ∂ 2ψ ⎞ ˆ ˆ ˆ = x⎜ − − − ⎟−y⎜ ⎟+ z⎜ ⎟ ⎝ ∂y∂z ∂z∂y ⎠ ⎝ ∂x∂z ∂z∂x ⎠ ⎝ ∂x∂y ∂y∂x ⎠ =0 Curl Property in Electrostatics Curl
E = −∇Φ ∇ × E = ∇ × ( −∇Φ ) = −∇ × ( ∇Φ ) =0 so ∇×E = 0 Example Example
Φ=
z q 4π ε 0 r [V ] q [C]
y x Find E from the point charge ⎡ ∂Φ ˆ 1 ∂Φ 1 ∂Φ ⎤ ˆ ˆ E = −∇Φ = − ⎢r +θ +φ r ∂θ r sin θ ∂φ ⎥ ⎣ ∂r ⎦ ∂Φ ˆ = −r ∂r ⎛ −q ⎞ ˆ⎜ = −r ⎟ 4π ε 0 r 2 ⎠ ⎝ ⎛q⎞ ˆ E = r⎜ ⎟ 4π ε 0 r 2 ⎠ ⎝ [ V/m] Example Example
z (0,0,z)
R Find: E (0,0,z)
ˆ E ( 0, 0, z ) = z Ez ( 0, 0, z )
y a x ρl0 [C/m] E = −∇Φ ∂Φ Ez = − ∂z Φ ( 0, 0, z ) = ρ 0a
2ε 0 z 2 + a 2 Example (cont.) Example
d Φ ( 0, 0, z ) ρ 0a d⎛ =− ⎜ Ez ( 0, 0, z ) = − ⎜ dz dz ⎝ 2ε 0 z 2 + a 2 ⎞ ⎟ ⎟ ⎠ ρ 0a ⎛ 1 ⎞ 2 2 − 32 Ez ( 0, 0, z ) = − ⎜ − ⎟ ( z + a ) ( 2z ) 2ε 0 ⎝ 2 ⎠
⎛ ρ 0a ⎜ +z Ez ( 0, 0, z ) = 2ε 0 ⎜ z 2 + a 2 3 2 ⎜( ) ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ [ V/m] Proof of Flux Line Property Proof
CV ⊥ E
r + dr dr r E proof: CV : Φ = constant dΦ = 0
so or CV : Φ = constant
The vector dr is a small change in position, moving along an equipotential contour. ∇Φ ⋅ dr = 0 −E ⋅ dr = 0 so E ⊥ dr Poisson’s Equation Poisson
∇ ⋅ D = ρv ∇ ⋅ ( ε 0E ) = ρv ∇ ⋅ ( −ε 0∇Φ ) = ρv ρv ∇ ⋅ ( ∇Φ ) = − ε0 Poisson’s Equation (cont.) Poisson
⎡ ∂Φ ∂Φ ∂Φ ⎤ ˆ ˆ ˆ ∇ ⋅ ( ∇Φ ) = ∇ ⋅ ⎢ x +y +z ∂x ∂y ∂z ⎥ ⎣ ⎦ ∂ 2Φ ∂ 2Φ ∂ 2Φ = 2+ 2+ 2 ∂x ∂y ∂z ∂ 2Φ ∂ 2Φ ∂ 2Φ Lap Φ ≡ 2 + 2 + 2 ∂x ∂y ∂z
Poisson’s Eq.: “Laplacian” ρv Lap Φ = − ε0 Poisson’s Equation (cont.) Poisson
Note: ∂ ∂ ∂ ˆ ˆ ˆ ∇=x +y +z ∂x ∂y ∂z ⎛ ∂2 ∂2 ∂2 ⎞ 2 ∇ ⋅∇ = ∇ = ⎜ 2 + 2 + 2 ⎟ ⎝ ∂x ∂y ∂z ⎠
“Laplacian operator” Lap Φ ≡ ∇ ⋅ ( ∇Φ ) = ∇ Φ
2 Poisson’s Equation Poisson
Hence ρv ∇ Φ=− ε0
2 “Poisson’s Equation” If ρv = 0
∇ 2Φ = 0
“Laplace’s Equation” then Electrostatic Triangle Electrostatic
ρv
ρv Φ=∫ dV ′ 4π ε 0 R V ˆ ρv R dV ′ E=∫ 2 4π ε 0 R V ∇ ⋅ (ε 0E ) = ρv
ρv ∇ Φ=− ε0
2 Φ E = −∇Φ
Φ (r ) = Φ ( R ) − ∫ E ⋅ d
R r E ECE 2317: Applied Electricity and Magnetism
Fall 2008
Prof. Jeffery T. Williams Dept. of Electrical & Computer Engineering University of Houston. Laplace’s equation and Boundary value problems
Notes prepared by the EM group, University of Houston.
Pierre Simon de Laplace Boundary Value Problem Boundary
(Laplace’s Equation)
Goal: Solve for the potential function inside of a region, given the value of the potential function on the boundary. ∇ 2 Φ ( x, y , z ) = 0 The Uniqueness Principle in electromagnetics assures us that: Φ ( x, y, z ) is unique Φ = Φ B ( x, y , z ) On boundary: Example Example V0 +  ε0 h x Goal: solve for Φ ( x, y, z ) Assume Φ ( x, y , z ) = Φ ( x ) Example (cont.) Example
∇ 2Φ = 0
∂ 2Φ ∂ 2Φ ∂ 2Φ + 2 + 2 =0 2 ∂x ∂y ∂z
Hence: ∂ 2Φ ( x ) =0 2 ∂x ∂Φ ( x ) = C1 ∂x Φ ( x ) = C1 x + C2 Solution: Example (cont.) Example
x = h: Φ (h) = 0 C1 ( h ) + C2 = 0 C2 = − hC1 ( reference point ) x = 0: Φ ( 0 ) = V0 C1 ( 0 ) + C2 = V0 C2 = V0
Hence: C2 = V0 C1 = − V0 h ⎛ V0 ⎞ Φ ( x, y, z ) = − ⎜ ⎟ x + V0 ⎝h⎠ [V ] Example (cont.) Example
Check: E = −∇Φ ∂Φ ( x ) ∂x ⎡ V0 ⎤ ˆ E = −x ⎢ − ⎥ ⎣ h⎦ ⎛ V0 ⎞ ˆ E = x⎜ ⎟ [ V/m] ⎝h⎠ ˆ = −x Previous method: V0 = ∫ E ⋅ d
Α h Β = ∫ Ex dx
0 = Ex h V0 Ex = h ⎛V ⎞ ˆ E = x⎜ 0 ⎟ ⎝h⎠ [ V/m] Example Example Φ = V0 V0 +  φ0 Φ=0 Goal: solve for Φ ( ρ , φ , z ) ∇ 2Φ = 0 1 ∂ ⎛ ∂Φ ⎞ 1 ∂ 2 Φ ∂ 2 Φ + 2 =0 ⎜ρ ⎟+ 2 2 ρ ∂ρ ⎝ ∂ρ ⎠ ρ ∂φ ∂z Example (cont.) Example
Assume Φ ( ρ , φ , z ) = Φ (φ ) ∂ 2Φ =0 2 ∂φ Φ = C1φ + C2 φ = 0: C1 ( 0 ) + C2 = 0 C2 = 0 φ = φ0 : Φ = V0 C1φ0 + C2 = V0 C1 = V0 φ0 Example (cont.) Example
⎛ V0 ⎞ Φ ( ρ , φ , z ) = ⎜ ⎟φ ⎝ φ0 ⎠
E = −∇Φ ⎡ ∂Φ 1 ∂Φ ∂Φ ⎤ ˆ ˆ ˆ = − ⎢ρ +φ +z ∂ρ ∂z ⎥ ρ ∂φ ⎣ ⎦ 1 ∂Φ ˆ = −φ ρ ∂φ ⎛ 1 ⎞ ⎛ V0 ⎞ ˆ E = −φ ⎜ ⎟ ⎜ ⎟ ⎝ ρ ⎠ ⎝ φ0 ⎠ [V] [ V/m] ECE 2317: Applied Electricity and Magnetism
Fall 2008
Prof. Jeffery T. Williams Dept. of Electrical & Computer Engineering University of Houston. Dielectrics
Notes prepared by the EM group, University of Houston. Dielectrics Dielectrics water “dipole” model: “Microscopically”: ˆ
“dipole moment”: H+ + + H+ H2O d +q  p = qd p = pˆ O q Dielectrics Dielectrics
+ + + + E
 x E + V  Dielectrics Dielectrics
Define: P = dipole moment/volume 1 P≡ ∆V ∑p
i =1 NV i ∆V NV = # of aligned dipoles
per unit volume Dielectrics: Bound Charge Dielectrics:
Assume nonuniform dielectric: ρvb x
For simplicity, the dipoles are shown perfectly aligned, with the number of aligned dipoles increasing with x. x A net volume charge density is created inside the material, called the “bound charge density” or “polarization charge density”. ρvb = “bound charge density”: due to molecules, it is not free to move ρv = “free charge density”: due to free charge inside the material:
it is free to move, in general Dielectrics: Bound Charge Dielectrics:
ρvb ρvb Total charge density inside the material: ρvtotal = ρv + ρvb
This total charge density may be viewed as being in free space. Dielectrics: Gauss’ Law Dielectrics:
∇ ⋅ (ε 0E ) = ρ
total v = ρ v + ρ vb We prefer to have only the freecharge density in the equation, since this is what is known. The goal is to calculate (and possibly eliminate) the boundcharge density term on the righthand side. Dielectrics: Gauss’ Law Dielectrics:
1 ρvb = Qb ∆V d⎞ d⎞ ⎛ ⎛ Qb = −qNV ⎜ x + ⎟ ∆V + q NV ⎜ x − ⎟ ∆V 2⎠ 2⎠ ⎝ ⎝
Hence ∆V ρvb = −q ⎢ NV ⎜ x + ⎟ − NV ⎜ x − ⎟ ⎥ 2 2
⎣ = − q∆NV ⎠ ⎠⎦
xd/2 x+d/2 ⎡ ⎛ ⎝ d⎞ ⎛ ⎝ d ⎞⎤ x NV = # of aligned dipoles
per unit volume Dielectrics: Gauss’ Law Dielectrics:
∆V ρvb = −q∆NV
1 = −q ∆Px p 1 = − ∆Px d ∂Px ρvb = − ∂x x xd/2 x+d/2 Hence 1 P≡ ∆V ∑p
i =1 NV i p = qd p = pˆ Dielectrics: Gauss’ Law Dielectrics:
In general or Hence, ∂Px ∂Py ∂Pz ρvb = − − − ∂x ∂y ∂z ρvb = −∇ ⋅ P ∇ ⋅ ( ε 0 E ) = ρ vtotal = ρ v − ∇ ⋅ P ∇ ⋅ (ε 0E + P ) = ρv Now lets define the electric flux density vector D as D = ε 0E + P
Thus, ∇ ⋅ D = ρv Dielectrics Dielectrics
Linear Material: P ∝ E = ε 0 χeE D = ε 0E + P = ε 0E + ε 0 χ eE = ε 0 (1 + χ e ) E Note: χe > 0 for most materials Electric flux density vector D: Define relative permittivity (dielectric constant): ε r ≡ 1 + χe
Then: ε = ε 0ε r D = ε 0ε r E = ε E Typical Linear Materials Typical
Teflon Water Styrofoam Quartz εr εr εr εr = 2.2 = 81 = 1.03 =5 Dielectrics: Gauss’ Law Dielectrics:
∇ ⋅ D = ρv
D = ε 0ε r E = ε E
Important conclusion: Gauss’ law works the same way inside a dielectric as it does in vacuum, with only the free charge density (i.e., the charge that is actually placed inside the material) being used on the righthand side. Example Example
Point Charge: Find D, E r r q b S a
Dielectric shell εr Gauss’ law: ˆ ∫ D ⋅ n dS = Q
S encl =q (The point charge q is the only free charge in the problem.) Example (cont.) Example
r r q b S a
Dielectric shell εr
⎡C/m 2 ⎤ ⎣ ⎦ ⎛q⎞ ˆ⎜ D=r 2⎟ ⎝ 4π r ⎠ ⎛q⎞ ˆ E = r⎜ 2⎟ ⎝ 4πε 0 r ⎠ ⎛ ⎞ q ˆ E = r⎜ ⎟ 4πε 0ε r r 2 ⎠ ⎝ [ V/m] [ V/m] r < a, r > b a<r<b Example (cont.) Example
Physically, why is the electric field weaker inside the dielectric? + + + q + + q
+  + + The alignment of the dipoles causes layers of bound surface charge. The electric field from these layers partially cancels that from the charge q. Homogeneous Dielectrics Homogeneous
ρvb = −∇ ⋅ P
= −∇ ⋅ ( ε 0 χ e E ) = −ε 0 χ e∇ ⋅ E ⎛D⎞ = −ε 0 χ e∇ ⋅ ⎜ ⎟ ε 0ε r ⎠ ⎝ ⎛1 = −ε 0 χ e ⎜ ⎝ ε 0ε r ⎛1 = −ε 0 χ e ⎜ ⎝ ε 0ε r =0 ⎞ ⎟∇ ⋅D ⎠ ⎞ ⎟ ρv ⎠
Assume a homogeneous dielectric with no free charge density inside ε ( x, y, z ) = constant ρvb = 0 Inside a homogeneous dielectric Bound charge only exists on the surface of a homogeneous dielectric. Homogeneous Dielectrics Homogeneous
Important note:
A homogeneous dielectric can always be modeled with layers of bound surface charge density. However, this is automatically accounted for when using the D vector. When using the D vector, the problem is solved using only the free charge density on the righthand side of Gauss’ law. ECE 2317: Applied Electricity and Magnetism
Fall 2008
Prof. Jeffery T. Williams Dept. of Electrical & Computer Engineering University of Houston. Dielectric breakdown
Notes prepared by the EM group, University of Houston. Dielectric Breakdown Dielectric
1 Arcing (spark) +  Ex Ex = Ec x Ec = 3 × 106 Air
Ec = “critical electric field” (air ionizes) [ V/m] [ V/m] [ V/m] Oil Ec = 15 × 106 Glass E = 30 × 106 c Dielectric Breakdown Dielectric
2 Corona Discharge +  E > Ec Power line Example Example
ε r = 2.0
Ec = 15 × 106 [V/m] εr a coaxial cable b b = 0.25 a = 0.5 [cm] [ mm] Find: Vmax (assume a safety factor (SF) = 2 ) ρlmax
Set ρlmax ⎛ ρ max ⎞ ˆ E = ρ⎜ ⎟ ⎝ 2π ε 0 ε r ρ ⎠ ρ =a ρ max Ec = 2π ε 0 ε r a Example Example
Thus ρ max = 2π ε 0 ε r a Ec
⎛ ρ max ⎞ ⎛a⎞ ˆ ˆ E = ρ⎜ ⎟ = ρ Ec ⎜ ⎟ ⎝ρ⎠ ⎝ 2π ε 0 ε r ρ ⎠ Vmax = VAB = ∫ E ⋅ d = ∫ Eρ d ρ
Α a Β b Hence ⎛a⎞ = + ∫ Ec ⎜ ⎟ d ρ ⎝ρ⎠ a
b = + Ec a ln ρ a
b or Vmax ⎛b⎞ = Ec a ln ⎜ ⎟ ⎝a⎠ Example Example
With safety factor:
SF max V ⎛1⎞ ⎛b⎞ =⎜ ⎟ Ec a ln ⎜ ⎟ ⎝ SF ⎠ ⎝a⎠ ⎛1⎞ = ⎜ ⎟ 15 × 106 ( 0.5 × 10−3 ) ln ( 5 ) ⎝2⎠ ( ) SF Vmax = 6035.4 [V ] Lightning Lightning Lightning Lightning Lightning Lightning
+++ ___ +++ ___ +++ ___ Most lightning comes from negative charges at the base of the thundercloud. About 10% comes from clouds with positive charges at the base. The base of the thundercloud is typically at an altitude of about 1,500 [m]. The top of the thundercloud may be at about 10,000 [m]. The voltage drop between the cloud base and ground is typically 150 [MV]. +++++++++++++++++++++++++++++++ Earth Lightning Lightning
A “stepped leader” begins descending from the cloud. It is a group of electrons that zigzags toward the earth. The charge is typically about 5 [C]. +++ ___ +++ ___ +++ ___ The stepped leader travels at an average speed of about 150 [km/s] (0.05% the speed of light). It travels about 50 [m] in about 1 [us] (about 10% the speed of light), pauses for about 50 [us], then continues. The stepped leader does not know where it will contact the ground until it is about 2050 [m] from the earth (40 [m] is typical). +++++++++++++++++++++++++++++++ Earth Lightning Lightning
+++ The stepped leader may branch as it descends (this is responsible for the forked appearance of some lighting). ___ +++ ___ +++ ___ +++++++++++++++++++++++++++++++ Earth Lightning Lightning
As the stepped leader descends, positively charged “streamers’ rise up from the earth to meet it. The streamers are more likely to come from conducting objects with sharp points. Streamers may be up to about 50 [m] in length + + +++ ___ +++ ___ +++ ___  + + + + + + + + + + + + +++++++++++++++++++++++++++++++ Earth Lightning Lightning
At a height of about 40 [m], a streamer meets with the stepped leader, and a conducting channel is formed to earth. +++ ___ +++ ___ +++ ___ + + + + + + + + +++++++++++++++++++++++++++++++ Earth Lightning Lightning
A powerful surge of current (“return stroke”) flows upward from the ground to the cloud. This is what is visible as the lightning bolt. The current surge travels upward at about 25% the speed of light. The charge in the branches is drained as the surge passes by. The average current is typically 10 to 100 [kA]. The return stroke lasts about 0.1 [ms] to 1 [ms]. +++ ___ +++ ___ +++ ___  current surge +++++++++++++++++++++++++++++++ Earth Lightning Lightning main channel branches The current stops after about 50 [ms]. A new leader, called the “dart leader” descends from the cloud along the previous path. The dart leader travels at about 2000 [km/s] (about 10 times faster than the stepped leader) without pausing. It carries about 1 [C]. Another return stroke occurs when the dart leader reaches the ground. Lightning Lightning
+++ ___ +++ ___ +++ ___ +++++++++++++++++++++++++++++++ Earth Lightning Lightning
The dartleader / returnstroke combination repeats typically 3 or 4 times. The entire lighting flask lasts typically 0.2 [s] (sometimes as long a 1 [s] or more). The total charge deposited on the ground is about 20 [C]. +++ ___ +++ ___ +++ ___ +++++++++++++++++++++++++++++++ Earth Lightning Rod Lightning
A lighting rod is a rod of metal that is well grounded, and ideally sharp at the end. It launches a good streamer (better than the surrounding points on the building) because of the high electric field near the tip. high E +++ ___ +++ ___ +++ ___ +++++++++++++++++++++++++++++++ Earth Lightning Safety Lightning
1. The best protection is to be inside of a closed metal structure such as an automobile (Faraday cage effect). If outside: • Make sure you are not the tallest object around. • Do not stand near the tallest object. • Do not carry metal poles or metal objects (golf clubs,etc.). 2. 3. If you must be in an open field, lie down in the lowest lying area, and cover your ears. Lightning Safety Lightning
The ground current near the strike may be very large. This causes an large electric field along the surface of the earth. It is possible to be electrocuted without being hit by the lightning. Best to keep your feet together! +++ ___ +++ ___ +++ ___ J =σE E=
V + E 1 σ J Earth Van de Graaff Generator Van
Electric field is high near sharp points: charges are free to jump on/off Faraday cage effect: no matter how much charge is placed on the dome, it goes to the outside and there is little field on the inside Example (classroom demo) Example
Van de Graaf: Find maximum V0 Qmax + + + + + + + + a Qmax = Ec 2 4π ε 0 a Qmax = 4π ε 0 a 2 Ec
r =a Er = Ec Qmax 4π ε 0 a 2 Ec V0 = = 4π ε 0 a 4π ε 0 a V0 = a Ec Example (cont.) Example
Assume [ m] EC = 3 × 106 [ V/m ] V0 = 300, 000 [ V ]
a = 0.1
V0 = Ec h 3 × 10 V0 = h= Ec 3 × 106 h = 0.1
5 V0 Maximum spark length: [ V/m]
h [V] [ m] (3.9 inches) ECE 2317: Applied Electricity and Magnetism
Fall 2008
Prof. Jeffery T. Williams Dept. of Electrical & Computer Engineering University of Houston. Boundary Conditions
Notes prepared by the EM group, University of Houston. Boundary Conditions for Dielectrics Boundary
z ε1 ε2 Tangential Component: A B x C ∫ E⋅d =0
only at the boundary ⇒ Ex1∆x z =0+ − Ex 2 ∆x z =0− = 0
Hence Ex1 = Ex 2 In general, Et 1 = Et 2 Boundary Conditions for Dielectrics Boundary
z ε1 ε2 A Ex1 Ex2 B x only at the boundary Tangential Component: z = 0+ : VAB ≈ Ex1∆x z = 0− : VAB ≈ Ex 2 ∆x Thus, Φ1 = Φ 2 Boundary Conditions for Dielectrics Boundary
Normal Component:
ρs
z Ez1
++++++++++ + ε1 ε2 Ez2 z Gauss’ Law: ∆S
++++++++++ + S (“pill box”) ˆ ∫ D ⋅ n ds = Q
S encl Boundary Conditions for Dielectrics Boundary
Dz1∆S − Dz 2 ∆S = ρ s ∆S Dz1 − Dz 2 = ρ s
In general, Dn1 − Dn 2 = ρ s ˆ n ⋅ ( D1 − D2 ) = ρ s Free charge density ˆ n
1 2 ˆ n points toward region 1 Boundary Conditions for Dielectrics Boundary
ˆ n ⋅ ( D1 − D2 ) = ρ s ˆ n ⋅ ( ( −ε1∇Φ1 ) − ( −ε 2∇Φ 2 ) ) = ρ s ∂Φ1 ∂Φ 2 −ε1 + ε2 = ρs ∂n ∂n
1 2 Free charge density ˆ n ˆ n points toward region 1 z Example Example
εtop = 2 ε0 + + + + + + + + + + + ++++ εbot = 3 ε0 E top ˆ ˆ ˆ = 2x − 3y + 4z ˆ ˆ ˆ Ebot = 2x − 3y − 3z
Find ρs
choose
1 2 ˆˆ n=z ˆ n (This establishes which region is region 1.) Example (cont.) Example (D ˆ ( D1 − D2 ) ⋅ n = ρ s
top The unit normal points towards region 1 −D bot ˆ )⋅z = ρ s Hence ρs = ε ε E
top 0r = ε 0 ( 2 )( 4 ) − ε 0 ( 3)( −3) ⎡C/m 2 ⎤ ⎣ ⎦ top z −ε ε bot 0r E bot z ρ s = 17ε 0 Example (cont.) Example
Alternative: pick ˆ ˆ n = −z
2 1 ˆ ˆ n = −z ˆ ( D1 − D2 ) ⋅ n = ρ s ˆ Dbot − Dtop ) ⋅ ( − z ) = ρ s ( (same result) Example Example
Draw flux lines going into a dielectric The electric field in the dielectric has the same horizontal component, but a smaller vertical component. q ε0 ε 0ε r Boundary Conditions for PEC Boundary
Region 2 is PEC: Et = 0
++++++ ˆ D ⋅ n = ρs PEC Et = 0
ˆ D ⋅ n = ρs
(The normal points outward.) Boundary Conditions for PEC Boundary PEC + + PEC + + + Electric lines of flux must end perpendicular to the boundary on a PEC. Summary Summary
Dielectric boundary PEC boundary Et 1 = Et 2 Et = 0 ˆ n ⋅ ( D1 − D2 ) = ρ s
ˆ n 1 +++++++
2 ˆ n ⋅ D = ρs
ˆ n
+++++++ PEC ε1 ε2 The normal points towards region 1. The normal points outward. ECE 2317: Applied Electricity and Magnetism
Fall 2008
Prof. Jeffery T. Williams Dept. of Electrical & Computer Engineering University of Houston. Boundary value problems with dielectrics
Notes prepared by the EM group, University of Houston.
Pierre Simon de Laplace Example Example
εr1 εr2
x=0 h1 h2 x=h V0 +  ∂ 2Φ ( x ) ∇ 2Φ = =0 2 ∂x Example (cont.)
εr1 εr2
x=0 h1 h2 x=h V0 +  Φ1 = c1 x + c2 Φ 2 = d1 x + d 2
x = 0: c2 = V0
d1h + d 2 = 0 Four unknowns, thus we need four boundary conditions. x = h: Example (cont.) Example
Hence: Φ 2 = d1 x − d1h = d1 ( x − h )
We need two more equations. (Use boundary conditions.)
Equivalent to continuous tangential E at the boundary Φ1 = c1 x + V0 BC #1 x = h1 : c1h1 + V0 = d1 ( h1 − h ) = − d1h2
or Φ1 = Φ 2 d1 = − 1 ( c1h1 + V0 ) h2 Example (cont.) Example
BC #2: Dx1 = Dx 2 ε r1 Ex1 = ε r 2 Ex 2
E = −∇Φ To calculate Ex , use Hence: ε r1 ( −c1 ) = ε r 2 ( −d1 ) Example (cont.) Example
Therefore εr2 εr2 c1 = d1 = ε r1 ε r1 ⎡1 ⎤ ⎢ − ( c1h1 + V0 ) ⎥ ⎢ h2 ⎥ ⎣ ⎦ or ⎛ h1 ε r 2 ⎞ V0 ε r 2 c1 ⎜1 + ⎟=− h2 ε r1 ⎝ h2 ε r1 ⎠ Hence: V0 ε r 2 − h2 ε r1 c1 = h1 ε r 2 1+ h2 ε r1 V0 − h2 d1 = h1 ε r 2 1+ h2 ε r1 Example (cont.)
V0
+  εr1 εr2 x=0 h1 h2 x=h Φ1 = −ε r 2V0 x + V0 h2ε r1 + h1ε r 2 −ε r1V0 Φ2 = ( x − h) h2ε r1 + h1ε r 2 ε r 2V0 ∂Φ1 ˆ ˆ E1 = − x= x h2ε r1 + h1ε r 2 ∂x ε r1V0 ∂Φ 2 ˆ ˆ E2 = − x= x h2ε r1 + h1ε r 2 ∂x ECE 2317: Applied Electricity and Magnetism
Fall 2008
Prof. Jeffery T. Williams Dept. of Electrical & Computer Engineering University of Houston. Capacitance
Notes prepared by the EM group, University of Houston. Capacitance Capacitance
A +Q
+ capacitor V
 εr
Q
B C [F] Q C= V [ F, Farad ] Note: the “A” conductor has the positive charge. Q = QA V = VAB 1 [ F] = 1 [ C/V ] Capacitance Capacitance
+Q
A ++ Q i
B i
+ ++ v (t) dQ i= dt d = ( Cv ) dt dv i=C dt Note: Q is the charge that flows from left to right, into plate A Capacitance Capacitance
So dv i (t ) = C dt
i i
+  i +  v (t) v (t) “passive sign convention” Example Example
A [m2] x εr Ex0 h [m] “parallel plate capacitor” Method #1 (start with E) Assume: Find C ˆ E = x Ex 0
bottom V = VAB = = Ex 0 h top E ⋅ d = ∫ Ex dx = ∫ Ex 0 dx
0 0 h h Example (cont.) Example
ρ sA
h [m] A [m2] x ˆ Q = ρ sA A = A D ⋅ n A ˆ = ε 0ε r A n A ⋅ E ˆ = ε 0ε r A x ⋅ E = ε 0ε r A E x = ε 0ε r A E x 0
C= Q ε 0ε r AEx 0 = V Ex 0 h +++++++++++ εr Hence ⎛ A⎞ εA C = ε 0ε r ⎜ ⎟ = ⎝h⎠ h [F ] Example (cont.) Example
Method #2 (start with ρs) +ρs0 x ρs0 ˆ D ⋅ n A = ρ sA = ρ s 0 ˆ ˆ ( x Dx ) ⋅ x = ρ s 0 Dx = ρ s 0 ρ sA = ρ s 0
Q = ρs0 A ε 0ε r E x = ρ s 0 ρs0 Ex = Ex 0 = ε 0ε r Example (cont.) Example
bottom V = VAB = = h Ex 0
Hence top ∫ E ⋅ d = ∫ Ex dx = ∫ Ex 0 dx
0 0 h h ⎛ ρs0 ⎞ V = h⎜ ⎟ ε 0ε r ⎠ ⎝
ρs0 A C= ⎛ ρs0 ⎞ h⎜ ⎟ ε 0ε r ⎠ ⎝ Therefore, ⎛ A⎞ εA C = ε 0ε r ⎜ ⎟ = ⎝h⎠ h [F ] Example Example
A [m2] εr = 6.0 (mica)
A = 100 [cm2] h = 0.1 [mm] εr h [m] ⎛ A⎞ C = ε 0ε r ⎜ ⎟ ⎝h⎠ ⎛ 100 × 10−4 ⎞ = ( 8.854 × 10−12 ) ( 6.0 ) ⎜ −3 ⎟ ⎝ 0.1× 10 ⎠ C = 0.005312 [ µ F ] = 5,312 [F ] [ pF ] Example Example
A Ex2 Ex1 εr2 εr1 D x0 h2 h1 B x Dx1 = Dx 2 = Dx 0
B. C. : ˆ D ⋅ n = ρs ˆ D2 ⋅ x = ρ sA Dx 2 = Dx 0 = ρ sA Q = ρ sA A = Dx 0 A Example (cont.) Example
V = Ex1h1 + Ex 2 h2 = Dx1 Dx 2 ε1 h1 + ε2 h2 Dx 0 ε1 h1 + Dx 0 ε2 h2 ⎡ h1 h2 ⎤ = Dx 0 ⎢ + ⎥ ⎣ ε1 ε 2 ⎦ Example (cont.) Example
Hence: So, C= 1 1 1 + C1 C2
C2 = Dx 0 A Q C= = V ⎛ h1 h2 ⎞ ⎜ + ⎟ Dx 0 ⎝ ε1 ε 2 ⎠ 1 A = = h1 h2 h1 h + +2 ε1 ε 2 ε1 A ε 2 A = 1 1 1 + ⎛ ε1 A ⎞ ⎛ ε 2 A ⎞ ⎜ ⎟⎜ ⎟ h1 ⎠ ⎝ h2 ⎠ ⎝ where C1 = ε1 A
h1 ε2 A
h2 11 1 =+ C C1 C2 Example Example
A1
++++++++++ A2 Dx1 εr1 ++++++ Ex0 εr2 Dx2 x x=h Ex1 = Ex 2 = Ex 0 V = Ex 0 h
ρ sA = Dx1 = ε1 Ex1 = ε1 Ex 0 1 ρ sA2 = Dx 2 = ε 2 Ex 2 = ε 2 Ex 0 Example (cont.) Example
Q = ρ s1 A1 + ρ s 2 A2 = ε1 Ex 0 A1 + ε 2 Ex 0 A2
Q ε1 A1 Ex 0 + ε 2 A2 Ex 0 C= = V Ex 0 h C= ε1 A1
h + ε 2 A2
h C = C1 + C2 Example Example
h = 1 [m] Find CL (capacitance / length) εr coaxial cable εr ⎛ ρ0 ⎞ ⎛ ρ0 ⎞ ˆ ˆ E=ρ ⎜ ⎟ ⎟ = ρ⎜ 2π ε ρ ⎠ 2π ε 0 ε r ρ ⎠ ⎝ ⎝
a b ρl0
ρl0 V = +∫ E ⋅ d
Α Β Q=ρ [C ] 0 (1m ) = ∫ Eρ d ρ
a b Example (cont.) Example
or ρ0 b1 V= ∫ ρ dρ 2π ε 0 ε r a ρ0 ⎛b⎞ ln ⎜ ⎟ = 2π ε 0 ε r ⎝ a ⎠
ρ 0 (1) Q CL = = V ⎛ ρ 0 ⎞ ⎛b⎞ ⎜ ⎟ ln ⎜ ⎟ ⎝ 2π ε 0 ε r ⎠ ⎝ a ⎠ Hence So 2π ε 0 ε r CL = ⎛b⎞ ln ⎜ ⎟ ⎝a⎠ [F/m] Example Example
h = 1 [m] Twolayer coax ε2 ε1 ε2
a c b Find CL ρl0
ρl0 ρ = a: ρ = c: ρ
−ρ 0 0 [C/m] [C/m] ε1 Example (cont.) Example
Gauss’s Law: ˆ ∫ D ⋅ nds = Q
s encl 2πρ h Dρ = ρ 0 h S
a c b ρ0 Dρ = 2πρ
ρl0
ρl0 ρ0 Eρ 1 = 2πε1 ρ ρ0 Eρ 2 = 2πε 2 ρ Example (cont.) Example
Q=ρ
0 (1m ) [ C]
c ε2
a c b ρl0
ρl0 V = ∫ E ⋅ d = ∫ Eρ d ρ
Α b a Β ε1 = ∫ Eρ 1 d ρ + ∫ Eρ 2 d ρ
a b c ρ0 ρ0 =∫ dρ + ∫ dρ 2π ε1 ρ 2π ε 2 ρ a b b c ρ 0 ⎛b⎞ ρ 0 ⎛c⎞ = ln ⎜ ⎟ + ln ⎜ ⎟ 2π ε1 ⎝ a ⎠ 2π ε 2 ⎝ b ⎠ Example (cont.) Example
CL = Q = V ρ 0 (1) ⎡1 1 ⎛b⎞ ⎛ c ⎞⎤ ρ 0⎢ ln ⎜ ⎟ + ln ⎜ ⎟ 2πε1 ⎝ a ⎠ 2πε 2 ⎝ b ⎠ ⎥ ⎣ ⎦ = 2π 1 ⎛b⎞ 1 ⎛c⎞ ln ⎜ ⎟ + ln ⎜ ⎟ ε1 ⎝ a ⎠ ε 2 ⎝ b ⎠ 1 1 1 1 1 ⎛b⎞ ⎛c⎞ = + ln ⎜ ⎟ + ln ⎜ ⎟ = CL 2πε1 ⎝ a ⎠ 2πε 2 ⎝ b ⎠ ⎛ ⎞⎛ ⎞ ⎜ 2πε ⎟ ⎜ 2πε ⎟ 1 2 ⎜ ⎟⎜ ⎟ b⎞⎟ ⎜ ⎛c⎞⎟ ⎜ ln ⎛ ⎜ ⎜ a ⎟ ⎟ ⎜ ln ⎜ b ⎟ ⎟ ⎝ ⎝ ⎠⎠ ⎝ ⎝ ⎠⎠ 1 1 1 = + CL CL1 CL 2 ε2
a c b ρl0
ρl0 2πε1 CL1 = ⎛b⎞ ln ⎜ ⎟ ⎝a⎠ CL 2 = 2πε 2 ⎛c⎞ ln ⎜ ⎟ ⎝b⎠ ε1 Stored Energy Stored
1 U E = ∫ ρ v Φ dV 2V ( Φ ( ∞ ) = 0) 1 1 2 U E = ∫ D ⋅ E dV = ∫ ε E dV 2V 2V Note: Please see the text book or supplementary notes for a derivation. Example Example
A
V0 +  [m2] h x εr Method #1 1 U E = ∫ D ⋅ E dV 2V ⎛V ⎞ ˆ E = x⎜ 0 ⎟ ⎝h⎠ ⎛V ⎞ ˆ D = ε 0ε r x ⎜ 0 ⎟ ⎝h⎠ 1 ⎛V ⎞ U E = ∫ ε 0ε r ⎜ 0 ⎟ dV 2V ⎝h⎠
2 Example Example
1 ⎛ V0 ⎞ U E = ε 0ε r ⎜ ⎟ 2 ⎝h⎠
2 ( Ah ) [ J ] Or, 1⎛ A⎞ 2 U E = ⎜ ε 0ε r ⎟ V0 2⎝ h⎠ Recall that A C = ε 0ε r h 1 U E = CV02 2 Hence Example (cont.) Example
Method #2 A [m2]
A +++++++++++++ B  εr h x 1 U E = ∫ ρv Φ dV 2V 1 = ∫ ρ s Φ dS 2S
(since we have surface charge in this problem) 1 1 = Φ A ∫ ρ s dS + Φ B ∫ ρ s dS 2 2 SA SB Example (cont.) Example
1 1 U E = Φ AQA + Φ B QB 2 2 1 1 = QA Φ A − QA Φ B 2 2 1 = Q (Φ A − ΦB ) 2 1 U E = QV0 2
1 U E = QV0 2 1 = ( CV0 ) V0 2 1 U E = CV02 2
A [m2]
A +++++++++++++ B  εr h x Example Example
a ε0 ρv0 [C/m3] Method #1 1 U E = ∫ D ⋅ E dV 2V Gauss’s Law: r<a r>a ⎛ ⎛4 ⎞⎞ ρv 0 ⎜ π r 3 ⎟ ⎟ ⎜ ⎝3 ⎠⎟ ˆ E = r⎜ 2 ⎜ 4πε 0 r ⎟ ⎜ ⎟ ⎝ ⎠ 4 3⎞ ⎛ ⎜ ρv 0 3 π a ⎟ ˆ E = r⎜ ⎟ 4πε 0 r 2 ⎟ ⎜ ⎜ ⎟ ⎝ ⎠ Example (cont.) Example
a UE = = = ε0
2 V ∫ ∫ E dV =
2 ε0
2 V Er2 dV ∫ ε0 ρv0 [C/m3] ε0
2 2π π ∞ 0 00 Er2 ( r ) r 2 sin θ dr dθ dφ ∫∫
∞ ε0
2 ( 2π )( 2 ) ∫ Er2 ( r ) r 2 dr
0 2 2 4 3⎞ 4 3⎞ ⎛ ⎛ πr ⎟ πa ⎟ a ρv 0 ∞ ρv 0 ⎜ ⎜ 2 2 3 3 U E = 2πε 0 ∫ ⎜ ⎟ r dr + 2πε 0 ∫ ⎜ ⎟ r dr 4πε 0 r 2 ⎟ 4πε 0 r 2 ⎟ a⎜ 0⎜ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ Example (cont.) Example
Result: ρv20 ⎛ 4π ⎞ 5 UE = ⎜ ⎟a ε 0 ⎝ 15 ⎠
⎛4 3⎞ Q = ρv 0 ⎜ π a ⎟ ⎝3 ⎠ ⎛3⎞ ρv 0 = Q ⎜ ⎟ 4π a 3 ⎠ ⎝
Note: Let: 1⎛Q UE = ⎜ a ⎝ ε0 2 ⎞⎛ 3 ⎞ ⎟⎜ ⎟ 20π ⎠ ⎠⎝ [J] UE → ∞ as a→0 ECE 2317: Applied Electricity and Magnetism
Fall 2008
Prof. Jeffery T. Williams Dept. of Electrical & Computer Engineering University of Houston. Image Theory
Notes prepared by the EM group, University of Houston. Uniqueness Theorem Uniqueness
S
ρv ( x, y, z ) (known) Φ = ΦB
Given: On boundary ρv ∇ Φ=− ε
2 Inside (correct D.E.) (correct B.C.) Φ = ΦB
Φ ( x, y , z ) On boundary Is unique Note: We can guess the solution, as long as we verify that the Poisson equation and the BC’s are correctly satisfied ! Example Example
PEC shell ε0 ρv = 0 V S ΦB
Φ ( x, y, z ) = V0 = V0 = constant Prove that E = 0 inside a hollow PEC shell (Faraday cage effect) Guess: Check: ( r ∈V ) ∇ 2 Φ = ∇ 2 (V0 ) = 0 in V Φ = V0 on S Therefore: Φ ( x, y, z ) = V0 is the correct solution. (Hence E = 0 ) Image Theory Image
z Φ(x,y,z) q h x PEC Image Theory Image
Image picture: z R1
Note: there is no ground plane in the image picture! r #1 h q R2 x h #2 z > 0: q Φ = Φ1 + Φ 2 −q = + 4πε 0 R1 4πε 0 R2 q Image Theory Image
z > 0: #1 V q Sh S = S0+ Sh x S0 #2 q r ∈V : ρv ρv1 2 ∇ Φ=− =− ε ε0 (correct D. E. in V) r ∈ S0 : Φ=0 since R1 = R2 since r =∞ r ∈ Sh : Φ = 0 (correct B. C. on S) Image Theory Image
z < 0: #1 q x V #2 q S ρv 2 ∇ Φ=− ε0
2 Wrong ! Incorrect charge in V ! Image Theory Image
z > 0: Φ= q 4πε 0 x + y + ( z − h )
2 2 2 + −q 4πε 0 x + y + ( z + h )
2 2 2 z < 0: z R1 #1 h q R2 r Φ=0 x h #2 q Example: HighVoltage Power Line Example:
V0 Highvoltage power line (radius a) h earth R Example: HighVoltage Power Line Example:
Linecharge approximation: ρ
h 0 earth R Example: HighVoltage Power Line
Image theory: ρ
h h Note: the earth is being modeled as a PEC.
0 −ρ 0 Note: the linecharge density can be found by forcing the voltage to be V0 at the radius of the wire. Example: HighVoltage Power Line
Image theory: ρ
h h A B 0 Set VAB = V0 −ρ 0 The point A may be selected to be the point at distance a below the line charge (the center of the power line) Linecharge formula: ⎛b⎞ ρ Φ= ln ⎜ ⎟ 2πε 0 ⎝ ρ ⎠ Note: b is the distance to the arbitrary reference point for the single linecharge solution. Example: HighVoltage Power Line Example:
Image theory: ρ
h h A B 0 Set VAB = V0 −ρ
ρ 0 ⎛ b ⎞ −ρ 0 ⎛ b ⎞ ln ⎜ ⎟ + ln ⎜ Φ ( A) = ⎟ 2πε 0 ⎝ a ⎠ 2πε 0 ⎝ 2h − a ⎠ ρ 0 ⎛ b ⎞ −ρ 0 ⎛ b ⎞ ln ⎜ ⎟ + ln ⎜ ⎟ = 0 Φ ( B) = 2πε 0 ⎝ h ⎠ 2πε 0 ⎝ h ⎠ 0 VAB = ρ 0 ⎛ 2h − a ⎞ ln ⎜ ⎟ 2πε 0 ⎝ a ⎠ Example: HighVoltage Power Line Example:
Image theory: ρ
h h 0 −ρ 0 Hence: 2πε 0 V0 2πε 0 V0 ≈ ρ0= 2h − a ⎞ ⎛ ⎛ 2h ⎞ ln ⎜ ⎟ ln ⎜ ⎟ ⎝a⎠ ⎝a⎠ Image Theory for Dielectric Region
Point charge over semiinfinite dielectric region: q h εr Image Theory for Dielectric Region
Solution for air region: q h h q´ ε0 ε0 q ' = −q εr −1 εr +1 Image Theory for Dielectric Region
Solution for dielectric region: q´´ h ε 0ε r ε 0ε r 2ε r q '' = q εr +1 Image Theory for Dielectric Region
Flux Plot
Note: the flux lines are straight lines in the dielectric region. q ε0 ε 0ε r ECE 2317: Applied Electricity and Magnetism
Fall 2008
Prof. Jeffery T. Williams Dept. of Electrical & Computer Engineering University of Houston. DC currents
Notes prepared by the EM group, University of Houston. DC Currents and Conductivity DC
Ohm’s Law
DC – Direct Currents J =σE
ChargeCurrent Formula ⎛∂ ⎞ = 0⎟ ⎜ ⎝ ∂t ⎠ Static (timeinvariant) form of Faraday’s Law still applies: ∇×E = 0 ⇒ E = −∇Φ J = ρv v Conductivities of Common Materials Conductivities
Material Silver Copper Gold Aluminum Nickel Seawater Germanium Silicon Glass Petroleum Oil Amber Quartz Conductivity [S/m] 6.17 x 107 5.8 x 107 4.1 x 107 3.53 x 107 1.38 x 107 4.0 2.2 1.56 x 103 1010  1014 1014 2.0x 1015 1.30 x 1018 Continuity Equation for DC Currents Continuity J
In general, ⎛∂ ⎞ = 0⎟ ⎜ ⎝ ∂t ⎠ ˆ n I out = ∫ S ˆ J ⋅ n dS
S ∂Q =− ∂t
For DC currents, I out = ∫ S ˆ J ⋅ n dS = 0 ∇⋅J = 0 Continuity Equation for DC Currents Continuity
I out = ∫ S ˆ J ⋅ n dS = 0 ⎛∂ ⎞ = 0⎟ ⎜ ⎝ ∂t ⎠ ˆ n S Kirchoff’s Current Law (KCL) Kirchoff
I out = ∫ S ˆ J ⋅ n dS = 0
i1 iN ∑i
n =1 N n =0
i2 i3 i4 tot iout = 0 S can be a “node” or S can be any closed surface. Electrostatics & DC Currents Electrostatics
Electrostatics DC Currents ∇×E = 0 ∇⋅D = 0 D = εE
E = −∇Φ ∇2Φ = 0
• Electric field is the same. ∇×E = 0 ∇⋅J = 0 J =σE •Assuming dielectric and conducting regions are the same shape, – Laplace’s equation applies in each region. – Same boundary conditions must be satisfied. Example: Electrostatic Capacitor Example:
+ h A V ε x V ∇ Φ = 0 ⇒ Φ = (h − x) h V εV ˆ E = −∇Φ = x ˆ dS = Q = ∫ ρ S dS = ∫ D ⋅ n A h S S h εV Q εA ˆ D = εE = x C= = h V h
2 Example: DC Current Resistor Example:
I
+ h A V σ x V ∇ Φ = 0 ⇒ Φ = (h − x) I = J ⋅ n dS = σ V A h ∫S ˆ h V 1 I σA ˆ E = −∇Φ = x [ S ,Siemens] G= = = h h RV σV h ˆ J =σE = x [Ω, Ohms] R= h σA
2 Electrostatic – DC Current Analogy
Electrostatics DC Currents E Φ → → → → →
→ → → E Φ ε
D= εE Q Q C= V
VAB = ∫ E ⋅ d
Α Β σ J =σE
I
I G= V VAB = ∫ E ⋅ d
Α Β Q= SA ˆ ∫ D ⋅ n ds I= SA ˆ ∫ J ⋅ n ds Example: TwoLayer Resistor Example:
I V + h2 h1 A σ2 σ1 x Find R Recall for twolayer capacitor
h2 h1 ε2 ε1 11 1 =+ C C1 C2
Where, C1 = ε1 A
h1 C2 = ε2 A
h2 Example: TwoLayer Resistor “cont.” Example:
I V + h2 h1 A σ2 σ1 x Thus, using analogy 1 1 1 + = R1 + R2 R= = G G1 G2
Where, 1 = G1 = R1 h1 σ1 A G2 = σ2A
h2 1 = R2 Joule’s Law Joule
A
++ ∆V + + B v ρv ++ ++ conducting body E
∆W = work (energy) supplied by E to a small volume of charge as it moves inside the conductor from point A to point B. This goes to heat! ∆W = ∆Q VAB ≈ ( ρ v ∆V ) ∫ E ⋅ d ≈ ( ρ v ∆V ) E ⋅ ∆ = ∆V ( ρ v ∆ ⎛ ∆⎞ ⋅ E = ∆V ⎜ ρ v ) ⎟ ⋅ E ∆t ⎝ ∆t ⎠
A B Joule’s Law Joule
⎛ ∆⎞ ∆W = ∆V ⎜ ρv ⋅ E ∆t = ∆ V ( ρ v v ) ⋅ E ∆ t ⎝ ∆t ⎠ ∆W = ∆V ρv v ⋅ E Power = P = ∆t = ∆V J ⋅ E
Power Density: Pd = J ⋅ E = σ E =
2 J 2 σ [W/m3 ] Dissipated Power : P = J ⋅ E dV [ W ]
V ∫ Joule’s Law Joule
I
V Resistor x +  h σ A P = ∫ J ⋅ E dV = J x Ex (A L)
V [W]
Hence, ⎛ I ⎞⎛V ⎞ = ⎜ ⎟ ⎜ ⎟ ( A L) ⎝ A ⎠⎝ L ⎠ = IV V P = VI = I R = R
2 2 Note: passive sign convention appears in the final result. ECE 2317: Applied Electricity and Magnetism
Fall 2008
Prof. Jeffery T. Williams Dept. of Electrical & Computer Engineering University of Houston. Laws of Magnetostatics Notes prepared by the EM group, University of Houston. Magnetic Field Magnetic
v N
εr q S
This experimental law defines the magnetic flux density vector B. Lorenz Force Law: F = q v×B In general, (with both E and B present): F = q (E + v × B) The units of B are Webers/m2 [Wb/m2] or Tesla [T]. Magnetic Gauss Law Magnetic
ˆ ∫ B ⋅ n ds = 0
S This is an experimental law.
z S (closed surface) B
N N
y x S B
Magnetic monopole (not possible) ! No net flux out ! Magnetic Gauss Law: Differential Form Magnetic
ˆ ∫ B ⋅ n dS = 0
S Apply the divergence theorem: V ∫ ( ∇ ⋅ B ) dV = 0 Since this applies for any volume, ∇⋅B = 0 Magnetic Field Magnetic
Define: H= 1 µ B=µH B H is called the “magnetic field,” [Amperes/m, A/m] B is the “magnetic flux density,” [Wb/m2, T] µ is the “magnetic permeability,” [Henries/m, H/m]
Freespace permeability µ = µ r µ0
Relative permeability µ0 = 4π ×10−7 [H/m] Ampere’s Law Ampere
C ∫ H⋅d = I encl = ˆ J ⋅ n ds ∫
S This is an experimental law. Iencl ˆ n S C
S and C are related by “RightHand Rule” Amperes’ Law: Differential Form Amperes
C ∫ H⋅d ˆ = I encl = ∫ J ⋅ n dS
S Using Stokes’ theorem ˆ ˆ ∫ (∇ × H) ⋅ n dS = ∫ J ⋅ n dS
S S Since this applies for any surface, ∇×H = J Maxwell’s Equations (Statics) Maxwell
∇ ⋅ D = ρv
Electrostatics electric Gauss’ law ∇×E = 0
Faraday’s law ∇⋅B = 0
Magnetostatics magnetic Gauss’ law ∇×H = J
Ampere’s law Constitutive Relations B=µH D=ε E Recall from Electrostatics Recall
∇ ⋅ D = ρv
Scalar potential ∇×E = 0 ρV ∇ Φ=− ε
2 D=ε E
in an infinite, homogeneous medium is Poisson’s Equation E from the solution of Poisson’s Equation (E = −∇Φ ) ρV ( r ') ρ S ( r ') ρl ( r ') E (r ) = ∫ R dV ' = ∫ R dS ' = ∫ R dl ' 2 2 2 4πε R 4πε R 4πε R V S C
Generalized Coulomb’s Equation Magnetostatics Magnetostatics
∇⋅B = 0 ∇×H = J ∇2 A = −µ J
Details are in the supplementary notes Magnetic vector potential B=µH H from the solution of this equation
in an infinite, homogeneous medium is H (r ) = ∫
V J ( r ') × R 4π R
2 dV ' = ∫
S J S ( r ') × R 4π R 2 Iˆ ×R dS ' = ∫ d′ 2 4π R C
BiotSavart Law BiotSavart Law Biot
H (r ) = ∫
V J ( r ') × R 4π R
2 dV ' dS ' z (x,y,z) R
dV´ =∫
S J S ( r ') × R 4π R 2 r r′ J
y Iˆ ×R =∫ d′ 2 4π R C
x R = r  r′ ˆ in direction of I
J dV ' = ( J dS ′ ) d ′ = I ˆ d ′ () Most Commonly used form of the Most BiotSavart Law
Iˆ ×R H (r ) = ∫ d′ 2 4π R C I
C dl´ ˆ′
R r′ r (x,y,z) Example: Infinite line of current Example:
z Find H
r´ R d′
I Iˆ ×R H (r ) = ∫ d′ 2 4π R C
ˆ d ′ = z dz ′ R = r − r′ ˆˆ = ρ ρ − z z′ R= ˆ ˆ ρ ρ − z′ z ρ
x y r ρ 2 + z ′2 R = ρ 2 + z ′2 Example: Infinite line of current ‘cont.d’ Example:
z Iˆ ×R H (r ) = ∫ d′ 2 4π R C
r´ R d′
I ⎡ ρ ρ − z′ z ˆ ˆ ˆ I z×⎢ ⎢ ρ 2 + z ′2 ⎣ H=∫ 4π ( ρ 2 + z ′2 ) C
y ⎤ ⎥ ⎥ ⎦ dz ′ ρ
x r I = 4π +∞ −∞ ∫ ˆ ρ φdz ′ (ρ 2 + z′ 2 ) 3 2 +∞ Iρ dz ′ ˆ∫ φ = 4π −∞ ρ 2 + z ′2 3 2 ( ) Example: Infinite line of current ‘cont.d’ Example:
z
+∞ Iρ dz ′ ˆ∫ H= φ 4π −∞ ρ 2 + z ′2 3 2 ( ) d′ r
´ I R ρ
x y Iρ z′ ˆ φ = 4π ρ 2 z ′2 + ρ 2 2 Iρ ˆ( 2) φ = 4π ρ ∞ =
−∞ Iρ ˆ φ 4π z′ z′ ∞ r ρ 2 1+ ρ2
z ′2
−∞ ⎛I⎞ ˆ H = φ⎜ ⎟ 2πρ ⎠ ⎝ [ A/m] Ampere’s Law Ampere
C ∫ H⋅d = = C ˆ ˆ ˆ ˆ ∫ ( φ Hφ ) ⋅ ( φ ρ dφ + ρ d ρ + z dz ) y C C ∫ H φ ρ dφ I
x ⎛I⎞ = ∫⎜ ⎟ ρ dφ 2πρ ⎠ 0⎝ = I 2π
2π 2π ∫ dφ =
0 I ( 2π ) 2π =I Ampere’s Law Ampere
y C I
x I ∫ H ⋅ d = ∫ 2π dφ C 0 =0 0 In this case the current is outside the closed path. Definition of 1 Amp of Current Definition
By definition of one Amp of current is such that in air: Bφ = 2 × 10−7
I [T]
Where, at ρ = 1 [ m]
I 2πρ ρ Bφ = µ0 Hφ = µ0 Example: Ring of current Example:
z H=
R r = (0,0,z) C ∫ Iˆ × R d′ 2 4π R a y ˆ = φ' ˆ d ′ = a dφ ′ ˆ ˆ R = z z + ρ ' ( −a ) R= ˆ ˆ z z − ρ'a z 2 + a2 ˆ ˆ ˆ aφ '× ( z z − ρ ' a ) z2 + a2 ) (
3 2 r´
x d′ I Find H R = z2 + a2 I H= 4π C ∫ dφ ′ Example: Ring of current Example:
z r= (0,0,z) Ia H= 4π
R 2π ∫
0 ˆ ˆ ρ ' z − a ( −z ) (z
Ia
2 2 + a2 ) 3 dφ ′ 2 =
a y 4π ( z + a Ia 2 ) 3 2 2π ⎡ 2π ⎤ ˆ ' dφ ′ + a ∫ z dφ ′ ⎥ ˆ ⎢z ∫ ρ 0 ⎣0 ⎦ H= r´
x d′ 4π ( z 2 + a 2 ) 3 ˆ ( a z 2π )
2 I ˆ H=z Ia 2 2 ( z 2 + a2 )
3 2 [ A/m] Example: Infinite sheet of current Example:
ˆ J S = z J sz [ A/m] z Find H y x Infinite sheet of current x top view y Example: Infinite sheet of current ‘cont.d’
ˆ JS = z JS z 0 [ A/m] z J S (r′) × R dS ′ H (r ) = ∫ 2 4π R S
y x ⎛ ˆ ˆ ˆ⎞ R yy − x′x − z ′z ⎟ ˆ J S × 2 = J S z0z × ⎜ ⎜ ⎡ y 2 + x′2 + z ′2 ⎤ 3/ 2 ⎟ R ⎦⎠ ⎝⎣ ˆ ˆ J S z 0 (− yx + x′y ) = 3/ 2 ⎡ y 2 + x′2 + z ′2 ⎤ ⎣ ⎦ dS ′ = dx′ dz ′ R = r − r′ ˆ ˆ ˆ = yy − ( x′x + z ′z ) ˆ ˆ ˆ yy − x′x − z ′z R= y 2 + x′ 2 + z ′ 2 R= y 2 + x′2 + z ′2 ˆ ˆ ˆ R R yy − x′x − z ′z = 3= 2 2 2 2 3/ 2 R R ⎡ y + x′ + z ′ ⎤ ⎣ ⎦ Example: Infinite sheet of current ‘cont.d’ Example:
ˆ JS = z JS z0 [ A/m] z J S (r′) × R dS ′ H (r ) = ∫ 2 4π R S
∞∞ =
y x y ⎣ + x′ + z ′ ⎤ ⎦ ˆ J S z 0 (− yx) ∞ ∞ 1 = ∫ −∞ ⎡ y 2 + x′2 + z′2 ⎤3/ 2 dx′dz′ ∫ 4π −∞ ⎣ ⎦ ˆ J S z 0 (− yx) ∞ 2dz ′ = ∫ ( y 2 + z ′2 ) 4π −∞
−∞ −∞ 2 2 ∞ ⎞ ˆ) ⎛ 2 J S z 0 (− yx −1 ⎛ z ′ ⎞ ⎜ tan ⎜ ⎟ ⎟ = ⎜y 4π ⎝ y ⎠ −∞ ⎟ ⎝ ⎠ ∫ ∫ 4π ⎡ y ˆ ˆ J S z 0 (− yx + x′y )
2 3/ 2 dx′dz ′ ⎧ JS z0 ˆ , y>0 ⎪−x ⎪ 2 H=⎨ ⎪ x J S z0 , y < 0 ˆ ⎪ 2 ⎩ Example: ParallelSheets of Current Example:
Find H everywhere
y w h I I
z x J bot s J top s ⎛I⎞ ˆ = z ⎜ ⎟ [ A/m ] ⎝ w⎠ ⎛ −I ⎞ ˆ = z ⎜ ⎟ [ A/m ] ⎝w⎠ Example: ParallelSheets of Current w
y Using superposition, h I I
J
bot sz x H=H bot +H top z
bot sz ⎧ bot ⎡ ⎛ J ⎞⎤ ˆ ⎪ 2H = 2 ⎢ x ⎜ − ⎟⎥ , 0 < y < h H=⎨ ⎣ ⎝ 2 ⎠⎦ ⎪ , otherwise ⎩0 ⎛I⎞ =⎜ ⎟ ⎝ w⎠ [ A/m ] ⎧ ⎛I⎞ ˆ ⎪− x ⎜ ⎟ [A/m], 0 < y < h H = ⎨ ⎝ w⎠ ⎪0 , otherwise ⎩ Using Ampere’s Law to Find H Using
For problems that have high degree of symmetry and the direction of the magnetic field can be determined, Ampere’s Law can be used by itself to find H C ∫ H⋅d = I encl 1) The “Amperian path” C must be a closed path. 2) The sign of Iencl is from the RH Rule. 3) Pick C in the direction of H (as much as possible). Example: Infinite Line of Current Example:
Calculate H
An infinite line current along the z axis. z Assume that H is in the φ direction I ˆ H = φ Hφ
r
x ρ
C
“Amperian path” y Example (cont.) Example
ˆ ˆ ∫ ( φ H φ ) ⋅ ( φ ρ dφ ) = I
2π C ∫ H⋅d = I encl
encl z I =I
ρ
C r
x y C ∫
0 H φ ρ dφ = I Hφ ρ ( 2π ) = I
Hφ = I 2πρ ⎛I⎞ ˆ⎜ H=φ ⎟ 2πρ ⎠ ⎝ [ A/m] Example (cont.) Example
2) Hz = 0 Hρ dρ cancels Hz dz = 0 To prove the other components are zero I
ρ h C
ρ=∞ −H z ( ρ ) h = 0 C ∫ H⋅d = I encl = 0 3) Hρ = 0 I Magnetic Gauss law: h ˆ ∫ B ⋅ n dS = 0 Bρ ( 2πρ h ) = 0
S
S Example: Coaxial Line Example:
y coaxial cable I
a b I a C r
x z
b
The outer jacket of the coax has a thickness of t = cb c
ρ<a Note: the permittivity of the material inside the coax does not matter here. I Jz = 2 πa ⎡ A/m 2 ⎤ ⎣ ⎦ b<ρ<c I Jz = − 2 π c − π b2 ⎡ A/m 2 ⎤ ⎣ ⎦ Example (cont.) Example
ˆ H = φ Hφ
y C ∫ H⋅d
φ
2π = I encl
a C
encl b C ˆ ˆ ∫ ( φ H ) ⋅ ( φ ρ dφ ) = I r x ρ ∫ Hφ dφ = I encl
0 c I encl Hφ = 2πρ This formula holds for any radius, as long as we get the Iencl correct. Example (cont.) Example
y ρ < a:
a <ρ < b: b <ρ < c: ⎛I⎞ I encl = J zA (πρ 2 ) = ⎜ 2 ⎟ πρ 2 ⎝πa ⎠ b a I encl = I I encl = I + J B z (πρ 2 −πb 2 ) C r x I =I− 2 πρ 2 − π b 2 ) 2( πc −πb
ρ > c: c I encl = I + ( − I ) = 0 I encl Hφ = 2πρ Note: There is no magnetic field outside of the coax (“shielding property”). Example: Solenoid Example:
Solenoid n = # turns/meter
a z µ = µ0 µ r
I Calculate Hz
ρ<a Hz C ∫ H⋅d = I encl H z h = I encl = I ( nh ) H z = nI
ˆ H = z ( nI ) C h ∞ [ A/m] Example (cont.) Example
ρ>a Hz C I encl = 0 Hzh = 0 Hz = 0 h ∞ ⎧z ( nI ) ⎪ˆ H=⎨ ⎪0 ⎩ [ A/m] , ρ<a , ρ >a B = µ0 µ r H , ρ < a Example (cont.) Example
The other components of the magnetic field are zero: 1) Hφ = 0 since I encl = 0 C Hφ 2πρ = I encl 2) Hρ = 0 from ∫ B ⋅ n ds = 0
S S Bρ 2πρ h = 0 Example: Infinite sheet of current Example:
ˆ J S = z J sz [ A/m] z Find H y x Infinite sheet of current x top view y Example (cont.) Example
Hxx Hx+ y By superposition: By symmetry: ˆ H = x Hx H x ( − y ) = −H x ( + y ) Hz = 0 Hy = 0 Example (cont.) Example
w x C
y + C ∫ H ⋅ d = I encl
bot Note: No contribution from the left and right edges (the edges are perpendicular to the field).
−w ∫ H⋅d = w w ∫
2 2 H x dx = − H x+ w 2 top ∫ H⋅d = −w ∫ H x dx = H x− w 2 Example (cont.)
− H x+ w + H x− w = J sz w − H x+ − H x+ = J sz 1 H = − J sz 2
+ x ⎧ ⎛ J sz ˆ ⎪x ⎜ − 2 ⎪⎝ H=⎨ ⎪x ⎛ + J sz ˆ⎜ ⎪⎝ 2 ⎩ ⎞ ⎟, ⎠ ⎞ ⎟, ⎠ y>0 y<0 Note: The magnetic field does not depend on y. ECE 2317: Applied Electricity and Magnetism
Fall 2008
Prof. Jeffery T. Williams Dept. of Electrical & Computer Engineering University of Houston. Magnetic Energy and Inductance
Notes prepared by the EM group, University of Houston. Magnetic Stored Energy Magnetic
UH = ∫
V 1 B ⋅ H dV 2 B = µH = µ0 µ r H
UH = ∫
V 1 2 µo µr H dV 2 Example Example
Ls
a z µ = µ0 µ r
I N turns 1 2 U H = ∫ µ0 µr ( nI ) dV 2 V
2⎤ ⎡1 = (π a Ls ) ⎢ µ0 µr ( nI ) ⎥ ⎣2 ⎦ 2 ˆ H = z ( nI ) , ρ < a = 0, ρ > a
N n= Ls 12 2⎛ N ⎞ = π a Ls µ0 µr I ⎜ ⎟ 2 ⎝ Ls ⎠ 1 2⎛ N ⎞ U H = µ0 µ rπ a ⎜ ⎟ I 2 2 ⎝ Ls ⎠
2 2 [J] Inductance Inductance
ˆ n
S I ˆ ψ = ∫ B ⋅ n dS
S [ Wb] ˆ n “right hand rule” based on I L≡ ψ
I [H ] NTurn Solenoid N Λ = total flux = Nψ ψ = flux through one turn
Λ Nψ L≡ = I I Ls
a Example Example
µ = µ0 µ r z I N turns I ψ = (π a 2 ) Bz = (π a 2 ) µ0 µ r H z = (π a 2 ) µ0 µr ( nI ) L = N (π a 2 ) µ0 µ r n N2 L= (π a 2 ) µ0 µ r Ls L=N ψ Find L [H] Example Example
Toroid N turns I µr µr
a Assume A = π a2 ˆ H = φ Hφ Example Example
C ∫ H⋅d = I encl
encl C C ˆ ∫ H ⋅ φ ρ dφ = I
2π ∫ H φ ρ dφ = I
0 encl N turns I
µr I encl Hφ = 2πρ I encl = − NI
⎛ − NI ⎞ ˆ⎜ H=φ ⎟ [A/m] ⎝ 2πρ ⎠ Example Example
N turns I
µr Nψ L= I N ≈ − Bφ ρ = R A I N = − A µ0 µ r H φ I ( ) ρ =R ⎛ −N ⎞ ˆ H = φ⎜ ⎟ [A/m] ⎝ 2πρ ⎠ N ⎛ NI ⎞ L = A µ0 µ r ⎜ ⎟ I 2π R ⎠ ⎝ ⎛ N2 ⎞ L = µ0 µr A ⎜ ⎟ [H] ⎝ 2π R ⎠ Example Example
coaxial cable I
a Hollow inner conductor I
b h Find L z I + z I S h Example (cont.) Example
I
+ I ˆ n
h S L= ψ
I ˆ ψ = ∫ B ⋅ n dS = ∫ Bφ dS = µ ∫ Hφ dS
S S S Example (cont.) Example
I
+ I ˆ n
h S ψ = µ ∫ H φ dS
S I b d ρ =µ h ψ = µ h ∫ Hφ d ρ = µ h ∫ [ln ρ ]a 2πρ 2π a a b b I Example (cont.) Example
I
a I
b h z I b ψ = µ h [ ln ρ ]a 2π I ⎛b⎞ ln ⎜ ⎟ = µh 2π ⎝ a ⎠ 1 ⎛b⎞ L = = µh ln ⎜ ⎟ I 2π ⎝ a ⎠ ψ µ0 µr ⎛ b ⎞ ln ⎜ ⎟ [H/m] L= 2π ⎝a⎠ Energy Formula for Inductor Energy
2U H L= 2 I 1 U H = ∫ B ⋅ H dV 2 V 1 L = 2 ∫ B ⋅ H dV IV Example Example
Ls
a Find L z µ = µ0 µ r
I N turns ⎛ N2 ⎞ 2 1 U H = µ0 µ rπ a 2 ⎜ ⎟I 2 ⎝ Ls ⎠ 2U H L= 2 I ⎛ N2 ⎞ L = µ0 µ rπ a 2 ⎜ ⎟ ⎝ Ls ⎠ [H] Example Example
Coax I
a L= 1 B ⋅ H dV 2∫ IV I
b h 1 2 = 2 µ0 µr ∫ H dV I V 1 = 2 µ0 µr ∫ Hφ2 dV I V 1 = 2 µ0 µ r ∫ I 0
h 2π b z ∫
0 ⎛I⎞ ∫ ⎜ 2πρ ⎟ ρ d ρ dφ dz ⎠ a⎝
b 2 Find L 1 ⎛1⎞ = µ0 µ r ⎜ ⎟ ( h )( 2π ) ∫ d ρ ρ ⎝ 2π ⎠ a 2 Example Example
Coax I
a 1 ⎛1⎞ L = µ0 µ r ⎜ ⎟ ( h )( 2π ) ∫ d ρ ρ ⎝ 2π ⎠ a
b 2 I
b h µ0 µ r h ⎛ b ⎞ ln ⎜ ⎟ L= 2π ⎝a⎠ [H] z L= µ0 µ r ⎛ b ⎞ ln ⎜ ⎟ 2π ⎝a⎠ [ H/m] Mutual Inductance Mutual
ˆ n1
I1 ˆ n2
ψ 21
I2 ˆ ψ 21 = ∫ B1 ⋅ n 2 ds
S2 M 21 = ψ 21
I1 M 12 = ψ 12
I2
Property: In general, In general, N 2ψ 21 M 21 = I1 N1ψ 12 M 12 = I2 M 12 = M 21 = M Example (cont.) Example
R2 R1 z I1 I2 Find M12 , M21
Λ12 N1ψ 12 M 12 = = I2 I2
Bz2
S1 ⎛ N2 ⎞ 2 M 12 = N1µ0 ⎜ ⎟ π R1 ⎝ Ls ⎠ ˆ ψ 12 = ∫ B 2 ⋅ n1 dS = ∫ Bz 2 dS
S1 [H] = Bz 2π R12 ⎛ N2 ⎞ 2 = µ0 ( n2 I 2 ) π R = µ0 I 2 ⎜ ⎟ π R1 ⎝ Ls ⎠
2 1 Example Example
R2 R1 z I1 I2 M 21 =
Bz1 Λ 21 N 2ψ 21 = I1 I1
S2 S1 ⎛ N1 ⎞ M 21 = N 2 µ0π R ⎜ ⎟ ⎝ Ls ⎠
2 1 ˆ ψ 21 = ∫ B1 ⋅ n 2 dS = ∫ Bz1 dS = π R12 Bz1 ⎛ N1 ⎞ = π R µ0 ( n1 I1 ) = π R µ0 ⎜ ⎟ I1 ⎝ Ls ⎠
2 1 2 1 Example (cont.) Example
R2 R1 z I1 I2 ⎛ N2 ⎞ 2 M 12 = N1µ0 ⎜ ⎟ π R1 ⎝ Ls ⎠
⎛ N1 ⎞ M 21 = N 2 µ0π R ⎜ ⎟ ⎝ Ls ⎠
2 1 [H] ⎛ π R12 ⎞ M 12 = M 21 = µ0 ⎜ ⎟ ( N1 N 2 ) ⎝ Ls ⎠ Force on Wire Force
F = q ( v × B)
B q v F charge: B wire: F = ∫ I d ×B
C I dl F I Example Example
z=0 x y h I2
1 z=L 2 I1 z Find Force on wire 2
ˆ F2 = ∫ I 2 ( z dz ) × B1
C I1 ⎞ ⎛ ˆ ˆ = ∫ I 2 ( z dz ) × ⎜ µ0 ( − y ) ⎟ 2π h ⎠ ⎝ C Example Example
I1 ⎞ ⎛ ˆ ˆ F2 = ∫ I 2 ( z dz ) × ⎜ µ0 ( − y ) ⎟ 2π h ⎠ ⎝ C I1 I 2 ˆ F2 = µ0 ( + x ) ∫ dz 2π h 0 L ⎛ I1 I 2 ⎞ ˆ F2 = x µ0 ⎜ ⎟L ⎝ 2π h ⎠ [ N] F2
z=0 x y h I2
1 z=L 2 I1 z F1 = −F2 Boundary Conditions Boundary
ˆ n
Ht1 Js µ1 = µ o µ r1 µ2 = µo µr 2 Ht2 The unit normal vector points towards region 1. ˆ n × ( H t1 − H t 2 ) = J s Bn1 = Bn 2 H t1 = H t 2 Note: If there is no surface current: Boundary Conditions Boundary
ˆ n
Ht1 Js µ1 = µ o µ r1
PEC
PEC Ht2=0 Assume zero magnetic field inside the PEC (true for any f > 0) ˆ n × Ht = J s Bn = 0 ...
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This note was uploaded on 10/29/2009 for the course ECE 2317 taught by Professor Staff during the Spring '08 term at University of Houston.
 Spring '08
 Staff

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