Transcript Document

Physics 712 – Electricity and Magnetism
Everyone Pick Up:
•Syllabus
•Two homework passes
Eric Carlson
“Eric”
“Professor Carlson”
Olin 306
Office Hours always
758-4994 (o) 407-6528 (c)
[email protected]
Materials
•Classical Electrodynamics
by John David Jackson
•Calculator
•Pencils or pens, paper
•Symbolic manipulation
http://users.wfu.edu/ecarlson/eandm/index.html
1/13
Dr. Carlson’s Approximate Schedule
9:00
10:00
11:00
12:00
1:00
2:00
3:00
4:00
5:00
Monday
PHY 712
PHY 742
PHY 109
research
Tuesday
research
office hour
office hour
office hour
Wednesday
PHY 712
Thursday
research
office hour
Friday
PHY 742
PHY 109
office hour
office hour
PHY 712
PHY 742
PHY 109
research
research
research
Free food
Collins Hall
colloquium
Collins Hall
Lab Meeting
•I will try to be in my office Tues. Thurs. 10-12
•When in doubt, call/email first
Reading Assignments / The Text
Classical Electrodynamics by J.D. Jackson, 3rd Edition
• I’m not writing my own textbook
• Contains useful information
• Reading assignments every day
ASSIGNMENTS
Day
Read
Homework
Today
none
none
Friday
1.1-1.4
0.1
Wednesday 1.5-1.8
1.1, 1.2
Homework
http://users.wfu.edu/ecarlson/quantum
•
•
•
•
•
•
•
•
Problems assigned as we go along
About one problems due every day
Homework is due at 10:00 on day it is due
Late homework penalty 20% per day
Two homework passes per semester
Working with other student is allowed
Seek my help when stuck
You should understand anything you turn in
ASSIGNMENTS
Day
Read
Homework
Today
none
none
Friday
1.1-1.4
0.1
Wednesday 1.5-1.8
1.1, 1.2
Attendance and Tests
Attendance
•I do not grade on attendance
•Attendance is expected
•Class participation is expected
•I take attendance every day
Tests
•Midterm will be from 10-12 on
March 4
•Final will be from 9-12 on April 29
Grades, Pandemic Plans
Percentage
Breakdown:
Homework 50%
Midterm
20%
Final
30%
Grade Assigned
94% A
77% C+
90% A73% C
87% B+
70% C83% B
<70% F
80% B-
Pandemic Plans
•If there is a catastrophic closing of
the university, we will attempt to
continue the class:
•Some curving
possible
Emergency contacts:
Web page
email
Cell: 336-407-6528
0A. Math
Coordinate Systems
•
•
•
•
•
We will generally work in three dimensions
A general coordinate in 3d will be denoted x
A general vector will be shown in bold face v
x  xxˆ  yyˆ  zzˆ
We will often work in Cartesian coordinates
2
2
2
Sometimes in
r

x

y

z
x  r sin  cos 
spherical coordinates,
  cos 1  z r 
y

r
sin

sin

related to Cartesian by
z  r cos 
  tan 1  y x 
• Coordinate x is
x  rrˆ
• Sometimes in
  x2  y 2
x   cos 
cylindrical coordinates,
y   sin 
  tan 1  y x 
related to Cartesian by
zz
zz
• Coordinate x is
x  ρˆ  zzˆ
Vector Identities
• Vectors can be combined using dot
a  b  axbx  a y by  az bz
products to make a scalar
• Vectors can be combined
a  b  xˆ  a y bz  az by   yˆ  az bx  axbz   zˆ  axby  a y bx 
using cross-products to
make a vector
• We often abbreviate
a2  a  a  ax2  ay2  az2
• Some vector identities:
a b  b a
• Symmetry/antisymmetry:
aa  0
a  b  b  a
• Triple scalar product:
a  b  c  b  c  a  c  a  b 
• Double cross-product:
a   b  c   b a  c   c a  b 
• These and many others in Jackson front cover
Derivatives in 3D
• The vector derivative  can be used for several types of derivatives:
• Gradient turns a scalar function
f
f
f
f  xˆ
 yˆ
 zˆ
into a vector function
x
y
z
• Divergence turns a vector
Ax Ay Az
function into a scalar function
A 


• Curl turns a vector
x
y
z
function into a
 Az Ay 
Ax Az   Ay Ax 

vector function
 A  xˆ 



  yˆ 

  zˆ 
z 
x   x y 
 z
 y
• There is also
the Laplacian, a second
2 f 2 f 2 f
2
derivative that can act on
 f  2  2  2
x
y
z
scalar or vector functions
• Each of these has more complicated forms in non-Cartesian coordinates
• See QM lecture notes or inside back cover of Jackson
Derivatives Product rule
• The product rule for derivatives:
d
df
dg
fg

g

f
 
dx
dx
dx
• Numerous 3D equivalents for products
• Gradient of product of scalars
  fg    f  g  f  g 
• Divergence of scalar times vector   fA   f  A  f   A 
• Curl of scalar times vector
• Divergence of a cross product
  fA    f   A  f   A 
  A  B     A   B  A   B 
• Each of these and many more in Jackson, front cover
Integrals in 3D
3
f
x
d
x


• 3d integrals of vector scalar functions will be common

• You should know how to handle these in any coordinate system
• Cartesian:



3
 f  x  d x   dx dy  dz f  x 

• Spherical:


2

0
2
0
1
0
1
f  x  d x   d  d
3



f  x  r 2 sin  dr
0

  d  d  cos    r 2 f  x  dr
• Cylindrical:
 f  x d x  
3
2
0
0



0
d  dz  f  x   d 
Fundamental Theorem of Calculus in 3D
• Fundamental theorem of calculus says:

b
a
df
dx  f  b   f  a 
dx
In general, in 3d, this theorem lets you do
one integral whenever you have an integral in 3d:
b
• Line integral:
 f  dl  f b   f a 
a
• Stokes’ Theorem:
  A  nˆ da   A  dl
Divergence Theorem:   A d x   A  nˆ da
Another theorem:
 f d x   fnˆ da
S
•
C
3
V
•
• And another theorem:
S
3
V
S

V
 A d 3x   nˆ  A da
S
• All these and more can be found on inside front cover of Jackson
Sample Problem 0.1
3.
Consider
the
vector
function
x/|x|
• Work in
x
rrˆ
rˆ

(a) Find the divergence for x  0.
3
spherical
3  2
r
r
x
(b) By integrating over a sphere
coordinates
ˆ
r
  1
• For x  0,
  2   r 2 2   0 around the origin, show that the
r
r  r 
divergence does not vanish there.
take divergence
• Tricky at x = 0 because everything is infinite there!
• Integrate over a sphere of radius R using the divergence theorem:
R
1
rˆ 
rˆ
1
3 
2
ˆ

da
d
x




r
da

4

R
 4
2 S
V  r 2  S r 2
2
R
R
• Since the integral is zero except at the origin, we must have   x3  4 3  x 
x
• You can generalize this where x is replaced by the
x  x
3
difference from an arbitrarily chosen origin x':


4

 x  x 
3
x  x
0B. Units
• Units is one of the most messed-up topics in electricity and magnetism
• We will use SI units throughout
• Fundamental units:
Distance
Time
Mass
Charge
meter
second
kilogram
coulomb
m
s
kg
C
• From these are derived lots of
non fundamental units:
• Equations of E and M sometimes
depends on choice of units!
Frequency
hertz
Force
newton
Energy
joule
Power
watt
Current
amp
Potential
volt
Resistance
ohm
Magnetic induction
tesla
Magnetic flux weber
Inductance
Henry
Hz
N
J
W
A
V

s–1
kgm/s2
Nm
J/s
C/s
J/C
V/A
T
Wb
H
kg/C/s
Tm2
Vs/A
1. Electrostatics
1A. Coulombs Law, El. Field, and Gauss’s Law
Coulomb’s Law
• Charges are measured in units called Coulombs
k1qqrˆ
• The force on a charge q at x from another charge q' at x':
F
2
x  x
• The unit vector points from x' to x rˆ  x  x
• We rewrite the unit vector as
x  x
1
• For reasons that will make some sense later, we rewrite constant k1 as k1 
4 0
• So we have Coulomb’s Law:
qq x  x
F
• For complicated reasons having
4 0 x  x 3
to do with unit definitions, the
constant 0 is known exactly:
 0  8.854 1012 N  C2 /m 2
• This constant is called the
permittivity of free space
Multiple Charges, and the Electric Field
• If there are several charges q'i,
you can add the forces:
• If you have a continuous distribution
of charges (x), you can integrate:
qqi x  xi
F
3
4

i
0 x  xi
F
q
4 0
qq x  x
F
4 0 x  x 3
3


x
d
x



x  x
3
x  x
• In the modern view, such “action at a distance” seems unnatural
• Instead, we claim that there is an
1
x  x
3
E x 
  x  d x
3

electric field caused by the other charges
4 0

xx
– Electric field has units N/C or V/m
• It is the electric field that
then causes the forces
F  qE
Gauss’s Law: Differential Version
x  x
E
  x  d x 
3

4 0
x  x
1
3
• Let’s find the divergence
of the electric field:
x  x
  E x 
  x  d x 
3

4 0

xx
• From four slides ago:
x  x
3


4

 x  x 
3
x  x
1
3
1
3
3
• We therefore have:   E  x   1



  x  d x 4  x  x     x 

4 0
0
• Gauss’s Law (differential version):
• Notice that this equation is local
 0 E  x     x 
Gauss’s Law: Integral Version
 0 E  x     x 
• Integrate this formula over   E  x  d 3x    x  d 3x
0 V

V
an arbitrary volume
• Use the divergence theorem:
 0  nˆ  E  x  da  q V 
S
– q(V) is the charge inside the volume V
• Integral of electric field over area is called electric flux
Why is it true?
• Consider a charge in a region
• Electric field from a charge inside a region
produces electric field lines
q
• All the field lines “escape” the region
somewhere
• Hence the total electric flux escaping must be
proportional to amount of charge in the region
Sample Problem 1.1 (1)
A charge q is at the center of a cylinder of radius r and height 2h. Find the electric
flux out of all sides of the cylinder, and check that it satisfies Gauss’s Law
• Let’s work in cylindrical coordinates
• Electric field is:
q  ρˆ  zzˆ
q x
q  ρˆ  zzˆ

 0E 
3 
3
4   2  z 2 3/2
4 x
4  ρˆ  zzˆ
E
q
• Do integral over top surface:  0  E  nˆ da 
top
4

q
4 0
2
r
 d 
0
0

h d 
2
h

2 3/2

• By symmetry, the
integral over the bottom
surface is the same
2 qh
4  2  h 2
r
0
ρ̂


z da
2
z
h
h
q
h
z
r

q
h
 1 

2
2
2
r h 
 0  E  nˆ da   0  E  nˆ da 
top

2 3/2
bot
E
ẑ

q
qh

2 2 r 2  h2
r
Sample Problem 1.1 (2)
A charge q is at the center of a cylinder of radius r and height 2h. Find the electric
flux out of all sides of the cylinder, and check that it satisfies Gauss’s Law
q  ρˆ  zzˆ
 0E 
4   2  z 2 3/2
q
qh
ˆ
 0  E  nda  
top
2 2 r 2  h2
• Do integral over lateral surface:
q
 da
q
 0  E  nˆ da 

lat
4    2  z 2 3/2
4
2 q
z

4 r 2  z 2
2
h
 rd 
0
h
r
h
h
 0  E  nˆ da 
lat
• Add in the top and bottom surfaces:
 0  E  nˆ da   0  E  nˆ da  2 0  E  nˆ da 
tot
lat
top
r
2
z
E
ρ̂

2 3/2
E
ẑ

h
h
q
h
z
r
dz
qh
r 2  h2
qh
r h
2
2
r
q
qh
r h
2
2
q
Using Gauss’s Law in Problems
 0  nˆ  E  x  da  q V 
S
Gauss’s Law can be used to solve three types of problems
• Total electric flux out of an enclosed region
– Simply calculate the total charge inside
• Electric flux out of one side of a symmetrical region
– Must first argue that the flux out of each side is the same
• Electric field in a highly symmetrical problem
– Must deduce direction and symmetry of electric field from other arguments
– Must define a Gaussian Surface to perform the calculation
– Generally use boxes, cylinders or spheres
Sample Problem 1.2
A line with uniform charge per unit length  passes through the long diagonal of
a cube of side a. What is the electric flux out of one face of the cube?
• The long diagonal of d  a2  a2  a2  a 3
the cube has a length
• The charge inside the cube is therefore
Q  d  a 3
• The total electric flux
 0  nˆ  E da  Q  a 3
out of the cube is
S
• If we rotate the cube 120 around the axis, the three faces at
one end will interchange
– So they must all have the same flux around them
• If we rotate the line of charge, the three faces at one
end will interchange with the three faces in back
– So front and back must be the same
a 3
1
• Therefore, all six
nˆ  E da   nˆ  E da 

one
face
faces have the same flux
6 0
6 S
Sample Problem 1.3
A sphere of radius R with total charge Q has its charge spread
uniformly over its volume. What is the electric field everywhere?
• By symmetry, electric field points directly away from the center
• By symmetry, electric field depends only on distance from origin
Outside the sphere:
E  x   E  r  rˆ
• Draw a larger sphere of radius r
• Charge inside this sphere is q(r) = Q
2


E
r
da
ˆ



4

r
0E r 
Q


E
x

n
da


0 S
• By Gauss’s Law,
0 S
Inside the sphere:
Q
E r  
• Draw a smaller sphere of radius r q  r   Q V  r   Qr 3 R 3
4 0 r 2
V  R
• Charge inside this sphere is only
• By Gauss’s Law,
Qr 3 R3   0  E  x   nˆ da   0  E  r  da  4 r 2 0 E  r 
S
S
• Final answer:
Qrˆ
E
4 0
 r 2 if r  R,
 3
if r  R .
rR
E r  
Qr
4 0 R 3
1B. Electric Potential
Curl of the Electric Field

x
0
• From homework problem 0.1:
3
x
• Generalize to origin at x':
x  x

0
3
x  x
• Consider the curl of the electric field:
1
x  x
1
x  x
3
3


E 
    x  d x
  x  d x  
3 
3

4 0
4 0
x  x

xx
• Using Stokes’ theorem, we can get
an integral version of this equation:
 E  dl  0
3

E
d
x0


S
 E  0
Electric Field: Discontinuity at a Boundary
Consider a surface (locally flat) with a surface charge 
• How does electric field change across the boundary?
A
• Consider a small thin box of area A
crossing the boundary
• Since it is small, assume E is constant over top surface and bottom surface
• Charge inside the box is A
• Use Gauss’s Law  A q

  E  x   nˆ da  Et  nˆ t A  Eb  nˆ b A  A  Et  Eb   nˆ t
on this small box
0
0

ˆ
 E   n 
• Consider a small loop of length L penetrating the surface
0
• Use the identity
0   E  dl  Et  xˆ L  Eb  xˆ L
• Ends are short, so
 E  xˆ  0

only include the lateral part
E  nˆ
0
• So the change in E across the boundary is
The Electric Potential
• In general, any function that has curl zero can be written as a gradient
 E  0
– Proven using Stokes’ Theorem
E  
• We therefore write:
–  is the potential (or electrostatic potential)
– Unit is volts (V)
It isn’t hard to find an expression for :
1
x  x
1
1
rˆ  x
• First note that


 
 2  3
3
x  x
• Generalize by shifting: x
r
x  x
r
x
• If we write:
  x 
1
4 0

  x  d 3x
x  x
• Then it follows that:
1
1
1
x  x
3
3


  x  
 x  d x 

  x  d x
3  E  x



4 0
x  x 4 0
x  x
Working with the Potential
E  
Why is potential useful?
• It is a scalar quantity – easier to work with
• It is useful when thinking about energy
– To be dealt with later
  x 
1
4 0

How can we compute it?
• Direct integration of charge density when possible
• We can integrate the electric field
• It satisfies the following differential equation:
   0  E   0   
 2    0 
Solving this equation is one of the main goals of the next couple chapters
  x  d 3x
x  x