Ben Gurion University of the Negev
www.bgu.ac.il/atomchip, www.bgu.ac.il/nanocenter
Physics 2 for Electrical Engineering
Lecturers: Daniel Rohrlich , Ron Folman
Teaching Assistants: Ben Yellin, Yoav Etzioni
Grader: Gad Afek
Week 12. Outlook – Introduction • method of images • Poynting
vector: examples • a paradox • transformers • Maxwell’s equations
in matter • reflection and refraction • polarized light • spherical
waves • interference
Source: Halliday, Resnick and Krane, 5th Edition, Chap. 38;
The Feynman Lectures on Physics, Chaps. 17, 21 and 27.
Introduction
In the previous weeks in this course, we have offered you a
basic understanding of classical electrodynamics, the theory
electric and magnetic fields and forces. This theory – which
Maxwell perfected – puts powerful tools in your hands. It does
not include quantum theory but it is already consistent with
Einstein’s special theory of relativity (although we have not
talked about relativity).
Introduction
In the previous weeks in this course, we have offered you a
basic understanding of classical electrodynamics, the theory
electric and magnetic fields and forces. This theory – which
Maxwell perfected – puts powerful tools in your hands. It does
not include quantum theory but it is already consistent with
Einstein’s special theory of relativity (although we have not
talked about relativity).
In this lecture, we hope to entertain you with examples of how
much you can already do with this theory, or can do with just
one or two more tricks, and examples of everyday phenomena
that you can now understand in detail.
Method of images
In electrostatics, E and B are constant in time;   E(r)  0,
so we can write E as a gradient: E(r)  V (r). Gauss’s law
then becomes
 (r)
2
 V (r ) 
.
0
Now suppose we have any function V0(r) with the property
that  2V0 (r)  0. Any linear combination of V(r) and V0(r)
will be another solution of Gauss’s law. Which combination is
the right one?
The answer depends on boundary conditions.
Method of images
We have solved electrostatic problems with a few charges, or
with a lot of symmetry. Here is a different sort of electrostatic
problem: A particle of charge q is at rest a distance R from an
infinite conducting plane. What
is the potential V(r) everywhere
to the right of the conductor?
Method of images
We have solved electrostatic problems with a few charges, or
with a lot of symmetry. Here is a different sort of electrostatic
problem: A particle of charge q is at rest a distance R from an
infinite conducting plane. What
z
is the potential V(r) everywhere
to the right of the conductor? It
can’t be just the potential of the
R
q x
particle…
Method of images
We have solved electrostatic problems with a few charges, or
with a lot of symmetry. Here is a different sort of electrostatic
problem: A particle of charge q is at rest a distance R from an
infinite conducting plane. What
z
is the potential V(r) everywhere
to the right of the conductor? It
can’t be just the potential of the
Equipotential
surface
R
q x
particle, because the plane is an
equipotential: V(0,y,z) = constant.
Method of images
Answer: Just put an “image” charge –q at on the x-axis at the
point (–R,0,0) and write down the potential V(r) for x > 0:
z
R
–q
R
q x
Method of images
Answer: Just put an “image” charge –q at on the x-axis at the
point (–R,0,0) and write down the potential V(r) for x > 0:
z
V ( x, y , z ) 

R
–q
q
4 0 [( x  R ) 2  y 2  z 2 ] 1 / 2
q
4 0 [( x  R )  y  z ]
2
2
2 1/ 2
R
q x
The Laplacian  2 of the “image”
potential vanishes for x > 0 and
also V(0,y,z) = 0!
.
Method of images
Answer: Just put an “image” charge –q at on the x-axis at the
point (–R,0,0) and write down the potential V(r) for x > 0:
z
V ( x, y , z ) 

Equipotential
surface
q
4 0 [( x  R ) 2  y 2  z 2 ] 1 / 2
q
4 0 [( x  R )  y  z ]
2
2
2 1/ 2
R
q x
The Laplacian  2 of the “image”
potential vanishes for x > 0 and
also V(0,y,z) = 0!
.
Method of images
This method also works for a hollow conducting sphere:
z
D
R
q x
Equipotential
surface
Method of images
This method also works for a hollow conducting sphere:
z
D
R'
q'
Equipotential
surface
R
q x
D
q'   q
R
D2
R' 
R
Poynting vector
The relation between energy and momentum in radiation:
E
k
B
Poynting vector
The relation between energy and momentum in radiation:
Consider electromagnetic radiation normal to a metal sheet of
size L2 and conductance σ. The electric field induces a current
density J = σE. The magnetic field is B perpendicular to J, and
like E lies in the metal sheet. It induces a force density J×B =
σE×B in the direction of the radiation. The power delivered by
the radiation to the metal sheet is I2R = (LJ)2/σ so the power per
unit surface area is J2/σ = JE while the force per unit surface
area is JB = JE/c. Although this result is for a special case, it
must correspond to the energy and momentum content of the
electromagnetic wave. Since the averaged Poynting vector Sav
is the energy density flux in radiation, the momentum density
flux in radiation – i.e. the radiation pressure – must be Sav/c.
Poynting vector
Example 1: A great amount of debris – asteroids, meteors, dust
– exists in interplanetary space. Although we might expect the
smallest dust particles to be of molecular size, very little of the
dust in our solar system is smaller than about 0.2 μm. Why?
Poynting vector
Example 1: A great amount of debris – asteroids, meteors, dust
– exists in interplanetary space. Although we might expect the
smallest dust particles to be of molecular size, very little of the
dust in our solar system is smaller than about 0.2 μm. Why?
The sun radiates energy at the rate P = 3.90 × 1026 W. Assume
each dust particle is spherical, with density ρ = 3.0 × 103 kg/m3
(density of the earth’s crust), and absorbs all energy incident on
it. The mass of the sun is M = 1.99 × 1030 kg, and Newton’s
gravitational constant is G = 6.67 × 10–11 m3/s2·kg. Assume the
distance of the dust particle from the sun is R.
Poynting vector
Example 1: A great amount of debris – asteroids, meteors, dust
– exists in interplanetary space. Although we might expect the
smallest dust particles to be of molecular size, very little of the
dust in our solar system is smaller than about 0.2 μm. Why?
The sun radiates energy at the rate P = 3.90 × 1026 W. Assume
each dust particle is spherical, with density ρ = 3.0 × 103 kg/m3
(density of the earth’s crust), and absorbs all energy incident on
it. The mass of the sun is M = 1.99 × 1030 kg, and Newton’s
gravitational constant is G = 6.67 × 10–11 m3/s2·kg. Assume the
distance of the dust particle from the sun is R. What is its
radius r if the gravitational attraction of the sun just balances
the radiation pressure from the sun?
Poynting vector
Example 1: A great amount of debris – asteroids, meteors, dust
– exists in interplanetary space. Although we might expect the
smallest dust particles to be of molecular size, very little of the
dust in our solar system is smaller than about 0.2 μm. Why?
Answer: The attraction of the sun is Fgrav 
GM  4 3 
r  .

2

R  3
The energy flux from the sun per unit area is the Poynting
vector S = P/4πR2. The radiation pressure S/c = (P/4πR2)/c if
the dust particle absorbs all the energy. The radiation force is
the cross-section of the particle times S/c: Frad = πr2(P/4πR2)/c.
GM  4 3 
If Frad = Fgrav then

r   so r = 3P/16πρcGM.

2
2

4R c R  3
.
r 2P
Poynting vector
GM  4 3 
r  .
Answer: The attraction of the sun is Fgrav  2 

R  3
The energy flux from the sun per unit area is the Poynting
vector S = P/4πR2. The radiation pressure S/c = (P/4πR2)/c if
the dust particle absorbs all the energy. The radiation force is
the cross-section of the particle times S/c: Frad = πr2(P/4πR2)/c.
GM  4 3 

r   so r = 3P/16πρcGM


4R 2c R 2  3
= 3(3.90 × 1026 W) / 16π (3.0 × 103 kg/m3) (3.00 × 108 m/s)
× (6.67 × 10–11 m3/s2·kg) (1.99 × 1030 kg)
If Frad = Fgrav then
= 0.19 μm.
r 2P
Poynting vector
GM  4 3 
r  .
Answer: The attraction of the sun is Fgrav  2 

R  3
The energy flux from the sun per unit area is the Poynting
vector S = P/4πR2. The radiation pressure S/c = (P/4πR2)/c if
the dust particle absorbs all the energy. The radiation force is
the cross-section of the particle times S/c: Frad = πr2(P/4πR2)/c.
GM  4 3 

r   so r = 3P/16πρcGM


4R 2c R 2  3
= 3(3.90 × 1026 W) / 16π (3.0 × 103 kg/m3) (3.00 × 108 m/s)
× (6.67 × 10–11 m3/s2·kg) (1.99 × 1030 kg)
If Frad = Fgrav then
r 2P
= 0.19 μm.
So why is interplanetary dust mostly larger than 0.2 μm?
Poynting vector
Example 2 (Halliday, Resnick and Krane, 5th Edition, Chap.
38, Prob. 12): The wire shown has length L, resistance R and
radius a, and carries a current I. (a) Show that the Poynting
vector S on the wire surface points inward. (b) Show that the
integral of S over the wire surface is –I2R.
B
a
I
Poynting vector
Example 2 (Halliday, Resnick and Krane, 5th Edition, Chap.
38, Prob. 12): The wire shown has length L, resistance R and
radius a, and carries a current I. (a) Show that the Poynting
vector S on the wire surface points inward. (b) Show that the
integral of S over the wire surface is –I2R.
Answer: (a) E is parallel to the wire, hence S = E×B/μ0 points
into the wire.
B
a
S
×
I
B
Poynting vector
Example 2 (Halliday, Resnick and Krane, 5th Edition, Chap.
38, Prob. 12): The wire shown has length L, resistance R and
radius a, and carries a current I. (a) Show that the Poynting
vector S on the wire surface points inward. (b) Show that the
integral of S over the wire surface is –I2R.
Answer: (b) E = V/L = IR/L and B = μ0I/2πa, thus S = I2R/2πaL
and ∫ S·dA = 2πaLS = –I2R.
B
a
S
×
I
B
dA
Poynting vector
Example 2 (Halliday, Resnick and Krane, 5th Edition, Chap.
38, Prob. 12): The wire shown has length L, resistance R and
radius a, and carries a current I. (a) Show that the Poynting
vector S on the wire surface points inward. (b) Show that the
integral of S over the wire surface is –I2R.
This looks strange at first – electromagnetic energy enters the
wire from the sides? But maybe it is not so strange – heat
energy leaves the wire from the sides….
B
a
S
×
I
B
dA
A paradox
Assume two uniformly (but oppositely)
charged cylinders, both infinitely long.
One fits inside the other but they have
almost the same radius r. Outside, a
distance L from their common axis, is
a particle of charge q at rest. The two
cylinders rotate at the same angular
speed but in opposite directions. But
mutual friction gradually stops them
from rotating, starting at t = 0.
L
q
A paradox
Assume two uniformly (but oppositely)
charged cylinders, both infinitely long.
One fits inside the other but they have
almost the same radius r. Outside, a
distance L from their common axis, is
a particle of charge q at rest. The two
cylinders rotate at the same angular
speed but in opposite directions. But
mutual friction gradually stops them
from rotating, starting at t = 0.
According to Faraday’s law, the decay
of the magnetic flux in the cylinders
creates an electric field that accelerates
the charged particle.
L
q
A paradox
Assume two uniformly (but oppositely)
charged cylinders, both infinitely long.
One fits inside the other but they have
almost the same radius r. Outside, a
distance L from their common axis, is
a particle of charge q at rest. The two
cylinders rotate at the same angular
speed but in opposite directions. But
mutual friction gradually stops them
from rotating, starting at t = 0.
According to Faraday’s law, the decay
of the magnetic flux in the cylinders
creates an electric field that accelerates
the charged particle.
L
q
A paradox
Let the magnitude of the magnetic field
inside the cylinders be B(t). The force
on the charged particle is qE(t) where
z
d
2 LE (t )   r
B(t ) ,
dt
2
hence the final momentum of the
charged particle is qr2B(0)/2L.
L
q
x
A paradox
Let’s integrate the initial Poynting
vector, assuming L >> r. We need
Ex(0,0,z) at the initial time:
E x (0,0, z ) 
qL
4 0 ( L  z )
2
2 3/ 2
z
,
so the integral of the Poynting vector is

qL r 2 B(0)
 4 0 0 ( L2  z 2 )3 / 2 dz ,
which reduces (using z = L tan θ) to
qr2B(0)/2ε0μ0L = qc2r2B(0)/2L.
L
q
x
A paradox
So the integral of the Poynting vector
S is qr2B(0)/2ε0μ0L = qc2r2B(0)/2L.
Now S/c is energy density; and if we
assume that S/c2 is momentum density,
then the electromagnetic momentum
was initially qr2B(0)/2L, exactly the
final mechanical momentum.
Momentum is conserved, after all!
z
L
q
x
Transformers
Power transmission lines lose I2R = IV of power on their way
to their destination. A way to reduce power losses is to reduce
I while increasing V. 350-kV power lines are common. At the
destination, where it is safer to use a smaller potential, a
transformer reduces V. How does a transformer work?
Get yours today!
Transformers
Power transmission lines lose I2R = IV of power on their way
to their destination. A way to reduce power losses is to reduce
I while increasing V. 350-kV power lines are common. At the
destination, where it is safer to use a smaller potential, a
transformer reduces V. How does a transformer work?
ΔV1
Primary
(input)
Soft iron
N1
N2
ΔV2
S
R2
Secondary
(output)
Transformers
With the switch S open, the potential across the primary coil
(with N1 windings) is ΔV1 = – N1dΦB/dt, by Faraday’s law. If
the coils link with same magnetic flux ΦB then the potential on
the secondary coil is ΔV2 = – N2dΦB/dt. The relation between
the potentials is ΔV1 = ΔV2 N1/N2. If N2 > N1 then ΔV2 > ΔV1.
This is a step-up transformer.
ΔV1
Primary
(input)
Soft iron
N1
N2
ΔV2
S
R2
Secondary
(output)
Transformers
With the switch S open, the potential across the primary coil
(with N1 windings) is ΔV1 = – N1dΦB/dt, by Faraday’s law. If
the coils link with same magnetic flux ΦB then the potential on
the secondary coil is ΔV2 = – N2dΦB/dt. The relation between
the potentials is ΔV1 = ΔV2 N1/N2. If N2 < N1 then ΔV2 < ΔV1.
This is a step-down transformer.
ΔV1
Primary
(input)
Soft iron
N1
N2
ΔV2
S
R2
Secondary
(output)
Transformers
With the switch S closed, the current in the secondary coil is
I2 = ΔV2 /R2.
In an ideal transformer, there are no losses and I1ΔV1 = I2 ΔV2 .
ΔV1
Primary
(input)
Soft iron
N1
N2
ΔV2
S
R2
Secondary
(output)
Transformers
Example: A power station in Beersheba delivers 20 MW of
power to a neighborhood 1 km away. The resistance in the
transmission line is 2.0 Ω and the industrial price of electricity
is ₪ 0.30/kWh. The power station produces the electricity at
22 kV but steps it up to 230 kV before transmission. (a) What
is the (approximate) daily transmission cost? (b) What would
be the cost without stepping up the potential?
Transformers
Example: A power station in Beersheba delivers 20 MW of
power to a neighborhood 1 km away. The resistance in the
transmission line is 2.0 Ω and the industrial price of electricity
is ₪ 0.30/kWh. The power station produces the electricity at
22 kV but steps it up to 230 kV before transmission. (a) What
is the (approximate) daily transmission cost? (b) What would
be the cost without stepping up the potential?
Answer: (a) We approximate using dc relations. The current is
power divided by potential: I = 20 MW / 230 kV = 87 A. The
power lost in the line is I2R = (87 A)2(2.0 Ω) = 15 kW; in one
day, the power loss is 360 kWh, worth ₪110.
Transformers
Example: A power station in Beersheba delivers 20 MW of
power to a neighborhood 1 km away. The resistance in the
transmission line is 2.0 Ω and the industrial price of electricity
is ₪ 0.30/kWh. The power station produces the electricity at
22 kV but steps it up to 230 kV before transmission. (a) What
is the (approximate) daily transmission cost? (b) What would
be the cost without stepping up the potential?
Answer: (b) We approximate using dc relations. The current
is power divided by potential: I = 20 MW / 22 kV = 910 A.
The power lost in the line is I2R = (910 A)2(2.0 Ω) = 1.65 MW;
in one day, the power loss is 40 MWh, worth ₪12,000.
Maxwell’s equations in matter
Here are Maxwell’s equations, including sources (charges,
currents) but not dielectric, diamagnetic, paramagnetic or
ferromagnetic materials:

E 
0
B  0
B
E  
t
E
  B   0 J   0 0
.
t
Maxwell’s equations in matter
Here are Maxwell’s equations, including sources (charges,
currents) and many dielectric, diamagnetic, paramagnetic or
ferromagnetic materials:

E 

B  0
B
E  
t
E
  B  J  
.
t
Maxwell’s equations in matter
We deduce the speed of electromagnetic waves in these
materials just as we deduced the speed in the vacuum. First,
we eliminate sources (charges, currents):
E  0
B  0
B
E  
t
E
  B  
.
t
Maxwell’s equations in matter
We find that the speed of electromagnetic waves in a medium
with electric permittivity ε and magnetic permeability μ is
1 /  .
B
0  E 
t
Faraday’s law
B 

0      E 


t


Vector identity


 (  B)
 (  E)   E 
t
Gauss’s law in a vacuum:
2
•E = 0
Ampère’s law, as
modified by Maxwell
2


E

E


2
2
  E   
   2   E
t 
t 
t
Maxwell’s equations in matter
The plane wave E(r,t) = E cos (k·r – ωt + δ) is often written as
a complex function:
E(r, t )  Eeik r it ,
where E is a constant vector that fixes the amplitude and the
polarization of the wave, k = 2π/λ is the wave number of the
wave, and ω is its angular frequency. (You can just ignore the
imaginary part of the function.) Substituting this plane
2

E  
wave into
 2E
 0, we obtain a solution if and
t
only if – k2 + μεω2 = 0, hence 1 /  = ω/k = f λ is the speed
2
of the wave in the material.
Maxwell’s equations in matter
Writing a plane wave as E(r,t) = Eeik·r – iωt is very convenient!
It allows us to write   E(r, t ) as ik  E(r, t ) , so we learn that E
is transverse in this case as well. The factor of i just indicates a
phase difference of π/2. It also allows us to write   E(r, t ) as
ik × E(r,t). Now from Faraday’s law we have

ik  E(r, t )    E(r, t )   B(r, t )  i B(r, t ) ,
t
hence B(r,t) = k × E(r,t)/ω.
Reflection and refraction
Now let’s consider what happens when an electromagnetic
wave crosses between two media, i.e. between two materials
with different permittivities and permeabilities. Let the
interface of the two materials be the plane x = 0, with a wave
Ei(r,t) incident on the left. There will be also a reflected wave
Er(r,t) on the left and a transmitted wave Et(r,t) on the right:
ki
θi
θt
θr
kr
x
kt
μiεi
μtεt
Reflection and refraction
On the x = 0 plane, the electric field must be continuous, which
means that Ei(0,y,z,t) + Er(0,y,z,t) = Et(0,y,z,t); or in complex
wave notation (after cancelling the common factor e– iωt):
Ei eik i (0, y, z )  E r eik r (0, y, z )  Et eik t (0, y, z ) .
ki
θi
θt
θr
kr
x
kt
μiεi
μtεt
Reflection and refraction
The equation Ei e
ik i (0, y , z )
 E r eik r (0, y, z )  Et eik t (0, y, z )
must hold for all y and all z. The only way it can hold for all y
and all z is if ki · (0,y,z) = kr · (0,y,z) = kt · (0,y,z), i.e. ki , kr and
kt have the same y- and z-components:
ki sin θi = kr sin θr = kt sin θt .
ki
θi
θt
θr
kr
x
kt
μiεi
μtεt
Reflection and refraction
The equation Ei e
ik i (0, y , z )
 E r eik r (0, y, z )  Et eik t (0, y, z )
must hold for all y and all z. The only way it can hold for all y
and all z is if ki · (0,y,z) = kr · (0,y,z) = kt · (0,y,z), i.e. ki , kr and
kt have the same y- and z-components:
ki sin θi = kr sin θr = kt sin θt .
Since ki = kr we have
angle of incidence =
angle of reflection!
ki
θi
θt
θr
kr
x
kt
μiεi
μtεt
Reflection and refraction
The equation Ei e
ik i (0, y , z )
 E r eik r (0, y, z )  Et eik t (0, y, z )
must hold for all y and all z. The only way it can hold for all y
and all z is if ki · (0,y,z) = kr · (0,y,z) = kt · (0,y,z), i.e. ki , kr and
kt have the same y- and z-components:
ki sin θi = kr sin θr = kt sin θt .
Snell’s law of
refraction!
ki
θi
θt
θr
kr
x
kt
μiεi
μtεt
Polarized light
Our plane wave solution E(r,t) = Eeik·r – iωt implies that the
electric field of a wave has a certain direction, perpendicular to
propagation vector k. Is this a prediction we can test?
Yes, with a polarizer! A polarizer has a polarization axis, and
passes only the component of light along the polarization axis.
We will not usually see an effect with one polarizer, because
most light is unpolarized – it is mixture of waves with E
pointing in any direction perpendicular to k. But with two
polarizers we see an effect: the first polarizer polarizes the
light emitted by a source, and as we turn the second polarizer
relative to the first, we may see all or part or none of the
polarized light.
Polarized light
Our plane wave solution E(r,t) = Eeik·r – iωt implies that the
electric field of a wave has a certain direction, perpendicular to
propagation vector k. Is this a prediction we can test?
Yes, with a polarizer! A polarizer has a polarization axis, and
passes only the component of light along the polarization axis.
Polarized light
A plane wave E(r,t) = Eeik·r – iωt is linearly polarized, but there
is also elliptically polarized light. Elliptically polarized light
(including circularly polarized light) reduces to a combination
of two linearly polarized waves, with orthogonal linear
polarizations, and a relative phase difference:
Polarized light
A plane wave E(r,t) = Eeik·r – iωt is linearly polarized, but there
is also elliptically polarized light. Elliptically polarized light
(including circularly polarized light) reduces to a combination
of two linearly polarized waves, with orthogonal linear
polarizations, and a relative phase difference:
Watch an animation!
Polarized light
A plane wave E(r,t) = Eeik·r – iωt is linearly polarized, but there
is also elliptically polarized light. Elliptically polarized light
(including circularly polarized light) reduces to a combination
of two linearly polarized waves, with orthogonal linear
polarizations, and a relative phase difference:
Watch an animation!
Spherical waves
We found plane-wave solutions to the wave equation, but don’t
assume these are the only solutions. Here a spherical wave:
B(r, t ) 
(t  r / c)
p (t  r / c)  (r / c) p
4 0c r
2 3
r ,
where p(t) = p0 sin ωt represents an electric dipole oscillating
on the z-axis at r = 0; p0 is a constant with dimensions of C·m,
pointing in the positive z-direction, and the dots denote time
derivatives. E(r,t) is a bit more complicated.

p
B
E
At large r, both B(r,t) and E(r,t) drop as 1/r.
Interference
Spherical-wave solutions can help us understand what happens
when a wave passes through a small hole in a screen. If the
wave passes only through the hole, then on the other side of the
screen it looks like a spherical wave with the hole as its source:
Watch this as an animation!
Interference
If the wave arrives at a screen with two holes and passes
through them, the result is an interference pattern where the
crests of one spherical wave and the troughs of the other
spherical wave add up to zero field:
Watch this as an animation!
Interference
There are regions of destructive interference, where waves
cancel each other,
L = difference in path
length = (n + ½) λ
d = hole separation
d
D = distance to screen
sin θn = L/d = (n + ½) λ/d
L
θn
Interference
and regions of constructive interference, where the waves add
in phase to produce a stronger oscillation.
L = difference in path
length = nλ
d = hole separation
d
D = distance to screen
sin θ'n = L/d = nλ/d
L
θ'n