Electricity
Rubbed amber attracts straw
(hlektronelectron) :
negative charge
Rubbed glass: positive
charge
Electrostatics
Stationary charges
Coulomb’s Law
Gauss’s Law
Maxwell’s Equations
Gauss’ Law for Electricity
Gauss’ Law for Magnetism
Faraday Induction Law
Ampere-Maxwell Law
Electromagnetic Waves
Magnetism
Natural magnet attracts iron
Compass
Moving charges
Electric Current
Current in a wire deflects
magnetic compass
Electromagnetism
Time-varying currents and
Time-varying charge distribution
→time-varying magnetic field
Faraday Induction Law
→time-varying electric field
Ampere-Maxwell Law
Magnetostatics
Steady currents
Time-independent
charge distribution
Biot-Savart Law
Ampere’s Law
Chapter 21 Electric Charge
Some properties of electric charges
i) Unit of electric charge: Coulomb(C ) = Ampere( A) Second( s )
ii) Charge is Quantized
q = ne
, e = 1.602 10 −19 C
n = 1, 2, 3,
e.g. The charge of a proton → e, The charge of an electron → −e
Note: Quarks have charges
e
2e
and
but they cannot be detected individually.
3
3
iii) Charge is conserved
Benjamin Franklin: Conservation of charge
Examples.
1.
238
U
92 protons
→
234
Th
90 protons
+
4
He
Note: 4 He is also known as particle
2 protons
2. e − + e + → + two-photon event of annihilation
3. → e − + e + pair production
Force between stationeary charges
i) Coulomb's Law (Charles Augustin Coulomb 1785)
F=
q1 q2
1
9
2
2
;
=
8.99
10
N
m
/
C
4e 0 r 2
4e 0
1
ii) Principle of Superposition
Individual electrostatic forces are not altered by each other.
n
The net force on q0 : Fnet = −
i =1
1 q0 qi
rˆi ( q0 at the origin)
2
4e 0 ri
iii)Shell theorem Electrostatic force on q
F =0
1 Qq
F=
rˆ
2
4e 0 r
q
r
Q
𝑄
𝑞
z
𝑞
Proof:
1) At an external point
𝑟Ԧ
Let the z-axis be in the direction of r .
𝛼 𝑟Ԧ − 𝑟′
Ԧ
𝑑𝑉′
dV ' = r ' sin ' dr ' d ' d '
2
𝜃′ 𝑟′
Ԧ
qr dV ' r − r '
dF =
(Coulomb's law)
2
4e 0 r − r ' r − r '
1
By symmetry, F = Fz kˆ and therefore
𝑅1
x
𝑅2
only the z-component of dF is of interest.
dFz = dF kˆ =
qr dV ' r − r ' ˆ
1 qr dV '
(
k
)
=
cos
2
2
4e 0 r − r ' r − r '
4e 0 r − r '
1
Let r − r ' = s
s 2 + r 2 − r '2
r '2 + r 2 − s 2
By cosine law, we have cos =
; cos ' =
2 sr
2r ' r
𝑄
y
1
F = Fz kˆ = kˆ dFz =
=
1
4e 0
qr kˆ
R2
R1
4e 0
cos
qr kˆ
V'
r −r'
2
r '2 cos sin '
0
0
r −r'
2
2
dV '
d ' d 'dr '
2 2
2
2
r' s + r − r'
ˆ
=
qr k 2
(sin ' d ') d 'dr '
R
0
0
1
4e 0
2 sr
s
r '2 + r 2 − s 2
r '2 + r 2 − s 2
Noting that cos ' =
d (cos ') = d (
)
2r ' r
2r ' r
sds
sin ' d ' =
r ' is considered constant
r'r
when calculating the integral in the bracket.
1
R2
2
s is a function of '.
F=
1
4e 0
qr kˆ
R2
R1
2
0
r + r ' r '2 s 2 + r 2 − r '2 sds
d 'dr '
r − r ' s 2
2 sr
r'r
F=
1
4e 0
qr kˆ
R2
R1
2
0
r + r ' r '2 s 2 + r 2 − r '2 sds
d 'dr '
r − r ' s 2
2 sr
r'r
2
2
r+r '
r
'
r
−
r
'
=
qr kˆ 2 (1 +
)
ds
d 'dr '
2
R
0
r
−
r
'
1
4e 0
s
2r
R2 2 r '
r+r '
r+r ' 1
1
r' 2
2
ˆ
=
qr k 2 ds + 2 ( r − r ' )
ds
d 'dr '
2
R
0
r
−
r
'
r
−
r
'
1
4e 0
2r
s
2r
1
R2
2
R2 2 r '
r'
1
1
ˆ
=
qr k 2 (2r ') + 2 ( r 2 − r '2 )(
−
) d 'dr '
R
0
1
4e 0
2r
r − r' r + r'
2r
R2 2 r '
1
ˆ
=
qr k 2 (2r '+ 2r ') d 'dr '
R1 0
4e 0
2r
1
1 4 qr ˆ R2 2
1 4 qr ( R23 − R13 ) ˆ
=
k r ' dr ' =
k
2
2
R
1
4e 0 r
4e 0
3r
=
q 4 3 4 3 ˆ
1 qQ ˆ
r
(
R
−
R
)
k
=
k
2
1
2
2
4e 0 r
3
3
4e 0 r
1
F=
1 qQ ˆ
k at an exterior point
2
4e 0 r
2) At an internal point
Consider an infinitesimal shell
1 qr dV1 '
1 qr dr ' da1
1 qr dr ' s12d
1 qr dr ' d
ˆ
ˆ
ˆ
dF1 =
e
=
e
=
e
=
eˆs1
s1
s1
s1
2
2
2
4e 0 s1
4e 0
s1
4e 0 s1 cos
4e 0 cos
1 qr dV2 '
1 qr dr ' da2
1 qr dr ' s22d
1 qr dr ' d
ˆ
ˆ
ˆ
dF2 =
e
=
e
=
e
=
eˆs2
s2
s2
s2
2
2
2
4e 0 s2
4e 0
s2
4e 0 s2 cos
4e 0 cos
Since eˆs1 = − eˆs2 dF = dF1 + dF2 = 0
𝑑𝑟′
F = dF = 0 F = 0 at an internal point
Note:
s
Angle = ;
r
r → radius of the circle,
s → arc length
S
Solid Angle = 2 ;
r
r → radius of the sphere,
S → sphere segment area
𝑠2
𝛼 𝑑𝛺
𝑞
𝑑𝛺 𝑠1
𝛼
𝛼
𝑠12 𝑑𝛺
𝑠12 𝑑𝛺
𝑑𝑎1 =
cos𝛼
Electric charges and materials:
Conductor → charge can move freely through conductors.
Insulators → charge cannot move freely through insulators.
Semiconductors → intermediate between conductors and insulators.
Superconductors → perfect conductors
Chapter 22 Electric Fields
Electric field
E=
F
;
q0
q0 : a positive test charge, F : the electrostatic force acting on the test charge
A. The electric field due to a point charge:
1 qq0
rˆ
2
F 4e 0 r
1 q
E= =
=
rˆ
2
q0
q0
4e 0 r
Ex. Calculate the electric field at the origin.
𝑞0
𝑞
y
30
30
x
E1 =
1 2Q
ˆ − 1 2Q sin 30 ˆj; E2 = 1 2Q cos 30iˆ − 1 2Q sin 30 ˆj
cos
30
i
4e 0 d 2
4e 0 d 2
4e 0 d 2
4e 0 d 2
E3 =
1 4Q
ˆ + 1 4Q sin 30 ˆj E = E1 + E2 + E3 = 1 8Q cos 30iˆ
cos
30
i
4e 0 d 2
4e 0 d 2
4e 0 d 2
B. The electric field due to an electric dipole
+q
An electric
dipole
d
Electric dipole moment: p = qd
−q
By the superposition principle, the electric field due to an electric dipole
is the vector sum of the electric fields due to + q and that due to − q.
z
P
(0,0, z )
Ex. Calculate the electric field at point P.
q
kˆ
4e 0 ( z − d ) 2
2
1
−q ˆ
2. The electric field due to − q : E( − ) =
k
d
4e 0 ( z + ) 2
2
1. The electric field due to + q : E( + ) =
1
+q
y
x
−q
By superposition principle,
1
1
q
1
1
−
]kˆ =
[
−
]kˆ
2
4e 0 ( z − d ) 2 ( z + d ) 2
4e 0 z (1 − d ) 2 (1 + d ) 2
2
2
2z
2z
q
d −2
d −2 ˆ
=
[(1
−
)
−
(1
+
) ]k
2
4e 0 z
2z
2z
E = E( + ) + E( − ) =
q
[
Note: Binomial Theorem
Expand (1 + x ) n in Taylor's series about x = 0.
(1 + x ) n = (1 + x ) n
x =0
+
n(1 + x ) n −1
1!
x =0
x+
n( n − 1)(1 + x ) n −2
2!
x =0
x2 +
n( n − 1) x 2
= 1 + nx +
+
2
d
Let n = −2 and x = ,
2z
d
d
d
d
d
d
we have [(1 − ) −2 − (1 + ) −2 ] = {[1 + + 3( ) 2 + ] − [1 − + 3( ) 2 +
2z
2z
z
2z
z
2z
d
2d
d3
=2 + =
+ O( 3 )
z
z
z
]}
d −2
d −2 ˆ
q
2d
d3 ˆ
E=
[(1 − ) − (1 + ) ]k =
[ + O ( 3 )]k
2
2
4e 0 z
2z
2z
4e 0 z z
z
q
qd
2e 0 z
3
kˆ if z
d.
Note that p = qdkˆ E =
p
2e 0 z
3
(z
d)
Note: When the point of interest P is off the z-axis, the electric field can be
more conveniently calculated using electric potential.
C. The electric field due to a continuous charge distribution
For continuous charge distributions
dq
l=
dl
linear charge density E = dE =
l
4e 0 l r − r 2 r − r
=
dq
dA
surface charge density E = dE =
r=
dq
dV
volume charge density E = dE =
A
V
l dl r − r
1
1
dA r − r
1
r dV r − r
4e 0 A r − r 2 r − r
4e 0 V r − r 2 r − r
Ex1. Electric field due to an infinitely long line of uniformly distributed charge
By symmetry, E = Erˆ.
1 l [( r / cos )d / cos ] dl
dE =
4e 0
( r / cos ) 2
l
dEr = dE cos =
cos d
4e 0 r
r/cos
r
l
l
2
E = rˆ dEr = rˆ
cos
d
=
rˆ
−
4e 0 r 2
2e 0 r
P dE
r
dE
dl=[(r/cos)d]/cos
Note:
(r/cos)d d→→
d
Ex 2. Electric field due to a circular ring of uniformly distributed charge
By symmetry, E = Ezˆ.
dE =
1 l ds
1
l ds
=
4e 0 r 2
4e 0 ( z 2 + R 2 )
1
l ds
1
l Rd
z
dE z = dE cos =
cos =
2
2
4e 0 ( z + R )
4e 0 ( z 2 + R 2 ) ( z 2 + R 2 )1/2
E = zˆ dE z = zˆ
E=
1
l zR
2
4e 0 ( z 2 + R 2 )3/2 0
(2 Rl ) z
1
qz
ˆ
z
=
zˆ
2
2 3/2
2
2 3/2
4e 0 ( z + R )
𝐸 4e 0 ( z + R )
d =
1
1
qz
zˆ
2
2 3/2
4e 0 ( z + R )
dE cos
Note:
If z
RE
1
qz
1 q
ˆ
z
=
zˆ
2 3/2
2
4e 0 ( z )
4e 0 z
𝑅2 + 𝑧 2
l
d𝐸
P
z
d
R
Ex 3. Electric field due to a uniformly charged circular arc of radius r
By symmetry, E = Eiˆ.
1 l ds
1 l rd
1 l d
dE =
=
=
4e 0 r 2
4e 0 r 2
4e 0 r
dE x = dE cos =
y
1 l
cos d
4e 0 r
r
1 l
ˆ
ˆ
E = −i dE x = −i
cos d
−
4e 0 r
l
l sin ˆ
= −iˆ
[sin ]+− = −
i
4e 0 r
2e 0 r
l sin ˆ
E=−
i
2e 0 r
P
rd
x
l
Ex 4. Electric field due to a uniformly charged disk of radius R
By symmetry, E = Ezˆ.
Consider the ring of radius r and infinitesimal thickness dr.
The charge of the ring is dq = dA = (2 r dr )
Recall E =
dE =
1
q
z
zˆ for a ring of radius R and charge q.
2
2
2
2 1/2
4e 0 z + R ( z + R )
dq
z
1 (2 rdr )
z
z
rdr
=
=
4e 0 z 2 + r 2 ( z 2 + r 2 )1/2 4e 0 z 2 + r 2 ( z 2 + r 2 )1/2 2e 0 ( z 2 + r 2 )3/2
1
z R
rdr
E = dE =
2e 0 0 ( z 2 + r 2 )3/2
Let X = z 2 + r 2 dX = 2rdr,
z2 + R2
z z + R dX
z
we have E =
=
−2 X
4e 0 z
( X )3/2 4e 0
z
2
2
1
−
2
2
2
z
E=
(1 −
) zˆ
2
2
2e 0
z +R
Note: When R → , E =
zˆ (infinite sheet)
2e 0
P
z
=
(1 −
)
2
2
2e 0
z +R
z
R
r
dr
A point charge in an electric field
F = qE (force on a point charge)
An initially stationary dipole in a (locally uniform) electric field
Fext ,net = qE + (−q ) E = 0 The dipole's center of mass does not move.
The center of mass (com) lies somewhere between + q and − q
Let the distance between + q and com be l.
The torque applied by the electric field on the dipole about its com is
t = (l cos , l sin , 0) ( qE , 0, 0) + (−(d − l ) cos , − ( d − l ) sin , 0) ( − qE , 0, 0)
= − qEl sin kˆ − qE (d − l ) sin kˆ = −qEd sin kˆ = −(qd ) E sin kˆ = − pE sin kˆ
= p E t = p E (torque on a dipole)
Note: The direction of positive z is chosen such that
y
the work done by the electric field on the dipole is
z
positive in the expression dW = t d . The angle
x
+q
decreases when the work is being done. Therefore d
is negative. A negative t to make dW positive can
be arranged by selecting a proper direction for the
positive z.
E
com l
d-l
−q
Potential energy of a dipole in electric field
dW = t d = − pE sin d
f
W = dW = − pE sin d = − pE [ − cos ]if = pE (cos f − cos i )
i
Electrostatic force is conservative W = −U = −[U ( f ) − U (i )]
pE (cos f − cos i ) = −[U ( f ) − U (i )]
Let f = 90, i = and U (90) = 0 U ( ) = − pE cos = − p E
U = −pE
y
z
x
+q
In summary,
t = pE
U = −pE
E
com l
d-l
−q
Chapter 23 Gauss’ Law
A. Coulomb's Law Gauss' Law
Consider a closed surface S enclosing a charge distribution (i.e. a Gauss' surface).
i) A surface element dA → magnitude: the infinitesimal area dA
direction: perpendicular to the surface and away from the interior of the surface.
ii) The surface element subtends a solid angle d at an interior point charge q.
s 2d = dA cos ; E =
1
q
(Coulomb's Law)
2
4e 0 s
E dA = E (dA) cos =
E dA =
S
4
q
4e 0
𝑑𝐴Ԧ
𝑠
1
q 2
q
s
d
=
d
2
4e 0 s
4e 0
d =
𝐸
𝐸2
𝒅
𝑑𝐴Ԧ2
𝑞
q
e0
𝐸1
𝑞′
d’
𝑑𝐴Ԧ1
𝑆
Note: The magnitude E dA depends on d and q only. (not on s neither .)
(iii) For an exterior point charge q, two surface elements dA1 and dA2 are intercepted
by the cone of solid angle d .
q
s1 d = −dA1 cos 1; E1 =
4e 0 s12
Note: s12d 0;cos 1 0
q
s2 2d = dA2 cos 2 ; E2 =
4e 0 s2 2
( cos 2 0)
1
2
1
𝑠
q
q
4e 0 s2
4e 0
2
(
s
d ) =
2
2
𝐸2
𝒅
q
q
2
E1 dA1 =
( − s1 d ) = −
d
2
4e 0 s1
4e 0
1
𝐸
1
E2 dA2 =
𝑑𝐴Ԧ
d
𝑑𝐴Ԧ2
𝑞
𝐸1
𝑞′
d’
𝑑𝐴Ԧ1
𝑆
E1 dA1 + E2 dA2 = 0
(iv) In the presence of multiple point charges, the total electric field E is the vector
sum of electric fields due to individual point charges. The contributions to the
outgoing flux of electric field E dA from exterior charges cancel. We have the
S
Gauss' law: E dA =
S
qenc
e0
where qenc is charge enclosed by the Gass' surface S .
A. Gauss' Law Coulomb's Law (for a statioary point charge)
Consider a sphere S with a stationary point charge q at its center.
By symmetry, E = Erˆ and E is independent of and .
E dA = E dA = 4 r 2 E
S
S
By Gauss' Law E dA =
S
4 r 2 E =
q
e0
E=
q
e0
1
q
1 q
ˆ
E
=
Er
=
rˆ (Coulomb's Law)
2
2
4e 0 r
4e 0 r
z
𝑞
x
S
r
y
Some applications of Gauss' law:
A. A charged isolated conductor E = 0 everywhere inside the cnductor
E dA =
S
4
q
4e 0
d =
qenc
e0
i) A solid conductor
E dA = 0 q
S
enc
=0
for any Gauss' surface inside the conductor.
There cannot be any charge anywhere inside the conducror.
All the amount of charge will be distributed on the surface
+
++ +
++
+
+
+
+
++
+
+
+
+
+
𝑆
+ + + +
++
of the conductor.
ii) A conductor with a cavity
For all Gauss' surfaces inside the conductor
and enclosing the cavity,
E dA = 0 qenc = 0
S
There is no net charge on the cavity wall.
Safe to stay in a car during lightning.
+
++ +
++
+
+
+
+
++
+
+
+
+
+
𝑆
+ + + +
++
B. The electric field just outside a charged conductor
The charges on the surface are stationary. E has no tangential comonent.
E is perpendicular to the conductor's surface.
Select a cylindrical Gauss' surface with two infinitesimal ends that are parallel
+
to the conductor's surface.
Gauss' Law EdA =
dA
E=
e0
e0
𝐸=0
C. Cylindrical Symmetry
Ex. An infinitely long line of uniformly distributed charge
By symmetry, E = Erˆ.
+ 𝑆
+
+
+ 𝑑𝐴 𝐸
+
+
𝜎
z
𝑆
𝑟
Select a cylindrical Gauss' surface with z axis as its central axis.
E dA = 2 rhE; q
enc
S
= hl
The Gauss' Law E dA =
S
ℎ
qenc
e0
E=
l
2e 0 r
𝜆
D. Planar Symmetry
Ex1. An infinite plane with uniform surface charge density .
By symmetry 1) E = Ezˆ and
S
𝐸
𝐴
𝜎
2) E is a constant on both ends of the cylinder.
Gauss' Law 2EA =
𝐸
A
E=
e0
2e 0
Ex 2. Two infinitely large conducting plates with uniform surface charge densities
and − , respectively.
A
Gauss' Law EA =
E=
e0
e0
S
𝐸
𝐸=0
𝐸=0
+𝜎
−𝜎
E. Spherical Symmetry
Ex. A uniformly charged shell
q
Gauss' surface : 1) inside the shell, e.g. S1
𝑆1
𝑆2
2) outside the shell e.g. S2
Gauss' Law: 1) inside the shell 4 r 2 E = 0 E = 0
2) outside the shell 4 r 2 E =
q
e0
E=
q
4e 0 r
2
(Note: shell theorem.)
Supplementary
qenc
Gauss' Law E dA =
e0
S
=
1
r dt
t
e
0
Divergence Theorem Edt = E dA
t
E =
t
S
r
(The differential form of Gauss' Law)
e0
𝑆
Chapter 24 Electric Potential
Electrostatic Force F = qE is conservative F = −U
f
Note: = iˆ + ˆj + kˆ ; U is the electric potential energy.
x
y
z
U
Define electric potential V : V =
i
q
𝑑𝑟Ԧ
𝑟Ԧ
𝑟Ԧ + 𝑑𝑟Ԧ
U
−U F
− V = −( ) =
= =E
q
q
q
O
f
f
f
ˆ )
ˆ V + ˆj V + kˆ V ) ( idx
ˆ + ˆjdy + kdz
E
dr
=
(
−
V
)
dr
=
−
(
i
i
i
i x y z
f V
f
V
V
= − (
dx +
dy +
dz ) = − dV = V ( ri ) − V ( rf )
i
i
x
y
z
r
Re-write ri as r0 and rf as r . Let V ( r0 ) = 0. We have V ( r ) = − E dr
r0
Note: V is often selected to be zero at 1) infinity or 2) a grounded conductor.
I. Electric potential due to a point charge q at the origin.
z
Let V = 0 at infinity. By symmetry, V ( r, , ) = V ( r )
q
q r 1
ˆ
ˆ
r
dr
r
=
−
dr
2
4e r 2
4e 0 r
0
r
V ( r ) = − E dr = −
=
r
1
1 1
1 q
( − )=
4e 0 r
4e 0 r
𝑞
r
y
q
x
II. Electric potential due to a group of point charges
n
n
qi
rˆ (a vector sum)
2 i
i =1 4e 0 ri
By superposition principle E = Ei =
i =1
n
qi
rˆi ) dr
2
i =1 4e 0 ri
V = − E dr = − (
n
P
1
1
ˆi i
qi ( 2 rˆi ) rdr
ri
i =1 4e 0
= −
n
1
P
ri
n
1
1 qi
=
= Vi (a scalar sum)
i =1 4e 0 ri
i =1
P
𝑟Ԧ1
𝑞1
𝑟Ԧ2
𝑞2
⋱
𝑟Ԧn
𝑞𝑛
Ex. Electric potential due to an electric dipole
Let V = 0 at infinity. By symmetry, V ( r, , ) = V ( r, )
V ( r, ) =
1
q
1 q
q 1 1
−
=
( − )
4e 0 r+ 4e 0 r− 4e 0 r+ r−
d
d
d
d
By cosine law r+ 2 = ( ) 2 + r 2 − 2( ) r cos ; r− 2 = ( ) 2 + r 2 − 2( ) r cos( − )
2
2
2
2
1
1
2
−
−
q
d2
d
2
2
V ( r, ) =
[( + r − rd cos ) 2 − ( + r + rd cos ) 2 ]
4e 0 4
4
1
1
2
−
−
q 1 d2 d
d
d
=
[( 2 − cos + 1) 2 − ( 2 + cos + 1) 2 ]
4e 0 r 4r
r
4r
r
z
q 1
1d
d2
1d
d2
=
{[1 +
cos + O ( 2 )] − [1 −
cos + O ( 2 )]}
4e 0 r
2r
r
2r
r
+q
(Taylor's expansions)
q 1 d
d2
=
[ cos + O ( 2 )]
4e 0 r r
r
=
p cos
1 p
=
rˆ
2
2
4e 0 r
4e 0 r
1
1 ( qd ) cos
(if r
2
4e 0
r
d)
𝜃
d
−q
𝑟+
𝑟
𝑟−
P
P
V ( r, ) =
p cos
; E = −V
2
4e 0 r
1
ˆ1 ˆ 1
+
+
r
r
r sin
1 ˆ 1
1 p cos
E = −V = −( rˆ +ˆ
+
)
r
r
r sin 4e 0 r 2
In sperical coordinates, = rˆ
p cos
1 p sin ˆ
ˆ
=
r
+
3
3
2e 0 r
4e 0 r
1
z
+q
𝜃
d
−q
𝑟+
𝑟
𝑟−
z
III. Electric potential due to a continuous charge distribution
Ex1. A line of charge
By the cosine law r2 = z 2 + r 2 − 2 rz cos
V ( r, ) =
Note
1
L
l dz
4e 0 0 r
dx
ax 2 + bx + c
=
l L
dz
4e 0 0 z 2 − (2r cos ) z + r 2
=
ln(2 a ( ax + bx + c ) + 2ax + b)
a
L
dz
2
𝜃
𝑟
l
l
V ( r, ) =
[ln(2 z 2 − (2 r cos ) z + r 2 + 2 z − 2 r cos )]zz ==0L
4e 0
2 L2 − (2 r cos ) L + r 2 + 2 L − 2 r cos
l
=
ln[
]
4e 0
2 r (1 − cos )
P
𝑟′
z
L
L2 − (2 r cos ) L + r 2 + L − r cos
l
=
ln[
]
4e 0
r (1 − cos )
l
L2 + d 2 + L)
Note: V ( d , ) =
ln[
]
2
4e 0
d
l
d
P
Ex 2. A uniformly charged disk of radius R
Consider the ring of radius r and infinitesimal thickness dr.
The charge of the ring is dq = dA = (2 r dr )
dV =
1
dq
4e 0
z +r
2
1 (2 rdr )
=
2
P
4e 0
z2 + r2
z
R
rdr
V = dV =
2e 0 0 ( z 2 + r 2 )1/2
R
r
Let X = z + r dX = 2rdr,
2
2
z2 + R2
z + R dX
we have V =
=
2
X
4e 0 z
( X )1/2 4e 0
z
2
2
1
2
=
2
2
( z 2 + R2 − z)
2e 0
dr
Electric Potential Energy of a System of Point Charges
Electric potential energy of a
system of fixed point charges
=
Work that must be done to
assemble the system
System
i.e. move the
charges one by one
from infinity
Work
1. one point charge q1
0=
V1 = 0
1
q1V1
2
2. two point charge q1 , q2
0 + q2 (
1
q2
1 q1
V1 =
;V2 =
4e 0 r21
4e 0 r12
1
q1
1
1
) = q1V1 + q2V2
4e 0 r21
2
2
3. three point charge q1 , q2 , q3
V1 =
V3 =
1
q
q
1 q1 q3
( 2 + 3 );V2 =
( + );
4e 0 r21 r31
4e 0 r12 r32
0 + q2 (
1
=
q q
( 1 + 2)
4e 0 r13 r23
n. n point charge q1 , q2 ,
Vi =
1
qj
4e 0 j i rji
, qn
1
q1
1 q1
1 q2
) + q3 (
+
)
4e 0 r12
4e 0 r13 4e 0 r23
1
1
1
q1V1 + q2V2 + q3V3
2
2
2
1 n
1 n
qV
qV
i i U =
i i
2 i =1
2 i =1
Consider a system of n point charges q1 ,q2
qn . The electric potential at the location
of point charge i is
i −1
n
i −1
Vi = Vi , j = Vi , j + Vi , j =
j i
where Vi , j =
j =1
1
j =i +1
qj
4e 0 ri , j
j =1
1
qj
4e 0 ri , j
n
+
j =i +1
1
qj
4e 0 ri , j
=
j i
qj
1
4e 0 ri , j
,
.
Note: ri , j = rj ,i qiVi , j =
1 qi q j
1 q j qi
1
=
= q jV j ,i = ( qV
i i , j + q jV j ,i )
4e 0 ri , j
4e 0 rj ,i
2
The charges are moved one by one from infinity to assemble the system.
i −1
The work needed to move the ith point charge is qV
i i, j
j =1
n
The total work required to assemble the entire system is
i −1
qV
i = 2 j =1
= ( q2V2,1 ) + ( q3V3,1 + q3V3,2 ) + ( q4V4,1 + q4V4,2 + q4V4,3 ) +
i i, j
( qnVn,1 + qnVn,2 +
n i −1
1 n i −1
1 n i −1
= ( qV
[ q jV j ,i + ( qV
i i , j + q jV j ,i ) =
i i , j )]
2 i =2 j =1
2 i =2 j =1
i = 2 j =1
+ qnVn,n −1 )
n i −1
1 n i −1
1 n i −1
( qV
[ q jV j ,i + ( qV
i i , j + q jV j ,i ) =
i i , j )]
2 i =2 j =1
2 i =2 j =1
i = 2 j =1
1
= [( q1V1,2 ) + ( q1V1,3 + q2V2,3 ) + ( q1V1,4 + q2V2,4 + q3V3,4 ) + ( q1V1,n + q2V2,n + + qn −1Vn −1,n )]
2
1
+ [( q2V2,1 ) + ( q3V3,1 + q3V3,2 ) + ( q4V4,1 + q4V4,2 + q4V4,3 ) + ( qnVn ,1 + qnVn ,2 + + qnVn ,n −1 )]
2
n
n
1 n
1 n
= q1V1, j + [ q2V2, j +
+
qV
+ qn −1Vn −1, j ]
i i, j +
2 j =2
2 j =3
j =i +1
j =n
2
1 1
+ [ q2V2, j + q3V3, j +
2 j =1
j =1
i −1
+ qiVi , j
+
j =1
1 n
1 n
= ( qi Vi , j ) = qV
i i
2 i =1
2
j i
i =1
1 n
Therefore, the potential energy of the system is U = qV
i i
2 i =1
1
For a continuous charge distribution U = V r dt
2 t
1 n −1
] + qnVn , j
2 j =1
Ex1. Three point charges: q1 = q, q2 = −4 q, q3 = 2 q
r1,2 = r2,3 = r3,1 = d
q2
1 q2
1 q3
1 ( −4 q + 2 q)
1 2q
V1 =
+
=
=−
4e 0 d 4e 0 d 4e 0
d
4e 0 d
1 q1
1 q3
1 ( q + 2q)
1 3q
V2 =
+
=
=
4e 0 d 4e 0 d 4e 0
d
4e 0 d
V3 =
1 q1
1 q2
1 ( q − 4q)
1 3q
+
=
=−
4e 0 d 4e 0 d 4e 0
d
4e 0 d
1 3
1
U = qV
=
( q1V1 + q2V2 + q3V3 )
i i
2 i =1
2
d
d
q1
1
1 2q
1 3q
1 3q
1 10q 2
= [ q( −
) + ( −4 q)
+ 2 q( −
)=−
2
4e 0 d
4e 0 d
4e 0 d
4e 0 d
d
q3
Ex 2. A sphere of radius R and uniform charge density r
(4 / 3) r 3 r
rr
rˆ
rR
rR
e0
3e 0
2
Gauss' Law 4 r E =
E= 3
3
(4 / 3) R r
R r rˆ
rR
rR
2
e0
3e 0 r
3
r
R R r
r r r
R2 r
r 2
2
ˆ
ˆ
ˆ
ˆ
V (r ) = − E dr = −
r
rdr
−
r
rdr
=
−
(
r
−
R
)
3e r 2
R 3e
3e 0 6e 0
0
0
r 2 r R2
=−
r +
6e 0
2e 0
1
1 R
r 2 r R2
U = V r dt = (−
r +
) r 4 r 2 dr
2 t
2 0 6e 0
2e 0
r
r 2 R 4
2 2
=−
(
r
−
3
R
r )dr
0
3e 0
dr
r
r 2 1 5
4 R 5 r 2
2 3 R
=−
[ r − R r ]0 =
3e 0 5
15e 0
Also, r =
R
2
Q
3Q
U
=
(4 / 3) R 3
20e 0 R
Q
Supplementary
qenc
Gauss' Law E dA =
e0
S
=
1
r dt
t
e
0
Divergence Theorem Edt = E dA
t
S
t
r
(The differential form of Gauss' Law)
e0
𝑆
r
r
E = −V E = − V = − 2V = 2V = −
(Poisson's Equation)
e0
e0
E =
2
2
2
Note: 2 = (iˆ + ˆj + kˆ ) (iˆ + ˆj + kˆ ) = 2 + 2 + 2
x
y
z
x
y
z
x
y
z
When r = 0 2V = 0 (Laplace's Equation)
*Uniqueness Theorem
Let V1 and V2 both satisfy Poisson's Equation
2V1 = −
r
r
and 2V2 = −
2 (V1 − V2 ) = 0
e0
e0
Noting that ( f f ) = (f ) (f ) + f ( f ) = f + f ( 2 f )
2
( f f )dt = f dt + f (2 f )dt
2
t
t
t
By divergence theorem ( f f )dt = ( f f ) dA
t
S
f dt = ( f f ) dA − f (2 f )dt
2
t
t
S
Let f = (V1 − V2 ), we have
2
(
V
−
V
)
d
t
=
[(
V
−
V
)
(
V
−
V
)]
dA
−
(
V
−
V
)[
1 2
1 2 1 2
1 2 (V1 − V2 )]dt
2
t
t
S
Since 2 (V1 − V2 ) = 0, if V1 = V2 on S then (V1 − V2 ) dt = 0.
2
t
Noting that (V1 − V2 ) 0, for (V1 − V2 ) dt to vanish, (V1 − V2 ) must be zero
2
2
t
and therefore V1 = V2 everywhere in t . Also , E = −V E1 = E2
and (V1 − V2 ) = 0 V1 − V2 = a constant. Since V1 − V2 = 0 on S, the constant is zero
V1 = V2 in t .
Summary: The uniqueness theorem states that the solution (V ) [and its gradient ()]
r
of Poisson's Equation ( V = − ) within a volume t is uniquely determined
e0
2
by the potential on the surface S enclosing that volume.
Chapter 25 Capacitance
Capacitor: A device in which electrical energy can be stored.
I. Isolated conductor
Consider an isolated conductor with charge q.
r=0
2V ( r ) = 0
𝐸=0
𝑉
1
𝑞
It stores an electrical energy U = qV .
2
Note: E = 0 inside the conductor so all charges have the same potential V .
Recall that the electric field just outside a charged conductor has magnitude E =
E ( r ) = −V ( r ) ( r ) = e 0 E ( r ) = e 0 −V ( r ) on the surface of the conductor.
If the charge of the conductor is changed to q such that V = aV on the surface,
V ( r ) = aV ( r ) is apparently a valid solution for the Laplace's equation outside
the conductor [i.e. 2V ( r ) = 0 2V ( r ) = a2V ( r ) = 0] in the space between
the conductor surface and infinity and also satisfies the boundary condition
(i.e. V ( r ) is aV on the conductor surface and zero at infinity).
.
e0
By uniqueness theorem V ( r ) = aV ( r ) [i.e. E ( r ) = aE ( r )] is unique.
We have ( r ) = e 0 −V ( r ) = ae 0 −V ( r ) = a ( r ) on the conductor surface.
q = dA = a dA = aq q V
S
S
Let q = CV . C is the capacitance of the isolated conductor.
1
1
q2
2
U = qV = CV =
2
2
2C
V ( r ) = aV ( r ) ?
V (r )
r=0
2V ( r ) = 0
𝐸=0
𝑉
E ( r ) = −V ( r )
r=0
𝐸=0
2 [aV ( r )] = 0 OK!
V’=a𝑉
uniqueness theorem
𝑞
E = −aV = aE
′
𝑞′
II. A capacitor of two conductors (called plates)
charged with + q and − q, respectively.
+
V+ − V− = − E dr
−
Let V = V+ − V− . If q is changed to q such that V = aV (i.e. V+ = aV+ , V− = aV− ),
V ( r ) = aV ( r ) is apparently a valid solution of Laplace's equation and satisfies the
boundary condition. By uniqueness theorem, E = −V ( r ) = − aV ( r ) = aE is unique.
= e 0 E = ae 0 E = a (on the surfaces of the two conductors ).
q = a dA = a dA = aq
S+
S+
Since V = aV q = aq, we have q V
Let q = C V C is the capacitance of the capacitor
𝑉+
+𝑞
r=0
V (r ) = 0
2
𝑉−𝑞
V is customarily written as V for historical reasons. q = CV
Electrical energy stored in the capacitor
1
1
1
1
qV+ + ( − q)V− = q(V+ − V− ) = qV
2
2
2
2
1
1
q2
2
U = qV = CV =
2
2
2C
Note: An isolated conductor can be viewed as
U=
a capacitor with a missing plate located at infinity.
To calculate the capacitance of a capacitor:
i) Calculate the electric field from charge distribution using Gauss' law.
r
S E dA = e 0 or E = e 0
q
ii) Calculate the potential difference
+
V = V+ − V− = − E dr
−
iii) C =
q
V
Ex1. A parallel-plate capacitor
i) EA =
q
e0
E=
q
e0 A
+
+𝑞
𝐴
q
qd
ˆ =
ii) V = − E dr = − ( −
zˆ ) zdz
−
0
e0 A
e0 A
iii) C =
−𝑞
d
𝑑
q e0 A
=
V
d
Ex 2. A cylindrical capacitor
i) E 2 rL =
q
e0
E=
+
q
2e 0 rL
q
q
b
ˆ =
ii) V = − E dr = − (
rˆ) rdr
ln
−
b 2e rL
2e 0 L a
0
iii) C =
q
2e 0 L
=
V ln(b / a )
a
𝑏
𝑎
+𝑞
−𝑞
L
Ex 3. A spherical capacitor
i) E 4 r 2 =
q
e0
E=
q
4e 0 r 2
+
a
q
−
b
4e 0 r
ii) V = − E dr = − (
ˆ =
rˆ) rdr
2
q
1 1
( − )
4e 0 a b
+𝑞
a
𝑏
q
4e 0
4e 0ab
iii) C = =
=
V ( 1 − 1 ) (b − a )
a b
−𝑞
Ex 4. An isolated conducting sphere
i) E 4 r 2 =
q
e0
E=
q
4e 0 r 2
+
a
q
−
4e 0 r
ii) V = − E dr = − (
iii) C =
+𝑞
q 4e 0
=
= 4e 0a
1
V
( )
a
a
ˆ =
rˆ) rdr
2
q
1 1
q
( − )=
4e 0 a
4e 0a
Capacitors in parallel
n
q = qi
i =1
q
Ci = i
V
+
n
−
𝑞n
𝐶n ⋯
𝑉
𝑞2
𝑞1
𝐶2
𝐶1
qi
n
n
q
q
Ceq = = i =1 = i = Ci
V
V
i =1 V
i =1
+𝑞
𝑉1 𝐶1
−𝑞
Capacitors in series
n
V = Vi
i =1
Ci =
q
q
Vi =
Vi
Ci
n
q
q
q
1
1
1
Ceq = = n
= n
= n
=
q
1
V
Ceq i =1 Ci
V
i
i =1
i =1 Ci
i =1 Ci
+
−
𝑉
+𝑞
𝑉2 𝐶2
−𝑞
⋮
+𝑞
𝑉n 𝐶n
−𝑞
Energy stored in an electric field.
+𝑞
𝐴
A. Energy stored in a parallel-plate capacitor
Recall E =
q
qd
e A
= ;V =
V = Ed ; C = 0
e0 A e0
e0 A
d
−𝑞
𝑑
for a parallel-plate capacitor.
1
1
1
1
( + q)(V+ ) + ( − q)(V− ) = q(V+ − V− ) = qV
2
2
2
2
1q 2 1
1 e 0 A qd 2 1 e 0 A
1
1
2
2
2
=
V = CV =
(
) =
( Ed ) = e 0 E ( Ad ) = e 0 E 2t
2V
2
2 d e0 A
2 d
2
2
Also, U =
dU 1
= e 0 E 2 (energy density)
dt 2
The energy stored in a parallel-plate capacitor can be alternatively viewed as energy
stored in the electric field with energy density
plates.
1
e 0 E 2 in the space between the two
2
B. General derivation of energy stored in an electric field.
r
; (VE ) = V E + V E
e0
1
1
e0
U = V r dt = V (e 0 E )dt = V Edt
Gauss' Law E =
t
2
=
e0
2
t
2
(VE )dt −
t
2
e0
t
t
V Edt
t
2
𝑆
By divergence theorem (VE )dt = VE dA.
t
S
Also, E = −V V Edt = − E Edt = − E 2dt
t
U =
e0
2
VE dA +
S
t
e0
2
2
E
dt =
t
t
e0
1
2
VE
dA
+
(
e
E
)dt
0
S
t
2
2
Let t → V → 0 at infinity
e0
2
r (r )
VE dA →
S
e0
VE dA = 0
2
1
1
U = ( e 0 E 2 )dt = ( e 0 E 2 )dt , t is the space where E 2 0
2
t 2
dU 1
= e 0 E 2 (energy density)
dt 2
Capacitor with a dielectric
Dielectric: insulating material e.g. plastic
1837 Michael Faraday C = k Cair , k 1
For a given q, V =
q
q
V
V V / k E0
=
= 0 ; V = Ed E = = 0
=
C k Cair k
d
d
k
+𝑞
+𝑞
𝐴
𝐴
−𝑞
dielectric
𝑑
𝑑
𝐶
𝐶𝑎𝑖𝑟
*In general, E =
E0
k
−𝑞
.
1
q
e.g. 1. A point charge inside a dielectric E =
rˆ
2
4e 0k r
2. The electric field just outside an isolated conductor imersed in a dielectric
E=
nˆ
ke 0
Question: Is Gauss' law still valid in the presence of dielectrics?
Answer: Yes
Question: Why?
Answer: Bound charges are generated on the dielectrics in the presence of
applied electric field. And, the total charge is the sum of free charge
and bound charge.
Theoretical model to explain the effect of reduced electric field in dielectric.
polar: permanent electric dipoles → aligned by external electric field
+electric field due to other dipoles
1. dielectric non-polar:electric dipole moments induced by external electric field
+electric field due to
other dipoles
2. It can be shown that the dipole moment per unit volume (electric polarization) P is
proportional to the total electric field (external + dipole-generated) E
in the dielectric: P = e 0 e E ( e is the dielectric's electric susceptibility)
3. Consider a volume of polarized dielectric with electric polarization P.
The potential dV at an exterior point r = ( x, y , z ) due to a volume element dt
Pdt r − r
at an interior point r = ( x, y , z) is dV =
2
4e 0 r − r r − r
1
[Recall the potential due to a dipole V ( r, ) =
𝑧
𝑑t′
𝑟′
p cos
1 p
=
rˆ]
2
2
4e 0 r
4e 0 r
1
𝑟Ԧ − 𝑟’
Ԧ
𝑟Ԧ
𝑦
𝑥
1
−
1
2
2
2
Note: (
) = (iˆ
+ ˆj
+ kˆ )[( x − x) + ( y − y ) + ( z − z ) ] 2
r − r
x
y
z
1
1
=−
[ −2( x − x)iˆ − 2( y − y ) ˆj − 2( z − z )kˆ]
2 ( ( x − x) 2 + ( y − y ) 2 + ( z − z ) 2 ) 3
=
=
1
( x − x)iˆ + ( y − y ) ˆj + ( z − z )kˆ
( ( x − x) 2 + ( y − y ) 2 + ( z − z ) 2 ) 2
( x − x) 2 + ( y − y ) 2 + ( z − z) 2
r − r
1 Pdt r − r
1
1
dV
=
=
P
(
) dt
2
2
4e 0 r − r r − r 4e 0
r − r
r − r r − r
1
Noting (
dV =
1
4e 0
V = dV =
1
1
1
P) = (
)P + (
) P
r − r
r − r
r − r
P (
1
1
1
1
1
1
) dt =
(
P )dt −
(
) Pdt
r − r
4e 0
r − r
4e 0 r − r
(
4e 0 t
1
1
− P
P )dt +
dt
t
r − r
4e 0
r − r
and by divergence theorem
1
(
4e 0 t
1
1
P nˆ
P )dt =
dA
S
r − r
4e 0
r − r
we have V =
=
P nˆ
1
− P
dA
+
dt
S
t
4e 0
r − r
4e 0
r − r
1
b
1
4e r − r
0
S
dA +
rb
1
dt
t
4e
r − r
0
b = P nˆ and r b = − P are the surface and volume bound charge distributions
induced by the electric field.
r r f + r b r f − P r f − e 0 e E
Gauss' Law E = =
=
=
e0
e0
e0
e0
(Note:P = e 0 e E ),
r f : free charge density e 0 (1 + e ) E = r f
rf
Define dielectric constant k = 1 + e . E =
e 0k
rf
E0
Noting that E0 =
in the absence of dielectric, we have k E = E0 E = .
e0
k
Define electric displacement D = e 0k E.
Gauss' law becomes D = r f ,
D dA = t Ddt = t r dt = q
S
f
f ,enc
Ex1. Parallel-plate capacitor.
DA = q D =
q
=f;
A
+𝑞
−𝑞𝑏
f
f
E
E=
=−
zˆ = ( −
zˆ ) / k = 0
e 0k
e 0k
e0
k
D
Note: P = e 0 e E = e 0 (k − 1) E
f
k −1
= e 0 (k − 1)( −
zˆ ) = −(
) f zˆ
e 0k
k
𝐴
𝑑
+𝑞𝑏
−𝑞
dielectric
𝐶
r b = − P = 0
k −1
k −1
ˆ
ˆ
−
(
)
z
z
=
−
(
) f upper conductor/dielectric interface
f
k
k
b = P nˆ =
−( k − 1) f zˆ ( − zˆ ) = ( k − 1) f lower conductor/dielectric interface
k
k
f
k −1
−
(
)
=
upper conductor/dielectric interface
f
f
k
k
f + b =
− + ( k − 1) = − f lower conductor/dielectric interface
f
f
k
k
Ex 2. A point charge q inside a dielectric.
q
;
2
4 r
D 4 r 2 = q D =
E=
D
e 0k
=
q
1 q
E0
ˆ
ˆ
r
=
(
r
)
/
k
=
4e 0 r 2k
4e 0 r 2
k
Note: P = e 0 e E = e 0 (k − 1) E
= e 0 (k − 1)(
q
k −1 q
ˆ
r
)
=
(
)
rˆ
2
2
4e 0 r k
k 4 r
1 ( r 2 Ar )
1 ( A sin )
1 A
In spherical coordinates, A = 2
+
+
r
r
r sin
r sin
1
k −1 q
1 k −1 q
r b = − P = 2 [ r 2 (
)
]
=
[(
) ]=0
2
2
r r
k 4 r
r r
k 4
k −1 q
k −1
ˆ
qb = lim( −4 r 2 rˆ P ) = lim[ −4 r 2 rˆ (
)
r
]
=
−
(
)q
2
r →0
r →0
k 4 r
k
k −1
q
q f + qb = q − (
)q =
k
k
Energy stored in an electric field in the presence of dielectrics.
In the presence of dielectrics, r = r f + r b . Again, we move the charges one by one
from infinity to assemble the system. Since the bound charges are simply induced on the
dielectrics , the energy for moving the bound charges from infinity should be excluded.
1
1
1
Therefore, U = V r dt − V r bdt = V r f dt .
2 t
2 t
2 t
1
By Gauss' law D = r f U = V Ddt . Noting (VD ) = V D + V D
2 t
1
1
U = (VD )dt − V Ddt .
2 t
2 t
r (r )
t
By divergence theorem (VD )dt = VD dA.
t
S
Also, E = −V V Ddt = − E (e 0k E )dt = −e 0k E 2dt
t
t
t
𝑆
1
1
1
1
2
VD
dA
+
(
e
k
E
)
d
t
;
Let
t
→
V
D
dA
→
0 D dA = 0
0
S
t
S
2
2
2
2
1
1
U = ( e 0k E 2 )dt = ( e 0k E 2 )dt , t is the space where E 2 0
2
t 2
dU 1
dU 1 2
= e 0k E 2 (energy density); Define e = e 0k
= eE
dt 2
dt 2
U =
Ex. A parallel-plate capacitor inserted with a dielectric.
1) The capacitor is charged with q :
q
D
q
E
E=
=
= 0
A
e 0k e 0k A k
DA = q D =
1 2
1
q 2
q 2d
q2
U = e E Ad = e 0k (
) Ad =
=
2
2
e 0k A
2e 0k A 2(ke 0 A / d )
e0 A
q2
q2
Note: i) C = k Cair = k (
) U =
=
d
2(ke 0 A / d ) 2C
ii) E =
E0
1
1
E
1 1
U
U = e E 2 Ad = e 0k ( 0 ) 2 Ad = ( e 0 E0 2 Ad ) = 0
k
2
2
k
k 2
k
+𝑞
+𝑞
𝐴
𝑈0 −𝑞
𝐸0
𝑑
𝐶𝑎𝑖𝑟
𝐴
𝑈
𝑑
dielectric
𝐶
𝐸
−𝑞
2) A voltage V is applied to the capacitor:
V
1 2
1
V 2
ke 0 AV 2 1 e 0 A 2
E = = E0 U = e E Ad = e 0k ( ) Ad =
= k(
)V
d
2
2
d
2d
2
d
e A
1 e A
1
Note: i) C = k Cair = k ( 0 ) U = k ( 0 )V 2 = CV 2
d
2
d
2
1
1
ii) E = E0 U = e E 2 Ad = e 0k E0 2 Ad
2
2
1
= k ( e 0 E0 2 Ad ) = kU 0
2
𝐴
𝐴
𝑉
𝑈0
+
−
𝐸0
𝑑
𝐶𝑎𝑖𝑟
𝑉
+
− 𝑑
𝑈
dielectric
𝑬
𝐶
Chapter 26 Current and Resistance
Definition of Current
dq charge pass through a surface in time dt
Current i =
dt an infinitesmal time interval
Current Density J
i
Current i =
J
dA
Note:
J
=
S
A
if the current is uniform across the surface
and parallel to dA.
Drift Velocity vd
i) Electrons move randomly in conductors with speeds~106 m / s.
ii) When a current exists, the randomly moving electrons tend to drift
with a velocity vd opposite to the direction of applied field.
vd ~ 10−5 − 10−4 m / s.
iii) J = ( ne)vd
i q / t ( ne) Al / t
l
Note: J = =
=
= ( ne) = ( ne)vd
A
A
A
t
𝑙
𝐴 q=(ne)Al
𝑡=0
𝑙
𝐴
𝑞
𝑡=𝑡
Ohm's Law
i) V = iR for some conducting devices
R : resistance
ii) E = r J for some materials
r : resistivity Note: conductivity =
1
r
J =E
*Dependence of r on temperature (empirical): r − r 0 = r 0 (T − T0 ), T0 = 293K
: temperature coefficient of resistivity
Consider a conducting device of length L, cross sectional area A and
resistance R made of a conducting material of resistivity r .
V = E L; i = JA
V EL
E L
By Ohm's law 1) R = =
= ( )( )
i
JA
J A
E
2) r =
J
L
R=r
A
𝑉
𝑖
𝐿
𝐴
A microscopic model for Ohm's law.
1) Conduction electrons are free to move in the sample of metal.
2) Electrons collide only with atoms of the metal (not with one another).
Let t be the average time between collisions.
eE
is the cceleration of electrons due to
m
the applied electric field (e : charge of electron, m : mass of electron).
The drift speed vd = at , where a =
vd =
eE
t
m
eE
ne 2t
m
J = nevd = ne t =
EE= 2 J
m
m
ne t
m
Let r = 2 . We have Ohm's law E = r J
ne t
Note: 1)For semiconductors, n increases with T r decreases with T .
2)For conductors, n ~ constant, t decreases with T r increases with T .
𝑉
Power in electric circuits.
𝑖
dq = idt
dU = dq V = idt V
𝑅
dU
P=
= iV (general)
dt
By Ohm's law V = iR , i =
V
R
V2
P=i R=
(special case) Resistive dissipation
R
2
Chapter 27 Circuits
i) Direct-current (dc) circuits: charge flows in one direction.
ii) Alternating-current (ac) circuits: charge flow reverses direction periodically.
Electrical Elements
i) emf devices : To provide an emf e to do work on charge carriers
creating a potential difference V .
+
Ɛ
Ɛ
𝑉
V = e − ir
dW Work done by the emf device on dq
−
emf e =
;
r : internal resistance
dq Charge passing through the emf device in time interval dt
Battery Generator
V = e (if r = 0)
ii) Resistor : Energy is thermally dissipated in a resistor as charge flows through it.
dq
; P = iV = i 2 R
dt
iii) Capacitor: To store electric energy.
V = iR = R
q
1
q2
2
V = ; U E = CV =
C
2
2C
iv) Inductor: To store magnetic energy.
V = e L = −L
2
di
d q
1
= − L 2 ; U B = Li 2
dt
dt
2
𝑅
𝐶
Capacitor
𝐿
Inductor
Resistor
Kirchhoff's Rules
i) Kirchhoff's loop rule (Kirchhoff's voltage law):
The algebraic sum of potential changes encountered in a complete traversal of any
loop of a circuit must be zero.
𝑖
Ex. Resistance in series
By Kirchhoff's loop rule, e − iR1 − iR2 −
iRn = 0
n
n
n
j =1
j =1
j =1
+
−
e − i R j = 0 i = e / R j Req = R j
𝑅1
𝑅2
Ɛ
⋮
𝑅n
ii) Kirchhoff's junction rule (Kirchhoff's current law):
The sum of current entering a junction (node) is equal to the sum of current leaving
that junction.
Ex1. Resistance in parallel
n
n
e =e
junction rule i = i j
i
=
j
=
1
Req
j =1 R j
loop rule e − i R = 0
j j
n
1
1
=
Req j =1 R j
+
−
𝑅1
𝑖1
Ɛ
𝑖
𝑅2 ⋯
𝑖2
𝑅n
𝑖𝑛
Ex 2. A multiloop circuit is plotted in the figure. Obtain the expressions for i1, i2 , i3
for given e1 , e 2 , R1 , R2 , R3.
junction rule i1 + i3 = i2
loop rule e1 − i1R1 + i3 R3 = 0
e 2 + i2 R2 + i3R3 = 0
R1e 2 + i1R1R2 + i3 R1 ( R2 + R3 ) = 0
R2e1 − i1R1R2 + i3 R2 R3 = 0
Ɛ2
Ɛ1
𝑅1
𝑅3
𝑅2
𝑖1
𝑖3
𝑖2
R2e1 + R1e 2
i3 = −
R1R2 + R2 R3 + R3 R1
i1 =
e1 + i3R3 = ( R2 + R3 )e1 − R3e 2
R1
i2 = i1 + i3 =
R1R2 + R2 R3 + R3 R1
R3e1 − ( R1 + R3 )e 2
R1R2 + R2 R3 + R3 R1
RC circuits
Kirchhoff's loop rule
𝑖
R
q
dq
q
e − iR − = 0 e − R − = 0
+
𝐶
C
dt
C
Ɛ
−
dq
1
e
+
q=
dt RC
R
q = qc + q p where qc is the complementary function and q p is a particular solution.
t
−
dqc
1
1 t
1
t
t
+
qc = 0; try qc = e e +
e = 0 = −
qc = Ke RC
dt RC
RC
RC
For a particular solution try q p = A (a constant)
1
e
1
e
+
qp =
A = A = Ce q p = Ce
dt RC
R
RC
R
dq p
q(t ) = qc (t ) + q p (t ) = Ke
−
t
RC
+ Ce
i) If q(0) = 0 K = −Ce q(t ) = Ce (1 − e
−
t
RC
) charging a capacitor
i) If e = 0 and q(0) = CV0 K = CV0 q( t ) = CV0e
−
t
RC
discharging a capacitor
q(t ) = Ce (1 − e
q(t ) = CV0e
t
−
RC
−
t
RC
) charging a capacitor
discharging a capacitor
Note:
𝑖
+
−
R
Ɛ
𝐶
Define capacitive time constant t = RC
t
q(t ) = Ce (1 − e t ) charging a capacitor
−
−
t
q(t ) = CV0e t discharging a capacitor
t
When i) t → e t → 0 q() = Ce the capacitor is fully charged
−
q( ) = 0 the capacitor is totally discharged
−
t
ii) t = t = RC e t = e −1
37% q(t ) = Ce 63% charging a capacitor
q(t ) = CV0 37% discharging a capacitor
Chapter 28 Magnetic Fields
A permanent magnet (natural or artificial ) produces a magnetic field B that
i) applies a magnetic force FB to a moving charged particle of charge q and
velocity v such that
FB = qv B (definition of magnetic field B ).
ii) deflects a magnetic compass (torque on a magnetic dipole)
Note:
1) Magnetic fields produced by permant magnets are due to
the spins of elementary particles.
2) Magnetic fields can also be prduced by electromagnets due to
electric current.
3) Lorentz force F = FE + FB = qE + qv B
Lorentz force F = FE + FB = qE + qv B
+
Applications
A. Cross Fields: E ⊥ B
Note: If v ⊥ B and v ⊥ E then qE ( qv B ).
B
E
I. Discovery of the electron (J. J. Thomson, 1897)
−
1. E = 0, B = 0, set y = 0
2. E 0, B = 0, measure y
− qE
qEL2
y=−
m
2mv 2
L
L = vt t =
v
2
m
EL
1
=−
2
y = a yt 2
q
2
yv
2
3. Maintain E and adjust B until y = 0
y
+
FE = − qE a y =
qE = qvB v =
B
E
−
Vacuum
pump
E
B
Such particles with a mass more than 1000 times
m
B 2 L2
=−
lighter than hydrogen are found in all matter.
q
2 yE
Discovery of the electrons.
y=0
II. The Hall effect (Edwin H. Hall, 1879)
B = Bkˆ; F = qv B
B
y
d
z
Carriers are subjected to the magnetic force and accumulte x
on one side of the bar until the magnetic force is balanced
V2
i
d
V1
by the electrostatic force due to carrier accumulation.
F = FE + FB = qE + qvd B = 0 E = − vd B
V
B
A
2
V = V2 − V1 = − E dr = − E d ˆj = (vd B ) ˆjd
1
1. Sign of carrier charge
i) negative carrier charge q = − e vd = − vd iˆ
V = ( vd B ) ˆjd = [( −vd iˆ) Bkˆ] ˆjd = − vd Bd [(iˆ kˆ) ˆj ] = vd Bd 0
ii) positive carrier charge q = e vd = vd iˆ
V = ( vd B ) ˆjd = [( vd iˆ) Bkˆ] ˆjd = vd Bd [(iˆ kˆ) ˆj ] = −vd Bd 0
+++++++++
−e
−−−−−−−−−
−−−−−−−−−
+e
+++++++++
2. Carrier concentration
eE = evd B vd =
V
ne V
E V /d
JBd (i / ld ) Bd
Bi
=
=
; J = (ne)vd =
n=
=
=
B
B
Bd
Bd
eV
eV
V le
l : thickness of the bar
A charged particle in a uniform magnetic field B = Bkˆ
F = qv B = q(v xiˆ + v y ˆj + v z kˆ) Bkˆ = qB ( −v x ˆj + v yiˆ + 0) = qB (v yiˆ − v x ˆj )
dv x ˆ dv y ˆ dv z
ˆ
+j
+k
)
Newton's 2nd Law F = ma = m(i
dt
dt
dt
Note F = qv B F ⊥ v v = v x2 + v 2y + v z2 is a constant.
d 2v x qB dv y
qB 2
dv x qB
=
=
−
(
) vx
dt 2
dt = m v y
m dt
m
2
d vy
qB
qB dv x
qB
dv y
=−
vx 2 = −
= − ( )2 v y
m
m dt
m
dt
dt
dv z
=
0
dt
v (t) = v a constant
z
z
d 2v x
qB 2
+
(
) v x =0
dt 2
m
2
d vy
qB 2
2 + ( ) v y =0
m
dt
v (t) = v
z
z
d 2v x
qB 2
qB 2
qB
t
2
+
(
)
v
=0
try
v
=
e
+
(
)
=
0
=
i
x
x
dt 2
m
m
m
qB
qB
v x (t ) = C1 exp(i
t ) + C2 exp( −i
t)
m
m
Select a coordinate system such that v x (0) = 0
v = v x2 + v 2y + v z2 v y (0) = v 2 − [v x (0)]2 − vz2 = v 2 − vz2
qB
v x (0) = 0 C2 = −C1 v x (t ) = 2iC1 sin( t )
m
dv
qB
m dv x
qB
Noting x =
v y v y (t ) =
= 2iC1 cos( t )
dt
m
qB dt
m
qB
2
2
v
(
t
)
=
v
−
v
sin(
t)
z
x
m
qB
v y (0) = v 2 − v z2 C1 = v 2 − v z2 / 2i v y (t ) = v 2 − v z2 cos( t )
m
v z (t) = v z
m
qB
2
2
qB
dx (t )
2
2
x
(
t
)
=
−
v
−
v
cos(
t ) + Cx
v
(
t
)
=
v
−
v
sin(
t
)
=
z
x
z
qB
m
m
dt
qB
dy (t )
m
qB
2
2
2
2
v
(
t
)
=
v
−
v
cos(
t
)
=
y
(
t
)
=
v
−
v
sin(
t) + Cy
y
z
z
m
dt
qB
m
dz (t )
z(t) = v z t + C z
v
(t)
=
v
=
z
z
dt
m
qB
2
2
m
x
(
t
)
=
−
v
−
v
cos(
t)
2
2
z
x
(0)
=
−
v
−
v
qB
m
z
qB
C x =0
m
qB
Let y (0) = 0
C y =0 y (t ) =
v 2 − v z2 sin( t )
qB
m
z(0) = 0
C =0
z
z(t) = v z t
m
Note: x 2 + y 2 =
v 2 − v z2 = a constant The charged particle follow a
qB
qB
m
2
2
helical path of radius
v − v z and angular speed =
.
qB
m
Period T=
2
=
2 m
mv
. If v z = 0 circular motion of radius r =
qB
qB
Magnetic Force on a Current-Carrying Wire
Consider a wire of length L carrying a current i in a uniform magnetic field B.
Let t be the time for charge carriers to travel a distance L.
The total charge moving in the wire q = it = i
L
qvd = iL
vd
FB = ( qi vd B ) = ( qi )vd B = qvd B = iL B
i
i
Note: For curved wire or non-uniform B, we can consider infinitesimal segments
of the wire. Each infinitesmal segment can be treated as straight and B is treated
as uniform within it. Therefore,
dFB = idL B
B
i
L
Torque on a current loop
FB = iL B F1 = − F3 ; F2 = − F4
i
z 𝜃 𝑛
ො
𝐹Ԧ4
F = F1 + F2 + F3 + F4 = 0
Note: F1 = F3 = iaB; F2 = F4 = ibB sin(
B
𝐹Ԧ1
1. Force exerted on the loop
2
i
− ) = ibB cos
𝑎
𝑏
y
x
i
i
𝐹Ԧ2
𝐹Ԧ3
2. Torque exerted on the loop
i) F2 and F4 zero net torque
b
b
b
b
ii) F1 and F3 torque t = sin F1 + sin F3 = sin iaB + sin iaB
2
2
2
2
= iabB sin (about y axis)
Define magnetic dipole moment = iabnˆ. We have t = B
The loop under the torque t tends to rotate so that nˆ becomes parallel to B.
Recall, for an electric dipole: t = p E U ( ) = − p E where U (90) = 0.
Analogously, t = B U ( ) = − B where U (90) = 0. magnetic potential energy
t 0 = (iab) B sin (about y axis)
=0 B sin = iA0 B sin
B
𝐹Ԧ1
i
𝐹Ԧ4
0 = iA0nˆ; t 0 = 0 B
𝑎
i
A0 : area of the rectangle
i
𝐹Ԧ2
𝑎
i
𝐹Ԧ1
i
𝑎
𝑏
i
𝑏
i
𝐹Ԧ2
𝐹Ԧ1
𝐹Ԧ1
i
𝐹Ԧ3
𝐹Ԧ4
i
i
𝐹Ԧ2
i
𝜃 𝑛ො𝐹Ԧ
3 i
𝑎
𝑏
2𝑎
𝐹Ԧ3
ො
z 𝜃𝑛
COM
i
𝐹Ԧ3
𝐹Ԧ4
𝐹Ԧ2
i
B
i
l
i
B
i
𝐹Ԧ4
=
i
y
x
𝜃 𝑛ො
i
𝐹Ԧ3
i
z 𝜃 𝑛
ො
𝐹Ԧ4
𝜃 𝑛ො
𝐹Ԧ4
𝑏
B
𝐹Ԧ1
B
𝐹Ԧ1
i
y
x
𝐹Ԧ2
2𝑏
i
𝐹Ԧ2
i
i
t = l sin i (2a) B + (b − l ) sin iaB + (2b − l ) sin iaB
= i (3ab) B sin == i (3 A0 ) B sin = [iA]B sin = B sin
= i (3 A0 )nˆ = iAnˆ
; A : area enclosed by the loop.
t = B
(about an axis through
the COM)
𝐹Ԧ3
Note:
1. A planar current loop of any shape can be constructed using infinite number of
infinitesimal rectangular current loops.
= iAnˆ
; A : area enclosed by the loop.
t = B
= NiAnˆ
2. For a coil of N turns
t = B
B
i
z 𝜃 𝑛
ො
A
y
x
i
Chapter 29 Magnetic Fields Due to Currents
𝑑 𝑠Ԧ
Current in a wire deflect magnetic compass.
Electric currents can produce magnetic fields.
0 ids rˆ
Biot-Savart Law: dB =
4 r 2
Ex1. A long straight wire
ids rˆ
0 ids sin
dB = 0
dB
=
4 r 2
4
r2
rd
rd
i
sin( + )
i
cos
cos
0 cos
2
= 0
=
4
r2
4
r2
id 0 id 0i cos d
= 0
=
=
4 r
4 r
4
R
0i 2
0i
0i
2
B = dB =
sin − =
cos d =
4 R − 2
4 R
2 R
2
r = rrˆ
P
i
𝑑 𝑠Ԧ
𝑟𝑑𝜙
𝑑 𝑠Ԧ 𝜃
r = rrˆ
𝜙
R
i
P
Note: Force between two parallel currents
Ba =
0ia
2 d
Fba = ib L Ba magnetic force on a length L of wire b
L
0 Liaib
Fba = ib LBa sin 90 =
2 d
d
i)If ia and ib are parallel attract each other
ii)If ia and ib are anti-parallel repel each other
𝑖𝑎
𝑖𝑏
Ex 2. A circular arc of wire
𝑑 𝑠Ԧ
0 ids rˆ
0 iRd sin 2
dB =
dB =
r = rrˆ
4 r 2
4
R2
i
𝜙 R
= 0 d
4 R
i
i
i
B = dB = 0 d = 0 ; For a full circle = 2 B = 0
4 R 0
4 R
2R
i
Ex 3. Along the axis of a circular current loop
ids rˆ
dB = 0
Note: ds ⊥ rˆ
2
4 r
ids iRd 0 iR
dB = 0 2 = 0
=
d
2
2
2
4 r
4 r
4 R + z
By symmetry, B = B
0i
R2
dB = dB cos = dB
=
d
2
2 3/2
2
2
4 ( R + z )
R +z
R
𝑧
dB
𝛼
𝑑 𝑠Ԧ
𝑅
dB
𝜃
0i
R2
0iR 2
0 (i R 2 )
0
B = dB =
d
=
=
=
0 4 ( R 2 + z 2 ) 3/2
2( R 2 + z 2 )3/2 2 ( R 2 + z 2 )3/2 2 ( R 2 + z 2 )3/2
0
0
If z R, B( z )
B
(
z
)
=
2 z 3
2 z 3
For a coil of N turns = Ni R 2 zˆ B( z ) = 0 3 is also valid.
2 z
2
For a current in a wire Biot-Savart Law: dB =
0 ids rˆ
4 r 2
For a continuous current distribution J , i = Jda ids = Jdads = J (dads ) = Jdt
0 J rˆ
0 J rˆ
dB =
dt B = dB =
dt t : the volume where J 0
2
2
t
4 r
4
r
Define vector potential A : B = A
1
1
Note: r = ( x − x)iˆ + ( y − y ) ˆj + ( z − z )kˆ =
r
( x − x) 2 + ( y − y ) 2 + ( z − z ) 2
1
1
1
( ) = −
[2( x − x)iˆ + 2( y − y ) ˆj + 2( z − z )kˆ]
r
2 [ ( x − x) 2 + ( y − y ) 2 + ( z − z ) 2 ]3
1
( x − x)iˆ + ( y − y ) ˆj + ( z − z )kˆ
1 r
rˆ
=−
=
−
=
−
( x − x) 2 + ( y − y ) 2 + ( z − z ) 2 ( x − x) 2 + ( y − y ) 2 + ( z − z ) 2
r2 r
r2
B=
0 J rˆ
0
1
0
1
d
t
=
−
[
J
(
)]
d
t
=
[
(
) J ]dt
2
t
t
t
4
r
4
r
4
r
0
1
1
[
(
J
)
−
( J )]dt [Note J = J ( x, y , z ) J = 0]
t
4
r
r
1
J
J
= 0 [ ( J )]dt = [ 0 dt ] = A A = 0 dt
4 t
r
4 t r
4 t r
=
A=
0 J
dt ; B = A
t
4
r
B = A = ( A) − 2 A in Coulomb gauge for steady states A = 0
B = − 2 A = − 2 [
0 J
0
2 1
d
t
]
=
−
[
J
( )dt ]
t
t
4
r
4
r
1
Note: To calculate ( ), we consider the following derivation:
r
Recall that in electrostatics E = −V
r
2
E
=
−
V
=
r
Gauss' law E =
(differential form)
e0
e0
Also, the electric potential at r due to a stationary point charge q at r is
2
V ( r − r ) =
1
q
and a single stationary point charge q at r
4e 0 r − r
corresponds to a charge density function r ( r ) = q ( r − r).
The Dirac delta function ( r − r) has the following properties:
1. ( r − r) = 0 if r r, ( r − r ) → if r = r ;
2. If t includes r ( r − r)dt = 1 and f ( r ) ( r − r )dt = f ( r )
t
t
r
1
q
q ( r − r )
1
2
− V = (
)=−
2 (
) = −4 ( r − r )
e0
4e 0 r − r
e0
r − r
2
0
0
2 1
2 1
J
(
)
d
t
=
−
J
( )dt
t
4
r
4
r
1
= − 0 J ( r)2 (
)dt = − 0 J ( r )[ −4 ( r − r )]dt = 0 J ( r ).
4
r − r
4
Therefore, B = −
We have the differential form of Ampere's circuital law B = 0 J
Consider a surface S enclosed by a closed curve C.
( B) da = J da = i where i is the total current enclosed by curve C.
By Stokes' theorem ( B ) da = B dr
B dr = i (The integral form of Ampere's circuital law)
0
S
0 enc
S
S
C
0 enc
enc
C
Note:
B =
iˆ
ˆj
kˆ
x
Bx
y
By
B B
B B
B B
= iˆ( z − y ) + ˆj ( x − z ) + kˆ( y − x )
z
y
z
z
x
x
y
Bz
ˆ
Consider a surface element parallel to the x -y plane: da = kdxdy
.
B y
Bx
B da = (
−
)dxdy
x
y
B ( x + dx, y , z ) − B y ( x, y , z ) Bx ( x, y + dy , z ) − Bx ( x, y , z )
=( y
−
)dxdy
dx
dy
= B y ( x + dx, y , z )dy + B y ( x, y , z )( −dy ) + Bx ( x, y + dy , z )( −dx ) + Bx ( x, y , z )dx
ˆ
= B ( x + dx, y , z ) ˆjdy + B ( x, y + dy , z ) iˆ( −dx ) + B ( x, y , z ) ˆj ( −dy ) + B ( x, y , z ) idx
= B ( x + dx, y , z ) ˆjdy + B ( x +dx, y + dy , z ) iˆ( −dx ) + B ( x, y + dy , z ) ˆj ( −dy )
ˆ + O[( dxi ) 2 ]
+ B ( x, y , z ) idx
𝑧
da
ˆj ( −dy )
iˆ( −dx )
iˆ( −dx )
ĵdy
𝑥
𝑦
Ex1. A long straight wire
0i
C B dr = 2 RB = 0ienc = 0i B = 2 R
Ex 2. Magnetic field inside a long straight wire with current
R
P
C
i
r 2 ir 2
ienc = i
= 2
2
R
R
ir 2
0ir
B
dr
=
2
rB
=
i
=
B
=
0 enc
0
C
R2
2 R 2
r2
i
2
Note: Let Aenc = r , ienc = i
=
Aenc = JAenc
2
2
R R
B
i
R
r
Solenoids and Toroids
I. A long solenoid
Outside: Br = 0 (from B da = 0), B =
S
0i
, Bz = 0
2 r
ℎ
Inside: Br = 0 (from B da = 0), B = 0
S
B dr = B h = i
C
z
0 enc
= 0inh (n : number of turns per unit length)
Bz = 0in
A solenoid provides a known uniform magnetic field B = 0in
Note:
V
A
parallel
plate
capacitor
provides
a
known
uniform
electric
field
E
=
d
II. A toroid
B dr = 2 rB = 0ienc = 0 Ni B =
C
0 Ni
2 r
𝑟
Chapter 30 Induction and Inductance
Magnetic flux through a surface S that is enclosed by a curve C :
B = B da (Note: B = BA is B is perpendicular to and uniform over S .)
S
Faraday's Law of Induction:
Induced emf e =
dW
dB
d
= E dr = −
= − B da (Integral Form)
C
dq
dt
dt S
By Stokes' theorem E dr = ( E ) da ; Noting −
C
S
d
B
B
da
=
(
−
) da
S
S
dt
t
B
(Differential form of Faraday's law)
t
( A)
A
A
Note: Since B = A E = −
= −
(E + ) = 0
t
t
t
A
A
By Stokes' theorem ( E + ) dr = [ ( E + )] da = 0 for any curve C
C
S
t
t
A
A
E+
= − ; In electrostatics,
= 0 and E = −V = V
t
t
A
E = −V −
, (V is also called scalar potential)
t
E = −
Generators
B
(Differential form of Faraday's law)
t
Recall the Lorentz force F = qE + qv B
E = −
F = q( E ) + q[ ( v B )] = − q(
B
) + q[ ( v B )]
t
B
( F ) da = − q ( ) da + q [ (v B )] da
S
S t
S
B
And by Stokes' theorem F dr = − q ( ) da + q ( v B ) dr
C
S t
C
B
W = F dr = − q ( ) da + q (v B ) dr
C
S t
C
W
B
e =
= − ( ) da + ( v B ) dr
S t
C
q
Ex1. A stationary loop of area S on the x - y plane in a time-varying magnetic
field B = B sin t kˆ.
0
B
d
dB
) da = − B da = −
= − SB0 cos t
S t
S
dt
dt
v = 0, B = SB0 sin t e = − (
B
e = − S ( ) da + C (v B) dr
t
Ex 2. A rectangular loop of width w and length l rotating
𝐹
𝜔
𝑧
𝑙
𝑤
about the x -axis with an angular speed in a constant
magnetic field B = B kˆ.
𝑥
𝐹
𝑦
B
1
1
= 0 e = ( v B ) dr = {[( w) ]B sin t} l + {[( w) ]B sin t} l
C
t
2
2
d
d "B "
= ( wl ) B sin t = SB sin t = − ( SB cos t ) = −
dt
dt
1
1
d "B "
For any axis : e = {[( w − y1 ) ]B sin t} l − {[( − w − y1 ) ]B sin t} l = −
2
2
dt
Ex 3. A rectangular loop of width w and length l lying on the x -y plane is moving along
the x -axis out of a constant magnetic field B = B kˆ. The speed of the loop is v and one
of the two sides of length w is already out of the magnetic field.
B
= 0 e = ( v B ) dr = ( vB sin 90) w = vBw
C
t
dx
d
d "B "
=
Bw = − [ Bw(l − x )] = −
dt
dt
dt
𝑤
𝐹
𝑧
𝑙
𝑦
𝑣
𝑥
Note:
1. Ex 2 and Ex 3 can be generalized to loops of all shapes. This can be easily seen
if we let the rectangular loops to be infinitesimal then use them to construct finite
loops of any shape.
2. The work done on charge carriers is then dissipated as thermal energy caused by
the resistance of the loops. Induction heating
3. If conducting plates are used to replace the loops eddy current
4. Lenz's law:
The magnetic field due to the induced current opposes the change in the magnetic
flux that induces the current. A convenient way to determine the direction of an
induced current in a loop. (curled -straight right-hand rule)
Inductors and Inductance
Inductor: A device that stores magnetic energy
e.g. A long solenoid of length l and cross sectional area A.
Ampere's law B = 0in
= N B = N ( BA) = ( nl )( 0in ) A = ( 0ln 2 A)i
Definition of self-inductance L =
(in general)
i
L = 0ln 2 A (long solenoid)
For any inductor = Li self-induced emf e L = −
d
di
= −L
dt
dt
RL circuits
𝑅
𝑖
𝐿
Ɛ
Kirchhoff's loop rule
e − iR − L di = 0 L di + Ri = e L = e − Ri − dt =
di
Ri − e
dt
dt
dt
di
L
R
R
− t
− t
t ln( Ri − e )
1
e
RC
− =
+ C e L = e ( Ri − e ) i (t ) = RC e L +
L
R
e R
R
−
L
e
Let A1 = RC and t L = , we have i (t ) = A1e t L +
e R
R
R
t
1
t
−
e
e
i) if i (0) = 0 A = − i (t ) = (1 − e t ) rise of current
L
1
R
R
−
t
ii) if e = 0 and i (0) = i0 A1 = i0 i (t ) = i0e t L decay of current
Energy Stored in a Magnetic Field
dB
di
For any current loop B = Li, Faraday's law e = −
= −L .
dt
dt
As a current is being built up from zero (di 0), as emf in the opposite direction
e = − L di is generated. Threfore, to deliver charge dq through the loop during time
dt
interval dt , work dW = L
dW = L
di
dq
dq has to be done. Note i = .
dt
dt
di
dq
dq = L di = Lidi
dt
dt
i
The work required to build up current i from zero is W = dW = Lidi =
0
This work is stored as magnetic energy U B =
For a system of many current loops U B =
1 2
Li in the magnetic field.
2
1
L ji j 2
2 j
1 2
Li
2
Ex1. A long solenoid of length l and cross sectional area A.
1
1
L = 0ln 2 A U B = Li 2 = 0ln 2 Ai 2
2
2
( 0in) 2
B2
1
B = 0in U B = u B lA =
lA =
lA = 0ln 2 Ai 2
2 0
2 0
2
dU B U B
B2
=
=
( for a long solenoid)
dt
lA 2 0
1
1
1
1
2
U B = L ji j = i j ( L ji j ) = i j ( B , j ) = i j ( B j da )
Sj
2 j
2 j
2 j
2 j
=
1
1
i
[
(
A
)
da
]
=
i j [ Aj dr ]
j
j
S
Cj
j
2 j
2 j
For continuous current distribution i j → Jda
i j [ Aj dr ] → ( Jda ) Aj dr = J A j da dr
Cj
UB =
Cj
Cj
1
1
[
J
A
da
dr
]
=
J Adt , t is the volumn where J 0.
j
C
t
j
2 j
2
By Ampere's law J =
1
0
B UB =
1
( B ) Adt
t
2 0
Noting ( B ) A = ( B A) + B ( A), we have
UB =
=
1
1
[ ( B A)dt + B ( A)dt ] =
[ ( B A) da + B ( B )dt ]
t
t
t
2 0
2 0 S
1
1
[ ( B A) da + B ( B )dt ] =
[0 + B 2dt ]
2 0
2 0
B2
dU B
B 2 magnetic energy
2
=
dt , t is the volumn where B 0. uB =
=
.
t 2
dt
2 0 density
0
Chapter 31 Electromagnetic Oscillations and Alternating Current
LC circuits
𝑖
Kirchhoff's loop rule
q
di
d dq
q
d 2q
1
− −L =0 L ( )+ =0 2 +
q=0
C
dt
dt dt
C
dt
LC
1
d 2q
2
Let =
, we have
+
q=0
2
dt
LC
C
To find two independent solutions for the basis of the solution space, try q = e t .
2e t + 2e t = 0 2 + 2 = 0 = i
q(t ) = Aeit + Be − it
A + B = q0
q(0) = q0
, Let
i0
dq
it
− it
i
(0)
=
i
A
−
B
=
0
i (t ) = dt = Aie − Bie
i
2
2
q0 i0
q0
i0
a = + −
A = 2 − i 2
A = aei
2 2
, Let
− i
q
i
B
=
ae
−
i
0
0
B = + i
0
=
arctan
2
2
q0
𝐿
q(t ) = aei (t + ) + ae − i (t + ) = 2a cos(t + )
Q = 2a
,
Let
dq
I = Q
i (t ) = dt = −2a sin(t + )
Q = q 2 + LCi 2
0
0
= 1
LC
q(t ) = Q cos(t + )
, where
−i0 LC
i
(
t
)
=
−
I
sin(
t
+
)
=
arctan
q0
q0 2
I =
+ i0 2
LC
Note: If i0 = 0 Q = q0 , = 0
Electrical and Magnetic Energy Oscillations
For an LC circuit
q(t ) = Q cos(t + )
Q
,
where
I
=
Q
=
LC
i (t ) = − I sin(t + )
Note:
q2 Q 2
Electrical Energy U E =
=
cos 2 (t + )
2C 2C
1
1
1
Q
Magnetic Energy U B = Li 2 = L[ − I sin(t + )]2 = L[ −
sin(t + )]2
2
2
2
LC
1 Q2
Q2
2
= L[
sin (t + )] =
sin 2 (t + )
2 LC
2C
Q2
Q2
Q2
2
2
UE + UB =
cos (t + ) +
sin (t + ) =
2C
2C
2C
q0 2 + LCi0 2 q0 2 1 2
=
=
+ Li0
2C
2C 2
Damped Oscilations in an RLC circuit
Kirchhoff's loop rule
𝑅
q
di
d dq
dq q
𝑖
− − iR − L = 0 L ( ) + R
+ =0
C
dt
dt dt
dt C
𝐿
C
2
d q R dq
1
2 +
+
q=0
dt
L dt LC
1
d 2 q R dq
2
Let =
, we have
+
+
q=0
2
dt
L dt
LC
To find two independent solutions for the basis of the solution space, try q = e t .
R t
R
2 t
2
e + e + e = 0 + + 2 = 0
L
L
R
R
− ( ) 2 − 4 2
R
R
R
L
= L
=−
i 2 − ( ) 2 , Let =
, = 2 − 2
2
2L
2L
2L
2 t
q(t ) = Ae − t eit + Be − t e − it =e − t (Aeit + Be − it )
dq
i
(
t
)
=
= − e − t (Aeit + Be − it )+e − t (Ai eit − Bi e − it )
dt
q(0) = q0
A + B = q0
Let
i
(0)
=
i
(i − ) A − (i + ) B = i0
0
1
i0
A
=
q
−
i
(
q
+
)
0
0
(i + ) A + (i + ) B = (i + ) q0
2
2
2
(i − ) A − (i + ) B = i0
B = 1 q0 + i ( q0 + i0 )
2
2
2
1 2
i0 2
q0 2 ( q0 + i0 ) 2
q0 +
) =
+
a = ( q0 ) + (
A = a ei
2
2
2
4
4 2
Let
− i
B
=
a
e
i
1
i
= arctan[ −(
q0 + 0 ) / q0 ] = arctan[ −( + 0 )]
2
2 2
q0
q(t ) = e − t (Aeit + Be − it ) = e − t a [ei (t + ) + e − i (t + ) ] = 2a e − t cos( t + )
Let Q = 2a,
we have q(t ) = Q e − t cos(t + ),
( q0 + i0 ) 2
R
R 2
i0
2
where Q = q0 +
,
=
,
=
−
(
)
,
=
arctan[
−
(
+
)]
2
2L
2L
q0
2
q 2 Q 2 −2 t
UE =
=
e cos2 ( t + ) electrical energy stored in C decays with time.
2C 2C
Alternating current (ac) and forced oscillations
emf induced in the loop of a generator e = e m sin d t
For an RLC circuit with a driving emf e = e m sin d t , Kirchhoff's loop rule:
q
di
d 2 q R dq
1
e
e m sin d t − − iR − L = 0 2 +
+
q = m sin d t
C
dt
dt
L dt LC
L
2
R
1
e
d
q
dq
m
Let =
, =
,A=
2 + 2
+ 2 q = A sin d t
2L
L
dt
dt
LC
q(t ) = qc (t ) + q p (t )
2
The complementary solution qc (t ) is the general solution of
𝑅
𝑖
𝐿
C
Ɛ
d q
dq
2
+
2
+
q = 0.
2
dt
dt
qc (t ) = Q e − t cos( t + ) (damped out with time)
d 2q
dq
The particular function q p (t ) is any solution of 2 + 2
+ 2 q = A sin d t.
dt
dt
To find a particular solution, try q p (t ) = D sin(d t − ) = Im[ Dei (d t − ) ]
d2
d
i ( d t − )
i ( d t − )
i ( d t − )
id t
2
Im[
De
]
+
2
Im[
De
]
+
Im
[
De
]
=
A
sin
t
=
Im[
Ae
]
d
2
dt
dt
d2
d
Im[ 2 Dei (d t − ) + 2 Dei (d t − ) + 2 Dei (d t − ) ] = Im[ Aeid t ]
dt
dt
Apparently, the above equation is satisfied if
d2
d
i ( d t − )
i ( d t − )
i ( d t − )
id t
2
De
+
2
De
+
De
=
Ae
dt 2
dt
−d 2 Dei (d t − ) + 2 id Dei (d t − ) + 2 Dei (d t − ) = Aeid t
−d 2 De − i + 2 id De − i + 2 De − i = A
−d 2 D cos + 2 d D sin + 2 D cos = A
2
2
D
sin
+
2
D
cos
−
D sin = 0
d
d
A
2
2
(
−
)
D
cos
+
2
D
sin
=
A
D
=
d
d
2
2
(
−
d ) cos + 2 d sin
−( 2 − d 2 )sin + 2 d cos = 0 = tan −1 2 d
( 2 − d 2 )
2 d
( 2 − d 2 )
2 d
−1
= tan
cos
=
,
sin
=
( 2 − d 2 )
( 2 − d 2 ) 2 + 4 2d 2
( 2 − d 2 ) 2 + 4 2d 2
A
D=
( 2 − d 2 ) 2
( − d ) + 4 d
2
2 2
2
2
+
4 2d 2
( 2 − d 2 ) 2 + 4 2d 2
=
A
( 2 − d 2 ) 2 + 4 2d 2
2 d
q p (t ) =
sin(d t − ), where = tan
2
2 2
2
2
2 − d 2
( − d ) + 4 d
A
At large t , q(t )
i (t ) =
=
q p (t ) =
−1
A
( − d ) + 4 d
2
2 2
2
2
sin(d t − )
dq(t )
Ad
=
cos(d t − )
2
2
2
2
2
dt
( − d ) + 4 d
(e m / L)d
2
cos(d t − ) =
em
cos(d t − )
1
1
R
2 2
2
(
− d L) 2 + R 2
(
− d ) + 2 d
d C
LC
L
2
( R / L)d
R
−1
= tan −1 2 d 2 = tan −1
=
tan
− d
(1 / LC ) − d 2
(1 / d C ) − d L
sin( + / 2)
cos
1
L − (1 / d C )
=
=−
+ = tan −1 d
2
cos( + / 2) − sin
tan
2
R
i (t ) = I cos(d t − ) = I sin(d t − ),
Note: tan( +
I=
)=
em
1
(
− d L) 2 + R 2
d C
, = +
2
= tan −1
d L − (
R
1
)
d C
Alternating current (ac) circuits
A. Three simple circuits (emf : e = e m sin d t )
1. A resistive load
Kirchhoff's loop rule: e − vR = 0
𝑅
vR = e = e m sin d t , vR = VR sin d t VR = e m
𝑖R
1
V
V
vR = R sin d t , iR = I R sin(d t − ) I R = R , = 0
R
R
R
VR = I R R, = 0
Ɛ
iR =
2. A capacitive load
Kirchhoff's loop rule: e − vC = 0
vC = e = e m sin d t , vC = VC sin d t VC = e m
qC = CvC
𝑖C
dqC
dv
= C C = d CVC cos d t = d CVC sin(d t + 90)
dt
dt
VC
iC = I C sin(d t − ) I C = d CVC =
, = −90
1 / d C
C
iC =
Let capacitive reactance X C =
1
VC = I C X C , = −90
d C
Ɛ
3. An inductive load
Kirchhoff's loop rule: e − vL = 0
vL = e = e m sin d t , vL = VL sin d t VL = e m
vL = L
𝑖L
𝐿
di
dt
Ɛ
1
VL
VL
VL
iL = di = vL dt = sin d tdt = −
cos d t =
sin(d t − 90)
L
L
d L
d L
iL = I L sin(d t − ) I L =
VL
, = +90
d L
𝑅
Let inductive reactance X L = d L VL = I L X L , = +90
B. The series RLC circuit
Neglect transient current. iR = iC = iL = i = I sin(d t − )
𝑖
𝐿
C
Kirchhoff's loop rule: e − vC − vR − vL = 0
e = vC + vR + vL
Ɛ
Note: i) iR = iC = iL = i = I sin(d t − ) iR , iC and iL are in phase with each another.
ii) i X and v X have the same angular frequency d
and a phase difference of 0, −90, or + 90.
as demonstrated in the three simple circuit section.
We will show that the method of phasors is a convenient way
for summing vR , vC , and vL .
Phasor A
A sin(t − )
Phasor:
Consider a sinusoidal function A sin(t − ). Phasor A
t −
e = vC + vR + vL represented in phasors:
Phasor VR is in phase with phasor I
Phasor VL is ahead of phasor I by 90
Phasor VL Phasor e m
Phasor VC is behind phasor I by 90
Note: Phasor e m represents e = e m sin d t
e m = VR + (VL − VC ) = ( IR) + ( IX L − IX C )
2
2
2
= ( IR ) 2 + ( I d L − I
I=
em
2
1 2
1 2
) = I 2 [ R 2 + (d L −
) ]
d C
d C
R 2 + (d L − 1 / d C ) 2
e
= m
phasor I
2
Z Impedance
𝜙
Phasor VR
t −
Phasor VC
1 2
−1 VL − VC
−1 X L − X C
Z = R + ( X L − X C ) = R + (d L −
) , = tan
= tan
d C
VR
R
2
2
2
In an RLC circuit, given emf e = e m sin d t ,
at large t , the current i( t ) = I sin(d t − ),
where I =
em
R + (d L − 1 / d C )
2
2
and = tan
−1
d L −
1
d C
𝑅
𝑖
R
1
Note: X L = d L, X C =
d C
If
𝐿
C
Ɛ
i) X L X C the circuit is more inductive than capacitive
0 I rotates behind e m
ii) X C X L the circuit is more capacitive than inductive
0 I rotates ahead of e m
iii) X L = X C I =
e m (maximum)
R
1
=
LC
the circuit is in resonance
d L =
1
d =
d C
= 0 I and e m rotate together.
Power in alternating-current circuits
i (t ) = I sin(d t − )
Note:
2
t +
d
t
𝑅
sin (d t − )dt =
2
t +
t
2
d
1 cos[2(d t − )]
{ −
}dt
2
2
𝑖
𝐿
C
2
sin(2d t − 2 ) t+
=
−[
]t =
−0=
d
4d
d
d
d
root − mean − square current :
I rms
=
t +
t
Ɛ
2
d
i 2dt
2 / d
=
2
t +
I2
2
d
sin
(d t − )dt =
t
2 / d
I2
I
=
2 / d d
2
Instantaneous rate of energy dissipation through R :
P (t ) = i 2 R = [ I sin(d t − )]2 R = I 2 R sin 2 (d t − )
Average rate of energy dissipation
t +
2
P (t )dt
=
=
d
Pavg
t
2 / d
Similarly, Vrms =
I 2 R t+ 2d
I 2R
1 2
2
2
sin
(
t
−
)
dt
=
=
I
R
=
I
d
rms R
t
2 / d
2 / d d 2
V
e
, e rms = m
2
2
𝑅
Note:
i) ac readings of multimeters: I rms ,Vrms , e rms
𝑖
e.g. household electrical outlet e rms = 120V
𝐿
C
maximum voltage of outlet V = 2 120V = 170V
ii) I =
em =
Z
em
R 2 + ( X L − X C )2
2
Pavg = I rms
R = I rms ( I rms R ) =
I rms =
e rms I
Z
I
e
e
= m / 2 = rms
Z
Z
2
rms R = e rms I rms
Ɛ
R
Z
X L − XC
R
and Z= R 2 + ( X L − X C ) 2 = cos
R
Z
R
We have Pavg = e rms I rms cos . The factor = cos is called power factor.
Z
recall tan =
Transformer
R
For resistive load as an example, Z = R power factor
=1
Z
R
Pavg = e rms I rms = e rms I rms = Vrms I rms
Z
To satisfy a given power requirement Pavg
Options:
2
i) high Vrms ; low I rms For transmission (to reduce Ohmiclosses Pcable = I rms
Rcable )
ii) low Vrms ; high I rms For household electrical devices
Tranformers are needed to raise and lower voltage for different requirements.
dB
dt
VP = e turn N P ; VS = e turn N S
emf per turn e turn =
VS = VP
NS
NP
(P : primary winding; S : secondary winding)
𝑁𝑃
𝑉𝑃
𝑁𝑆
𝑉𝑆
𝑅
NP
VP
= IP
By conservation of energy I PVP = I SVS I S = I P
NS
VS
IS =
NV
N
VS
IP = S IS = S S =
NPR
NP
R
NS
)
N V
NP
= ( S )2 P
NP R
NPR
N S (VP
VP
IP =
N
[( P ) 2 R ]
NS
Req = (
NP 2
) R load resistance seen by the generator
NS
𝑁𝑃
𝑉𝑃
𝑁𝑆
𝑉𝑆
𝑅
Chapter 32 Maxwell’s Equations; Magnetism of Matter
r
A. Gauss' Law for Electric Fields: E da =
E =
S
e0
e0
qenc
E da = 0, if there is no electric charge (electric monopole) in the volume
S
enclosed by a closed surface S.
B. No magnetic monopole has ever been observed.
It is assumed that magnetic monopoles do not exist.
Gauss' Law for Magnetic Fields: B da = 0 B = 0
S
C. In electrostatics: electrostatic force is conservative Echarge = −V
Echarge dr = 0 Echarge = 0
C
For non-steady states: Faraday's Law of Induction Einduced dr = −
C
Einduced = −
B
t
E dr = Echarge dr + Einduced dr = −
C
C
C
dB
B
E = −
dt
t
dB
dt
D. In magnetostatics: Ampere's Law Bcurrent dr = 0ienc
C
Bcurrent = 0 J
dE
For non-steady states: Maxwell's Law of Induction Binduced dr =0 e 0
C
dt
E
Binduced = 0e 0
t
Ampere-Maxwell Law
dE
B
dr
=
B
dr
+
B
dr
=
i
+
0 enc
0 e0
C
C current
C induced
dt
dE
(Note: displacement current id ,enc = e 0
)
dt
E
B = 0 J + 0e 0
t
Ex. Induced magnetic field in a circular parallel-plate capacitor
For the circular Amperean loop: i) By symmetry B dr = 2 rB; ii) ienc = 0;
C
d
q
r 2 dq r 2 inside the
2
e 0 dt ( r R 2e ) = R 2 dt = R 2 i capacitor
q
dE
0
iii) E = =
;
iv)
i
=
e
=
d ,enc
0
e 0 R 2e 0
dt
e d ( R 2 q ) = dq = i outside the
0
capacitor
dt
R 2e 0
dt
Ampere-Maxwell Law B dr = 0ienc + 0id ,enc
C
r 2 inside the
0i inside the
0 R 2 i capacitor
2 R 2 r capacitor
2 rB = 0 +
B=
i outside the
0i outside the
0 capacitor
2 r capacitor
−𝑞
+𝑞
𝑖
r
𝑅
𝑖
Electricity
Rubbed amber attracts straw
(hlektronelectron) :
negative charge
Rubbed glass: positive
charge
Electrostatics
Stationary charges
Coulomb’s Law
Gauss’s Law
Maxwell’s Equations
Gauss’ Law for Electricity
Gauss’ Law for Magnetism
Faraday Induction Law
Ampere-Maxwell Law
Electromagnetic Waves
Magnetism
Natural magnet attracts iron
Compass
Moving charges
Electric Current
Current in a wire deflects
magnetic compass
Electromagnetism
Time-varying currents and
Time-varying charge distribution
→time-varying magnetic field
Faraday Induction Law
→time-varying electric field
Ampere-Maxwell Law
Magnetostatics
Steady currents
Time-independent
charge distribution
Biot-Savart Law
Ampere’s Law
Maxwell's Equations
Gauss' Law for Electricity
E da =
qenc
S
e0
Gauss' Law for Magnetism B da = 0
E =
r
e0
B = 0
S
dB
dt
B
t
Faraday's Law
E dr = −
Ampere-Maxwell Law
E
dE
B
dr
=
i
+
e
B
=
J
+
e
0 enc
0 0
0
0 0
C
dt
t
C
Potentials
A
scalar potential V E = −V −
t
vector potential A
B = A
Force
Lorentz force F = qE + qv B
E = −
Magnetism and Electrons
Consider the loop model for electron orbits
(A non-quatum derivation to obtain the relation between
𝐿𝑜𝑟𝑏
orbital angular momentum Lorband
orbital magnetic dipole moment orb .)
Lorb = rmv
e
evr
2
=
iA
=
(
)(
r
)
=
orb
2 r / v
2
𝐴
e
orb =
Lorb
2m
Electrons have a negative charge. orb = −
e
2m
Note:
En =
E1
, n = 1, 2,3 ; Lorb =
2
n
𝑖
𝜇Ԧ𝑜𝑟𝑏
e
Lorb
2m
From quantum mechanics, Lorb, z = ml , ml = −l , −l + 1,
orb, z = − ml
𝑟
l (l + 1), l = 0,1,
,n −1
,0,
, l − 1, l
𝑚
−𝑒
𝑣
Spin magnetic dipole moment
spin angular momentum (spin) S
spin magnetic dipole moment s
Note: g 2 1
e
s = −
gS , g : g-factor
2m
explanantion--i) Classical: the charge distribution (that gives rise to s ) is different from
the mass distribution (that gives rise to S ).
ii) Quantum mechanical: A relativistic correction for Schroedinger's eqution
when reduced from Dirac equation.
From quantum mechanics, S z = ms , ms = − s, − s + 1,
1
1
1
ms = , S z =
2
2
2
e
e
; B =
(Bohr magneton) s , z = B
2m
2m
For electrons s =
s , z =
Note:
S=
, s − 1, s
s( s + 1) =
3
2
Magnetic Materials
In a solid material:
i) orbital magnetic dipole moment and spin magnetic moment of an
electron combine vectorially. magnetic dipole moment of an electron.
ii) magnetic dipole moments of all electrons in an atom combine vectorially.
magnetic dipole moment of an atom.
iii) magnetic dipole moments of all atoms in a sample of materials combined
vectorially. 1. Diamagnetism: feeble (easy to be masked by para- or
ferromagnetism), exhibited by all materials.
Bext
, C : Curie constant
T
3. Ferromagnetism: Hysteresis loop, Domains.
2. Paramagnetism: Curie's law M = C
Consider a planar loop that carries a current i. The area enclosed by the loop is A.
1
C 2 r dr = nˆ dA = Anˆ,
1
r dr
2
1
1
ˆ
The magnetic dipole moment of the loop = iAn = i
r dr = r idr
C 2
2 C
1
1
1
1
= r ( Jda )dr = r Jda dr = r Jdt = r Jdt
2 C
2 C
2 t
2 t
0
J ( r)
Recall A( r ) =
dt
t
4
r −r
Note: infinitesmal area dA =
𝑧
1
Expand
in Taylor's series about r ,
r −r
1
Note f ( r ) = [( r − r0 ) ]n f ( r ) r = r ;
0
n =0 n !
1
r − r
(
)=−
3
r − r
r − r
1
1
r
r2
1 r r
we have
= + ( − r) ( − 3 ) + O ( 3 ) + 3 if r
r − r r
r
r
r
r
r
𝑖
r
𝑥
r
dA =
1
r dr
2
𝑦
dr
A( r ) =
0
J ( r)
0
J ( r )
J ( r )
d
t
=
[
d
t
+
( r r )dt ]
3
t
t
t
4
r − r
4
r
r
J ( r)
1
1
i
Note that
dt = J ( r)dt = [ J ( r )da ]dr = dr = 0
t
r
r t
r t
r C
0 J ( r)
0
A( r ) =
(
r
r
)
d
t
=
J ( r )( r rˆ)dt
3
2 t
t
4
r
4 r
To find the dependence of the vector potential A( r ) on the magnetic moment ,
J y
J x
J
we note: ( xy J )dt = ( y J x + xy
+ xJ y + xy
+ xy z )dt
t
t
x
y
z
J x J y J z
= ( y J x + xJ y )dt + xy (
+
+
)dt
t
t
x
y
z
= ( y J x + xJ y )dt + xy ( J )dt
t
t
J x
J
2 J y
Also, ( x J )dt = (2 xJ x + x
+ x
+ x2 z )dt
t
t
x
y
z
J y J z
2 J x
= (2 xJ x )dt + x (
+
+
)dt
t
t
x
y
z
2
= 2 xJ x dt + x2 ( J )dt
t
t
2
( xy J )dt =
xy J da = ( xy J )dt = xy J da = 0
t
S
J = 0 at infinity
2
2
2
2
t ( x J )dt = S x J da = ( x J )dt = x J da = 0
r
r
Equation of continuity J +
= 0. In steady states
= 0. J = 0
t
t
( y J x + xJ y )dt + xy ( J )dt = 0 y J x dt = − xJ y dt
t
t
t
t
2 xJ x dt + x2 ( J )dt = 0 xJ x dt = 0
t
t
t
zJ y dt = − y J z dt ,
xJ z dt = − zJ x dt
t
t
t
t
Similarly,
and t y J y dt = t zJ z dt = 0
1
1
= r Jdt = [iˆ( y J z − zJ y ) + ˆj ( z J x − xJ z ) + kˆ( xJ y − y J x )]dt
2 t
2 t
ˆ J ]dt
ˆ J z + ˆjzJ x + kx
= [iy
y
t
ˆ J ) (iˆrˆ + ˆjrˆ + kr
ˆ ˆ )]dt
ˆ J z + ˆjzJ x + kx
rˆ = [(iy
y
x
y
z
t
ˆ ˆ ) + ( ˆjz J ir
ˆ ˆ ) + (kx
ˆ J ir
ˆ J ˆjrˆ ]dt
ˆ J z ˆjrˆy + iy
ˆ J z kr
ˆˆx + ˆjz J x kr
ˆˆx + kx
= [(iy
z
x
z
y
y
y
t
= [( y J z rˆy kˆ − y J z rˆz ˆj ) + ( − z J x rˆx kˆ + z J x rˆziˆ) + ( x J y rˆx ˆj − x J y rˆyiˆ)]dt
t
= [( y J z rˆy − zJ x rˆx + 0)kˆ + ( zJ x rˆz − xJ y rˆy + 0)iˆ + ( xJ y rˆx − y J z rˆz + 0) ˆj ]dt
t
= [( y J z rˆy + xJ z rˆx + zJ z rˆz )kˆ + ( z J x rˆz + y J x rˆy + xJ x rˆx )iˆ + ( xJ y rˆx + z J y rˆz + y J y rˆy ) ˆj ]dt
t
= [( xrˆx + y rˆy + zrˆz ) J xiˆ + ( xrˆx + y rˆy + zrˆz ) J y ˆj + ( xrˆx + y rˆy + zrˆz ) J z kˆ]dt
t
= [( r rˆ) J xiˆ + ( r rˆ) J y ˆj + ( r rˆ) J z kˆ]dt = J ( r )( r rˆ)dt
t
t
A( r ) =
0
0
0 r
ˆ
ˆ
J
(
r
)(
r
r
)
d
t
=
r
=
4 r 2 t
4 r 2
4 r 3
(vector potential at r due to a magnetic dipole at the origin.)
The vector potential at r , A( r ), due to a volume t of magnetic material with
magnetization M ( r) (magnetic dipole moment per unit volume) is
A( r ) =
0
M ( r) ( r − r )
0
1
d
t
=
M
(
r
)
(
)dt
3
t
t
4
4
r − r
r − r
0
M ( r)
M ( r)
=
[
−
]dt
t
4
r − r
r − r
=
0 M ( r)
0
M ( r)
d
t
−
dt
t
t
4
r − r
4
r − r
Consider del identities B ( A C ) = ( B A) C ; ( A B ) = B ( A) − A ( B )
If B is a constant vector ( A B ) = B ( A) ( A B )dt = B ( A)dt
t
t
By divergence theorem ( A B )dt = ( A B ) da = − ( B A) da
t
S
S
Use the 1st identidy and let C = da − ( B A) da = B ( − A da )
S
S
B ( A)dt = B ( − A da ) ( A)dt = − A da
t
S
t
S
0
M ( r)
0
M ( r )
0 M ( r) nˆ
And, −
dt = −
[−
da ] =
da
4 t
r − r
4 S r − r
4 S r − r
A( r ) =
0 M ( r)
0 M ( r) nˆ
d
t
+
da
t
S
4
r − r
4
r − r
Compared to the vector potential due to free currents:A( r ) =
0 J f ( r)
dt , we have
t
4
r − r
J b ( r) = M ( r) volume bound current density, and
K b ( r) = M ( r ) nˆ surface bound current density.
Note: Recall the bound charge densities r b = − P and b = P nˆ
Maxwell's Equations in Matter
In matter
Polarization P : Electric dipole moment per unit volumn
Magnetization M : Magnetic dipole moment per unit volumn
Bound charge density r b = − P
Bound current density J b = M
Polarization current density J P :
r b
r
P
P
= 0 J P = − b = − ( − P ) = ( ) J P =
t
t
t
t
t
r = r f + rb = r f − P
JP +
J = J f + Jb + J P = J f + M +
P
t
r r f + rb r f − P
Gauss' Law E = =
=
(e 0 E + P ) = r f
e0
e0
e0
Let D = e 0 E + P D = r f
E
P
E
= 0 ( J f + M + ) + 0e 0
t
t
t
B
(e 0 E + P )
D
( − M) = J f +
= Jf +
0
t
t
Ampere-Maxwell Law B = 0 J + 0e 0
D
Let H =
− M H = J f +
0
t
B
For linear materials
P = e 0 e E ; D = e E ; e = e 0 (1 + e ) = e 0k
M = m H ; B = H ; = 0 (1 + m )
D = rf
B = 0
E = −
B
t
H = J f +
D
t
Chapter 33 Electromagnetic Waves
Maxwell's Equations
E =
r
;
e0
B = 0;
E = −
B
;
t
B = 0 J + 0e 0
E
t
B
E
J
2E
( E ) = ( − ) = − ( B ) = − ( 0 J + 0e 0
) = − 0
− 0e 0 2
t
t
t
t
t
t
Del Identity ( E ) = ( E ) − 2 E ( E ) = (
r
) − 2 E
e0
J
2E
r
2 E r
J
2
2
− 0
− 0e 0 2 = ( ) − E E − 0e 0 2 =
+ 0
t
t
e0
t
e0
t
( E )
2B
Similarly, ( B ) = 0 J + 0e 0
= 0 J − 0e 0 2
t
t
2B
2
2
2
=( B ) − B = − B B − 0e 0 2 = − 0 J
t
In vacuum, for distant source, r = 0 and J = 0
2
2
E
(
r
,
t
)
B( r , t )
2
2 E ( r , t ) − 0e 0
=
0
;
B
(
r
,
t
)
−
e
=0
0 0
2
2
t
t
Note: The wave equations are not only for "waves".
All electric fields are required to satisfy the wave equations for E.
For example the electrostatic field due to a point charge q at the origin :
E=
q
4e 0 r
2
rˆ
1
x
y
z
ˆ
ˆj +
=
iˆ +
k
2
2
2
2
2
2
2
2
2
2
2
2
4e 0 ( x + y + z ) x + y + z
x +y +z
x + y + z
q
q
x
y
z
ˆ
ˆ+
ˆj +
=
i
k
3
3
3
4e 0 2
2
2 2
2
2
2 2
2
2
2 2
(x + y + z )
(x + y + z )
(x + y + z )
2 Ex
q
1
3
x
Note : 2 =
−
(2 x)
3
5
x
4e 0 x 2
2 2
2
2 2
2
2 2
(x + y + z )
(x + y + z )
2
q
1
3x
=
−
3
5
4e 0 x 2
2
2 2
2
2
2 2
(x + y + z )
(x + y + z )
1
6x
3
−
(2
x
)
−
5
5
2
2
2
2 2
2
2
2 2
(x + y + z )
(x + y + z )
q
=
4e 0 5
3x 2
+
(2
x
)
7
2 2
2
2 2
(x + y + z )
3
q
3x
6x
15 x
=
−
−
+
5
5
7
4e 0
2
2
2 2
2
2
2 2
2
2
2 2
(x + y + z )
(x + y + z )
(x + y + z )
q
9x
15 x 3
=
−
+
5
7
4e 0
2
2
2 2
2
2
2 2
(x + y + z )
(x + y + z )
2 Ex
q 3
x
q
3
xy
=
−
=
−
(2
y
)
5
5
4e 0 y
y 2
4e 0 y 2 2
2
2 2
2
2
2 2
(x + y + z )
(x + y + z )
q
3x
5
3 xy
=
−
+
(2
y
)
5
7
4e 0
2 2
2
2
2 2
2
2 2
(x + y + z )
(x + y + z )
q
3x
15 xy 2
=
−
+
5
7
4e 0
2
2
2 2
2
2
2 2
(x + y + z )
(x + y + z )
Similarly,
2 Ex
q
3x
15 xz 2
=
−
+
5
7
z 2 4e 0
2
2
2 2
2
2
2 2
(x + y + z )
(x + y + z )
2 Ex 2 Ex 2 Ex
Ex =
+
+
2
2
x
y
z 2
2
9x
15 x 3
3x
+
−
−
5
7
5
2
2
2
2
2
2
2
2
2
q (x + y + z )2 (x + y + z )2 (x + y + z )2
=
2
2
4e 0
15 xy
3x
15 xz
+
−
+
7
5
7
( x2 + y 2 + z 2 ) 2 ( x2 + y 2 + z 2 ) 2 ( x2 + y 2 + z 2 ) 2
q
15 x
15 x 3
15 xy 2
15 xz 2
=
−
+
+
+
5
7
7
7
4e 0
2
2
2 2
2
2
2 2
2
2
2 2
2
2
2 2
(x + y + z )
(x + y + z )
(x + y + z )
(x + y + z )
2
2
2
q
15 x
15 x( x + y + z )
=
−
+
=0
5
7
4e 0
2
2
2 2
( x 2 + y 2 + z 2 ) 2
(x + y + z )
Similarly,
Ey =
2
2 Ey
x
2
+
2 Ey
y
2
+
2 Ey
z
2
=0
2
2
2
E
E
Ez
z
z
2 Ez =
+
+
=0
2
2
2
x
y
z
And 2 E = (iˆ 2 E + ˆj 2 E + kˆ 2 E ) = 0
x
y
z
2 E
Since, for electrostaic fields, E is independent of time t 2 =0
t
2 E (r , t )
2
We have E (r , t ) − 0e 0
=0
2
t
2
2 Ex (r , t )
=0
E x ( r , t ) − 0e 0
2
t
2
2
E y (r , t )
E
(
r
,
t
)
2
2 E ( r , t ) − 0e 0
=
0
E
(
r
,
t
)
−
e
=0
y
0 0
2
2
t
t
2
2 Ez (r , t )
=0
E z ( r , t ) − 0e 0
2
t
2 Ex (r , t )
2 Ex 2 Ex 2 Ex
2 Ex
E x ( r , t ) − 0e 0
=0
+
+
− 0e 0 2 = 0
2
2
2
2
t
x
y
z
t
Separation of variables E x ( x, y, z, t ) = X ( x )Y ( y ) Z ( z )T (t )
2
d2X
d 2Y
d 2Z
d 2T
YZT
+ XZT 2 + XYT 2 − 0e 0 XYZ 2 = 0
2
dx
dy
dz
dt
1 d 2 X 1 d 2Y 1 d 2 Z
1 d 2T
+
+
− 0e 0
=0
2
2
2
2
X dx
Y dy
Z dz
T dt
1 d2X
d2X
ik x x
− ik x x
2
2
=
−
k
+
k
X
=
0
X
(
x
)
=
A
e
+
B
e
x
x
x
x
X dx 2
2
dx
d 2Y
1 d 2Y
ik y y
− ik y y
2
2
=
−
k
+
k
Y
=
0
Y
(
y
)
=
A
e
+
B
e
y
y
y
y
Y dy 2
dy 2
1 d 2 Z
d 2Z
ik z z
− ik z z
2
2
=
−
k
+
k
Z
=
0
Z
(
z
)
=
A
e
+
B
e
z
z
z
z
2
2
Z
dz
dz
1 d 2T
d 2T
2
2
it
− it
=
−
+
T
=
0
T
(
t
)
=
A
e
+
B
e
t
t
2
2
T
dt
dt
1
k 2 + k 2 + k 2 − e 2 = 0
=
=c
y
z
0 0
x
2
2
2
0e 0
kx + k y + kz
Ex ( x, y, z , t ) = X ( x)Y ( y ) Z ( z )T (t )
[ ik x + ik y y + ik z z −it ]
Apparently, Ex (r , t ) = E0, x e x
= E0, x ei ( k r −t ) , where k = (k x , k y , k z ),
is a valid solution. Note that the condition
k
=
1
0e 0
= c has to be satisfied.
Apply the same derivation for Ex (r , t ) to E y ( r , t ) and Ez (r , t ), we obtain a solution
for E (r , t ) that is of very special interest: E (r , t ) = E0 ei ( k r −t ) a plane wave solution.
(E and E0 are complex. The physical field Re[ E ] should be used for non-linear cases.)
Note: for a plane wave E ( r , t ) = E0ei ( k r −t )
B
t
Faraday's Law E = −
E0, x e
iˆ
ˆj
kˆ
x
y
z
i ( k x x + k y y + k z z −t )
E0, y e
i ( k x x + k y y + k z z −t )
E0, z e
B
B
B
= −iˆ x − ˆj y − kˆ z
t
t
t
i ( k x x + k y y + k z z −t )
ky
Bx
kz
i ( k r −t )
i ( k r −t )
−
=
ik
E
e
−
ik
E
e
B
=
E
−
E y + Cx (r )
y 0, z
z 0, y
x
z
t
k
k
B y
−
= −ik x E0, z ei ( k r −t ) + ik z E0, x ei ( k r −t ) B y = − x E z + z E x + C y ( r )
t
ky
Bz
kx
i ( k r −t )
i ( k r −t )
− ik y E0, x e
Bz = E y − E x + C z ( r )
− t = ik x E0, y e
B=
1
k E + C (r ) =
k E0
Let C ( r ) = 0. B( r , t ) =
ei ( k r −t ) + C ( r )
k E0
ei ( k r −t ) = B0ei ( k r −t ) B0 =
k E0
and B =
k E
r
Gauss' Law for Electricity E ( r , t ) = = 0
e0
i ( k x + k y + k z −t )
i ( k x + k y + k z −t )
i ( k x + k y + k z −t )
[ E0, x e x y z
] + [ E0, y e x y z
] + [ E0, z e x y z
]=0
x
y
z
iE0, x k x e
i ( k x x + k y y + k z z −t )
+ iE0, y k y e
( E0, x k x + E0, y k y + E0, z k z )[ie
i ( k x x + k y y + k z z −t )
i ( k x x + k y y + k z z −t )
+ iE0, z k z e
i ( k x x + k y y + k z z −t )
=0
i ( k x x + k y y + k z z −t )
]=0
] = ( E0 k )[ie
E0 k = 0 E 0 ⊥ k E ⊥ k
Recall B =
k E
B ⊥ k , B ⊥ E , k is in the direction of E B
E
Also E ⊥ k k E = k E B =
= . ; =
B k
k
kE
E
= c.
=c
B
0e 0
1
In summary,
E ( r , t ) = E0ei ( k r −t )
for the plane E.M. wave
, the real parts are the physical fields.
i ( k r −t )
B ( r , t ) = B0e
E ⊥ k , B ⊥ k , B ⊥ E , k is in the direction of E B
k E E
1
B
=
;
=
c
=
B
0e 0
Energy Transport and Poynting Vector
Poynting vector S =
S =
=
1
0
1
0
1
0
( E B) =
[ B (−
E B (John Henry Poynting)
1
0
[ B ( E ) − E ( B )]
B
E
1
B
E
) − E ( 0 J + 0e 0
)] =
B (− ) − e 0 E
−EJ
t
t
0
t
t
1 ( B B) 1 ( E E )
B2 1
=−
− e0
−EJ = − [
+ e0E 2 ] − E J
2 0
t
2
t
t 2 0 2
dU B dU E
u
u
=− [
+
]− E J = − − E J S +
= −E J
t dt
dt
t
t
u
r
If J = 0 S +
= 0 compared to J +
=0
t
t
S is the energy current density of the electromagnetic wave
( r E + r v B ) dr
dt
work done per unit time by the electromagnetic field
Note E J = E r v = ( r E + r v B ) v =
on electric charge per unit volume
E ( r , t ) = E0 cos( k r − t )
E
1
For a plane electromagnetic field,
=c=
B
e 0 0
B ( r , t ) = B0 cos( k r − t )
Note: S contains non-linear term of the fields, complex waves cannot be used.
E , E0 , B, B0 are all real.
E2
B2
B2 1 E 2
B 2 1 e 0 0 E 2
S=
EB =
=
=c
= c(
+
) = c(
+
)
2
0
0
c 0
0
2 0 2 c 0
2 0 2 0
1
EB
B2 1
= c(
+ e0E 2 )
2 0 2
E0 2
The intensity of the wave I = Savg =
[ E ]avg =
[cos 2 ( k r − t )]avg
c 0
c 0
1
2
E0 2
=
2 c 0
E0
1 2
c 2
e0 2
B 2 rms
Note Erms =
I=
E rms =
B rms = c[ E rms +
]
c 0
0
2
2 0
2
Radiation Pressure
Electromagnetic wave Photons
special relativity E 2 = p 2c 2 + m 2c 4
E
c
During time interval t , the energy of the photons an area A is bombarded with is
for photons m = 0 E = pc p =
U IAt
U = IAt. The total momentum of the photons is
=
.
c
c
U IAt
i) Total absorption: the momentum change of the object p =
=
c
c
F
p
IAt
I
radiation pressure Pr = = ( ) / A = [(
) / t ] / A =
A
t
c
c
ii) Total reflection back along path: the momentum change of the object
p =
2U 2 IAt
=
c
c
F
p
2 IAt
2I
radiation pressure Pr = = ( ) / A = [(
) / t ] / A =
A
t
c
c
Polarization: directions of the electric field oscillations.
Ex.
E ( r , t ) = ( E0 ˆj )ei ( kx −t )
vertically polarized. x − y plane is the plane of oscillation.
i
(
kx
−
t
)
B ( r , t ) = ( B0kˆ)e
E ( r , t ) = ( E0kˆ)ei ( kx −t )
horizontally polarized. z − x plane is the plane of oscillation.
i
(
kx
−
t
)
B ( r , t ) = ( − B0 ˆj )e
Light of other polarization
a linear combination of vertically and horizontally polarized light.
Unpolarized light polarized randomly.
(a statistical ensemble of vertically and horizontally polarized light.)
Polarizing sheet: aligned long molecules embedded in plastic.
An electric field component (a) parallel to the polarizing direction → pass
(b) perpendicular to the polarizing direction → absorbed
1
For an unpolarized light passing through a polarizing sheet: I = I 0 (one-half rule)
2
For a polarized light passing through a polarizing sheet:
2
E = E0 cos I = I 0 cos2 (cosine-squared rule) recall I = Erms
/ c 0
Electromagnetic waves in matters
For linear materials
[ P = e 0 e E ; e = e 0 (1 + e ) = e 0k ; D = e E ] ;[ M = m H ; = 0 (1 + m ); B = H ]
Maxwell's equations
rf
D
=
r
e
E
=
r
E
=
f
f
e
H = J + D B = J + (e E ) B = J + e E
f
f
f
t
t
t
rf
B
E
E =
;
B = 0;
E = −
;
B = J f + e
(in matters)
e
t
t
compare (r f , J f , , e ) (r , J , 0 , e 0 )
E =
r
;
e0
B = 0;
E = −
B
;
t
B = 0 J + 0e 0
E
(in vacuum)
t
2 E (r , t )
2 B( r , t )
2
Recall r = 0 and J = 0 E ( r , t ) − 0e 0
= 0 ; B( r , t ) − 0e 0
=0
2
2
t
t
2 E (r , t )
2 B( r , t )
2
2
r f = 0 and J f = 0 E ( r , t ) − e
= 0 ; B( r , t ) − e
=0
2
2
t
t
(in matters)
2
2 E (r , t )
2 B(r , t )
2
E (r , t ) − e
= 0 ; B(r , t ) − e
=0
2
2
t
t
E (r , t ) = E0ei ( k r −t )
1
c
Plane-wave solutions:
;
=
=v=
i ( k r −t )
k
n
e
B (r , t ) = B0 e
2
Boundary conditions for E , D, B, H at the interface of two media
Consider a curve C and a cylindrical closed surface S in which → 0
i) Faraday's law E dr = −
C
dB
; B = 0 Et ,1l − Et ,2l = 0 Et ,1 = Et ,2
dt
ii) Gauss' law for electricity D da = q f ,enc Dn ,1 A − Dn ,2 A = f A Dn ,1 − Dn ,2 = f
S
iii) Gauss' law for magnetism B da = 0 Bn ,1 A − Bn ,2 A = 0 Bn ,1 = Bn ,2
S
d
D da ; D da = 0
C
S
S
dt
H t ,1 l − H t ,2 l = (K f nˆ ) l for all loops, where nˆ is the normal vector
iv) Ampere-Maxwell law H dr = i f +
from midium 2 to midium1.
H t ,1 − H t ,2 = K f nˆ
1
2
𝑙
S’
𝐶
1
𝛿
2
𝐴
𝛿
𝑆
Reflection and Refraction (applying boundary conditions on phase)
An incident electromagnetic wavel, travelig in medium 1 (1 , e1 ) towards medium 2
(2 , e 2 ), is partially reflected by the interface between the two media (the x -y plane)
and partially transmitted into medium 2. The electric field can be written as
E0ei ( k r −t ) + E0ei ( k r −t ) z 0
1
c
E=
Note: v =
=
i ( k r − t )
e n
z0
E0e
If f = 0 and K f = 0, the boundary conditions become:
e1En ,1 = e 2 En ,2 ; Et ,1 = Et ,2 ; Bn ,1 = Bn ,2 ;
Bt ,1
1
=
Bt ,2
𝑘
medium 1
medium 2
2
𝜃𝜃
″
𝑘″
𝑥
𝜃′ 𝑘′
Boundary conditions must hold for all points rI on the interface for all time. 𝑧
k rI − t = k rI − t = k rI − t , Noting rI = ( x , y ,0)
k x x + k y y − t = k xx + k y y − t = k x x + k y y − t for all x , y , and t
k , k ,and k lie in the same plane (plane of incidence).
k x = k x = k x
k y = k y = k y = 0
k y = k y = k y
Let it be the x -z plane.
k x = k sin = k x = k sin = k x = k sin
ck ck ck
n2
= =
=
=
k = k; k = k
n1 n2
n1
n1
n2
k
=
k
;
k
=
k
n1
k sin = k sin = k sin
k sin = k sin = Law of Reflection
n2
k
sin
=
k sin n1 sin = n2 sin Law of Refraction (Snell's Law)
n1
Total Internal Reflection
Snell's Law n1 sin = n2 sin increases with
Note: If n1 n2 then .
Let = 90 when = C n1 sin C = n2 sin 90 = n2
critical angle C = sin −1
n2
n1
There is no refracted ray for angles of incidence larger than critical angle C .
Total internal reflection.
Polarization by Reflection
Note:
E0ei ( k r −t ) + E0ei ( k r −t ) z 0
E=
(recall = = )
i ( k r −t )
z0
E0e
TE polarization (transverse electric): electric field perpendicular to
the plane of incidence (x -z plane).
TM polarization (transverse magnetic): magnetic field perpendicular to
the plane of incidence (x-z plane).
A. For TE polarization
E0ei ( k r −t ) = E0 ˆjei ( k r −t ) ; E0ei ( k r −t ) = E0 ˆjei ( k r −t ) ; E0ei ( k r −t ) = E0 ˆjei ( k r −t )
(B=
k E
c ck
; k = k ; = ; n = = )
v
1 ˆ
ˆ cos ) ˆjE ei ( k r −t ) = E n1 ( kˆ sin − iˆ cos )ei ( k r −t )
(ik sin + kk
0
0
c
1 ˆ
ˆ cos ) ˆjE ei ( k r −t ) = E n2 ( kˆ sin − iˆ cos )ei ( k r −t )
sin + kk
B0ei ( k r −t ) = (ik
0
0
c
1 ˆ
ˆ cos ) ˆjE ei ( k r −t ) = E n1 ( kˆ sin + iˆ cos )ei ( k r −t )
B0ei ( k r −t ) = (ik
sin − kk
0
0
c
B0ei ( k r −t ) =
E0 ˆjei ( k r −t ) + E0 ˆjei ( k r −t ) z 0
E=
i ( k r −t )
z0
E0 ˆje
n1 ˆ
n1 ˆ
i ( k r −t )
i ( k r −t )
ˆ
ˆ
E
(
k
sin
−
i
cos
)
e
+
E
(
k
sin
+
i
cos
)
e
z0
0
0 c
c
B=
E0 n2 ( kˆ sin − iˆ cos )ei ( k r −t )
z0
c
Boundary conditions
[Note: We have previously shown that the incident, reflected, and refracted waves
all have the same phase at the interface. i.e. when r = ( x, y,0)]
e1En ,1 = e 2 En ,2 e1 0 = e 2 0
Et ,1 = Et ,2 E0 + E0 = E0
Bn ,1 = Bn ,2 E0
Bt ,1
1
=
Bt ,2
2
n1
n
n
sin + E0 1 sin = E0 2 sin n1 sin ( E0 + E0) = n2 sin E0
c
c
c
Snell's law n1 sin = n2 sin E0 + E0 = E0
− E0
n1
n
n
cos + E0 1 cos − E0 2 cos
n ( E − E0)
n E
c
c
c
=
1 0
cos = 2 0 cos
1
2
1
2
E0 + E0 = E0
n1 ( E0 − E0)
n2 ( E0 + E0)
cos
=
cos
n
(
E
−
E
)
n
E
1 0
0
2 0
cos =
cos
1
2
1
2
n12 cos − n2 1 cos
2n12 cos
E0 =
E0 ; E0 = E0 + E 0 =
E0
n12 cos + n2 1 cos
n12 cos + n2 1 cos
sin
, we have
sin
sin
n
cos
−
n
cos
1
1
n12 cos − n2 1 cos
sin cos − cos sin
sin
E0 =
E0
E0 =
E0
sin
n12 cos + n2 1 cos
sin cos + sin cos
n1 cos + n1
cos
sin
sin( − )
sin( − )
=
E0 E0 =
E0 Fresnel's equation for TE polarization
sin( + )
sin( + )
If 1 2 0 , noting n2 = n1
B. For TM polarization
Note:B =
k E
and E ⊥ k , B ⊥ k , B ⊥ E B =
kE
=
E nE
=
v
c
B0ei ( k r −t ) + B0ei ( k r −t ) z 0
B=
i ( k r −t )
z0
B0e
n
n
n
B0ei ( k r −t ) = 1 E0 ˆjei ( k r −t ) ; B0ei ( k r −t ) = 2 E0 ˆjei ( k r −t ) ; B0ei ( k r −t ) = 1 E0 ˆjei ( k r −t )
c
c
c
c ck
( E is in the B k direction; k = k ; = ; n = = )
v
E0ei ( k r −t ) = E0ei ( k r −t ) [ ˆj (iˆ sin + kˆ cos )] = E0 ( − kˆ sin + iˆ cos )ei ( k r −t )
E0ei ( k r −t ) = E0ei ( k r −t ) [ ˆj (iˆ sin + kˆ cos )] = E0 ( −kˆ sin + iˆ cos )ei ( k r −t )
E ei ( k r −t ) = E ei ( k r −t ) [ ˆj (iˆ sin − kˆ cos )] = E ( −kˆ sin − iˆ cos )ei ( k r −t )
0
0
0
E0 ( −kˆ sin + iˆ cos )ei ( k r −t ) + E0( −kˆ sin − iˆ cos )ei ( k r −t ) z 0
E=
i ( k r −t )
z0
E0 ( −kˆ sin + iˆ cos )e
n1 ˆ i ( k r −t ) n1 ˆ i ( k r −t )
+ E0 je
z0
c E0 je
c
B=
n2 E0 ˆjei ( k r −t )
z0
c
Boundary conditions
[Note: We have previously shown that the incident, reflected, and refracted waves
all have the same phase at the interface. i.e. when r = ( x, y,0)]
e1En ,1 = e 2 En ,2 e1[ E0 ( − sin ) + E0( − sin )] = e 2 [ E0 ( − sin )]
e1 ( E0 + E0)sin = e 2 E0 sin
e1
n1
( E0 + E0)( n1 sin ) =
e2
n2
e11
e 2 2
n12
n2 2
( E0 + E0) =
E0
( E0 + E0) =
E
2
2
n11
n2 2
n11c
n2 2c
n1
( E0 + E0) =
n2
E
1
2
Et ,1 = Et ,2 E0 cos − E0 cos =E0 cos ( E0 − E0) cos =E0 cos
E0 ( n2 sin )
Bn ,1 = Bn ,2 0 = 0
n1
n2
(
E
+
E
)
E0
0
0
Bt ,1 Bt ,2
n1 ( E0 + E0) n2
c
c
=
=
=
E
1
2
1
2
1
2
0
n2
n1
(
E
+
E
)
=
E
n12
0
0
(
E
−
E
)
cos
=
( E0 + E0)0 cos
1
2
0
0
n2 1
( E − E ) cos =E cos
0
0
0
n12
cos
n cos − n12 cos
n2 1
E0 =
E0 = 2 1
E0
n12
n2 1 cos + n12 cos
cos +
cos
n2 1
cos
cos
2n12 cos
2n12 cos
E0 =
( E0 − E0 ) =
E0 =
E0
cos
cos n2 1 cos + n12 cos
n2 1 cos + n12 cos
cos −
sin
, we have
sin
sin
n
cos − n1 cos
1
n2 1 cos − n12 cos
E0 =
E0 sin
E0
sin
n2 1 cos + n12 cos
n1
cos + n1 cos
sin
If 1 2 0 , noting n2 = n1
sin cos − sin cos
E0
sin cos + sin cos
sin cos (sin 2 + cos 2 ) − sin cos (sin 2 + cos 2 )
=
E0
2
2
2
2
sin cos (sin + cos ) + sin cos (sin + cos )
sin sin (cos sin − cos sin ) + cos cos (sin cos − sin cos )
=
E0
sin sin (cos sin + cos sin ) + cos cos (sin cos + sin cos )
sin sin sin( − ) + cos cos sin( − )
=
E0
sin sin sin( + ) + cos cos sin( + )
(cos cos − sin sin ) sin( − )
cos( + ) sin( − )
=
E0 =
E0
(cos cos + sin sin ) sin( + )
cos( − ) sin( + )
sin( − ) / cos( − )
tan( − )
=
E0 =
E0
sin( + ) / cos( + )
tan( + )
tan( − )
E0 =
E0 Fresnel's equation for TM polarization
tan( + )
=
sin( − )
E0 Fresnel's equation for TE polarization
sin( + )
tan( − )
E0 =
E0 Fresnel's equation for TM polarization
tan( + )
E0 =
Note: i) When + = 90 tan( + ) → E0 = 0
TM component is not reflected only reflect TE component
The incident angle is called Brewster's angle B
ii) n1 sin B = n2 sin(90 − B ) = n2 cos B tan B =
In air n1 1 B = tan −1 n
n2
n
B = tan −1 2
n1
n1
Chapter 38 Photons and Matter Waves
A. Electromagnetic waves are particles, too!
Photoelectric Effect
hn = K max + = K max + hn 0 ; n : frequency of incident light; : work function;
K max : kinetic energy of the most energetic electrons; n 0 : cut-off frequency.
or = K max + = K max + 0
Einstein: Photons with energy quanta hn to explain photoelectric effect.
photon energy: E = hn = ;
E hn hn h
=
=
=
c
c ln l
Note: It was shown in 1969 that quantum mechanics can explain photoelectric
photon momentum: E 2 = c 2 p 2 + m 2c 4 ; m = 0 p =
effect without the concept of photons.(time-dependent harmonic perturbation)
Compton Effect:
(experimentally proved that momentum and energy are transferred via photons)
l = l + l ;
l : incident-x -ray wavelength, l : scattered-x-ray wavelength , l : Compton shift.
Note: Classically, electrons should oscillate at the same frequency as the incident wave
(driven oscillations) and should emit waves at this same frequency.
Electromagnetic wave Probability wave of photons
Interference Probability fringes of photons
B. Electrons (particles) are also waves (matter waves)
de Broglie wavelength for matter (1924):
h
p
Two-slit interference of electrons (1927 Davisson, Germer; Thomson):
l=
An experimental proof for matter waves.
Wave equation for matter waves? Schroedinger
Schroedinger Equation
Plausibility Arguments:
i) de Broglie-Einstein Postulates
h
2
2 p
l
=
k
=
=
p
=
p E
i( x− t )
p
l
h
i ( kx −t )
sinusoidal matter wave : Ae
= Ae
E
E = =
p2
ii) E =
+U
2m
iii) If 1 and 2 are two solutions, then = c1 1 + c2 2 is also a solution.
(linear space of )
iv) For free particles, U ( x, t ) is constant. F = −
k =
p
and =
E
U
= 0 p and E are constant.
x
are constant. The wave function of free particles can be
written as a sinusoidal traveling wave Ae
i ( kx −t )
= Ae
p E
i( x− t )
.
p E
p E
p E
p E
i( x− t )
i( x− t )
i( x− t )
i( x− t )
1. From (iv) we note ( − i
) Ae
= pAe
; (i
) Ae
= EAe
x
t
Define momentum operator pˆ = −i
and total energy operator i
.
x
t
p2
2. From (ii) E =
+ U , the corresponding total energy Hamiltonian operator is
2m
2
2
2
ˆ
p
Hˆ =
+U = −
+U
2
2m
2m x
3. For all potential functions, let Hˆ ( x, t ) = i
( x, t ), we have the
t
2
2 ( x, t )
Schroedinger Equation −
+
U
(
x
,
t
)
=
i
( x, t )
2m x 2
t
Note: i) For free particles ( x, t ) = Ae
p E
i( x− t )
, Schroedinger Equation shows that
p2
(
+ U ) ( x, t ) = E ( x, t ).
2m
ii) In 3-D pˆ = −i . The Schroedinger Equation becomes
−
( r , t ) + U ( r , t ) = i
(r , t )
2m
t
2
2
Born's postulate:
For a wave function ( r , t ) of a matter wave,
2
is a probability density.
dt is the probability that measurement of the particle's position at the time t
2
finds it in the volume dt about r .
Four basic postulates of quantum mechanics
Postulate I: To any observable (physical quantity) A, there corresponds an operator
Aˆ such that measurement of A yields values a which are eigenvalues of Aˆ .
Aˆ a = a a ; a is the eigenfunction of Aˆ corresponding to the eigenvalue a.
2
ˆ
p
e.g. momentum pˆ = −i , total energy Hamiltonian Hˆ =
+ V ( r ).
2m
Postulate II: Measurement of A that yields the value a leaves the system in the state
, which is the eigenfunction of Aˆ corresponding to the eigenvalue a.
a
Postulate III: The state of a system can be represented by a state function (wave
function) ( r , t ) which contains all information regarding the state of the system.
The expectation value of any observable O for the system in state ( r , t ) is
O = *Oˆ dt .
Note that *r dt = r * dt = r dt . According to Born's postulate,
2
2
is
2
the probability density. r dt = r = *rˆ dt The position operator rˆ = r
Postulate IV: The state function ( r , t ) of the system develops in time according to the
ˆ
time-dependent Schroedinger equation H ( r , t ) = i
(r , t ).
t
Note:
1 for i = j
i) Dirac Notation: = * dt ; Kronecker delta: i , j =
0 for i j
ii) If the state function is a linear combination of eigenfunctions of Aˆ ,
= cia , where Aˆa = aia and a a = i , j
i
i
i
i
j
i
we have A = * Aˆ dt = Aˆ =
*
ˆ c =
c
A
c
i ai j a j i c j a j ai a j
i
= c*i c j a j i , j = c*i ci ai = ci ai ci
2
i
j
i
j
2
i
j
is the probability that measurement of Aˆ
i
on a system in the state of finds a value of ai and leaves the system in the state of a j .
iii) Hermitian operators: For any operator Aˆ , its Hermitian adjoint Aˆ † is defined
such that Aˆ † = Aˆ for all and in the Hilbert space. If Aˆ † = Aˆ ,
i.e. Aˆ = Aˆ then Aˆ is called a Hermitian operator.
If Aˆ is Hermitian, for any eigenfunction n of Aˆ , we have Aˆ n = an n where an is
the corresponding eigenvalue. Aˆ n n = n Aˆ n an* =an
eigenvalues of Hermitian operators are real.
Note: Measurements of all physical observables yield real values. Therefore, the
operators corresponding to all physical observables are Hermitian.
iv) The commutator between two operators Aˆ and Bˆ is defined as
ˆ ˆ − BA
ˆˆ
[ Aˆ ,Bˆ ] = AB
) − [ x ( −i
) + ( −i
x )] = i
x
x
x
If [ Aˆ ,Bˆ ] = 0, Aˆ and Bˆ are said to commute and A and B compatible with each other.
ˆ ˆ − px
ˆ ˆ = x ( −i
e.g. [ xˆ , pˆ ] = xp
Note: Let i and j be eigenfunctions of a Hermitian operator Aˆ
Aˆi = aii and Aˆ j = a j j
Aˆi j = i Aˆ j ai* i j = a j i j
Since ai ' s are real, we have ai* i j = ai i j = a j i j
(ai − a j ) i j = 0
(i) i j ai − a j 0 i j = 0
(ii) i = j i j = i i = i*i dt = i dt = 1
2
i j = i , j (orthogonal eigenfunctions)
Robertson-Schroedinger Relation
Consider two Hermitian operators Aˆ , and Bˆ . Suppose [ Aˆ , Bˆ ] = Cˆ . For any state
of the system, the expectation value of A is A = Aˆ = * Aˆ dt .
The uncertainty of A is defined as
A =
=
=
( A − A )2 =
( Aˆ − A ) 2 =
* ˆ2
[ A − 2 A Aˆ + A ] dt =
2
2
2
Aˆ 2 − 2 A + A =
* ˆ
2
(
A
−
A
)
dt
* ˆ2
* ˆ
A
d
t
−
2
A
A dt + A
2
Aˆ 2 − A
Similarly, the uncertainty of B
B =
( B − B )2 =
2
Bˆ 2 − B
Define Dˆ = Aˆ − A ; Eˆ = Bˆ − B Note that Dˆ and Eˆ are both Hermitian
ˆ ˆ − ED
ˆ ˆ = ( Aˆ − A )( Bˆ − B ) − ( Bˆ − B )( Aˆ − A )
[ Dˆ , Eˆ ] = DE
ˆ ˆ − Aˆ B − A Bˆ + A B − BA
ˆ ˆ + Bˆ A + B Aˆ − B A
= AB
ˆ ˆ − BA
ˆ ˆ = [ Aˆ , Bˆ ] = Cˆ
= AB
2
*
dt
( AB ) 2 = Dˆ 2 Eˆ 2 = Dˆ Dˆ
Eˆ Eˆ
By Cauchy-Schwartz inequality:
For any two vectors F and G, (F F )(G G ) ( F G ) 2
( AB ) = Dˆ Dˆ
Eˆ Eˆ Dˆ Eˆ
2
2
ˆ ˆ
= DE
2
1 ˆˆ ˆˆ
1 ˆˆ ˆˆ
1 ˆˆ ˆˆ
1 ˆ ˆ
ˆ
ˆ
Note: DE = ( DE + ED ) + ( DE − ED ) = ( DE + ED ) + [ D, E ]
2
2
2
2
ˆ ˆ = 1 ( DE
ˆ ˆ + ED
ˆ ˆ ) + 1 [ Dˆ , Eˆ ]
DE
2
2
1
ˆ ˆ + ED
ˆ ˆ ) = 1 [ DE
ˆ ˆ + ED
ˆ ˆ ] = 1 [ DE
ˆ ˆ + DE
ˆ ˆ ]
( DE
2
2
2
*
1
ˆ ˆ + DE
ˆ ˆ ] = Re{ DE
ˆ ˆ }
= [ DE
2
1
1
ˆ ˆ − ED
ˆ ˆ ) = i Im{ DE
ˆ ˆ }
Similarly, [ Dˆ , Eˆ ] = ( DE
2
2
ˆ ˆ
( AB ) DE
2
1
= [ Dˆ , Eˆ ]
4
2
2
2
2
ˆ ˆ } + Im{ Dˆ Eˆ } Im{ DE
ˆ ˆ }
= Re{ DE
1
= Cˆ
4
2
=
2
1
1
C AB
C
4
2
2
Heisenberg's Uncertainty Principle
ˆ ˆ − px
ˆ ˆ = x ( −i
Recall [ xˆ , pˆ ] = xp
Therefore,
1
i =
2
2
In 3-D, we have
x p
x p x
y p y
z p z
2
2
2
) − [ x ( −i
) + ( −i
x )] = i
x
x
x
Heisenberg's Uncertainty Principle
1
i = 0
2
2
If p = 0 then x = .
[ xˆ , pˆ ] = i xp
EX . For a free particle
( x, t ) = Ae
p E
i ( x− t )
, an eigenstate of momentum.
p E
p E
i( x− t )
i( x− t )
ˆ
i.e. pAe
= −i
Ae
= pAe
p = 0
x
by Born's postulate:
p E
i ( x− t )
The probability density ( x, t ) = A * e
2
p E
−i ( x− t )
Ae
p E
i( x− t )
= A * A = A2
The probability density is constant everywhere. x =
In one dimension, for time-independent potential V ( x ) the Schroedinger eqaution is
2 ( x, t )
−
+
V
(
x
)
(
x
,
t
)
=
i
( x, t )
2
2m x
t
To solve this equation, let ( x, t ) = ( x )T (t ). (separation of variables)
2
2 [ ( x )T (t )]
−
+ V ( x )[ ( x )T (t )] = i
[ ( x )T (t )]
2m
x 2
t
2
d 2 ( x )
dT (t )
T (t )[ −
+ V ( x ) ( x )] = ( x )[i
]
2m dx 2
dt
2
T (t )
d 2 ( x )
( x)
dT (t )
[−
+ V ( x ) ( x )] =
[i
]
( x )T (t ) 2m dx 2
( x )T (t )
dt
2
2
1
d 2 ( x )
1
dT (t )
[−
+ V ( x ) ( x )] =
[i
]
( x ) 2m dx 2
T (t )
dt
2
2
1
d 2 ( x )
d2
( x ) [ − 2m dx 2 + V ( x ) ( x )] = E [ − 2m dx 2 + V ( x )] ( x ) = E ( x )
E
1 [i dT (t ) ] = E dT = −i E dt ln(T ) = −i E t + C T (t ) = A e − i t
t
t
T (t )
dt
T
2
time-independent
d2
ˆ ( x ) = E ( x )
[−
+
V
(
x
)]
(
x
)
=
E
(
x
)
or
H
Schroedinger Equation
2m dx 2
2
d2
[−
+ V ( x )] ( x ) = E ( x )
2m dx 2
2
d2
For free particles V ( x ) = V0 (constant) [ −
+ V0 ] ( x ) = E ( x)
2m dx 2
2 2
d 2 ( x ) 2m( E − V0 )
2m( E − V0 )
k
2
+
( x ) = 0 Let
= k or E − V0 =
2
2
dx 2
2m
2m( E − V0 )
d 2 ( x )
2
ikx
− ikx
+
k
(
x
)
=
0
(
x
)
=
A
e
+
B
e
where
k
=
dx 2
E
−i t
E
Recall T (t ) = At e . Let
= T (t ) = At e − it
( x, t ) = ( x )T (t ) = Aei ( kx −t ) + Be − i ( kx +t )
Note: If V0 E k is imaginary k = ik , where k =
( x ) = Ae −k x + Bek x
2m(V0 − E )
is real.
Barrier Tunneling
𝑉𝑏
𝐸
𝐼
𝐼𝐼𝐼
𝐼𝐼
𝑥
𝐿
0 x 0 or x L
V ( x) =
. Electrons are emitted from left.
Vb 0 x L
For region I: I ( x ) = AI eikx + BI e − ikx , k =
For region II: II ( x ) = AII e
−k x
kx
2mE
+ BII e , k =
For region III: III ( x ) = AIII eikx
2m(Vb − E )
Boundary conditions:
I (0) = II (0) AI + BI = AII + BII
I (0) = II (0) ik ( AI − BI ) = −k ( AII − BII )
II ( L) = III ( L) AII e −k L + BII ek L = AIII eikL
( L) −k AII e −k L + k BII ek L = ikAIII eikL
II ( L) = III
(k + ik )eikL
(k − ik )eikL
BII =
AIII ; AII =
AIII
kL
−k L
2k e
2k e
ik ( AII + BII ) − k ( AII − BII )
AI =
2ik
(k − ik )eikL (k + ik )eikL
(k − ik )eikL (k + ik )eikL
ik [
+
] − k[
−
]
−k L
kL
−k L
kL
2k e
2k e
2k e
2k e
=
AIII
2ik
−(k − ik ) 2 (k + ik ) 2
+
ikL
2
2
−k L
kL
e
−
(
k
−
ik
)
(
k
+
ik
)
ikL
2k e
= 2k e
e AIII =
[
+
] AIII
−k L
kL
2ik
4ikk
e
e
ieikL −(k − ik ) 2 (k + ik ) 2
ieikL
2 kL
2 −k L
=−
[
+
]
A
=
[(
k
−
ik
)
e
−
(
k
+
ik
)
e ] AIII
III
−k L
kL
4kk
e
e
4kk
AI
2
ikL
−ie − ikL
2 kL
2 −k L
* ie
=
[(k + ik ) e − (k − ik ) e ] AIII
[(k − ik ) 2 ek L − (k + ik ) 2 e −k L ] AIII
4kk
4kk
1
2
2 kL
2 −k L
2 kL
2 −k L
[(
k
+
ik
)
e
−
(
k
−
ik
)
e
][(
k
−
ik
)
e
−
(
k
+
ik
)
e
]
A
III
16k 2k 2
1
2
2
2 2 2k L
2
2 2 −2k L
4
4
=
[(
k
+
k
)
e
+
(
k
+
k
)
e
−
(
k
+
ik
)
−
(
k
−
ik
)
]
A
III
16k 2k 2
1
2
2
2 2 2k L
2
2 2 −2k L
4
2 2
4
=
[(
k
+
k
)
e
+
(
k
+
k
)
e
−
(2
k
−
12
k
k
+
2
k
)]
A
III
16k 2k 2
1
2
2
2 2 2k L
2
2 2 −2k L
4
2 2
4
= {1 +
[(
k
+
k
)
e
+
(
k
+
k
)
e
−
(2
k
+
4
k
k
+
2
k
)]}
A
III
16k 2k 2
1
2
2
2 2 2k L
2
2 2 −2k L
2
2 2
= {1 +
[(
k
+
k
)
e
+
(
k
+
k
)
e
−
2(
k
+
k
)
]}
A
III
16k 2k 2
(k 2 + k 2 ) 2 2k L
2
−2 k L
= [1 +
(
e
+
e
−
2)]
A
III
16k 2k 2
k 2 + k 2 2 kL 2
k 2 + k 2 2 k L −k L 2
2
2
−k L 2
k L −k L
= {1 + (
) [( e ) + ( e ) − 2e e ]} AIII = [1 + (
) ( e − e ) ] AIII
4kk
4kk
1 x
1 k 2 + k2 2
2
2
−x
Note: sinh( x ) = ( e − e ) AI = [1 + (
) sinh 2 (k L)] AIII
2
4
kk
=
The transmission coefficient is:
T=
AIII
2
AIII
=
2
2
1
=
1 k 2 + k2 2
1 k 2 + k2 2
2
2
[1 + (
) sinh (k L)] AIII
1+ (
) sinh 2 (k L)
4
kk
4
kk
2m(Vb − E )
2mE
2mVb
2m
Recall k =
;k=
k 2 + k2 =
;
k
k
=
E (Vb − E )
2
2
T=
AI
1
2m(Vb − E )
1
Vb2
1+
sinh 2 (
L)
4 E (Vb − E )
If Vb
E sinh (
2
2m(Vb − E )
1
L) = [ ( e
2
2 m (Vb − E )
L
−e
−
2 m (Vb − E )
L
1 2
)] e
4
2 m (Vb − E )
2
2 m (Vb − E )
L
2m(Vb − E )
Vb2
Vb
1
Vb2
Vb 2
2
;
1+
sinh (
L)
e
E (Vb − E ) E
4 E (Vb − E )
16 E
16 E −2
T
e
Vb
2 m (Vb − E )
L
e −2 k L
L
Chapter 39 More About Matter Waves
2
d2
Recall [ −
+ V ( x )] ( x ) = E ( x )
2
2m dx
For free particles V ( x ) = V0 (constant)
d 2 ( x )
2
ikx
− ikx
i) If V0 E
+
k
(
x
)
=
0
(
x
)
=
A
e
+
B
e
where k =
2
dx
d 2 ( x )
2
−k x
kx
ii) If V0 E
−
k
(
x
)
=
0
(
x
)
=
A
e
+
B
e
, where k =
2
dx
𝑉0
An electron in a finite well
𝐼
𝐼𝐼
2m( E − V0 )
2m(V0 − E )
𝐼𝐼𝐼
V0 x 0 or x L
𝐸
V =
0 0 x L
𝑥
𝐿
Region I: I ( x ) = Aek x (The coefficient of e −k x , that blows up as x → −, has to be zero.)
Region II: II ( x ) = Beikx + Ce − ikx
Region III: III ( x ) = De −k x (The coefficient of ek x , that blows up as x → , has to be zero.)
k=
2mE
;k=
2m(V0 − E )
Boundary conditions
I (0) = II (0) A = B + C
k + ik
k − ik
B
=
A
;
C
=
−
A
(0)
=
(0)
A
k
=
ik
(
B
−
C
)
2ik
2ik
I
II
II ( L) = III ( L) BeikL + Ce − ikL = De −k L
ikL
− ikL
−k L
( L) ikBe − ikCe = −k De
II ( L) = III
k + ik (k +ik ) L k − ik (k −ik ) L
D
=
[
e
−
e
]A
2ik
2ik
D = −[ k + ik e(k +ik ) L + k − ik e(k −ik ) L ] A
2k
2k
k + ik (k +ik ) L k − ik (k −ik ) L
k + ik (k +ik ) L k − ik (k −ik ) L
e
−
e
=−
e
−
e
2ik
2ik
2k
2k
k + ik ikL k − ik − ikL
k + ik ikL k − ik − ikL
e −
e =−
e −
e
ik
ik
k
k
(k + ik ) 2 eikL = (k − ik ) 2 e − ikL k − ik = (k + ik )eikL
Note: 2ik = (k + ik ) − (k − ik ) = (k + ik ) − [ (k + ik )eikL ] = (1 eikL )(k + ik )
= (e
− ik
L
2
e
ik
L
2
)e
ik
L
2
(k + ik )
L
− ik
2ik
A=
B = (e 2
k + ik
kx
I ( x ) = Ae
= (e
− ik
e
L
2
ik
e
L
2
ik
) Be
L
2
ik
L
2
ik
L
2 kx
) Be e
k − ik
(k + ik )eikL
C=−
B=
B = eikL B
k + ik
k + ik
II ( x ) = Be + Ce
ikx
ik
L
L
ik ( x − )
2
2
= Be e
e
ik
L
2
− ikx
Be
= Be
L
− ik ( x − )
2
ikx
ikL
− ikx
L
2
L
ik ( x − )
2
e Be
ik
= Be [e
D = BeikL ek L + Ce − ikL ek L = BeikL ek L
ik
L
2 kL
=Be e ( e
ik
III ( x ) = De
L
2
−k x
e
− ik
L
2
= Be
L
2
(e
ik
L
2
e
L
L
ik ( x − )
2
2
= Be e
e
L
− ik ( x − )
2
− ik
L
2
) ek L e − k x
e
ik
L
2
Be
L
− ik ( x − )
2
]
eikL Be −ikL ek L = BeikL ek L
)
ik
ik
Bek L = Bek L (eikL 1)
L
L
ik
ik
L
L kx
k
x
2
2
I ( x ) = −2iBe sin( k 2 )e
I ( x ) = 2 Be cos( k 2 )e
L
L
ik
ik
L
L
2
2
(
x
)
=
2
iBe
sin[
k
(
x
−
)]
(odd)
or
(
x
)
=
2
Be
cos[
k
(
x
−
)] ( even)
II
II
2
2
L
L
ik
ik
L
L −k ( x − L )
−k ( x − L )
2
2
(
x
)
=
2
iBe
sin(
k
)
e
(
x
)
=
2
Be
cos(
k
)e
III
III
2
2
k − ik = (k + ik )eikL ; Let = tan −1 ( k / k ) k + ik = k 2 + k 2 ei ; k − ik = k 2 + k 2 e − i
e − i = ei eikL e − i e
− ik
L
2
= ei e
ik
L
2
e
L
− i ( + k )
2
= e
L
i ( + k )
2
L
L
L
L
k
sin(
+
k
)
=
0
+
k
=
n
=
tan(
n
−
k
)
=
−
tan
k
k
2
2
2
2
cos( + k L ) = 0 + k L = ( n + 1 )
k = tan[( n + 1 ) − k L ] = cot k L
k
2
2
2
2
2
2
L 2
L 2 mL2V0
L
L
Note ( k ) +(k ) =
Let
K
=
k
and
=
k
2
2
2 2
2
2
= − K cot K
odd
mL2V0
2
2
Plot K + =
in
K
plane
with
for
the
states to find
2
2
= K tan K
even
their quantized eigenenergies. (obtain K → k → E )
An electron in a infinitely deep potential energy well
2
d2
x 0 or x L [ − 2m dx 2 + ] ( x ) = E ( x ) ( x ) = 0
V =
2
2
d
( x)
ikx
− ikx
0 0 x L −
=
E
(
x
)
(
x
)
=
Ae
+
Be
;k=
2
2m dx
Boundary conditions:
2mE
(0) = 0 A + B = 0 B = − A ( x ) = Aeikx + Be − ikx = A(eikx − e −ikx ) = 2iA sin kx
n
( L) = 0 2iA sin kL = 0 kL = n k =
where n = 1, 2,3
( x ) = 2iA sin
n
n
x = C sin
x
L
L
L
Normalization
n
n
2 L
2 n
(
x
)
(
x
)
dx
=
1
(
C
sin
x
)(
C
sin
x
)
dx
=
C
sin
xdx
0
0
0
L
L
L
L
*
L
*
2
C L
1 1
2n
= C ( − cos
x )dx =
=1 C =
0 2
2
L
2
2
eigenstates n ( x ) =
L
2
L
2 2
2 2
2
n
k
2
sin
x quantized eigenenergies En =
=
n
2
L
L
2m 2mL
Note:
2
C L
2 i
i)
=1 C =
e A wavefunction in quantum mechanics is determined
2
L
only to within a constant phase factor. That is, ( x ) and ei ( x ) represent exactly
the same physical state of the system. Therefore, we select = 0 for simplicity.
2 2
k2
2
ii) En =
=
n , n = 1, 2,3,
2
2m 2mL
2
n
(If n = 0, n ( x ) =
sin
x = 0. The particle does not exist.)
L
L
2
E1 =
2
2
0 zero-point energy for a confined particle.
2mL2
(A confined particle can never be at rest.)
While a free particle can have any energy including zero, a confined particle
has discrete energies and its ground state energy has to be larger than zero.
2
2
iii) The zero-point energy
2
2
2
2mL
, as well as energy difference between two states
2
2
(
n
−
n
) increases with deceasing L.
2
1
2
2mL
A Three-Dimensional Box
0 0 x Lx and 0 y Ly and 0 z Lz
V ( x, y , z ) =
everywhere else
Time-independent Schroedinger Eqaution
2
2
2
[−
2 + V ( x, y, z )] ( x, y, z ) = E ( x, y, z ); 2 = 2 + 2 + 2
2m
x y
z
Separation of variables ( x, y, z ) = X ( x)Y ( y ) Z ( z )
2
2
2
d2X
d 2Y
d 2Z
1 d 2 X 1 d 2Y 1 d 2 Z
−
[YZ
+ XZ 2 + XY 2 ] = EXYZ −
[
+
+
]= E
2
2
2
2
2m
dx
dy
dz
2m X dx
Y dy
Z dz
2
2 2
d2X
n
2
2
x
− 2m dx 2 = Ex X ; X (0) = X ( Lx ) = 0 X ( x) =
sin
x; Ex ,nx =
n
2 x
L
L
2
mL
x
x
x
2
2
d Y
2 2
−
= E yY ; Y (0) = Y ( Ly ) = 0
n y
2
2
2
m
dy
Y
(
y
)
=
sin
y
;
E
=
n2
y ,ny
2 y
Ly
Ly
2mLy
2
2
d Z
−
= Ez Z ; Z (0) = Z ( Lz ) = 0
2 2
2
n
2
2
2m dz
Z ( z) =
sin z z; Ez ,nz =
n
2 z
E +E +E =E
L
L
2
mL
z
z
z
y
z
x
2
2 2
n y
n x
nz
nz 2
8
nx 2 n y
nx , n y , nz ( x, y , z ) =
sin
x sin
y sin
z ; En x , n y , n z =
( 2 + 2 + 2)
Lx Ly Lz
Lx
Ly
Lz
2m Lx
Ly
Lz
n ,n ,n ( x , y , z ) =
x
y
z
2
2 2
8
n
n
n
nx 2 n y
nz 2
sin
x sin
y sin
z ; E n x ,n y ,n z =
( 2 + 2 + 2)
Lx Ly Lz
Lx
Ly
Lz
2m Lx
Ly
Lz
If Enx ,ny ,nz = Enx ,n y ,nz nx ,ny ,nz and nx ,n y ,nz are degenerate.
The Bohr Model of the Hydrogen Atom
Johann Balmer: Wavelengths emitted or absorbed by a hydrogen atom
1
1
1
= R ( 2 − 2 ), n = 3, 4,5 and 6.
l
2
n
Bohr (1913)
1. circular orbits for electrons with center at the nucleus
h
(de Broglie wavelength)
p
h
h
2 r = nl = n L = rp = n
=n
p
2
2. Angular momentum of the electron is quantized. L = n , n = 1, 2,3
L = rp; l =
1 e2
v2
On the other hand −
= m( − )
2
4e 0 r
r
2
4e 0
4e 0
4e 0 2
2
2
2 4e 0
2
r=
(
rmv
)
=
(
rp
)
=
L
=
n
=
n
a
me 2
me 2
me 2
me 2
4e 0 2
Radius is quantized. Bohr radius a =
me 2
1 e2
v2
4e 0 2 e 0h 2
2
Noting −
= m( − ); L = n ; r = n a; a =
=
; n = 1, 2,3
2
2
2
4e 0 r
r
me
me
p2
1 e2
p2
v2
L2
L2
L2
E = K +U =
+ (−
)=
+ [m( − ) r ] =
− 2 =−
2
2m
4e 0 r
2m
r
2mr
mr
2mr 2
2
2
n2 2
m 2e4
1
me 4
1
me 4
=−
=−
=−
= 2 (−
) = 2 (− 2 2 )
2
2
2 2
2
2 2 4
2 2 2
2m( n a )
2mn a
2mn 16 e 0
n
32 e 0
n
8e 0 h
1
me 4
me 4
En = 2 ( − 2 2 ) Energy is quantized.
13.6eV
2 2
n
8e 0 h
8e 0 h
me 4
1
1
c
E = Enhigh − Enlow = − 2 2 ( 2 − 2 ) emitted as light. E = hn = h
8e 0 h nhigh nlow
l
me 4
1
1
c
1
me 4
1
1
− 2 2( 2 − 2 )=h =
(
−
)
2 3
2
2
8e 0 h nhigh nlow
l
l 8e 0 h c nlow nhigh
me 4
Note that Rydberg constant R =
and let nlow = 2, nhigh = n
2 3
8e 0 h c
1
1
= R ( 2 − 2 ) Bohr model reproduces Balmer's equation.
l
2
n
Note: Bohr model is not correct. But Bohr radius is a good estimate
1
for the size of particles and quasi-particles.
Schroedinger Equation and Hydrogen Atom
Time-independent Schroedinger Equation in 3D
−
2
2m
2 ( r ) + U ( r ) = E ( r )
1 e2
For an electron in hydrogen atom, U ( r ) = −
.
4e 0 r
For convenience, the sperical coordinate system in which
1 2
1 1
1 2
=
r+ 2(
sin
+ 2
) is adopted.
2
2
r r
r sin
sin
The time-independent Schroedinger equation becomes
2
1 2
1 1
1 2
1 e2
{−
[
r+ 2(
sin
+ 2
)] −
} ( r, , ) = E ( r, , )
2
2
2m r r
r sin
sin
4e 0 r
2
Separation of variables ( r, , ) = R( r ) ( )( )
2
1 d2
1 e2
( ) ( )[ −
r−
]R ( r )
2
2m r dr
4e 0 r
1 1
1 2
− R( r )
(
sin
+ 2
)( ) ( ) = ER( r )( ) ( )
2
2
2m r sin
sin
2
2
1
1 d2
1 e2
[−
r−
]R ( r )
2
R ( r ) 2m r dr
4e 0 r
1
1
1
1 2
−
(
sin
+ 2
)( ) ( ) = E
2
2
2m r ( ) ( ) sin
sin
2
2
1
1 d2
1 e2
2
{r
[−
r
−
]
R
(
r
)
−
Er
}
2
R ( r ) 2m r dr
4e 0 r
2
1
1
1 2
−{
(
sin
+ 2
)( ) ( )} = 0
2
2m ( ) ( ) sin
sin
2
2
1
1 d2
1 e2
2
2
r
[
−
r
−
]
R
(
r
)
−
Er
=C
2
R ( r ) 2m r dr
4e 0 r
2
1
1
1 2
(
sin
+ 2
)( ) ( ) = C
2
2m ( ) ( ) sin
sin
− 2l (l + 1)
1 d2
l (l + 1) 2m 1 e 2 2m
Let the constant C =
[
r−
+ 2
+ 2 E ]R ( r ) = 0
2
2
2m
r dr
r
4e 0 r
hcR E1
2r
− r / na 2 r l 2 l +1
=
,
R
:
Rydberg
const
ant.
R
(
r
)
=
A
e
(
)
L
(
);
nl
nl
n −l −1
2
2
n
n
na
na
2r
a : Bohr radius; L2nl−+l1−1 ( ) : associated Laguerre polynomials; Anl : normalization constant
na
En = −
1
1
1 2
(
sin
+ 2
)( ) ( ) = C = −
2
2m ( ) ( ) sin
sin
2
2
l (l + 1)
2m
1
1 2
(
sin
+ 2
)( ) ( ) = −l (l + 1)( ) ( )
2
sin
sin
1
1 d
d
1 d2
2
2
sin (
sin
)( ) +
(
)
=
−
l
(
l
+
1)
sin
2
( )
sin d
d
( ) d
1
1 d
d
1 d2
2
2
{
sin (
sin
)( ) + l (l + 1) sin } +
( ) = 0
2
( )
sin d
d
( ) d
1 d
d
1
2
2
sin
(
sin
)
(
)
+
l
(
l
+
1)
sin
= C
( )
sin d
d
2
1
d
( ) = −C
2
( ) d
Let the constant C = m 2 .
1 d2
d 2 ( )
2
2
im
− im
(
)
=
−
C
=
−
m
+
m
(
)
=
0
(
)
=
A
e
+
B
e
( ) d 2
d 2
( + 2 ) = ( ) m is an integer. m = 0,1, 2
1
1 d
d
sin 2 (
sin
)( ) + l (l + 1) sin 2 = C = m 2
( )
sin d
d
1 d
d ( )
m2
[sin
] + [l (l + 1) − 2 ]( ) = 0
sin d
d
sin
( ) = Pl m (cos ) associated Legendre functions
Yl m ( , ) =
2l + 1 (l − m )! m
Pl (cos )eim spherical harmonics
4 (l + m )!
Finally, the nomalized solutions of the time-independent Schroedinger equation
2 ( n − l − 1)! − r / na 2 r l 2 l +1 2 r m
nlm ( r, , ) =
e
( ) Ln −l −1 ( )Yl ( , )
na
2
n
[(
n
+
l
)!]
na
na
2
En = −
hcR E1
= 2
2
n
n
Note:
i) En is determned only by the principal quantum number n.
ii) n = 1, 2,3,
; l = 0,1, 2
n − 1; m = −l , −l + 1,
, l − 1, + l
iii) l = 0 ( s ), 1 (p ), 2(d )
e. g. 1s: n = 1, l = 0; 2 p : n = 2, l = 1; 3d : n = 3, l = 2
Matrix Mechanics (Heisenberg, Born, Jordan)
State fuction = aii , where i ' s are elements of a basis of the Hilbert space.
i
a1
a
2
an
Fij = i Fˆ j → matrix elements in {i } representation.
Operator Fˆ
1 Fˆ 1
Fˆ
1
2
n Fˆ 1
1 Fˆ 2
2 Fˆ 2
n Fˆ 2
1 Fˆ n
2 Fˆ n
n Fˆ n
If = aii and = Fˆ
i
1 Fˆ 1
1 Fˆ 2
1 Fˆ n
a1
a1
a Fˆ
ˆ
ˆ a
F
F
2
2
1
2
2
2
n
=
2
a
n Fˆ
ˆ
ˆ an
F
F
1
n
2
n
n
n
Note:
If Gˆ i = gii i.e. i ' s are eigenfunctions of Gˆ with eigenvalues gi ' s ,
then i Gˆ j = g j i j = g j ij .
gi if i = j
Gij = g j ij =
if i j
0
0
g1 0
0 g
0
2
Gˆ is diagonal
Gˆ =
0
0
g
n
0
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