Download ELECTRODYNAMICS—lecture notes second semester 2004 Ora Entin-Wohlman

ELECTRODYNAMICS—lecture notes
second semester 2004
Ora Entin-Wohlman
References:
1. J. D. Jackson, “Classical Electrodynamics”, Wiley.
2. G. B. Arfken, "Mathematical methods for Physicists",
Academic Press.
1
Summary of vector analysis
r
r
∂F
∂F
∂F
∇F = xˆ
+ yˆ
+ zˆ
1. Gradient, ∇ ,
∂z
∂x
∂y
r r
a. ∇F (r ) is perpendicular to the surface of constant F that
r
the point r .
r
r
r dF
b. The gradient of F (r ) : ∇F =
r dr
includes
r r ∂Vx ∂V y ∂Vz
r
∇
⋅V =
+
+
∇
⋅
,
2. Divergence,
∂x
∂y
∂z
r r
a. ∇ ⋅ V is the flux per unit volume through an infinitesimal closed
surface,
r r
lim 1 r
∇ ⋅V =
V ⋅ nˆ da
v → 0 v ∫S
xˆ
r
r
∂
r
∇
×
V
=
3. Curl (rotor), ∇ × ,
∂x
Vx
yˆ
∂
∂y
Vy
zˆ
∂
∂z
Vz
a. The n̂ component of the rotor is related to its circulation by
r r
lim 1 r r
nˆ ⋅ ∇ × V =
V ⋅ dl
S → 0 S C∫
where n̂ is normal to the plane where C lies.
r
r
Exercise: Given the vector V = r F (r ) , find F (r ) so that
r r r r
∇ ⋅V = ∇ ×V = 0 .
∂2F ∂2F ∂2F
+
+ 2
4. Div.grad=Laplacian, ∇ , ∇ F =
∂x 2 ∂y 2
∂z
2
2
2
r r
5. Rot × grad =0, ∇ × ∇F = 0
r r r
6. Div.rot=0, ∇ ⋅ ∇ × V = 0
r
r
r
r r r
r
2
7. Rot × rot, ∇ × (∇ × V ) = ∇(∇ ⋅ V ) − ∇ V
8. Gauss’s theorem
r
rr r
∫ V ⋅ nˆda = ∫ dr ∇ ⋅ V
S
V
Important consequence:
r
rr r 1
r r r ⎧− 4π
∫ dr ∇ ⋅ ∇ r = − ∫ dr ∇ ⋅ r 3 = ⎨⎩ 0
r
r
r
r
Exercise: show that when V = A × P, where A is a constant vector, Gauss’s
r
rr r
ˆ
n
×
P
da
=
d
r
∇ × P . Hint: use
theorem gives ∫
∫
S
V
r r
r r
r r
r r r r r r r r r
nˆ ⋅ A × P = −nˆ ⋅ P × A = −nˆ × P ⋅ A and ∇ ⋅ A × P = P ⋅ ∇ × A − A ⋅ ∇ × P .
9. Stoke’s theorem
r r
r r
V
⋅
d
l
=
∇
∫
∫ × V ⋅ nˆda
S
r
r
r
Exercise: show that when V = AF , where A is a constant vector,
r
r
ˆ
n
×
∇
Fda
=
Fd
l . Hint: use
Stoke’s theorem gives ∫
∫
S
r
r
r
r r
r
∇ × ( AF ) = F∇ × A + (∇F ) × A .
3
Electrostatics
1. Coulomb’s law
r r
r
The electric field, E (r ) , at point r , due to a collection of point-charges,
r
qi , located at ri , is
r r
n
r r
r − ri
E (r ) = ∑ qi r r 3
| r − ri |
i =1
When those charges can be described as a charge-density (of dimension
r
charge per unit volume), ρ (r ' ) , then
r r
r r
r r r − r'
E (r ) = ∫ dr ' ρ (r ' ) r r 3
| r − r '|
The two expressions coincide for
n
r
r r
ρ (r ) = ∑ qi δ (r − ri )
i =1
The delta-function
Dirac delta-function obeys the following in one-dimension:
δ ( x − a) = 0, x ≠ a
a+
∫ dxδ ( x − a) = 1
a−
and has the properties
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∫ dxf ( x)δ ( x − a) = f (a)
δ ( f ( x) ) = ∑
i
1
df
dx
∫ dxf ( x)δ ' ( x) = − f ' (a)
δ ( x − xi )
xi
∞
Exercise: calculate ∫ dxf ( x )δ ( ax + b) =
−∞
1
b
f (− ) .
|a|
a
In more than one dimension, the delta-function becomes a product. In
Cartesian coordinates
r r
δ (r − R) = δ ( x − X )δ ( y − Y )δ ( z − Z )
In spherical coordinates
r r
δ (r − r ' ) =
Exercise: calculate
r
r
1
δ (r − r ' )δ (cosθ − cosθ ' )δ (ϕ − ϕ ' )
r2
r
r r
dxdyF
(
r
)
δ
(
A
r
+ B ), where A is a two by two
∫
matrix, B and r are two-dimensional vectors. Written explicitly, we
have
r
dxdyF
(
r
)δ ( A11 x + A12 y + Bx )δ ( A21 x + A22 y + B y ) .
∫
r
r
r
r
−1
Hence, Ar = r ' and A r ' = r . It remains to note that
∂x
∂x '
dxdy =
∂y
∂x '
∂x
dx ' dy '
r
∂y '
1
−1
dx ' dy ' =
F (− A B ) .
∂y
| det A | , to obtain | det A |
∂y '
5
Examples:
1. Gaussian representation of a one-dimensional delta-function, the limit
α → 0 of
⎡ ⎛ x ⎞2 ⎤
δ ( x) =
exp ⎢− ⎜ ⎟ ⎥
α π
⎣⎢ ⎝ α ⎠ ⎦⎥
1
2. Gaussian representation of a three-dimensional delta-function, the
limit α → 0 of
⎛ 1 ⎞
δ (r ) = ⎜ ⎟
⎝ 2π ⎠
3/ 2
(
)
⎡ 1
⎤
exp
x2 + y2 + z2 ⎥
−
3
2
⎢
α
⎣ 2α
⎦
1
3. Other one-dimensional delta-functions: (a) Lorentzian--the limit α → 0
1
1
;
of δ ( x) =
απ 1 + ( x / α ) 2
(b) of
sin( x / α )
1
δ ( x) =
=
πx
2π
1/ α
ixt
dte
∫
−1 / α
(c) and of
δ ( x) =
1
α
e −( x / α ) , x > 0
4. From Gauss’s theorem,
δ (r ) = −
1 2⎛ 1 ⎞
∇ ⎜ ⎟
4π
⎝r⎠
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2. Gauss’s law
n
r
r r
ˆ
E
⋅
n
da
=
4
π
q
=
4
π
d
r
∑ i
∫
∫ ρ (r )
i =1
S
S
V
n
V
q
E
r r
r r r
r
r r r r − r'
∇ ⋅ E ( r ) = ∫ dr ' ρ (r ' )∇ ⋅ r r 3 = 4πρ ( r )
| r − r '|
r r r
∇ × E (r ) = 0
explanation:
r rr
r
∇ ⋅ 3 = 4πδ ( r )
r
and
r rr
∇× 3 = 0
r
Example: Calculate the electric field of an infinite cylinder, of radius a,
charged with a constant charge density λ per unit length. We use
cylindrical coordinates, ρ , ϕ , z . By symmetry, the field is only along the
ρ̂ -direction. Consider a volume of length L around the cylinder. By
Gauss’s law,
⎛Q⎞
⎛λ⎞
EL 2πρ = 4π ⎜ ⎟ Lπρ 2 = 4π ⎜ ⎟ Lπρ 2 , ρ < a ,
⎝ SL ⎠
⎝S⎠
⎛Q⎞
⎛λ⎞
EL 2πρ = 4π ⎜ ⎟ Lπa 2 = 4π ⎜ ⎟ Lπa 2 , ρ > a ,
⎝ SL ⎠
⎝S⎠
giving
r
2λ
E ( ρ ) = ρˆ 2 ρ , ρ < a ,
a
7
r
2λ
E ( ρ ) = ρˆ
,
ρ > a.
ρ
3. scalar potential
r r
r r ρ (rr ' )
r r
E (r ) = −∇ ∫ dr ' r r ≡ −∇Φ (r ),
| r − r '|
r
r
r ρ (r ' )
Φ (r ) = ∫ dr ' r r
| r − r '|
r
r1
r r1
r rr
r
r
2 1
= −∇ ⋅ 3 = −4πδ (r )
since ∇ = − 3 ⇒ ∇ ⋅ ∇ ≡ ∇
r
r
r
r
r
r r
∫ E ⋅ dl = −(Φ B − Φ A ),
B
Thus:
A
(work per charge)
r r
E
∫ ⋅ dl = 0
Exercise: Find the potential of a uniformly charged straight wire of
length 2l . The charge per unit length is λ . Hint:
∫
dx
1 + x2
= ln( x + 1 + x 2 ) .
Let the wire be along the z-axis, such that the x-y plane cuts it at its
middle point. Then for points on the x-y plane we have
l
Φ( ρ ) = λ ∫
−l
dz '
ρ + z'
2
2
= λ ln
l + l2 + ρ 2
− l + l2 + ρ 2
For points in the x-z plane
8
l
Φ ( x,0, z ) = λ ∫
−l
dz '
ρ + ( z − z' )
2
2
= λ ln
z + l + ( z + l) 2 + x 2
z − l + ( z − l) 2 + x 2
By symmetry, the potential at any point is
Φ ( x, y, z ) = λ ln
z + l + ( z + l) 2 + x 2 + y 2
z − l + ( z − l) 2 + x 2 + y 2
.
The equi-potential surfaces are ellipsoids. Consider points on the x-z
plane. An ellipse there is given by
x2 z 2
+
=1
a2 c2
The sum of the distances from the points (0, l) and (0,−l) should be
constant for points on the ellipse. Namely,
z2
z2
2
2
a (1 − 2 ) + ( z − l) + a (1 − 2 ) + ( z + l) 2 = const .
c
c
2
For z=0 this gives 2 a + l and for z=c (namely, x=0) it gives 2c.
2
2
2
2
2
Therefore, c = a + l . Inserting into the result for Φ ( x,0, z ) , we find
that the equi-potential curves are
Φ 0 = λ ln
c+l
.
c−l
4. surface charge distributions
n
E2
r
σ (r )
E1
9
(
)
r
r
E 2 − E1 ⋅ nˆ = 4πσ
discontinuity in the normal component of the
electric field. The tangential components are continuous.
Exercise: The electric field of a uniformly charged (infinite) plane, of
charge σ per unit area. By symmetry, (for a plane perpendicular to the
z-axis),
r r ⎧ zˆE ( z ),
E (r ) = ⎨
⎩− zˆE (| z |),
z > 0,
z < 0.
Then, applying Gauss’s law for a cylinder along the z-direction, the
electric field on each side of the plane is
discontinuity as given above.
E ( z ) = 2πσ
,with the
5. Poisson and Laplace equations
When there is charge density, the potential satisfies the Poisson
equation:
∇ 2 Φ = −4πρ . Otherwise, it satisfies the Laplace equation:
∇ 2Φ = 0 .
Example: Let us find the charge distribution giving rise to the scalar
potential
e −α r ⎛ α r ⎞
Φ (r ) = q
⎜1 + ⎟
r ⎝
2 ⎠
Let us the Laplacian of this potential (in Cartesian coordinates)
10
∂r ∂Φ
∂
x
= q F (r ),
Φ(r ) =
∂x ∂r
∂x
r
α α2 ⎞
−αr ⎛ 1
⎟⎟
F (r ) = −e ⎜⎜ 2 + +
r
r
2
⎝
⎠
⎛ F (r ) ⎛ x ⎞ 2 ∂
⎞
∂2
∂ x
⎜
⎟
Φ
=
=
+
(
r
)
q
F
(
r
)
q
F
(
r
)
⎜ ⎟
⎜ r
⎟
∂x 2
∂x r
∂
r
r
⎝ ⎠
⎝
⎠
Therefore,
∇ 2 Φ(r ) =
q −αr 3
e α and
2
ρ ( r ) = − 2πqe −α r α 3
6. Electrostatic potential energy
The potential energy of a collection of point charges is:
qi q j
1
W = ∑∑ r r
2 i j | ri − r j |
The potential energy of continuous charge distribution is:
r
v
r r ρ (r ' ) ρ (r ) 1 r r
r
r r
r
1
1
W = ∫ ∫ dr ' dr r r = ∫ dr ρ (r )Φ(r ) = − ∫ dr Φ(r )∇ 2 Φ(r )
2
| r − r '|
2
8π
W=
r r r 2 1
r 2
1
d
r
r
d
r
E
∇
Φ
=
|
(
)
|
8π ∫
8π ∫
Exercise: Find the electrostatic energy of a uniformly charged sphere
of radius a. Let us first find the electric field, using Gauss’s law. By
r r
symmetry, E ( r ) = rˆE (r ). Then,
11
⎧
4πa 3
r>a
⎪⎪ 4π 3 ρ = 4πQ,
2
E ( r ) 4πr = ⎨
3
3
⎪4π 4πr ρ = 4πQ r , r < a
⎪⎩
3
a3
The energy is then
∞
a
2
⎤ 3 Q2
4π 2 ⎡
2 r
2 1
W=
Q ⎢ ∫ drr 6 + ∫ drr 4 ⎥ =
8π
a
r ⎦ 5 a
a
⎣0
7. Conductors
a.
b.
c.
d.
The
The
The
The
electric field is zero inside a conductor (at equilibrium).
potential inside a conductor is constant.
charge density inside a conductor is zero.
electric field on the surface of a conductor satisfies:
where
r
Et = 0
En = 4πσ ,
σ is the charge density on the surface.
Example: the electric field of a conducting sphere, of radius a, carrying
r r
E
(r ) = rˆE (r ) . The Laplace equation is then
a charge q. By symmetry,
(everywhere, except on the surface)
∇ 2Φ =
1 d ⎛ 2 dΦ ⎞
⎜r
⎟ = 0,
r 2 dr ⎝ dr ⎠
with
Φ ( a ) = Φ 0 , Φ ( ∞ ) = 0.
12
The solution of this Laplace equation is in general
Φ(r ) =
A
+B
r
The boundary conditions require A = aΦ 0 ,
B = 0. Thus the potential is
a
a
Φ (r ) = Φ 0 , and the electric field is E (r ) = 2 Φ 0 , from which we find
r
r
q
q
Φ 0 = 4πσa = 4π
a
=
.
4πa 2
a
Example: a grounded sphere, of radius a, in the presence of an external
r r
field, E (r ) = E0 zˆ . The boundary conditions on the potential are then
Φ (a, θ ) = 0, Φ (r → ∞, θ ) = − E0 z = − E0 r cos θ .
The Laplace equation is then
1 ∂2
1
∂ ⎛
∂Φ ⎞
(
)
sin
r
θ
∇ Φ (r , θ ) =
Φ
+
⎟ =0.
⎜
r ∂r 2
r 2 sin θ ∂θ ⎝
∂θ ⎠
2
1 ∂2
1 ∂ ⎛ 2 ∂Φ ⎞
(rΦ) = 2
[Note:
⎟ .] The solution which includes just terms
⎜r
2
r ∂r
r ∂r ⎝ ∂r ⎠
linear in cosθ is
Φ (r ,θ ) =
A
C cos θ
+B+
+ Dr cos θ .
r
r2
[To check this, it is convenient to put µ = cosθ .] B is just a constant, so
we take it to be zero. Also, D = − E0 . In addition, A=0 because the
electric field for cosθ = 0 [the plane z=0] has opposite signs above and
below the plane z=0. Finally,
13
Φ ( a, θ ) =
leading to
C cos θ
− E0 a cos θ = 0,
a2
C = E0 a 3 . The solution for the potential is therefore
a 3 cos θ
Φ ( r , θ ) = E0
− E0 r cos θ ,
r2
which yields for the electric field
⎛ 2a 3 ⎞
Er (r , θ ) = E0 ⎜⎜1 + 3 ⎟⎟ cos θ
r ⎠
⎝
⎛ a3 ⎞
Eθ (r , θ ) = − E0 ⎜⎜1 − 3 ⎟⎟ sin θ
⎝ r ⎠
The surface charge density on the sphere is
σ (θ ) = E0
and the total surface charge is
3
cos θ ,
4π
π
∫ dθ sin θσ (θ ) = 0 .
0
This leads to a dipole-like behavior: the second terms in the result for
the electric field can be put in the form
with
r
r
r
3rˆ( p ⋅ rˆ) − p
E (r ,θ ) =
,
r3
r
p = E0 a 3 rˆ cos θ − θˆ sin θ = E0 a 3 zˆ , since
(
)
rˆ = xˆ sin θ cos ϕ + yˆ sin θ sin ϕ + zˆ cos θ ,
θˆ = xˆ cos θ cos ϕ + yˆ cos θ sin ϕ − zˆ sin θ ,
ϕˆ = − xˆ sin θ sin ϕ + yˆ sin θ cos ϕ .
14