Download Physics 2. Electromagnetism 1 Fields Lecture 1. Vector and tensor analysis

Physics 2. Electromagnetism
Lecture 1. Vector and tensor analysis
1
1.1
Fields
Scalar field
A scalar field is a function U (r, t). Field means that in each point of space r there is a number U (r)
given. Scalar means that this number does not depend on the choice of coordinates. If we choose
new coordinates x0i = x0i (xj ) this number does not change although the functional dependence does
change: U 0 (x0i ) = U (xj ).
Example 1.1.
Let U = x2 + y 2 and we rotate the coordinates: x0 = x cos θ + y sin θ, y 0 =
−x sin θ + y cos θ. In order to find U 0 we only have to substitute x and y in the original expression,
which gives U 0 = x0 2 + y 0 2 .
Example 1.2.
Let U = x2 and we stretch the coordinate x0 = Cx, C = const. Then U 0 (x0 ) =
U (x) = x2 = (x0 /C)2 = x0 2 /C 2 , that is, U 0 = x0 2 /C 2 .
1.2
Vector field
A vector field is a vector function A(r, t). Field means that in each point of n-dimensional space r
there are n numbers Ai (r), i = 1, . . . , n given. Vector means that these numbers change in a very
special way when we change the coordinates. Namely, let us choose new coordinates x0i = x0i (xj ). We
can also express the old coordinates in terms of the new ones: xi = xi (x0j ). The n number Ai are
components of a vector (or simply a vector) if
A0i
X ∂x0
∂x0i
i
=
Aj ≡
Aj .
∂xj
∂xj
j
(1)
In the right hand side of (1) we used the Einstein’s rule: if an index is repeated in the expression
a summation over this index is implied. It should be understood that A0i depend on x0j , while Ai
depend on xj .
Example 1.3.
L et Ax = x2 , Ay = xy + y 2 , and we rotate the coordinates: x0 = x cos θ + y sin θ,
1
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Physics 2. Electromagnetism
Lecture 1
y 0 = −x sin θ + y cos θ. According to (1) we have
A0x =
∂x0
∂x0
Ax +
Ay
∂x
∂y
(2)
= cos θx2 + sin θ(xy + y 2 )
(3)
= cos θ(x0 cos θ − y 0 sin θ)2 + sin θ((x0 cos θ − y 0 sin θ)(x0 sin θ + y 0 cos θ)
+ (x0 sin θ + y 0 cos θ)2 )
(4)
Simplify and calculate A0y respectively.
Example 1.4.
L et Ax = x2 and x0 = Cx (one-dimensional case). According to our rule A0x =
(∂x0 /∂x)Ax = Cx2 = C(x0 /C)2 = x0 2 /C. Compare to U 0 = x0 2 /C 2 !
2
Tensor field
Let us recall that a (second rank) tensor is a two-indices construction Tij , i = 1, . . . , n, j = 1, . . . , n
(n2 numbers). In accordance with the previous definition the coordinate transformation x0i = x0i (xj )
results in the following transformation of the tensor:
Tij0 (x0 )
∂x0i ∂x0j
=
Tkl (x)
∂xk ∂xl
(5)
(Einstein’s summation rule applied !).
In a more general way, a construction Ti1 i2 ...is with s indices (ns numbers) is a s-rank tensor if
the coordinate transformation x0i = x0i (xj ) results in the transformation
T 0 i1 . . . is (x0 ) =
∂x0i1
∂x0
. . . is Tj1 ...js (x).
∂xj1
∂xjs
(6)
In other words, we first have to perform all differentiations, all summations, and finally substitute
xi = xi (x0j ).
Example 2.1.
Let Txx = y 2 + z 2 , Txy = Tyx = −xy, Tyy = x2 + z 2 , Txz = Tzx = −xz,
Tyz = Tzy = −yz, Tzz = x2 + y 2 (this is taken from the functional form of the inertia tensor
Tij = r2 δij − xi xj ). The coordinate rotation x0 = x cos θ + y sin θ, y 0 = −x sin θ + y cos θ, z 0 = z will
give
0
Tzz
= Tzz ,
(7)
2
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Physics 2. Electromagnetism
Lecture 1
0
= cos θTxz + sin θTyz
Txz
(8)
= − cos θxz − sin θyz = −x0 z 0 ,
0
= cos θ cos θTxx + 2 cos θ sin θTxy + sin θ sin θTyy
Txx
= cos2 θ(y 2 + z 2 ) − 2 cos θ sin θxy + sin2 θ(x2 + z 2 )
2
2
= y0 + z0 ,
(9)
(10)
(11)
(12)
0
Txy
= − cos θ sin θTxx + cos θ cos θTxy − sin θ sin θTyx + sin θ cos θTyy
(13)
= −x2 cos θ sin θ − xy(cos2 θ − sin2 θ) + y 2 cos θ sin θ = −x0 y 0
(14)
Find other components.
3
Vector analysis
In the previous course (Physics 1) we defined the vector differential operator ∇ which can be applied
to a scalar to give a vector (gradient) grad U = ∇U , to a vector to give a scalar (divergence)
div A = ∇ · A, or to a vector to give a new vector (curl or rotor) rot A = ∇ × A. Here we consider
these operations in more detail.
3.1
Gradient
In the past gradient was defined as (grad U )i = (∂U/∂xi ). Such definition is good for Cartesian
coordinates but would be not quite useful in curvilinear coordinates. Let us consider, for example,
spherical coordinates x = r sin θ cos φ, y = sin θ sin φ, z = r cos θ. According to the naive definition,
we would have (grad U )r = (∂U/∂r), (grad U )θ = (∂U/∂θ), and (grad U )φ = (∂U/∂φ). Let us now
recall that the first and the most important use of the gradient is to get the force from the potential
energy F = − grad U . According to the above definition we would have Fr and Fθ in different units
(if Fr is in N, then Fθ is in N·m), which is not physical. Thus, we have to give a more successful
definition of the gradient. The proper definition is given in terms of scalars:
grad U · dr = dU,
(15)
grad U = (grad U )x x̂ + (grad U )y ŷ + (grad U )z ẑ,
(16)
dr = dxx̂ + dyŷ + dzẑ,
(17)
where dU is the full differential.
3.1.1
Cartesian coordinates
In the cartesian coordinates we have
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Physics 2. Electromagnetism
Lecture 1
dU =
∂U
∂U
∂U
dx +
dy +
dz,
∂x
∂y
∂dz
(18)
so that
(grad U )x dx + (grad U )y dy + (grad U )z dz =
and therefore
(grad U )x =
∂U
,
∂x
(grad U )y =
∂U
,
∂y
∂U
∂U
∂U
dx +
dy +
dz,
∂x
∂y
∂dz
(grad U )y =
∂U
,
∂y
(19)
(20)
as usual.
3.1.2
Spherical coordinates
Let us consider now spherical coordinates, where
grad U = (grad U )r r̂ + (grad U )θ θ̂ + (grad U )φ φ̂,
(21)
dr = drr̂ + rdθθ̂ + r sin θdφφ̂,
∂U
∂U
∂U
dr +
dθ +
dφ.
dU =
∂r
∂θ
∂φ
(22)
(23)
Putting al this together we get
(grad U )r dr + (grad U )θ rdθ + (grad U )φ r sin θdφ =
and therefore
(grad Ur ) =
∂U
,
∂r
(grad U )θ =
1 ∂U
,
r ∂θ
∂U
∂U
∂U
dr +
dθ +
dφ,
∂r
∂θ
∂φ
(grad U )φ =
1 ∂U
.
r sin θ ∂φ
(24)
(25)
Now all components of the force would be measures in the same units.
Show that in the cylindrical coordinates x = r cos φ, y = r sin φ, z, the gradient is
(grad Ur ) =
3.2
∂U
,
∂r
(grad U )φ =
1 ∂U
,
r ∂φ
(grad U )z =
∂U
.
∂z
(26)
Divergence
Divergence was defined earlier as div A = (∂Ax /∂x) + (∂Ay /∂y) + (∂Az /∂z). As we have seen above
definitions in Cartesian coordinates are not general enough and we have to think of a generalization.
For this purpose let us consider a small (infinitesimal) volume dV and it surface. In each point of the
surface we define an outward directed (unit) normal vector n̂ and the a vector field flux dΦ = A · n̂dS
H
H
(here dS is the infinitesimal area). Let dΦ = A · n̂dS be the flux from the volume dV in the
outward direction. Then we define
div AdV = A · n̂dS.
(27)
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Physics 2. Electromagnetism
3.2.1
Lecture 1
Cartesian coordinates
We again start the analysis with the Cartesian coordinates. The corresponding infinitesimal volume
will be a cube with the coordinates (x, y, z), (x+dx, y, z), (x, y+dy, z), (x, y, z+dz), (x+dx, y+dy, z),
(x + dx, y, z + dz), (x, y + dy, z + dz), (x + dx, y + dy, z + dz), so that the volume is dV = dxdydz. Let
us consider the flux from inside the cube through the sides parallel to the x − y plane. The area of
each of these sides is dxdy, while the normal is ẑ at the side passing through z +dz and −ẑ at the side
passing through z. Thus, the flux will be Az [x, y, z + dz]dxdy − Az [x, y, z]dxdy = (∂Az /∂z)dxdydz
(from now on we shall use square brackets to denote functional dependence). In the same way (SHOW
!) the flux through other sides will give (∂Ax /∂x)dxdydz and (∂Ay /∂y)dxdydz, so that we have
div Adxdydz = (
∂Ax ∂Ay ∂Az
+
+
)dxdydz,
∂x
∂y
∂z
(28)
which gives usual Cartesian divergence.
3.2.2
General orthogonal curvilinear coordinates
For the following it is more convenient to consider general orthogonal curvilinear coordinates (x1 , x2 , x3 ),
such that the distance in these coordinates is given by
ds2 =
X
h2i dx2i .
(29)
i
The coefficients hi depend on the coordinates, in general. The vector dr is written as follows:
dr =
X
hi dxi êi ,
(30)
i
where êi · eˆj = δij .
1. Cartesian coordinates: x1 = x, x2 = y, x3 = z, hi = 1 for i = 1, 2, 3, eˆ1 = x̂, eˆ2 = ŷ, eˆ3 = ẑ.
2. Spherical coordinates: x1 = r, x2 = θ, x3 = φ, h1 = 1, h2 = r, h3 = r sin θ, eˆ1 = r̂, eˆ2 = θ̂,
eˆ3 = φ̂.
3. Cylindrical coordinates: x1 = r, x2 = θ, x3 = z, h1 = 1, h2 = r, h3 = 1, eˆ1 = r̂, eˆ2 = θ̂, eˆ3 = ẑ.
Now let us consider the infinitesimal volume dV which is between (x1 , x2 , x3 ) and (x1 + dx1 , x2 +
dx2 , x3 + dx3 ), so that dV = h1 h2 h3 dx1 dx2 dx3 . Let us calculate the flux across the sides “parallel” to x1 − x2 “plane”: A3 [x1 , x2 , x3 + dx3 ]dS[x3 + dx3 ] − A3 [x1 , x2 , x3 ]dS[x3 ]. The area dS =
(h1 dx1 )(h2 dx2 ), where h1 and h2 depend on x3 too. Therefore, dΦ3 = (h1 h2 A3 )[x3 + dx3 ]dx1 dx2 −
(h1 h2 A3 )[x3 ]dx1 dx2 = (∂h1 h2 A3 /∂x3 )dx1 dx2 dx3 . Calculating in the same way we have
dΦ =
X ∂h1 h2 h3 Ai /hi
i
∂xi
dx1 dx2 dx3 = div Ah1 h2 h3 dx1 dx2 dx3 ,
(31)
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Physics 2. Electromagnetism
Lecture 1
so that
1 X ∂h1 h2 h3 Ai /hi
.
h1 h2 h3 i
∂xi
P
1. Cartesian coordinates: all hi = 1 therefore div A = i (∂Ai /∂xi ).
2. Spherical coordinates: substituting the corresponding hi we have h1 h2 h3 = r2 sin θ and
div A =
div A =
1 ∂ sin θAθ
1 ∂Aφ
1 ∂r2 Ar
+
+
.
r2 ∂r
r sin θ
∂θ
r sin θ ∂φ
(32)
(33)
Derive div A in cylindrical coordinates.
4
Rotor
The previous Cartesian definition of rot A was as follows:
(rot A)i = ijk
∂Ak
∂xj
(34)
(Einstein summation rule applied !). Again we need a generalization, which requires definition of a
circulation.
4.1
Circulation
Let us consider a closed path which encloses some surface. Let us define together a normal n̂ to
the surface (which changes with the position on the surface) and the direction of the path l̂ (which
changes along the path). The unit vector l̂ is always tangential to the path and its direction is always
in the direction of our “movement” along this path (we do not turn back). The two unit vectors n̂
and l̂ must be chosen consistently, so that n̂ × l̂ always points inside the path. Circulation of the
vector A along the path is defined as
I
Ψ=
A · l̂dl,
(35)
where l is the length.
4.2
Rotor in general
Let us consider and infinitesimal dS area with the normal n̂ (since the area is infinitesimal the
direction of the normal does not change) enclosed in an infinitesimal path. Then rotor is defined as
follows:
I
rot A · n̂dS = A · l̂dl.
(36)
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Physics 2. Electromagnetism
4.2.1
Lecture 1
Cartesian coordinates
Let us choose a small rectangle (x, y), (x + dx, y), (x + dx, y + dy), (x, y + dy). If we move along
the rectangle in this order, then the normal is n̂ = ẑ. The circulation will be dΨ = Ax [x, y]dx +
Ay [x + dx, y]dy − Ax [x, y + dy]dx − Ay [x, y]dy = ((∂Ay /∂x) − (∂Ax /∂y))dxdy. Comparing with
dΨ = (rot A)z dxdy we find
∂Ay ∂Ax
−
(37)
(rot A)z =
∂x
∂y
that is, the usual expression.
4.2.2
General orthogonal curvilinear coordinates
Let the “rectangle” be (x1 , x2 ), (x1 +dx1 , x2 ), (x1 +dx1 , x2 +dx2 ), (x1 , x2 +dx2 ). Then the circulation
is calculated as follows (dl1 = h1 dx1 , dl2 = h2 dx2 , dS = h1 h2 dx1 dx2 , n̂ = eˆ3 ):
dΨ = (A1 h1 )[x1 , x2 ]dx1 + (A2 h2 )[x1 + dx1 , x2 ]dx2
− (A1 h1 )[x1 , x2 + dx2 ]dx1 − (A2 h2 )[x1 , x2 ]dx2
∂h2 A2 ∂h1 A1
=(
−
)dx1 dx2 .
∂x1
∂x2
(38)
Comparison with dΨ = (rot A)3 h1 h2 dx1 dx2 gives
1
(rot A)3 =
h1 h2
∂h2 A2 ∂h1 A1
−
∂x1
∂x2
.
(39)
Similar relations can be derived for other components. The order 1,2,3 should be conserved.
Spherical coordinates:
1
∂r sin θAφ ∂rAθ
(rot A)r = 2
−
,
r sin θ
∂θ
∂φ
∂Ar ∂r sin θAφ
1
−
(rot Aθ =
,
r sin θ ∂φ
∂r
1 ∂rAθ ∂Ar
(rot A)φ =
−
r
∂r
∂θ
(40)
(41)
(42)
Calculate rotor in cylindrical coordinates.
5
Gauss-Stokes-Green’s theorems
The two theorems result immediately from the definitions of the divergence and rotor. Let us consider
H
a finite (not infinitesimal) volume and the integral div AdV . We can divide the volume into many
H
H
P
small volumes so that div AdV = i (div AdV )i , where div AdVi = i A · n̂dS for each small volume
7
Physics 2. Electromagnetism
dVi . Thus
Lecture 1
I
div AdV =
XI
i
I
A · n̂dS =
A · n̂dS
(43)
i
where in the last integral the integration is carried out only over the outer surface, since the integral
over the surfaces, which belong to two pieces simultaneously, vanish because the corresponding
normals are pointing in the opposite directions, while the areas are the same.
In a similar way (prove) we find
I
I
rot A · n̂dS =
6
A · l̂dl.
(44)
Second order operations
We can apply ∇ once again to get the following second order operations: div grad U , grad div A,
rot rot A. The two other operations vanish identically: rot grad U ≡ 0 and div rot A ≡ 0 (prove).
The operator div grad acting on a scalar is usually called a Laplacian and denoted ∆.
P
Exercise. Show that in the Cartesian coordinates ∆U = i (∂ 2 U/∂x2i ).
Exercise. Show that in the spherical coordinates
1 ∂
∆U = 2
r ∂r
7
r
2 ∂U
∂r
1
∂
+ 2
r sin θ ∂θ
∂U
sin θ
∂θ
+
1
∂2U
.
r2 sin2 θ ∂φ2
(45)
What can you do with tensors
In what follows we apply the Einstein’s rule of summation over the repeated indices.
Let Ai be a vector and Tij be a tensor. Then Ai Ai is a scalar, Tij Aj is a vector, Tij Ai Aj is a
scalar, Tii (summation !) is a scalar, Tij Tjk is a tensor, etc.
There are two special tensors. The Kronecker delta δij is defined as follows:

1 i = j,
δij =
0 i =
6 j.
(46)
The following equalities are easily derived: δii = 3, δij Aj = Ai , δij Tjk = Tik , etc.
The Levy-Civita fully antisymmetric tensor ijk is defined as follows 123 = 1 and ijk = −jik =
−ikj .
8
Physics 2. Electromagnetism
8
Lecture 1
Area and volume in curvilinear coordinates
An infinitesimal area built on dxi and dxj will be dS = hi dxi hj dxj (no summation over i and j !).
Example 8.1.
Cylindrical coordinates. In the plane perpendicular to the axis z = const, so that
we have an area built on dr and dφ, with hr = 1, hφ = r: dS = rdrdφ. On the cylindrical surface
r = const and dS = hz dzhφ dφ = rdzdφ.
Example 8.2. Spherical coordinates. On a sphere r = const and dS = hθ dθhφ dφ = rdθr sin θdφ =
r2 sin θdθdφ.
An infinitesimal volume is dV = h1 h2 h3 dx1 dx2 dx3 .
Example 8.3.
Cylindrical coordinates: dV = rdrdθdz. Spherical coordinates: dV = r2 sin θdrdθdφ.
9
Physics 2. Electromagnetism
Lecture 2. Electric field
1
Electric charge and Coulomb law
If two particles can interact electromagnetically, we say they both have electric charge. Let two
particles have electric charges q1 and q2 and are in the positions r1 and r2 . Then the force acting on
the charge q2 from the charge q1 is
F1→2 =
kq1 q2 (r2 − r1 )
.
|r2 − r1 |3
(1)
Eq. (1) is known as the Coulomb law. Obviously, F1→2 = −F2→1 . Here k is some universal world
constant, whose value depends on the choice of units. The force looks similar to the gravitational
force. The important difference is that the electric charge can be positive and negative. As is clear
from (1) two charges of the same sign repel each other, while two charges of the opposite sign attract
each other.
1.1
Charge properties
• There is an elementary charge (proton charge) e = 1.6 · 10−19 C. The electron charge is −e. All
other charges are multiples of e, that is, q = Ze, where Z = 0, ±1, ±2, . . .. We shall usually
deal with charges q e.
• Charge is conserved, that is, no charge can disappear or appear.
• Not all particles possess charge. Particles or bodies without charge are called (electrically)
neutral.
2
Superposition principle
The superposition principle states that if there are a number of charges, qi , in the positions ri ,
i = 1, . . . , n, then the force acting on the charge q in the position r is the vector sum of the forces
1
Physics 2. Electromagnetism
Lecture 2
acting on q from each other charge:
F=
X kqqi (r − ri )
|r − ri |3
i
3
.
(2)
Electric field
Let us fix a system of charges qi in the positions ri and start measuring the force which acts on a
test charge q in different positions. We call this charge a test charge since we assume that it does
not change the positions of other charges. The force as a function of r is given by (2). It is seen that
the only parameter which belongs to the test charge is q which appears in all terms. In other words,
the force can be written as F = qE, where the vector E depends on r but does not depend on the
test particle at all. Thus, E[r] exists even without a test particle and is a field. This is called electric
field. The electric field of the system of charges is
E ==
X kqi (r − ri )
i
3.1
|r − ri |3
.
(3)
Charge distribution
Not all charges are point charges. Indeed, is a body is charge to a charge which is much larger
than the proton charge, it means that there are many charged particles present and distributed in
the body, so that it is more convenient to think of it as a continuous charge distribution with some
volume density ρ = dq/dV (charge per unit volume,) surface density σ = dq/dA (charge per unit
area), or linear density λ = da/dl (charge per unit length). In this case we divide the body into
small pieces. Let one such piece has the charge dq and its position is r0 . Then we can treat it as a
point charge and its electric field would be
dE =
so that the total electric field is
Z
E=
kdq(r − r0 )
,
|r − r0 |3
kdq(r − r0 )
,
|r − r0 |3
(4)
(5)
where the integral is taken over the whole charge.
Try to study the following examples without looking at the figures (in the end of the
lecture - when I have time). It is very instructive to learn to deal with such issues using
only formulae. Look at the figures only if you fail to understand.
Example 3.1.
Thin ring of the radius R is homogeneously charged with the charge q. What is
the electric field on the ring axis? Let us choose coordinates so that the origin is in the center of the
ring, and z axis is perpendicular to the ring plane (which is, clearly, x−y plane). The ring equation is
2
Physics 2. Electromagnetism
Lecture 2
given by r = R in polar coordinates, or x2 +y 2 = R2 in Cartesian coordinates. Each point on the ring
is characterized by the polar angle φ0 so that its Cartesian coordinates are r0 = (R cos φ0 , R sin φ0 , 0).
An infinitesimal part is cut out by the infinitesimal angle dφ0 and is of the length dl = Rdφ0 . The
linear charge density is λ = q/2πR, and the infinitesimal charge is now
dq = λdl =
q
dφ0 .
2π
The coordinates of a point on the ring axis are r = (0, 0, z), so that r − r0 = (−R cos φ0 , −R sin φ0 , z),
√
and |r − r0 | = R2 + z 2 . Now all this has to be substituted into (5):

−R cos φ0


dφ0  −R sin φ0 
z

E=
2π(R2
kq
+ z 2 )3/2
2π
Z
0
(6)
which gives Ex = Ey = 0 and
Ez =
(R2
kqz
.
+ z 2 )3/2
(7)
Limiting cases: At z = 0 (in the ring center) Ez = 0. When |z| R we get Ez = kqz/|z|3 - Coulomb
field.
Example 3.2.
A homogeneous disk of the radius R and charge q is given. What is the electric
field on the disk axis ? The choice of the coordinates is as in the previous example. The difference
is that we have now to split the disk into infinitesimal areas dA = r0 dr0 dφ0 with the position vector
r0 = (r0 cos φ0 , r0 sin φ0 , 0). The infinitesimal charge will be dq = σdA, where σ = q/πR2 is the surface
charge density. Now eq. (6) will be substituted by the following

E=
kq
πR2
Z
0
R

r0 dr0 

Z
2π
dφ0
0
(r0 2
1
+ z 2 )3/2

−r0 cos φ0
 0

 −r sin φ0 
z
(8)
Again the only nonzero component is
2kq
Ez = 2
R
z
z
−√
2
|z|
R + z2
.
(9)
Check that if R |z| we again get the Coulomb field Ez = kqz/|z|3 . The limit |z| R is more
interesting since in this case the disk becomes an infinite charged plane with the charge q → ∞ and
R → ∞ but the charge density σ = q/πR2 < ∞, and the electric field Ez = 2πσ sign z (sign z = 1
when z > 0 and sign z = −1 when z < 0).
Example 3.3.
A thin sphere of the radius R is homogeneously charged with the charge q. What
3
Physics 2. Electromagnetism
Lecture 2
is the electric field in an arbitrary point ? We choose the coordinates so that the origin is in the
sphere center, and z axis passes through the point in which we are looking for the electric field.
In other words, r = (0, 0, z). Each point on the sphere is given by two spherical angles, so that
r0 = (R sin θ0 cos φ0 , R sin θ0 sin φ0 , R cos θ0 ). The surface charge density will be σ = q/4πR2 and the
infinitesimal area is dA = R2 sin θ0 dθ0 dφ0 . Thus, we have (check that Ex = Ey = 0)
Z
π
0
sin θ dθ
Ez =
0
Z
0
0
2π
kq(z − R cos θ0 )
dφ
4π[(z − R cos θ0 )2 + R2 sin2 θ0 ]3/2
0
.
(10)
Changing the variable µ = cos θ0 we obtain
Ez =
kq
2
Z
1
dµ
−1
(z 2
z − Rµ
− 2zRµ + R2 )3/2

kq sign z/|z|3 , |z| > R
=
0,
|z| < R
(11)
Thus, the field outside the sphere is the Coulomb field, the field inside vanishes.
Example 3.4.
A rod of the length l is homogeneously charged with the linear charge density λ.
Find the electric field. We choose the coordinates so that the origin is in the middle of the rod and
the rod is along z axis. Thus, the coordinates of the rod points are r0 = (0, 0, z 0 ), −l/2 < z 0 < l/2.
The infinitesimal charge dq = λdz 0 . For simplicity, let us choose r = (x, 0, z) (otherwise we can
p
always rotate the coordinates in the x − y plane to do so). Then |r − r0 | = x2 + (z − z 0 )2 , and we
have
Z
l/2
Ex =
−l/2
kλdz 0 x
(x2 + (z − z 0 )2 )3/2
!
kλ
l/2 − z
l/2 + z
p
+p
,
2
2
2
x
x + (l/2 − z)
x + (l/2 + z)2
Z l/2
kλdz 0 (z − z 0 )
Ez =
2
0 2 3/2
−l/2 (x + (z − z ) )
!
1
1
= kλ p
−p
.
x2 + (l/2 + z)2
x2 + (l/2 − z)2
=
(12)
(13)
It is of interest to consider the case l → ∞ (infinite rod), which gives Ex = 2λ/x, Ez = 0.
Example 3.5.
A hollow cylinder (outer radius a, inner radius b < a, length l) is homogeneously
charged with the (volume) charge density ρ. Find the electric field in the plane perpendicular to the
cylinder axis and passing through the axis middle. Let us choose r = (x, 0, z) (in the plane of interest
z = 0) and the cylinder axis along z axis, with the coordinate origin in the middle. The charge is
distributed homogeneously inside the volume limited by −l/2 ≤ z 0 ≤ l/2, b ≤ r0 ≤ a (cylindrical
4
Physics 2. Electromagnetism
Lecture 2
coordinates). The infinitesimal charge dq = ρr0 dr0 dφ0 dz 0 , and r0 = (r0 cos φ0 , r0 sin φ0 , z 0 ), so that
0
0
02
0 2
2
0
0 2
R ≡ |r − r | = (x − r cos φ ) + r sin φ + (z − z )
1/2
q
= x2 − 2xr0 cos φ0 + r0 2 + z 0 2 .
Now
Z
a
0
Ex =
r dr
0
Z
l/2
dz
0
−l/2
2π
b
a
2π
Z
dφ0
0
kρ(x − r0 cos φ0 )
R3
klρ(x − r0 cos φ0 )
p
(x2 − 2xr0 cos φ0 + r0 2 ) x2 − 2xr0 cos φ0 + r0 2 + l2 /4
b
0
Z l/2
Z 2π
Z a
kρr0 sin φ0
0
0
0
r dr
dz
dφ0
Ey = −
= 0,
R3
b
−l/2
0
Z a
Z l/2
Z 2π
kρ(z − z 0 )
0
0
0
Ez = −
r dr
dz
dφ0
R3
b
−l/2
0
Z a
Z l/2
Z 2π
kρz
0
0
0
=−
r dr
dz
dφ0 3 .
R
b
−l/2
0
Z
=
r0 dr0
Z
dφ0
(14)
(15)
(16)
The integrals are rather complicated. Let is consider what happens when l → ∞ (infinite cylinder).
Physically this corresponds to the calculation of the electric field of a long cylinder a, b l near the
cylinder surface |x| l and middle plane |z| l. In this case it is worth to calculate first
Z
l/2
I≡
−l/2
2
dz 0
= 2
,
3
0
R
x − 2xr cos φ0 + r0 2
so that
a
2π
2kρ(x − r0 cos φ0 )
,
x2 − 2xr0 cos φ0 + r0 2
b
0
Z a
Z 2π
2kρz
0
0
Ez =
r dr
dφ0 2
.
x − 2xr0 cos φ0 + r0 2
b
0
Z
Ex =
0
r dr
0
Z
dφ0
(17)
(18)
The expressions are still complicated. We will not calculate the integrals here. There is a more
efficient method which we shall learn in the next lecture.
A
Integral calculation technique
Let us consider the integral
Z
1
I(z) =
dµ
−1
Notice that
(z 2
(z 2
z − Rµ
− 2zRµ + R2 )3/2
z − Rµ
d
1
=−
2
3/2
2
− 2zRµ + R )
dz (z − 2zRµ + R2 )1/2
5
Physics 2. Electromagnetism
Lecture 2
then
Z
1
d
1
dµ
2
dz (z − 2zRµ + R2 )1/2
−1
Z 1
1
d
dµ
=−
dz −1 (z 2 − 2zRµ + R2 )1/2
d (z 2 − 2zRµ + R2 )1/2 µ=1
|µ=−1
=
dz
zR
d |z − R| − |z + R|
=
dz
zR



− z2 , z > R


d
=
− 2 , |z| < R
 R
dz 

2,
z < −R
z

2


,
z>R

 z2
= 0,
|z| < R



− 2 , z < −R
I(z) = −
z2
6
Physics 2. Electromagnetism
Lecture 3. Gauss law
1
Derivation
Let us consider the electric field of a point charge q. Let us choose the coordinates so that the
point charge be in the origin, then we have E = kqr̂/r2 (in spherical coordinates). Let us consider
a closed surface S which does not enclose the charge and calculate the flux across the surface Φ =
H
H
E · dS = E · n̂dS (we will often use the short notation for convenience). According to the general
Gauss-Stokes theorem we have
I
I
div EdV = 0.
(1)
E · dS =
Φ=
V
S
The last equality follows from
1 ∂r2 Er
= 0.
(2)
r2 ∂r
Thus, the electric flux across a closed surface vanishes is there is no charge inside it.
Let us now consider the flux across a closed surface when the charge is inside. In this case the
direct application of the Gauss-Stokes theorem is impossible since E and div E ar poorly defined in
r = 0, which is inside the volume of integration. Instead, let us build a small sphere P around the
charge which is completely inside our surface S. Then we get a new closed volume which is limited
by S and P (we have cut out the small sphere from the volume limited by S). The charge is outside
this volume so that
I
I
ΦS−P =
E · dS −
E · dS = 0
(3)
div E =
S
P
(signs because of the normals). Therefore, we have
I
I
E · dS =
ΦS =
S
Z
E · dS = 4π
P
r2 dr
kq
= 4πkq.
r2
(4)
Summarizing the above we get

4πkq, the charge is inside S,
E · dS =
0,
S
the charge is outside
I
1
(5)
Physics 2. Electromagnetism
Lecture 3
The superposition principle immediately gives for arbitrary charge distribution
I
E · dS = 4πkqin ,
(6)
S
where qin is the charge which is inside S. Taking into account the definition of the volume charge
density ρ = dq/dV we can write
I
I
E · dS =
4πkρdV,
(7)
S
V
and from the general Gauss-Stokes theorem one finds
div E = 4πkρ
(8)
which is the first Maxwell equation.
2
Applications
Practical applications of the Gauss law are usually related to symmetric systems where we can make
reasonable guess about the behavior of the electric field.
2.1
Spherical symmetry
Let the charge density depend on r only (spherical coordinates). We can therefore expect that
E = E(r)r̂. Let us choose a sphere with the radius r as a Gaussian surface. Then
I
I
E · dS =
S
I
dS = 4πr2 E(r).
E · r̂dS = E(r)
(9)
S
H
On the other hand, S E · dS = 4πkq(r) where q(r) is the charge inside the sphere with the radius r.
If the charge density is known, one has
Z
q(r) = 4π
r
2
ρ(r0 )r0 dr0 .
(10)
0
It is instructive to show that the same can be found from the Maxwell equation (8). Indeed, in
spherical coordinates we have
1 dr2 E
= 4πkρ
(11)
r2 dr
which has the solution
Z
4πk r
2
ρ(r0 )r0 dr0 + C],
(12)
E= 2 [
r
0
where the constant C = 0 since E = 0 when ρ = 0 (see below delta-function).
2
Physics 2. Electromagnetism
Lecture 3
If ρ = 0 for r > a then outside r = a the electric field is
E(r) =
where
kQ
,
r2
a
Z
(13)
2
ρ(r0 )r0 dr0 .
Q = 4π
(14)
0
This is correct even when a → 0 while Q remains constant (point charge).
2.2
Cylindrical symmetry
Let the charge density depend on r only (cylindrical coordinates !). We can therefore expect that
E = E(r)r̂. Let us choose a Gaussian surface as a cylinder with the radius r and length l. The flux
through the surfaces perpendicular to z vanishes (since n̂ = ẑ ⊥ r̂) and
I
I
E · dS =
S
On the other hand,
H
S
I
E · r̂dS = E(r)
dS = 2πrlE(r).
(15)
S
E · dS = 4πkq(r), where
Z
r
ρ(r0 )r0 dr0 ,
q(r) = 2πl
(16)
0
and we obtain
Rr
ρ(r0 )r0 dr0
.
E(r) =
r
Again this can be obtained from (8) which in the cylindrically symmetric case reads
4πk
0
1 drE
= 4πkρ.
r dr
(17)
(18)
If ρ = 0 for r > a then
E(r) =
2kλ
r
for r > 0, where the effective linear charge density λ = 2π
a → 0 but λ remains finite (thin infinite rod).
2.3
(19)
Ra
0
ρ(r0 )r0 dr0 . This is correct even when
Plane symmetry
Let ρ = ρ(z) and does not depend on x and y. We can expect that E = E(z)ẑ. We shall also assume
that ρ = 0 when |z| > d, that is, there is a charged layer with a finite width, so that we can find a
place outside the charged volume. Since we can choose the direction of z axis arbitrarily, we have
the boundary conditions in the form Ez (−∞) = −Ez (∞). Choosing a Gaussian surface as a cylinder
3
Physics 2. Electromagnetism
Lecture 3
whose axis is parallel to z and which is limited by z1 < z 0 < z2 , we get
Z
I
z2
E · dS = (Ez (z2 ) − Ez (z1 ))S = 4πkS
ρ(z 0 )dz 0
(20)
z1
S
where S is the area of the cylinder base. Therefore,
z2
Z
Ez (z2 ) − Ez (z1 ) = 4πk
ρ(z 0 )dz 0 .
(21)
z1
Using the boundary conditions we get
∞
Z
ρ(z 0 )dz 0
Ez (∞) − Ez (−∞) = 2Ez (∞) = 4πk
−∞
(22)
d
Z
0
ρ(z )dz
= 4πk
0
−d
Substituting in (21) z2 = z, z1 = ∞, and Ez (∞) from (22) one has
d
Z
0
Ez (z) = 2πk
Z
0
z
ρ(z 0 )dz 0
ρ(z )dz + 4πk
−d
(23)
∞
Another approach
Using (8) one finds
dEz
=ρ
dz
(24)
which gives
Z
z
ρ(z 0 )dz 0 .
Ez (z) = Ez (−∞) +
(25)
−∞
The condition Ez (∞) = −Ez (−∞) gives
Z
∞
2Ez (−∞) = −
ρ(z 0 )dz 0
−∞
so that eventually
Ez (z) =
1
2
Z
z
0
0
Z
ρ(z )dz −
∞
∞
0
ρ(z )dz
0
.
(26)
z
which is the same as above (CHECK !).
4
Physics 2. Electromagnetism
Lecture 4. Potential
1
Potential energy
The Coulomb force by a non-moving point charge is a central force. As we already know (see
Physics 1. Mechanics) any central force is conservative (or potential which is the same). From the
superposition principle it follows that the force produced by any distribution of non-moving charges
on a test charge q is conservative. Such force is F = qE where E is the electric field produced by our
system of charges and does not depend on q. Since this force is potential, it can be represented as
F = − grad U , where U is the test particle potential energy (this definition is not rigorous enough and
we’ll make it more rigorous later). Dividing this relation by q we get E = − grad φ, where φ = U/q
does not depend on q either and is called electric potential. From the definition we get immediately:
R2
φ[r2 ] − φ[r1 ] = − 1 E · r0 .
Let us know have a closer look at the potential energy. Let a point charge q1 be held non-moving
in r1 and let us bring a second charge q2 from the infinity (infinite distance) into the position r2 .
The electric field produced by q1 in arbitrary r is
E=
kq1 (r − r1 )
.
|r − r1 |3
(1)
According to the definition of the potential, we can directly calculate
φ(r) − φ(r0 ) =
kq1
kq1
−
,
|r − r1 | |r0 − r1 |
(2)
where r0 is some (arbitrarily chosen) reference point. In this case it is convenient to choose |r0 −r1 | →
∞ and φ(r0 ) = 0, so that
kq1
φ(r) =
.
(3)
|r − r1 |
Now the potential energy when q2 is brought to r2 is
U = qφ =
kq1 q2
,
R12
(4)
where R12 = |r2 − r1 |. This potential energy is the interaction energy, that is, belongs to the system
of two charges and not to a single charge.
1
Physics 2. Electromagnetism
Lecture 4
Let us know bring another charge, q3 , into r3 . The superposition principle guarantees that
F3 = F1→3 + F2→3 = qE1 + qE2 = qE, so that the potential energy of the third charge is U3 = qφ
where φ is the potential produced by the two charges, q1 and q2 ,
φ=
kq2
kq1
+
.
R13 R23
(5)
More rigorously, U3 is the potential energy of three charges q1 , q2 , and q3 , minus the potential energy
of two charges, q1 and q2 .
Let us generalize what we have learnt so far. Let there be a system of charges, qi in ri , and let
Ri = |r − ri | and Rij = |ri − rj |. Then the potential produced by the system of charges in r will be
X kqi
φ(r) =
Ri
i
.
(6)
If we bring a test charge q into r its potential energy will be Uq = qφ. The potential energy of the
system itself is
U=
X kqi qj
Rij
X kqi qj
i>j
=
1
2
=
1
2
Rij
i6=j
X
(7)
qi φi .
i
Here φi is the potential produced in ri by all charges except qi itself,
φi =
X kqj
j6=i
Rij
.
(8)
The generalization onto continuous charge distributions is obvious (it is assumed that all charges
are confined within a finite volume):
Z
φ(r) =
kdq 0
,
|r − r0 |
while the potential energy
U=
1
2
Z
0
0
φ dq =
1
2
Z
kdq 0 dq 00
.
|r0 − r00 |
(9)
(10)
2
Physics 2. Electromagnetism
2
Lecture 4
Examples
Example 2.1.
Let a thing ring of the radius R be homogeneously charged, total charge being q.
The potential on the ring axis at the distance z above the ring plane can be found with the same
choice of the coordinates as in Lecture 2:
Z
kqdϕ
kq
√
φ=
=√
.
(11)
2π R2 + z 2
R2 + z 2
The electric field on the axis can be found by differentiating
kqz
∂φ
.
=
2
∂z
(R + z 2 )3/2
Ez = −
(12)
Example 2.2.
Let a disk of the radius R be homogeneously charged with the surface charge
density σ. The potential on the disk axis is
Z
φ(z) =
R
0
r dr
0
0
Z
2π
dϕ0 √
0
√
kσ
= 2πkσ( R2 + z 2 − |z|).
r02 + z 2
(13)
Example 2.3.
Let us consider a homogeneously charged solid sphere (volume charge density ρ,
radius R). The Gauss law gives for the electric field Er

4πr3 ρ/3, r < R,
Q(r) =
4πR3 ρ/3, r > R
4πr2 Er = 4πkQ(r),
so that

4πkρr/3,
r < R,
Er =
4πkR3 ρ/3r2 , r > R.
(14)
(15)
Now the potential can be found as follows

4πkR3 ρ/3r,
r > R,
φ(r) = −
E(r0 )dr0 =
2πkR2 ρ − 2πkρr2 /3, r < R.
∞
Z
r
(16)
The potential energy of the sphere can be found (write in detail) as
U=
1
2
Z
· 4πρ
0
R
3kQ2
r dr φ(r ) =
,
5R
02
0
0
(17)
where Q is the total charge of the sphere.
3
Physics 2. Electromagnetism
Lecture 4
The same can be more easily calculated in the following way. Let us imagine that we are building
the sphere layer by layer. Let a sphere of the radius r is already built which has the charge q(r) =
4πρr3 /3. The potential on the surface of this sphere is φ(r) = kq(r)/r. By adding a layer with the
charge dq we add the potential energy dU = φdq, so that the total potential energy in the end will
be
Z
Z
Z Q
qdq
U = dU = φdq =
,
(18)
r
0
where r has to be expressed in terms of q: q/Q = (r/R)3 . Let us change variables as follows:
q/Q = τ , then
Z
3kQ2
kQ2 1 2/3
τ dτ =
.
(19)
U=
R 0
5R
Example 2.4.
Let there be two charges +q and −q in the positions d/2 and −d/2, respectively.
Let us find the potential of this system, electric dipole, at large distances, r d. The exact potential
is
kq
kq
φ(r) =
−
.
(20)
|r − d/2| |r + d/2|
In order to make use of the condition r d we shall write
−1
|r − d/2|
−1/2 1
= r − r · d + d /4
≈
r
2
2
r·d
1+
2r2
(21)
where Taylor expansion is used in the last approximation. Using similar approximation for the other
absolute value (do that yourself !) we find
φ(r) =
kqd · r
.
r3
(22)
kp · r
.
r3
(23)
Let us define the dipole moment as p ≡ qd, then
φ(r) =
Exercise. Find the electric field of the dipole. Use E = − grad φ.
3
Laplace and Poisson equations
From E = − grad φ and div E = 4πkρ one immediately gets the Poisson equation
div grad φ = ∇2 φ = ∆φ = −4πkρ.
(24)
4
Physics 2. Electromagnetism
Lecture 4
The counterpart of this equation free from charges is known as the Laplace equation
∆φ = 0.
4
(25)
Electric field energy
Let us consider once again the potential energy of a charge distribution
U=
1
2
Z
0
0
φ dq =
I
1
2
φ0 ρ0 dV,
(26)
V
where we assume that all charges are confined within a finite volume (ρ = 0 outside V ). We shall
use the following relations:
div(φE) = φ div E + E · grad φ = 4πkρφ − E2 .
(27)
Substituting into (26) we obtain
1
U=
8πk
I
I
div(φE)dV +
V
V
E2
dV.
8πk
(28)
The first integral can be transformed according to the Gauss-Stokes theorem:
I
I
φE · dS.
div(φE)dV =
(29)
S
V
If S becomes infinitely large,
I
|
I
φE · dS| ≤ | max φr→∞ | · |
S
E · dS| → 0,
(30)
S
so that we obtain
I
U=
V
E2
dV.
8πk
(31)
E2
.
8πk
(32)
Defining energy density as u = dU/dV we get
u=
Example 4.1.
Potential energy of a homogeneously charged sphere (we calculated it earlier) can
be found using (32) and

4πkρr/3,
r<R
E = r̂ ·
(33)
4πkρR3 /3r2 , r > R.
5
Physics 2. Electromagnetism
Lecture 4
Indeed,
∞
E2
U = 4π
r2 dr
8πk
0
Z ∞
2 Z R 4
4πR3 ρ
r
1
k
=2
dr +
dr
6
2
3
0 R
R r
3kQ2
=
5R
Z
(34)
Example 4.2. Potential energy of a thin sphere. If the charge of the sphere is q and the radius is
R, the potential on the sphere is φ = kq/R, that is, constant all over the sphere. Thus, the potential
energy
Z
Z
kq
kq 2
0
0
1
.
(35)
U = 2 φ dq =
dq 0 =
2R
2R
We can arrive at the same result calculating
Z
∞
U = 4π
R
r2 dr
E2
,
8πk
(36)
where E = kq/r2 .
Example 4.3.
Dipole in a slowly varying potential. A dipole is a couple of charges (+q and −q)
which are bound to each other. Let the center of the dipole be at r and the charges are at r + d/2
and r − d/2. The potential energy of this dipole in the external potential (not including the energy
of their mutual attraction) is
U = qφ[r + d/2] − qφ[r − d/2].
(37)
If the potential φ(r) does not change substantially on the distance d, we may Taylor expand to get
φ[r + d/2] − φ[r − d/2] = d · grad U = −d · E
(38)
so that the potential energy of the dipole is
U = −qd · E = −p · E.
(39)
The force which would act on a dipole in the electric field E is
F = grad U = grad(p · E) = p grad Ek ,
(40)
where Ek = E · p̂ is the electric field in the direction of the dipole vector. Let is consider the simplest
6
Physics 2. Electromagnetism
Lecture 4
case where p k E k x̂ and Ex depends only on x. Then we get
Fx = p(dEx /dx),
(41)
that is, the dipole is attracted into the region where E increases.
Let us now consider the case where the electric field is constant. Then the total force is zero, but
P
the torque N = i ri × Fi = (r + d/2) × qE − (r − d/2) × qE = qd × E = p × E.
5
Dipole moment in general
Let us consider the potential of a charge distribution in a volume at distances much larger than the
size of the volume:
Z
kρ(r0 )dV 0
,
(42)
φ(r) =
|r − r0 |
where |r0 | |r|. Using the Taylor expansion
0 −1
|r − r |
02
2
0 −1/2
= (r + r − 2r · r )
1
r · r0
≈ (1 + 2 )
r
r
we get
Z
Z
k
kr
0
0
φ=
ρ(r )dV + 3 · ρ(r0 )r0 dV 0
r
r
kq kr · p
+
.
=
r
r3
Here
Z
q=
(43)
(44)
ρ(r0 )dV 0
(45)
r0 ρ(r0 )dV 0
(46)
is the total charge inside the volume, while
Z
p=
is the dipole moment.
7
Physics 2. Electromagnetism
Intermezzo 1. Field lines and equipotential surfaces
Field lines and equipotential surfaces are used for visual representation of the electrostatic field
structure. Field lines are curves such that in each curve point the electric field is tangent to the
curve, in other words, the unit tangent vector is given by dr/dl = E/|E| (here l denotes the curve
length). Field lines are denser when the field is stronger. The electric field lines can start in the
positive charge or at infinity, and end in the negative charge or at infinity. They cannot be closed.
Example 0.1.
F ield lines of a point charge. For simplicity we choose the coordinate origin in the
charge, then the electric field is E = kqr̂/r 2 , which gives (in spherical coordinates)
r̂
dr
dθ
dϕ
+ θ̂r + ϕ̂r sin θ
= r̂,
dl
dl
dl
(1)
that is, the field lines are given by θ = const, ϕ = const, l = r, that is, are radial.
Example 0.2.
L et two identical charges q are in the positions r1 = (a, 0, 0) and r2 = (−a, 0, 0).
We shall consider the field lines in the x − y plane. The electric field is E =
i=1,2 kq(r − ri )/|r − ri |
3
,
that is,
Ex =
Ey =
kq(x − a)
y 2)3/2
+
((x − a)2 + y 2)3/2
+
((x − a)2 +
kqy
kq(x + a)
((x + a)2 + y 2 )3/2
kqy
((x + a)2 + y 2 )3/2
,
(2)
,
(3)
and the field line equation should be derived from
Ey
dy
=
.
dx
Ex
(4)
Draw qualitative structure of field lines in this case. Draw field lines for the qase of
opposite charges.
Equipotential surfaces are those at which φ = const. Taking differential from the both sides of
this equation we get grad φ · dr = 0, which means that E = − grad φ ⊥ dr, that is, the electric field
1
Physics 2. Electromagnetism
Intermezzo 1
3
2
1
0
-1
-2
-3
-3
-2
-1
0
1
2
3
Figure 1: Equipotential surfaces for identical charges.
is perpendicular to the equipotential surfaces.
Example 0.3.
F or a point charge φ = kq/r so that equipotential surfaces are spheres r = const.
Example 0.4.
F or the two identical charges (as above)
φ= kq
(x − a)2 + y 2
+
kq
(x + a)2 + y 2
,
(5)
so that equipotential surfaces are given by
1
1
+
= const.
(x − a)2 + y 2
(x + a)2 + y 2
(6)
Respectively, for two opposite charges equipotential surfaces will be given by
1
1
−
= const.
(x − a)2 + y 2
(x + a)2 + y 2
(7)
2
Physics 2. Electromagnetism
Intermezzo 1
3
2
1
0
-1
-2
-3
-3
-2
-1
0
1
2
3
Figure 2: Equipotential surfaces for opposite charges.
3
Physics 2. Electromagnetism
Intermezzo 2. Conductors in electrostatic fields.
1
Conductors
In insulators charges almost cannot move from their places, while in conductors charges can move
freely. The last feature immediately means that in a static equilibrium there can be no electric field
inside a conductor. Indeed, suppose E = 0 inside, then a charge q would experience force F = qE
and would be accelerated with the acceleration a = F/m = (q/m)E. In the equilibrium a = 0 so is
the electric field E = 0.
Since E = 0 inside a conductor, div E = 0 = 4πkρ, so that ρ = 0, which means that there cannot
be a net charge inside a conductor. If a conductor is charged, all the charge is on its surface.
Let us now assume that there is a cavity in a conductor which is charged. Where is the charge ?
Is it on the outer surface, on the cavity surface, or on both ? Let us build a closed envelope which
is completely inside the conductor and encloses the cavity. According to the Gauss law, the electric
field flux across this envelope is proportional to the total charge inside. Since the electric field is zero
on this envelope, the flux is zero too, so is the total charge inside it. Since we know that there are
no charges inside the conductor, the total charge on the cavity surface is zero also (although there
may be some distribution, that is, in each point on the cavity surface the surface charge density may
be nonzero). Thus, the whole charge is on the outer surface of the conductor.
On the surface of a conductor a charge cannot move along the normal outside, so that there may
be nonzero electric field just outside the conductor provided it is directed along the normal to the
surface. The surface itself becomes a discontinuity: the electric field is nonzero on the outside of the
surface but vanishes inside. Let the surface charge density on the surface be σ (this charge density
may vary on the surface). In order to calculate the electric field just above the surface let is first
consider an auxiliary problem. Let there is a slab charge distribution, that is, ρ = ρ(z) for z1 ≤ z < z2
and zero outside this layer. Since everything depends on z only we expect that the electric field be
directed along z too, so that E = (0, 0, Ez ). Applying the divergence equation div E = 4πkρ we get
dEz /dz = 4πkρ(z).
1
(1)
Physics 2. Electromagnetism
Intermezzo 2
Integrating the both sides, we obtain
Ez (z = z2 ) − Ez (z = z1 ) = 4πk
Let us now assume that z1 = − → 0−, z2 = → 0+, but
z2
z1
0+
0−
ρ(z )dz .
(2)
ρdz → σ (surface density). We find
Ez [0+] − Ez [0−] = 4πkσ,
(3)
or, taking into account that z is in the direction of the normal to the plane, we can write this in a
more general way:
En+ − En− = 4πkσ.
(4)
We say that the plane with a nonzero surface charge density is a discontinuity and (4) is the conditions
on the discontinuity or matching conditions.
Now we apply (4) to a small part of the conductor surface, which can be approximately considered
to be planar. At the inner side of the surface En− = 0 so that we have
En+ = 4πkσ.
(5)
This relation gives the local electric field at each point of the conductor surface on the outside.
Since the electric field is perpendicular to the surface, the conductor surface itself is an equipotential surface.
2
Physics 2. Electromagnetism
Lecture 5. Capacitance and capacitors
1
Capacitance
Let us have a conductor which is initially at the potential φ = 0, and let us charge it with a charge
q. As a result of charging the conductor is now at a new potential φ ∝ q. The ratio C = q/φ
shows the ability of the conductor acquire charges without substantial changing its potential and is
called capacitance. In fact, since an arbitrary constant can be always added to a potential, we need a
“global” reference point at which the potential would always remain zero. In order to find a physical
body which could play the role of such reference point, let us first consider a conducting sphere of the
radius R. If it is charge with the charge q, its potential is φ = kq/R and the capacitance is C = R/k.
Thus, a sphere with a very large radius would have a very big capacitance and its potential would
be always remain almost the same. This is the reason why Earth is taken as this reference body.
In what follows we shall use the fact that the Earth potential is always zero. A conductor is called
“grounded” if it is brought to the contact with Earth so that their potentials are equal and both
equal zero.
In so doing we have already calculated the capacitance of a spherical conductor C = R/k.
Capacitance of more complicated shapes is difficult to calculate, since it requires calculation of the
electric field distribution around the conducting body, a problem which can be solved analytically
only in a very limiting number of cases. Therefore, it is usually done numerically or experimentally.
Anyway, if we know the capacitance C of a given conductor, then given its potential φ we can find
the charge on it q = Cφ. Alternatively, we can calculate the potential from the charge as φ = P q,
where P = 1/C is the potential coefficient. Let us now assume that we bring another conductor close
to the charged one. The second conductors distorts the electric field in such a way that the potential
on the first conductor changes, in general. On the other hand, the second conductor acquires induced
potential. Thus, conductors affect each other in such a way that their potentials change even when
the charges do not, so that we can no longer write φi = Pi qi for each conductor separately but have
to take into account the mutual influence by writing
φi =
X
Pij qj ,
(1)
Cij φj ,
(2)
j
qi =
X
j
1
Physics 2. Electromagnetism
Lecture 5
Comment: Matrices Cij and Pij are mutually inverse.
2
Capacitors
A capacitor (in its simplest form) is a couple of conductors with the zero total charge, so that if one
conductor has the charge q, the other has −q. Let us denote φ+ the potential of the conductor with
the charge q and φ− the potential of the other. Then from (1) we have
φ+ = (P11 − P12 )q,
(3)
φ− = (P21 − P22 )q,
(4)
V = φ+ − φ− = (P11 + P22 − P12 − P21 )q ≡ q/C,
(5)
and eventually
where we call V = φ+ − φ− potential difference or voltage, and C is the capacitance of the capacitor,
that is, of the pair of conductors and not a single conductor as earlier.
If we bring another conductor it can affect the potentials of the first two and thus change the
relation between the charge q and voltage V . It is not what we want from the practical point of
view. Indeed, it would not be very convenient if the capacity depended not only on the construction
built specially for charge storage but also on anything around. Therefore, we would like to reduce
the effects of other conductors to a minimum. We do that by making such systems which “contain”
almost all their electric field “inside” the system itself, so that any other body does not feel the
electric field of the capacitor, and therefore cannot disturb it.
Example 2.1.
Parallel-plate capacitor is built of two parallel conducting plates with the area S
each. The distance d between the plates must be much smaller than the dimensions of the plates
(length and width) so that the field between the plates be very close to the field produced by two
infinite plains. Let the charge on one of them be q, hence the surface charge density σ = q/S. It
will produce the electric field E+ = 2πkσ directed from the plane. The other will have the charge
−q and the surface charge density −σ and will produce the field E− = −2πkσ directed towards the
plane. Together they will produce the field E = 4πkσ between the plates, directed from q to −q,
R
and zero field outside. The potential difference V = − E · dr = 4πkσd = Q/(S/4πkd), so that the
capacitance is C = S/4πkd.
Example 2.2.
Spherical capacitor consists of two concentric spheres with the radii a and b,
a < b. Let the inner sphere have the charge q and the outer one have the charge −q. The electric
field is zero everywhere except between the spheres where it is E = kq/r 2 . Thus, the voltage
Rb
V = a (kq/r 2 )dr = kq(1/a − 1/b), so that the capacitance is C = ab/k(b − a).
2
Physics 2. Electromagnetism
Lecture 5
Exercise Calculate capacitance of a cylindrical capacitor.
3
Energy of a capacitor
Let a capacitor is charged to the voltage V and, therefore, has the charge q = CV on the positively
charged plate (we call it plate in the analogy with the parallel-place capacitor despite that the
actual shape can be different). Let as add an infinitesimal charge dq on the positively charged
plate, and, respectively, −dq on the negatively charged plate. The addition of energy would be
dU = dq · φ+ + (−dq) · φ− = dq · V = qdq/C. Integrating this relation we get
U(q) =
Z
0
q
q ′ dq ′/C =
CV 2
q2
=
,
2C
2
(6)
where we have taken into account that the energy of discharged capacitor (q = 0) is zero.
R
Exercise Show that the capacitor energy can be also found as U = V udV , where u = E 2 /8πk
is the energy density of the electric field and V is the volume of the capacitor (that part where
the electric field is nonzero). Use E = 4πkσ = const for a parallel-plate capacitor (V = Sd) and
E = kq/r 2 for a spherical capacitor (dV = 4πr ′ 2 dr ′ ).
4
Capacitor “circuits”
Capacitors are used to store energy due to charge separation. In practice, usually more than one
capacitor are combined in one system for various reasons (scheme requirement, flexibility of control,
or just technical limitations). In many cases we are interested to know what are voltages and charges
on all capacitors if some known external voltage is applied to the whole system. Alternatively, we
may be asked to substitute the whole system by a single effective capacitor whose capacity must
be found. These tasks are extremely easy and require only some routine technical work (which
sometimes may be quite long if a scheme is complicated) and do not require from us any art or new
ideas. One has to take into account the following “rules”:
• All plates connected to each other by a conducting wire are at the same potential. This is
just a special case of earlier conclusion that all points of a conductor have the same potential:
several conducting plates connected by conducting wires actually are parts of one conductor.
• Plates which are not connected to any external source cannot gain or loose charge, that is,
the sum of the charges on the plates which are connected only to each other remains constant.
This is just the charge conservation.
3
Physics 2. Electromagnetism
Lecture 5
• Only on the plates connected to external sources can the charge change. However, the total
charge on all plates, positively and negatively charged, is always zero. This follows from our
requirement that there will be no net charge, otherwise there will be electric field outside the
capacitors.
• The potential difference between the two points does not depend on the path. This is just
rephrasing of the potential character of the electric field.
Example 4.1.
Capacitors are said to be connected in series is the “positive” plate of the preceding
capacitor is connected to the “negative” plate of the next. Thus for two neighboring capacitors, ith
and i + 1th we have: qi + (−q)i+1 = 0, that is, on every capacitors have the same charge q. If we
substituted the whole series with a single capacitor it would also have the charge q. In the same
P
P
time, the voltage on the whole series is just the sum V = i Vi = i q/Ci. Since q = Cef f V , we get
P
1/Cef f = i 1/Ci .
Example 4.2.
Capacitors are connected in parallel if all “positive” plates are connected together
and all “negative” plates are connected together, so that on each capacitors is the same voltage V
P
P
(which is also the voltage on the effective capacitor). At the same time, q = i qi = i Ci V , so that
P
Cef f = i Ci .
Most connections are neither in series not in parallel and require usage of the above given rules.
+
C1 , q1
+
C2 , q2
V
Figure 1: Connection is series: V = V1 + V2 , q = q1 = q2
Example 4.3.
Connection which is neither in parallel nor in series: how to treat.
4
Physics 2. Electromagnetism
Lecture 5
C1 , q1
C2 , q2
Figure 2: Connection in parallel: Q = q1 + q2 , V = V1 = V2 .
+
+
C1 , q1
C2 , q2
C5 , q5
+
+
+
C3 , q3
C4 , q4
Figure 3: Complicated connection.
5
Physics 2. Electromagnetism
Lecture 5
For this connection we have to write:
q = q1 + q2 ,
(7)
− q1 − q5 + q3 = 0,
(8)
− q2 + q5 + q4 = 0,
q1
q3
+
= V,
C1 C3
q4
q2
+
= V,
C2 C4
q2
q5
q3
+
+
= V.
C2 C5 C3
(9)
(10)
(11)
(12)
Thus, we have 6 equations for 6 unknowns q, q1 , . . . , q5 , which have to be expressed in terms of
C1 , . . . , C5 and V . In the end we get the effective capacitance as follows C = q/V .
Example 4.4.
Let two capacitors C1 and C2 be charged with q1 and q2 and then connected as
shown in the figure.
+
C1 , q1
−
C2 , q2
+
C1 , q1′
C2 , q2′
+
Figure 4: Connection of charged capacitors.
In this case charge conservation requires q1′ + q2′ = q1 − q2 , while the voltage on the two capacitors
is the same: q1′ /C1 = q2′ /C2 . We have two equations for two unknowns.
5
Dielectric - very briefly
If a dielectric is inserted into electric field it becomes polarized (we shall learn that later) and reduces
the electric field by the factor 1/ε, that is, the electric field in the dielectric E = Ev /ε, where Ev is
the field which would be measured without dielectric, and ε is the dielectric constant.
Example 5.1.
Let us consider a parallel-plate capacitor with a dielectric inside. If the charge
density is σ the field becomes E = 4πkσ/ε so that the voltage is V = 4πkqd/Sε and the capacitance
C=
εS
4πkd
(13)
6
Physics 2. Electromagnetism
Lecture 5
The Gauss law will be written as follows:
I
εE · dS = 4πkQin
(14)
S
where ε in the integrand takes into account that different materials can be inside the volume limited
by S, and Qin refers to external charges (not related to microscopic charges - see Lecture 11).
6
Proof of Cij = Cji
Let us consider two immobile conductors, so that C11 , C12 , C21 , and C22 are constant. Let us start
charging the first conductor keeping φ2 = 0. Then the work which has to be done is
W1 =
Z
φ1 dq1 =
Z
φ1 C11 dφ1 = 12 C11 φ21
Let us now start charging the second conductor keeping φ1 = const. Now
W2 =
Z
φ1 dq1 +
Z
φ2 dq2 =
Z
C12 φ1 dφ2 +
Z
C22 φ2 dφ2
= C12 φ1 φ2 + 21 C22 φ22
The sum U = W1 + W2 = 12 C11 φ21 + C12 φ1 φ2 + 21 C22 φ22 is the final energy of the two conductors.
However, we could charge first the second conductor and only after that the first conductor, so that
U = 21 C11 φ21 + C21 φ1 φ2 + 21 C22 φ22 as well. Since the potential energy does not depend on the history,
which means C12 = C21 . Since this can be done for any pair of conductors, we have Cij = Cji.
7
Physics 2. Electromagnetism
Lecture 6. Electric currents, resistivity, etc.
1
Electric current
Let us assume that there is some concentration (number of particles per unit volume) ns of charges
qs , and they are moving with the velocity vs (the index s shows that we speaking about the charges
of the same kind s, e.g. about electrons). Let us choose a small oriented area dS = dS · n̂ and
calculate the charge dq which passes through this area during infinitesimal time dt. It is clear that
all charges which are initially at the distance less than vs,n dt = vs · n̂dt will cross the area. The total
number of charge crossing the area during time dt would be them dN = ns dSvs,n dt = ns vs · n̂dSdt
and the total charge crossing this area would be
dQ = (qs ns vs) · dSdt ≡ Js · dS.
(1)
The ratio IdS = dQ/dt is called the electric current crossing the area dS, while Js = qs ns vs is called
the current density.
The expression for the current density can and have to be generalized. Indeed, there may be
more than one kinds of charges (carriers), which simply would mean that we have to add summation
over s. Moreover, not all particles must move with the same velocity, and their velocity can change.
This means that we have use the average velocity of the particles, and this average can be varying
in space and time. Thus, the most general expression for the current density would be
J=
X
qs ns v¯s,
(2)
s
where the bar denotes averaging.
The expression for the current can be also generalized. If we wish to know what current flows
across an arbitrary, not infinitesimal, surface S we have to use the integral
I=
Z
J · dS,
(3)
S
and the charge crossing this area is related to the current by the same relation I = dQ/dt.
H
Let us now consider the the current crossing a closed surface outwards, I = S J · dS. Since
this integral gives the total charge leaving the volume limited by the surface S, the charge inside
1
Physics 2. Electromagnetism
Lecture 6
should decrease by the same value because of the charge conservation, that is, dq/dt = −I, where
H
q = V ρdV is the (time varying) charge inside the volume V . We have now:
I =−
I
J · dS
I
d
=−
ρdV
dt V
I
∂ρ
=−
dV
∂t
I V
=
div J dV (Gauss theorem)
S
(4)
V
and therefore
∂ρ
= 0.
(5)
∂t
Eq. (5) is the differential form of the charge conservation law.
The state (and, respectively, the current) is stationary if the charge density is time independent,
div J +
(∂r/∂t) = 0. Not that this is not a static state where charges do not move at all.
2
Resistivity
Let us consider a homogeneous conducting material with charge carriers (concentration n and charge
q) which can move freely, and apply a homogeneous electric field E = const. Every charge will be
accelerated by this field, so that v = v0 + qEt/m (where m is the mass of the charged particle,
and v0 is a random initial velocity), so that J = nq v̄ = nqEt/m and growth indefinitely. We know
that it is not the case, which means that there are some other forces acting on charged particle and
decelerating them so that their velocity remains finite. Indeed, charged particles collide with other
particles (of other kinds) which are present in the system (e.g. electrons collide with ions).
Let us consider a simple model of a conductor. Let the current carrying particles collide with
ions from time to time, and the velocity of the current carrying particle just after the collision uc is
random. Then at some moment the particle would have the velocity
v = uc +
qE
tc ,
m
(6)
where tc is the time passed since the last collision. The current density will be
J = nq v̄ = nq[ūc +
qE ¯
tc ].
m
(7)
The first term vanishes since uc is randomly distributed, while τ = t¯c is just the average time between
the collisions, so that J = (nq 2 τ /m)E. Summarizing all this and generalizing onto several kinds or
2
Physics 2. Electromagnetism
Lecture 6
carriers, we obtain
J = σE,
(8)
P
where σ = s ns qs2 τs /ms is the electrical conductivity of the conducting material and depends only
on the material properties. Eq. (8) is the differential form of the Ohm’s law. It is often written as
follows E = ρJ where ρ = 1/σ is the resistivity.
3
Resistive dissipation
A force acting on a moving particle produces work. The electric force power (work in unit time) is
P = F · v = qE · v. Applying this to the charge carriers we get the instantaneous power
P = qs E · (us +
qs Ets
).
ms
(9)
The power which is dissipated (is lost and goes into heat) in the unit volume is obtained by multiplying
(9) by ns , averaging, and summing over s:
X
dP
qs Eτs
=
ns qs E ·
= σE 2 ,
dV
m
s
s
(10)
where V denotes volume.
4
Ohm’s law
Let us consider a resistor which is an appliance made of a conductor. Let have a an elongated shape
with the constant cross-section area of S and a length l, and we assume that a homogeneous electric
field E is applied in the direction along the resistor. If the conductor conductivity σ is known then
we can find the current density J = σE and the total current crossing the cross-section of the resistor
is I = JS = σSE. On the other hand, presence of the electric field implies the potential difference
(or voltage) along the resistor (the potential is higher “upstream”, that is, decreases in the direction
of the current flow): V = El. Combining these two expressions we come to the Ohm’s law V = IR,
where the resistance R = l/σS = ρl/S. This Ohm’s law is valid even for more complicated resistors
which do not have a cylindrical shape, since for at least not very high and more or less stationary
currents the current density distribution is the same and only proportional to the applied voltage.
The relation R = ρl/S is, however, only for resistors with constant cross-section area. For other
shapes different relations hold which have to be calculated in each case.
The total power dissipated in a homogeneous resistor is
P =
V2
dP
· V = σE 2 Sl =
= I 2 R.
dV
R
(11)
3
Physics 2. Electromagnetism
5
Lecture 6
Circuits
The following rules are known as Kirchhoff’s laws and actually are representations of the charge and
energy conservation:
• In each node the sum of incoming currents is equal to the sum of the outgoing currents
P
Iout .
P
Iin =
• When moving along the resistor in the direction of the current the voltage drops by V = IR.
• The total voltage drop in a closed (sub)circuit is zero.
5.1
Simple connections
Example 5.1.
Connection in series (see Fig. 1). Same current I passes through both resistors,
the total voltage V = V1 + V2 = I(R1 + R2 ) = IRef f , so that Ref f = R1 + R2 .
↓I
↓ R1 , I1
V
↓ R2 , I2
Figure 1: Connection is series: V = V1 + V2 , q = q1 = q2
Example 5.2.
Connection in parallel (see Fig. 2). Same voltage V is on both resistors while the
current splits I = I1 + I2 , so that V /Ref f = V /R1 + V /R2 and 1/Ref f = 1/R1 + 1/R2 .
4
Physics 2. Electromagnetism
Lecture 6
↓ R1 , I1
↓ R2 , I2
Figure 2: Connection in parallel: I = I1 + I2 , V = V1 = V2 .
Example 5.3.
Connection which is neither in parallel nor in series: how to treat.
↓ R1 , I1
→ R5 , I5
↓ R3 , I3
↓ R2 , I2
↓ R4 , I4
Figure 3: Complicated connection
Choosing current directions as in Fig. 3 we can write
I = I1 + I2 ,
(12)
I1 = I3 + I5 ,
(13)
I2 = I4 − I5 ,
(14)
V = I1 R1 + I3 R3 ,
(15)
V = I2 R2 + I4 R4 ,
(16)
V = I1 R1 + I5 R5 + I4 R4 .
(17)
From this equation we have to find I as a function of V and finally Ref f = V /I.
5
Physics 2. Electromagnetism
5.2
Lecture 6
EMF
Since there is ohmic (resistive) dissipation in any resistor, the energy is being lost continuously. Thus,
a source of energy is needed. This source is called emf and works on a “chemical energy” which is
nothing but the energy of electric interaction between positive and negative charges inside molecules,
that is, local electric fields. Let us consider an imaginary battery consisting of two electrodes and
electrolyte (Fig.4).
A+ C −
B+ C −
A+ C −
B+ C −
A+
B+
A+
B+
−−−−−
+++++
+++++
−−−−−
+++++
−−−−−
A
B
−−−−−
+++++
+
−
A
+
B
−
−
Figure 4: A battery (emf).
The electrolyte in this case is a water solution of two chemicals AC and BC which can dissolve
into positive (A+ and B + ) and negative (C − ) ions. Let us not assume that at the level electrode the
positive A+ tend to leave the solution and enter the region above the electrode which is filled with
another material able to absorb A. At the other electrode B + tend to leave the region above the
electrode (which is B rich) and join the solution but as ions, leaving behind electrons. Thus, at each
electrode two thin boundary layers are formed where the average (or macroscopic, or self-consistent)
electric field is directed against the motion of positive charges (the charges are dragged against the
average force by local forces). This can happen if the new situation is energetically preferred, that
is, has a lower energy. Thus, two potential jumps occur at the two electrodes, so that there is a
potential difference between the left (positive, anode) and the right (negative, cathode) one. If we
connect them by a conductor electrons start moving from the cathode to the anode (that is, current is
flowing from anode to cathode). Once these electrons tend to reduced the charges on both electrodes,
more ions go out and into and maintain the potential difference, until there are no more ions in the
solution and/or the enriched new electrode regions.
6
Physics 2. Electromagnetism
5.3
Lecture 6
EMF and resistor
We will consider a battery as a source of the voltage (emf) with its one internal resistivity. When
we cross the emf from − to + the potential increases by E. Let us assume that the emf is connected
to a resistor so that a current is flowing in the circuit from + to − (see Fig. 5). Here we neglect the
internal resistivity of the battery.
↓ R, I
E
Figure 5: EMF and resistor
Let us now count the potential jumps (from now one we shall count them with +) and potential
drops (with −). Moving in the direction of the current we cross the battery with the potential jump
UE = E and then experience the potential drop UR = −IR at the resistor. The total voltage on the
closed circuit U = UE + UR = E − IR = 0, so that I = E/R.
Example 5.4.
A more complicated circuit is shown in Fig. 6.
↑ R1 , I1
↓ R3 , I3
↑ R2 , I2
E1
E3
E2
Figure 6: Circuit
To start with we choose (arbitrarily) the direction of currents. For the currents we now have:
I3 = I1 + I2 . Let us now consider the closed path including E1 , R1 , E3 and R3 and follow the path in
the direction of the current I1 :
E1 − I1 R1 − E3 − I3 R3 = 0.
(18)
For the path including E2 , R2 , E3 , and R3 we will follow the direction of I2 (quite arbitrarily):
E2 − I2 R2 − E3 − I3 R3 = 0.
(19)
7
Physics 2. Electromagnetism
Lecture 6
Adding the equation for the currents we get three linear equations for three variables (currents).
6
RC circuit
This is the first circuit with variable current.
S
R, I ↓
E
+
C, q
Figure 7: RC circuit
Let us denote the charge and current and choose the polarity is shown in Fig. 7. Then dq/dt = I
and E − IR − q/C = 0. Thus, we have
q̇ +
E
1
q= .
RC
R
(20)
We can use the same equation (20) for charging and discharging. The difference is that during
discharging we have to put E = 0. The general solution of (20) is
q = CE + Ae−t/RC ,
(21)
where A is a constant which has to be determined from the initial conditions q = q0 when t = 0, so
that A = q0 − CE. Eventually we get
q = CE(1 − e−t/RC ) + q0 e−t/RC .
(22)
Respectively,
q
= E(1 − e−t/RC ) + V0 e−t/RC ,
C
E
V0
I = q̇ = ( − )e−t/RC
R
R
V =
V0 = q0 /C,
(23)
(24)
Charging: q0 = 0 so that V = E(1 − e−t/RC ).
8
Physics 2. Electromagnetism
Lecture 6
q,I
2
1.5
1
0.5
0.5
1
1.5
2
2.5
3
t
Figure 8: Charging: charge (red) and current (blue)
Discharge: E = 0 so that V = V0 e−t/RC .
q,I
2
1.5
1
0.5
0.5
1
1.5
2
2.5
3
t
Figure 9: Discharge: charge (red) and current (blue)
9
Physics 2. Electromagnetism
Lecture 7. Relativity and magnetic field
1
Relativity reminder
Let two frames S and S 0 are moving with the relative velocity V0 = V0 n̂. Then Lorentz transformations read:
n̂ · dr = γ0 (n̂ · dr 0 + V0 dt0 )
dt = γ0 (dt0 + V0 n̂ · dr 0 /c2 )
0
dr⊥ = dr⊥
(1)
(2)
(3)
γ0 = (1 − V02 /c2 )−1/2
(4)
V0 = V0 n̂
(5)
For simplicity of notation we shall choose n̂ = x̂, so that n̂ · r = x and r⊥ = (y, z). Then one has
dx = γ0 (dx0 + V0 dt0 )
(6)
dt = γ0 (dt0 + V0 dx0 /c2 )
(7)
dy = dy 0 ,
(8)
dz = dz 0
γ0 = (1 − V02 /c2 )−1/2
V0 = V0 x̂
(9)
(10)
Let both observers, S and S 0 , observer the same moving particle. The particle motion in the
frame S will be described by r(t) and v(t) = dr/dt. Respectively, in the frame S 0 the particle
trajectory and velocity will be given as r 0 (t0 ) and v 0 (t0 ) = dr 0 /dt0 .
Using the relations between x, y, z, t and x0 , y 0 , z 0 , t0 one immediately gets the following velocity
transformation rules:
dx
vx0 + V0
=
dt
1 + V0 vx0 /c2
vy0
dy
vy =
=
dt
γ0 (1 + V0 vx0 /c2 )
vx =
1
(11)
(12)
Physics 2. Electromagnetism
Lecture 7
vk =
v⊥ =
vk0 + V0
(13)
1 + V0 vk0 /c2
0
v⊥
γ0 (1 + V0 vk0 /c2 )
(14)
In addition to the velocity v we shall define the particle Lorentz factor γ = (1 − v 2 /c2 )−1/2 . This
is NOT the Lorentz factor for the frame transformation, γ belongs to the particle (like v) and also
changes when we move into another frame, γ 0 6= γ.
Let us now recollect the non-relativistic momentum is defined as
p = mv = m
dr
dt
(15)
where t is invariant (the same in both frame - NON-relativistic !). Special relativity states that
coordinates and time change when we switch frames, and this definition is not appropriate. We have
to find some invariant time to differentiate with. For these purposes we can choose the proper time
dτ 2 = dt2 − dr 2 /c2 = dt2 (1 − v 2 /c2 ) = inv
so that
dτ =
dt0
dt
= 0
γ
γ
(16)
(17)
Now the proper momentum definition will look as follows:
p=m
dr
dp
= mγ
= mγv
dτ
dt
(18)
It is easy to see that in the non-relativistic limit γ → 1 and the new momentum reduces to the
well-known non-relativistic expression p = mv.
Since time also changes it makes sense to ask what is the physical sense of the construction
K = mc2 (dt/dτ ) = mc2 γ. In the non-relativistic limit Taylor expansion gives
K ≈ mc2 +
mv 2
2
(19)
that is, we recover the usual expression for the kinetic energy with the addition of the rest energy.
Thus, we identify K with the particle energy. It is easy to see that
K 2 − p2 c2 = m2 c4 = inv
(20)
Now we can see that the momentum p = m(dr/dτ ) and energy K = mc2 (dt/dτ ) should transform
2
Physics 2. Electromagnetism
Lecture 7
similarly to the coordinates and time, namely,
pk = γ0 (p0k + V0 K 0 /c2 )
(21)
K = γ0 (K 0 + V0 p0k )
(22)
p⊥ = p0⊥
(23)
where k and ⊥ are with respect to n̂. Since p = mvγ and K = mc2 γ, this can be written also as
follows
γvk = γ0 γ 0 (vk0 + V0 )
γ = γ0 γ 0 (1 + V0 vk0 /c2 )
0
γv⊥ = γ 0 v⊥
(24)
(25)
(26)
It is easy to derive from here the rules for the velocity transformation which we obtained earlier in
a more usual way.
In order to be able to relate all this to the electric and magnetic field we have to know the
transformation rules for the force. The non-relativistic definition of force is
F =
dp
dt
(27)
which means that the proper relativistic definition will be
dp
= Fγ
dτ
(28)
dK
dK
=γ
= γF · v = f · v
dτ
dt
(29)
f=
It is of interest to mention that
f0 =
Since we are differentiating with respect to the invariant proper time, f , f0 should transform
similarly to p, K, that is,
fk = γ0 (fk0 + V0 f00 /c2 )
(30)
f0 = γ0 (f00 + V0 fk0 )
(31)
f⊥ = f⊥0
(32)
or, for the usual force
γFk = γ0 γ 0 (Fk0 + V0 (F 0 · v 0 )/c2 )
γ(F · v) = γ0 γ 0 ((F 0 · v 0 ) + V0 Fk0 )
(33)
(34)
3
Physics 2. Electromagnetism
Lecture 7
γF⊥ = γ 0 F⊥0
(35)
Do not forget that γ and γ 0 belong to the particle when viewed from two different frames, while γ0
is related to the relative velocity of the two frames.
2
Magnetic field
Let in the frame S 0 a particle experiences only usual electric field F 0 = qE 0 , so that
f 0 = γqE 0 ,
f00 = γ 0 qE 0 · v 0
(36)
0
= 0, so that
Let us first consider the case E⊥0 = 0, v⊥
γFk = γ0 γ 0 qEk0 (1 + V0 vk0 /c2 )
(37)
Fk = qEk0
(38)
or
and since Ek = Fk /q (if independent of the particle parameters) we may identify Ek = Ek0 .
0
Let now Ek0 = 0 and v⊥
= 0, so that
γF⊥ = γ 0 qE⊥0
(39)
γ 0 = γγ0 (1 − V0 vk /c2 )
(40)
F⊥ = γ0 qE⊥0 − γ0 qE 0 V0 vk /c2
(41)
Since
we get
The second term depends on the particle velocity and cannot be electric force.
3
Lorentz force
Let us introduce magnetic field via the Lorentz force
F = qE + αqv × B
(42)
and try to find the transformation law. In doing so we have to write down
F 0 = qE 0 + αqv 0 × B 0
(43)
4
Physics 2. Electromagnetism
Lecture 7
and use the transformation rules for the forces, taking into account that they should be satisfied for
any particle velocity v. Calculations are a little bit long but eventually one will have
Bk = Bk0
Ek = Ek0
0
)
E⊥ = γ0 (E⊥0 − αV0 × B⊥
0
+ V0 × E⊥0 /c2 )
αB⊥ = γ0 (αB⊥
4
∗
(44)
(45)
(46)
Derivation of the field transformation
Let V0 = v0 x̂. Then, from the transformation rules for the force one has
γFx = γ0 γ 0 (Fx0 + V0 (F 0 · v 0 /c2 )
γFy = γ 0 Fy0 ,
γFz = γ 0 Fz0
γ(F · v) = γ0 γ 0 ((F 0 · v 0 ) + V0 Fx0 )
where
F = q(E + αv × B)
F 0 = q(E 0 + αv 0 × B 0 )
Since the transformation laws for the fields should depend only on the relative velocity of the two
frames and be independent of the particle velocity, we may choose v as we wish to make calculations
easier. Let us choose v = 0 so that v 0 = −V0 = −V0 x̂, γ = 1, γ 0 = γ0 , and one has
Fx0 = qEx0
Fx = qEx ,
F⊥ = qE⊥ ,
F · v = 0,
0
F⊥0 = q(E⊥0 − αV0 × B⊥
)
F 0 · v 0 = −qV0 Ex0
Substituting all this into the force transformation expressions one immediately finds (k is in xdirection)
Ek = Ek0
0
E⊥ = γ0 (E⊥0 − αV0 × B⊥
)
Inverting these expressions, we get
E⊥0 = γ0 (E⊥ + αV0 × B⊥ )
5
Physics 2. Electromagnetism
Lecture 7
or
γ0 αV0 × B⊥ = E⊥0 − γ0 E⊥
0
= E⊥0 − γ02 (E⊥0 − αV0 × B⊥
)
=−
γ02 V02 0
0
E⊥ + γ02 αV0 × B⊥
c2
Multiplying by ×V0 we get
0
αB⊥ = γ0 (αB⊥
+ V0 × E⊥0 /c2 )
If we now notice that the transformation is symmetric for E and B we have to conclude that
Bx = Bx0 .
6
Physics 2. Electromagnetism
Lecture 8. Magnetic field
1
Magnetic force and magnetic field
Magnetic field B can be recognized by the force acting on a moving particle FB = qv × B. The
total (electric and magnetic) force on a particle (Lorentz force) is:
F = q(E + αv × B),
(1)
where α = 1 in SI and α = 1/c in CGS.
If we wish to know the magnetic force on an infinitesimal volume dV we just sum all the forces
acting on all charged particles inside this volume:
FB = α
X
i
qi vi × B = α
X
s
ns qs v¯s × B = αJ × BdV,
(2)
P
P
where i is taken over all individual particles, while s is taken over all sorts of particles (species),
and v¯s is the average velocity of the species s, while ns is its number density.
If we want to know the force on a small piece of wire, J dV = Idl we find
dF = αIdl × B.
2
(3)
Laws of the magnetic field
The below laws can be, in principle, proven using the Lorentz transformations. At present we,
however, will take them as experimental facts.
2.1
Ampere’s law
Ampere’s law says that for any closed contour L and chosen path direction
I
L
B · dr = 4πKI,
1
(4)
Physics 2. Electromagnetism
Lecture 8
where I is the total current flowing across the surface S enclosed by L, and the directions of the path
along L and the normal to S are in mutual agreement, as usual. The current is positive if it flows in
the positive direction of the normal, and negative otherwise.
Taking into account that
I
I=
S
and the Stokes theorem
I
L
B · dr =
J · dS,
I
S
rot B · dS
we immediately obtain
rot B = 4πKJ ..
(5)
The constant K depends on the choice of units of measurements, in SI it is usually written as
K = µ0 /4π, and the Gaussian system K = 1/c (c is the speed of light).
2.2
No magnetic charges
As we know the equation (first Maxwell equation)
div E = 4πρ
states that electric charge is the source of the electric field. The corresponding equation for the
magnetic field reads
div B = 0.
(6)
and means that there is no magnetic charge. In this case the magnetic field can be written as
B = rot A,
(7)
where A is the vector potential.
In Cartesian coordinates, substituting (7) into (5) we get
rot rot A = grad div A − ∆A = 4πKJ
(8)
∆≡
(9)
∂2
∂2
∂2
+
+
.
∂x2 ∂y 2 ∂z 2
We shall use this equation for the calculation of the vector potential.
2
Physics 2. Electromagnetism
2.3
Lecture 8
Gauge invariance and Bio-Savart law
If A is a solution of (8) then A + grad Λ, where Λ is an arbitrary scalar field, is also a solution since
rot grad Λ ≡ 0. Choosing properly Λ we can achieve div A = 0 so that (8) takes the form
∆A = −4πKJ .
(10)
∆Ai = −4πKJi
(11)
Rewriting this in the component form
we see that the equation for each component is the same as the equation for the electric potential
(Poisson equation)
∆φ = −4πkρ.
Thus, the solutions should be similar too, that is, Since the solution of the Poisson equation is
φ(r) =
Z
kρ(r ′)dV
|r − r ′ |
the vector potential should take the following form
A[r] =
Z
KJ [r ′]dV ′
.
|r − r ′|
(12)
The magnetic field B = rot A be easily found using the relation rot(Af ) = ∇ × (Af ) = f rot A +
grad f × A:
B=K
Z
1
× J dV = K
grad
|r − r ′|
Z
J × (r − r ′)dV
.
|r − r ′|3
(13)
Exercise Derive (13).
For a wire current J V = Idl one has
B=K
Z
Idl × (r − r ′)
.
|r − r ′ |3
(14)
which is called Bio-Savart law.
3
Application of Lorentz force
As was found earlier the force which acts on a infinitesimal current carrying wire with the length dl
is
dF = αIdl × B,
(15)
3
Physics 2. Electromagnetism
Lecture 8
where the vector |dl| = dl and is directed along the current. The force of a finite wire is
F = αI
Z
dl × B = αI
Z
dr × B,
(16)
where we substituted dl = dr taking into account that the integral is actually a path integral.
Example 3.1.
integral, so that
Constant magnetic field. In this case the magnetic field can be taken out of the
F = αI(
Z
dr) × B = αI(r2 − r1 ) × B,
(17)
where r1 and r2 are the positions of the beginning and the end of the wire. One important conclusion
is if the current is closed, r2 = r1 , the force vanishes: there is no magnetic force on a closed current
in a homogeneous magnetic field.
Example 3.2.
Force on a small closed current loop. Let us now consider a closed current loop
but in a spatially varying magnetic field. We shall assume that the loop is small and the magnetic
field is slowly varying so that we can Taylor expand the magnetic field relative to the “center of the
loop”. Let us write the force as follows:
I
dr ′ × B[r ′]
(18)
I
ǫijk dx′j Bk [x′ ].
(19)
F = αI
L
or
Fi = αI
L
We now Taylor expand the magnetic field
Bk [x′ ] = Bk [x] + (x′l − xl )
∂Bk
∂Bk
= Bk + Xl
.
∂xl
∂xl
(20)
Substituting this into integral (19) we get
∂Bk
ǫijk dXj (Bk + Xl
)
∂xl
L
I
I
∂Bk
ǫijk Xl dXj
= αIBk ǫijk dXj + αI
∂xl L
L
Fi = αI
I
(21)
The first integral vanishes, the second one can be written as
I
Xl dXj = ǫljm n̂m ∆S,
(22)
L
where ∆S is the area of the small loop, and n̂ is the normal to the loop.
4
Physics 2. Electromagnetism
Lecture 8
Quasi-proof. Let us consider a small circle in x − y plane and define the path direction so that
x = r cos ϕ, y = r sin ϕ, 0 ≤ ϕ < 2π. The normal then will be n̂ = ẑ. For l = 1, j = 2 we get
I
xdy =
L
Z
2π
r cos ϕ(r cos ϕdϕ) = πr 2 = ǫ312 nz ∆S.
0
Using (22) in (21) we get
∂Bk
ǫijk ǫljm∆S n̂m
∂xl
∂Bk
∆S n̂m (δil δkm − δim δkl )
= αI
∂xl
∂Bk
= αI
∆S n̂k ,
∂xi
Fi = αI
(23)
where we used (∂Bk /∂xk ) = div B = 0. The combination m = αI∆Sn is called magnetic moment
(and the loop is a magnetic dipole) so that the force can be now written as follows:
F = grad(m · B).
(24)
Recalling the analogy with the electric dipole:
F = − grad Up ,
Up = −p · E,
(25)
we see that it is possible to identify Um = −m · B as a potential energy of the magnetic dipole.
Example 3.3.
Torque on a small closed loop. Since we already established that such loop is a
magnetic dipole and found that the forces are similar we may expect that the torque will be of the
form
N = m × B.
(26)
Indeed, the torque on an infinitesimal part of a loop is dN = r × dF , so that the total torque on a
closed loop is
N=
I
L
r × dF = αI
I
L
r × (dr × B)
(27)
5
Physics 2. Electromagnetism
Lecture 8
or
Ni = αIǫijk
I
xj ǫklm dxl Bm
I
= αIǫijk Bm ǫklm xj dxl
L
L
= αISǫijk Bm ǫklm ǫjla na
(28)
= αISǫijk Bm na (δkj δma − δka δim )
= ǫijk (αISj )Bk = ǫijk mj Bk ,
that is, exactly what is expected.
4
Application of Bio-Savart law
Example 4.1.
Magnetic field of an infinite straight current. Let us choose the coordinates so that
the current is on z axis and point towards positive direction, dl = dz ′ ẑ, and r ′ = (0, 0, z ′). Using
our freedom in choosing the coordinate frame we also choose the coordinates of the point in which
we are looking for the magnetic field, as r = (r, 0, 0), where r is the distance from the wire. Thus,
√
r − r ′ = (r, 0, −z ′ ), |r − r ′| = r 2 + z ′ 2 , and dl × (r − r ′ ) = dz ′ (0, r, 0). Combining all this we have
Bx = Bz = 0,
Z +∞
dz ′ r
2KI
.
(29)
By = KI
2 3/2 =
2
′
r
−∞ (r + z )
Example 4.2. Magnetic field of a circular loop. Let the loop be in x−y plane, then the loop point
positions are r ′ = r r̂ (cylindrical coordinates), and dl = rdϕ′ ϕ̂. We shall find the magnetic field on
√
the loop axis, at r = z ẑ, so that r−r ′ = z ẑ−r r̂, |r−r ′| = r 2 + z 2 , and dl×(r−r ′ ) = rdϕ′(z r̂+r ẑ).
Now
Z 2π
rdϕ′ (z r̂ + r ẑ)
B = KI
(r 2 + z 2 )3/2
0
2Kπr 2 I
(30)
= 2
ẑ
(r + z 2 )3/2
2Km
=
2
α(r + z 2 )3/2
where m = απr 2 I ẑ.
Example 4.3.
Magnetic field of a small loop - general case. Here we illustrate usage of the vector
6
Physics 2. Electromagnetism
Lecture 8
potential which has to be calculated as follows:
A=K
I
L
Idr ′
,
|r − r ′|
(31)
where L is the closed loop. We are interested in the magnetic field (and therefore vector potential)
at large distances r ≡ |r| ≫ |r ′| so that we Taylor expand
1
1
= (1 + r · r ′/r 2 ),
′
|r − r |
r
(32)
and substitute into (31):
A = KI
I
L
dr ′
+
r
I
L
(r · r ′)dr ′
.
r3
(33)
The first integral is zero, the second one can be written as follows:
I
KIxj
Ai =
x′j dx′i
r3
L
Kxj
= 3 ǫjik mk
r
(34)
and finally
Km×r
.
α r3
Now the magnetic field is (see vector analysis problem 1.5)
A=
K 3r(m · r) − r 2 m
.
B = rot A =
α
r5
(35)
(36)
Example 4.4.
Interaction of two straight currents. Let two currents, I1 and I2 be z directed and
at the distance d from each other (I1 and I2 can be positive and negative as well). Let us choose
coordinates so that I1 passes through the coordinate origin, while I2 passes through r = d (cylindrical
coordinates). Then I1 produced magnetic field at the position of I2 is B1 = 2KI1 ϕ̂/d. The force
which acts on the part of the wire of the length l is F2 = αI2 lẑ × B1 = −2αKI1 I2 lr̂/d. The two
currents attract each other if I1 I2 > 0 and repel each other if I1 I2 < 0.
5
Application of the Ampere law
Example 5.1. Magnetic field of an infinite straight wire. Because of the symmetry we can expect
that (in cylindrical coordinates) B = B(r)ϕ̂ so that we choose a circle as an integration contour.
7
Physics 2. Electromagnetism
According to the Ampere law
Lecture 8
I
L
B · dr = 2πrB = 4πKI
(37)
and we immediately have B = 2KI/r.
Example 5.2.
Magnetic field of a cylindrical current distribution. Let the current is distributed
so that the current density J = J(r)ẑ. Then the Ampere contours will be circles, and as above
I
so that
L
B · dr = 2πrB = 4πKIin = 4πK(2π
4πK
B=
r
Z
Z
r
J(r ′ )r ′dr ′ ),
(38)
0
r
J(r ′ )r ′ dr ′
(39)
0
If J = const for r < a and J = 0 for r > a we get

2KJr,
r<a
B=
2Kπa2 J/r, r > a.
Example 5.3.
in series.
(40)
Magnetic field of a coil. A coil is just a number of identical circular loops connected
The more loops are connected and the denser is their distribution, the stronger is the magnetic
field inside the coil and the weaker is the magnetic field outside. When the coil is sufficiently long
and the number of loops of the unit length n is sufficiently large, we may neglect the edge effects and
consider the magnetic field inside the coil as homogeneous and directed along the coil axis. Then,
H
choosing a contour as shown in the figure, we get B · dr = Bl where l is the length of the side in
H
inside the coil. Ampere’s law gives B · dr = 4πKIin where Iin is the current across the rectangular
area enclosed by the contour, Iin = Inl (I being the current in the coil, that is, the current flowing
8
Physics 2. Electromagnetism
Lecture 8
in each loop), so that
Bl = 4πKInl → B = 4πKIn
(41)
Example 5.4.
Current sheet. Assume a current is flowing in the y − z plane in z direction. Let
the surface current density be J . Let us break the current sheet into narrow striped of the width
dy. Each stripe has the current dI = J dy flowing in z direction. Such infinite straight current would
produce a magnetic field whose direction is found according to the right hand rule.
l
y
x
Because of the symmetry, we can expect that the current sheet produces the magnetic field
B = B ŷ for x > 0 and B = −B ŷ for x < 0. Then, choosing a path (blue in the figure) we
H
find B · dr = 2Bl = 4πKIin = 4πKJ l and B = 2πKJ . Thus, we have B = 2πKJ ŷ sign(x),
where sign(x) = x/|x|, sign(x) = 1 for x > 0 and sign(x) = −1 for x < 0. Compare this with the
corresponding expression for the electric field of an infinite charged plane with a given surface charge
density E = 2πσ x̂ sign(x).
6
Use of the Maxwell equation and vector potential
Here we provide examples of the use of
rot B = 4πKJ
B = rot A
Uniform magnetic field Let B = B ẑ. In Cartesian coordinates
∂Ay ∂Ax
−
=B
∂x
∂y
∂Az ∂Ay
−
=0
∂y
∂z
∂Ax ∂Az
−
=0
∂z
∂x
9
Physics 2. Electromagnetism
Lecture 8
One possible choice is Ay = Az = 0, then Ay = Bx. Another possible choice is Ay =
Ax = − 12 By, that is, A = 21 B × r.
1
Bx,
2
Cylindrical symmetry 1 Let J = Jz (ρ)ẑ then
1 ∂Bz
∂ρBϕ
=0
−
ρ ∂ϕ
∂z
∂Bρ ∂Bz
−
=0
∂z
∂ρ
1 ∂ρBϕ ∂Bρ
= 4πKJz
−
ρ
∂ρ
∂ϕ
Because of the symmetry there is no dependence on z and ϕ, so that Bz = const and
Z ρ
1
′ ′
′
4πK
J(ρ )ρ dρ + C
Bϕ =
ρ
0
where C = const and describes the magnetic field of an infinite wire.
A step further:
1 ∂Az
∂ρAϕ
=0
−
ρ ∂ϕ
∂z
∂Aρ ∂Az
−
= Bϕ
∂z
∂ρ
1 ∂ρAϕ ∂Aρ
=0
−
ρ
∂ρ
∂ϕ
or
∂Az
= Bϕ ⇒ Az =
∂ρ
Z
ρ
Bϕ (ρ′ )dρ′
0
and other components are zero. This is only one of possible solutions.
Cylindrical symmetry 2 Let J = Jz (ρ)ϕ̂ then
∂ρBϕ
1 ∂Bz
=0
−
ρ ∂ϕ
∂z
∂Bρ ∂Bz
−
= 4πKJϕ
∂z
∂ρ
1 ∂ρBϕ ∂Bρ
=0
−
ρ
∂ρ
∂ϕ
Again (∂/∂z) = 0, (∂/∂ϕ) = 0, and
∂Bz
= −4πKJϕ ⇒ Bz = −4πK
∂ρ
Z
ρ
Jϕ ρ′ dρ′
∞
10
Physics 2. Electromagnetism
Lecture 8
(explain the boundary condition B = 0 at ρ → ∞).
Respectively,
1 ∂ρAϕ
= Bz
ρ ∂ρ
Z
1 ρ
Bz (ρ′ )ρ′ dρ′
Aϕ =
ρ
(what about boundary conditions ? ).
Current sheet Let J = Jy (x)ŷ then
∂By
∂Bz
−
=0
∂y
∂z
∂Bx ∂Bz
−
= 4πKJy
∂z
∂x
∂Bx
∂By
−
=0
∂x
∂y
Since (∂/∂y) = 0 and (∂/∂z) = 0, one has
Bz (x2 ) − Bz (x1 ) = −
Z
x2
Jy (x′ )dx′
x1
If all current is in the layer 0 < x < d we have
Bz (d) − By (0) = −
Z
0
d
Jy dx ≡ −Iy
Respectively, we find
Bz =
∂Az
Z∂x
x
Ax =
Bz (x′ )dx′
x0
where x0 is arbitrary.
7
Magnetic properties - very briefly
Similar to dielectrics, magnetic materials change the magnetic field inside the material. However,
most magnetic materials enhance the magnetic field, so that B = µB0 , where B0 is the magnetic
field which would be produced in the vacuum and µ is the magnetic characteristic of the material
(magnetic permeability). Thus, magnetic field in a coil with a magnetic material inside would be
11
Physics 2. Electromagnetism
Lecture 8
B = 4πKµIn. Respectively, the Ampere law takes the form
1
B · dr = 4πKIin
µ
I
(42)
where 1/µ under the integral takes into account that the path can go through different materials,
and Iin is the external current.
8
Magnetic moment in general
Let us consider the vector potential which is produce by currents flowing in a finite volume. We are
interested in the vector potential at distances much larger than the size of the volume. Thus, the
Bio-Savart law gives
Z
KJ (r ′ )dV ′
A(r) =
(43)
r − r ′|
where |r ′| ≪ |r|. Using the Taylor expansion
1
r · r′
2
|r − r ′|−1 = (r 2 + r ′ − 2r · r ′)−1/2 ≈ (1 + 2 )
r
r
we get
The integral Qm ≡
K
A=
r
R
Z
K
J (r )dV + 3
r
′
′
Z
J (r · r ′ )dV ′ .
(44)
J dV ′ can be called magnetic charge for steady (time-independent) currents.
Let Let us first prove that Qm = 0. For this purpose is it sufficient to prove that Qm · a = 0 for
arbitrary constant vector a. Notice it is possible to write a = grad f , f = a · r ′ and therefore we have
R
to prove J · grad f dV ′ = 0. The last integral can be rewritten using div(f J ) = J · grad f + f div J .
The second term vanishes since for steady currents div J = −ρ̇ = 0. Therefore, we have
a·
Z
J dV =
Z
=
Z
′
J · grad f dV ′
′
div(f J )dV =
I
S
f J · dS ′ = 0
The last integral vanishes since we can choose the surface S enclosing volume V completely, so that
J = 0 at this surface. Thus, Qm · a = 0 for arbitrary constant vector a, which means Qm = 0.
The second term in (44) is more convenient to deal with in index representation:
Z
′
Mi = J (r · r )dV
′
i
= Ji
Z
xj x′j dV ′ .
(45)
12
Physics 2. Electromagnetism
Lecture 8
Let is write Mi = 21 xj (Lij + Kij ), where
Lij =
Z
(Ji x′j + Jj x′i )dV ′ ,
(46)
Kij =
Z
(Ji x′j − Jj x′i )dV ′ ,
(47)
Let us first show that Lij = 0. In order to do that let us consider the scalar L̃ = ai bj Lij where ai
and bj are arbitrary constant vectors. Similar to above, we have
L̃ =
Z
[(J · grad f )g + (J · grad g)f ]dV ′
=
Z
J · grad(f g)dV ′
=
I
S
div(J f g) · dS −
(48)
Z
f g div J dV ′ = 0
where f = a · r ′, g = b · r ′.
For Kij let us notice that (r ′ × J )k = ǫklm x′l Jm . Multiplying this relation by ǫijk we have
ǫijk (r ′ × J )k = ǫijk ǫklm x′l Jm
= (δil δjm − δim δjl )x′l Jm
(49)
= x′i Jj − x′j Ji
Comparing this with (47) we find
Kij = −
and finally
Z
ǫijk ǫijk (r ′ × J )k dV ′
K
Ai = − 3 ǫijk xj (
r
The integral
m=α
Z
Z
r ′ × J dV ′ )k
r ′ × J dV ′
(50)
(51)
(52)
is the magnetic moment of the current system, so that
A=
Km×r
α r3
(53)
13
Physics 2. Electromagnetism
9
Lecture 8
Angular momentum and magnetic moment
Let a charged particle with the charge q and mass M move on a circle with the radius r and angular
velocity ω. The angular momentum of the particle is
L = r × p = r × M(ω × r) = Mr 2 ω
(54)
The magnetic moment that the particle produces is
m = αIS ω̂
(55)
where S = πr 2 and I = q/T = qω/2π. Thus,
m = αqr 2 ω = (αq/m)L
(56)
14
Physics 2. Electromagnetism
Lecture 9. Electromagnetic induction.
1
Conductor moving in a magnetic field
We have seen in the past that motion of charges creates magnetic fields. Here we shall see that
electric fields can be produced due to the motion of conducting bodies in a magnetic field. Let us
consider a simple situation where there is a homogeneous magnetic field B = const in the space and
a conducting bar (directed along n̂ ∦ B) is moving with the velocity v ∦ B, n̂. On each charged
particle in the bar the magnetic force FB = αqv × B is acting. This force has a nonzero component
along the bar Fk = n̂ · F = αq n̂ · (v × B), which makes the charges move along the bar until there
is a charge separation at the two ends causing a parallel electric field, such that the total parallel
force Fk = qEk + αq n̂ · (v × B) = 0. Thus, motion of a conductor in the magnetic field led to
the generation of the electric field Ek = −αn̂ · (v × B). In a more general way, whenever charges
can freely move, in the equilibrium E + αv × B = 0. Yet in these cases the electric field is due to
the magnetic field and charge separation. If we look at this phenomenon in the reference frame of
the moving body we’ll find that electric field appears in this frame (which is the result of Lorentz
transformation of the electric and magnetic fields). This electric field is also homogeneous.
2
Inhomogeneous field
Let now analyze two inertial systems, S and S ′ , such that S ′ moves relative to S with the velocity
V0 = V0 x̂. Let the magnetic field in the frame S be B = Bz (x)ẑ and the electric field is absent.
Then the magnetic field in the moving frame will be B ′ = γ0 Bz (x) while the electric field will be
E ′ = γ0 V0 × B
= −γ0 V0 Bz (x)/cŷ
(1)
(2)
where x = γ0 (x′ + V0 t′ ). Let us calculate now rot E ′ :
∂
(γ0 V0 Bz (x)/c)ẑ
∂x′
1 ∂ ′
Bz
=−
c ∂t′
rot E ′ = −
1
(3)
(4)
Physics 2. Electromagnetism
Lecture 9
Thus, a) the electric field is not potential, and b) rot E ′ is related to (∂B/∂t′ ).
3
Electromagnetic induction
It appears that moving a conductor in a magnetic field is not the only way to produce electric
fields from magnetic fields. Time-variable magnetic produces electric field of a non-electrostatic
nature. This phenomenon is called electromagnetic induction. Although electromagnetic induction
was historically formulated in an integral (non-local) form, we will start with the local expression of
this law (which is nothing but the induction Maxwell equation). This law is written as follows:
rot E = −α
∂B
.
∂t
(5)
Even a quick glance at (5) shows that the obtained electric field is not similar to the electrostatic
field, in the sense that it cannot be written as E = − grad φ. Indeed, in this case rot E ≡ 0. On the
contrary, since B = rot A, we get
rot E = −α
∂
rot A = rot(−αȦ)
∂t
so that we can write
E = −αȦ,
(6)
E = −αȦ − grad φ,
(7)
or, more generally,
Since the induced electric field is not potential, its circulation
H
L
E · dr 6= 0, in general.
Equation (5) relates the variations of the electric and magnetic field in a point (locally), while
traditionally the induction law is formulated for circuits or closed contours. Let us choose a contour
L (may be changing, i.e. time-dependent), which enclosed the surface S. We shall assume that the
path direction and surface normal directions are properly defined. Let us consider the circulation
H
H
E
·dr.
If
we
move
a
charge
along
this
path
the
work
of
the
electric
forces
would
be
W
=
q
E ·dr.
L
L
However, W/q is nothing but emf, so that we can write:
E=
=
I
E · dr
IL
rot E · dS
I
∂B
· dS
= −α
S ∂t
(8)
S
It is, however, only a part of the emf. The other part appears due to the motion of the contour itself.
Indeed, in any part dr of the contour moving with the velocity v an electric field E = −αv × B is
2
Physics 2. Electromagnetism
Lecture 9
produced (see previous section). Therefore, the additional emf is
I
I
E · dr = −α
L
(v × B) · dr = −α
L
I
(dr × v) · B
L
Combining these two parts of the emf we shall have
E = −α
I
S
∂B
· dS − α
∂t
I
(dr × v) · B.
The last expression can be simplified if we denote ΦB =
H
S
B · dS and notice that
I
∂B
· dSdt +
B · dS
dΦB =
∂t
I dS
I S
∂B
· dSdt + B · (dr × v)dt
L
S ∂t
I
(9)
L
(10)
so that we eventually get
d
Φb .
(11)
dt
It has to be understood that (11) describes the combination of two effects: a) generation of electric
fields by time-dependent magnetic fields, and b) transformation of magnetic fields into electric fields
E = −α
in a moving frame.
4
Total electric field
The direct consequence of the induction equation (5) is the expression for the total electric field
E = − grad φ − Ȧ.
(12)
Taking into account the expression for the magnetic field
B = rot A
(13)
we can easily come to the conclusion that the scalar potential φ and vector potential A are not
uniquely determined. Indeed, we always can perform the transformation
A → A + grad Λ,
φ → φ − Λ̇
(14)
where Λ is an arbitrary function of time and coordinates. This transformation leaves E and B
unchanged (invariant), hence the name gauge invariance.
3
Physics 2. Electromagnetism
5
Lecture 9
Electromagnetic induction - Faraday and Lenz laws
The law of electromagnetic induction in the integral form, (11), is called Faraday law. The minus in
this expression if the Lenz law and has a very deep physical meaning. Let us consider a conducting
wire made in a closed contour and let us assume that magnetic field lines cross it in some direction. We
may choose the normal so that the angle between the normal and the magnetic field ∠(B n̂) < π/2.
H
Then the magnetic flux ΦB = S B · dS > 0. Let the magnetic field increase. This causes the flux
increase. The resulting emf causes a current in the conductor which produces its own magnetic field
whose direction is opposite to the direction of the initial magnetic field. Thus, the induced magnetic
field is acting to reduce its increase. It is easy to check that in general the response of the emf (and
induced current) is to reduce the changes in the system.
Example 5.1.
to a resistor R.
An ideally conducting rod is sliding on the two ideally conducting rails, connected
⇐v
+
R
The rod length is l, the velocity is v, and the homogeneous magnetic field B is directed into the
plane of the figure. The change of the magnetic flux during the time dt is dΦB = BdS = Bvldt (v
does not have to be constant !). Here we chose the normal in the direction of the magnetic field, so
that the positive direction at the contour (formed by the rails, the rod, and the resistor) is clockwise.
The induced emf is E = −αdΦb /dt = −αBvl. The sign − means that the emf is causing current
counterclockwise. The current is I = E/R = −αBvl/R and causes its own electric field which goes
out of the figure plane towards, thus adding a negative magnetic flux and reducing the flux change.
6
Inductance
Let us consider a coil of the length l, loop density n (number of loops on the unit length), and the
cross-section area S. We already know that the magnetic field produced by this coil inside itself is
B = 4πKIn. This field produces a magnetic flux across one loop Φ0 = BS = 4πKInS. Since there
are N = nl loops the total magnetic flux is ΦB = LI/α, where L = 4απKn2 S does not depend
on current and is called inductance. If the current changes the flux changes also which causes the
˙ Thus, changes of the current in the coil cause an inductive
inductive emf E = −αdΦB /dt = −LI.
emf in the same coil.
In a more general way, let us assume that there are several coils (or other closed current loops)
4
Physics 2. Electromagnetism
Lecture 9
with currents Ii flowing. The magnetic field is produced by all currents, and according to the
P
superposition principle, B = i Bi , where Bi is produced by Ii . Thus, for any coil i the magnetic
H
P H
flux Φi = i B · dS = j i Bj · dS. Since Bj ∝ Ij one finds
αΦi =
X
Lij Ij
(15)
j
where Lij is called mutual inductance (compare with the mutual capacity Cij ). Respectively, the
inductive emf will be
X
Lij I˙j
(16)
Ei = −
j
Let us now assume that the currents are changing by dIi during time dt. The work which has
been done in order to ensure such current change is
dW = −
X
Ei dIidt =
i
X
Lij Ij dIi
(17)
ij
Integrating this expression from initially zero current to final Ii we get the energy of the coils in the
following form
X
Lij Ii Ij
(18)
U = 21
ij
7
7.1
Proof of Lij = Lji
First method
This method is indirect and does not require calculation of Lij . It is based on the energy consideration
and is similar to what has been done for capacity (see Lecture 5). Let is consider two coils, so that
L11 , L12 , L21 , and L22 are fixed. Let us change the current in the first coil keeping I2 = 0. The work
which has been done is
W1 = −
Z
E∞ I1 dt =
Z
L11 I1 dI1 = 21 L11 I12
Now we change the current in the second coil keeping I1 = const so that
W2 = −
Z
E∞ I1 dt −
Z
E∈ I2 dt =
= L12 I1 I2 + 21 L22 I22
Z
L12 I1 dI2 +
Z
L22 I2 dI2
The total work is the final energy of the two coils
U = 21 L11 I12 + L12 I1 I2 + 12 L22 I22
5
Physics 2. Electromagnetism
Lecture 9
Inverting the order of the current change we get
U = 21 L11 I12 + L21 I1 I2 + 12 L22 I22
which means that L12 = L21 . Since the arguments are valid for each pair of coils, we have Lij = Lji .
7.2
Direct method
Let us use the definition Φ2 = L21 I1 and calculated
I
I
B1 · dS =
Φ2 =
S2
A1 · dr2
L2
where B1 is the magnetic field produced by current I1 and A1 is the corresponding vector-potential,
B1 = rot A1 . In the above expression we used the Stokes theorem.
Using Bio-Savart law we have
I
I1 dr1
A 1 = µ0
L1 |r2 − r1 |
and substituting this into the expression for the flux we get
L21
Φ2
=α
= µ0
I1
I
L1
I
L2
dr1 · dr2
|r2 − r1 |
(19)
The obtained expression does not depend on the order of indices so that L12 = L21 .
8
RL circuit
Let us consider a circuit consisting of a resistor R, inductance L, and constant emf E (which can be
zero too).
S
R, I ↓
E
L, I ↓
The same current I flows through the resistor and the coil. However, while the resistor is a passive
˙ so that the circuit equation will have the
element, the coil generates the inductive emf EL = −LI,
6
Physics 2. Electromagnetism
Lecture 9
form
X
E = IR ⇒ I˙ + I(R/L) = E/L
(20)
The solution of this equation with the initial condition I = I0 at t = 0 is
I=
E
[1 − exp(−t/τL )] + I0 exp(−t/τL )
R
(21)
where τL = L/R (compare with τC = RC for RC circuit).
Switching on:
In this case I0 = 0 (inductance prevent immediate change of current), and we
have
E
[1 − exp(−t/τL )]
(22)
R
Finally the current stops changing when t → ∞ and the coil does not affect anything. The final
I=
current is E/R. During all this time the power of the external emf is Pext = EI. From this power
the part I 2 R is dissipated into heat, the rest is obviously stored in the energy of the coil. The power
of the coil emf
Pl = −EL I = −
dUL
d LI 2
=−
dt 2
dt
From here we have the inductance energy
LI 2
UL =
2
(23)
It can be easily verified (do that !) that during the ”charging” half of the emf work goes into the
coil and half goes into heat.
Switching off In this case E = 0 and I = I0 exp(−t/τL ). It can be easily verified that the whole
coil energy finally goes into heat.
9
Magnetic energy
We have already calculated the magnetic field B = 4πKIn and the inductance L = 4απKn2 Sl of
a long coil (such that all the magnetic field can be considered to be inside the coil). Thus, the coil
energy is
LI 2
αB 2 Sl
αB 2
UL =
=
=
V
(24)
2
8πK
8πK
where V = Sl is the volume of the coil. Thus, the magnetic field energy density is
uB =
dU
αB 2
=
dV
8πK
(25)
7
Physics 2. Electromagnetism
Lecture 9
which is very similar to the electric field energy density uE = E 2 /8πk. The similarity is not coincidental as we shall see later.
RLC circuit
10
Let us consider the three elements, resistor R, capacitor C, and inductance L connected in series.
R, I →
−
C, +
L, I ↓
We choose the capacitor polarity so that the current enters the positive plate, and I = q̇. Thus,
we have
q/C + IR − E = q/C + Rq̇ + Lq̈ = 0.
(26)
We shall start analysis with the simplest case R = 0 (LC circuit). In this case we have
1
q
LC
(27)
q = Aeiω0 t ,
(28)
q̈ = −
and the solution is
p
where ω0 = 1/LC. Thus, the charge (and current) are oscillating harmonically. WARNING: in
order to get a physical solution we have to take the real part of (28), that is, the solution is
q = Re (A exp(iω0 t)) .
(29)
Now it is easy to substitute q = A exp(pt) into (26) to get
R
p2 + ( )p + ω02 = 0,
L
(30)
that is,
p=−
s
1
±i
2τ
ω02 −
1
4τL2
(31)
where τL = L/R and we assumed that ω0 τL > 1/2.
8
Physics 2. Electromagnetism
Lecture 9
The solution is
q = Re Ae−t/2τL exp(iω ′t)
where ω ′ =
11
p
ω02 − 1/4τL2 . These are damped oscillations.
(32)
AC current
Let us now add an external AC emf of the form E = Re E0 exp(iωt) (see Figure).
R, I →
−
L, I ↓
C, +
∼
Then the circuit voltage equation would be
q/C + IR + LI˙ = E
(33)
and we can expect that the solution be of the form I = Re(I0 exp(iωt)) (ω is the frequency of the
˙ For the current
external source). Let us define the ”voltages” UC = q/C, UR = IR, UL = LI.
I ∝ exp(iωt) we have q = I/iω, I˙ = iωI. Thus, one has
UC = (1/iωC)IC ,
UR = RIR ,
UL = (iωL)IL
(34)
It is widely accepted to write Us = Xs Is , where s = R, C, L, and X are called impedances. In some
sense impedances behave like resistances. For example, in our case we can just write
XI = E
(35)
where X = XL + XC + XR = R + i(ωL − 1/ωC). Respectively, we have
E0
exp(iφ0 )
I0 = E0 /X = p
R2 + (ωL − 1/ωC)2
1/ωC − ωL
φ0 = arctan
R
Which are forced oscillations. The current is maximum if ω =
case φ0 = 0.
(36)
p
1/LC = ω0 (resonance). In this
9
Physics 2. Electromagnetism
Lecture 9
Exercise Show that the average power P = hRe(E) Re Ii = E0 I0 cos φ0 /2.
10
Physics 2. Electromagnetism
Lecture 10. Maxwell equations and electromagnetic waves
1
Maxwell equations
We already know several Maxwell equations:
div E = 4πkρ
(1)
which says that electric charge is the source of the electric field and is equivalent to the Coulomb
law.
div B = 0
(2)
which says that there are no magnetic charges.
rot B = 4πKJ
(3)
which says that magnetic field is produced by currents.
rot E = −α
∂B
∂t
(4)
which is the induction law, saying that a time-varying magnetic field produces electric fields. This
has to be completed with the charge conservation
div J +
∂ρ
=0
∂t
(5)
There are some discrepancies though. Indeed, there is certain asymmetry in (3) and (4): the second
one predicts generation of electric fields but time-varying magnetic fields, while the first one does
not even mentions electric fields. On the other hand, we know that under Lorentz transformations
electric field goes into magnetic field and the opposite is true also. Second, applying div to (3) we
get div J = 0 and not (5). This means that (3) lacks a term. Let us rewrite (3) as follows:
rot B = 4πKJ + X
1
Physics 2. Electromagnetism
Lecture 10
where X is the lacking term to be found. Applying div again we get
4πK div J + div X = 0 → div X = 4πK
=
∂ρ
∂t
K
∂E
K ∂
div E =
div
k ∂t
k
∂t
and finally
K ∂E
k ∂t
Substituting this into (3) we get the complete Maxwell equation in the form
X=
rot B =
K ∂E
+ 4πKJ
k ∂t
(6)
1
(∂E/∂t) is called displacement current.
The term 4πk
In order to see whether this ”current” is real let us recall the behavior of a parallel-plate capacitor
in a circuit. If a current I enters the capacitor, the charge on the plate changes: q̇ = I. The
electric field inside is E = 4πkq/S, and the corresponding ”displacement current” density would
be JE = (1/4πk)Ė = q̇/S. The total displacement current IE = JS = I. Thus, the displacement
current in this case simply closes the circuit playing the role of the conductivity current for the
capacitor.
2
Energy balance of the fields
We know that the magnetic field work on charged particle is zero, since the magnetic force FB =
αqv × B ⊥ v. On the other hand, the electric field power in the unit volume
dPE
=E·J
dV
(7)
Let us multiply (6) by E/4πK and add to (4) multiplied by −B/4πK. We obtain
∂
1
[E rot B − B rot E] =
4πK
∂t
E2
αB 2
+
8πk 8πK
+E·J
(8)
Now let us use the relation (PROVE !)
and obtain
div(E × B) = B · rot E − E · rot B
(9)
∂u
+ div J = −E · J
∂t
(10)
2
Physics 2. Electromagnetism
Lecture 10
where
u=
E2
αB 2
+
8πk 8πK
(11)
is the electromagnetic field energy,
1
E ×B
4πK
and is the Pointing vector - energy flux density.
J =
3
(12)
Electromagnetic waves
Maxwell equations have solutions even in the absence of charges and currents. Indeed, let us assume
ρ = 0, J = 0. Then we have
K ∂E
k ∂t
∂B
rot E = −α
∂t
rot B =
(13)
(14)
For simplicity let us assume that E and B depend only on X = x − V t, where V has to be found.
Then rot → x̂ × (d/dX) and (∂/∂t) = −V (d/dX). We find
d
VK
(x̂ × B +
E) = 0,
dX
k
d
(x̂ × E − V αB) = 0
dX
(15)
(16)
As a special solution we can choose, for example, Ey arbitrarily, then Bz = (V K/k)Ey , and V 2 =
p
k/αK. Since both k and K are universal constants, so is c = kα/K - the speed of light. Thus, the
perturbations propagate with the constant velocity in x direction. This wave is called a plane wave.
3.1
Monochromatic waves
A special kind of a plane wane is a monochromatic wave of the form
E = E0 exp(ik · r − iωt),
(17)
B = B0 exp(ik · r − iωt).
(18)
Substitution into Maxwell equations with rot → ik× and (∂/∂t) → −iω gives
B0 = k × E0 /ω,
(19)
E0 = −(K/kαω)k × B0 ,
(20)
from which we get ω 2 = k 2 c2 and B0 = k̂ × E0 /c. The wave has definite frequency ω and wavevector
3
Physics 2. Electromagnetism
Lecture 10
k. The last one is related to the wavelength λ = 2π/k.
4
Physics 2. Electromagnetism
Lecture 11. Electromagnetic fields in medium
1
Maxwell equations
Let us rewrite once again the Maxwell equations we already know (here I will write them in the CGS
- Gaussian form):
div E = 4πρ,
(1)
div B = 0,
(2)
1 ∂B
,
c ∂t
1 ∂E 4π
+
J.
rot B =
c ∂t
c
rot E = −
(3)
(4)
In these equations ρ and j are exact charged and current densities, that is, all the microcharges and
currents have to be included if we wish to find the exact fields. Let us assume that we wish to
know the electric field in a medium consisting of a large number of identical dipoles with q and −q
separated by the distance d (so that the dipole moment p = qd). If we are interested in the exact
electric field in each point, in the vicinity of each charge q or −q the field would be greatly affected by
this charge and much less by other charges, so that the electric field would be very inhomogeneous.
Moreover, since all dipoles are in permanent thermal motion, this field would be also fast varying
in time. However, this is not what we usually want to know. What we wish to know is the electric
field which is averaged over times much greater than the typical time of the change of the dipole
orientation, and over volumes which contain large (macroscopic) number of dipoles. In this case we
loose all information about the field inhomogeneities (fluctuations) but become able to describe the
macroscopic field using very limited number of medium characteristics.
In order to see what happens let us assume that for some reason (we shall speak about these
reasons later) the dipoles become ordered, that is, the preferred direction has been chosen, so that the
average dipole momentum of a dipole is p. If the number of dipoles in the unit volume (concentration)
is n = dN/dV , the average dipole momentum density (polarization) would be P = np. The total
dipole momentum of the volume V = Sl limited by two ”plates” with the area S (we choose the
plates perpendicular to P) and distance l between them, is ptot = PSl. Let us now imagine that
these two plates are like a capacitor with the surface densities σ and −σ. In order that the total
1
Physics 2. Electromagnetism
Lecture 11
dipole momentum be what we have already calculated the surface charge density should be so that
σSl = P Sl, that is, σ = P . The electric field inside this ”capacitor” is E = 4πσ = 4πP . Taking
into account that the electric field direction is from the positive ”plate” to the negative one, that
is, in the direction opposite to P we get EP = −4πP. Thus, a polarized dielectric produces its own
electric field which is directed opposite to the polarization.
Let us consider now what happens with the magnetic field. Let there there are magnetic dipoles
with the average magnetic moment m and concentration n. The volume density of the magnetic
moment (magnetization) would be M = nm, and the magnetic moment of the same V = SL as above
would be just mtot = MSL. In order to find out what magnetic field is produced by this magnetic
moment inside it we shall compare is with a coil. A coil with the cross-section area S, length l,
number of turns per unit length n, and current I produces the magnetic field of B = 4πIn. Magnetic
moment of one turn is IS and the total magnetic moment is mtot = ISnl. Now the magnetic field and
the magnetic moment point in the same direction so we can write B = 4πmtot /SL. When comparing
with the above medium we come to the conclusion that the magnetic field which is produced is
BM = 4πM.
Now we are ready to analyze how the polarization and magnetization are produced.
2
Dielectrics and polarization
We will be considering simple models.
2.1
No dipoles initially
Let us assume that the dielectric consists of molecules are not polarized initially. Let each molecule
be a couple of two charges, q and −q, which are connected by a spring with the spring constant κ.
When an electric field E is applied the spring force has to balance the electric force, κx = qE. Since
the force on the positive charge is in the direction of the field, a dipole is produced with p = qE/κ.
The polarization then would be P = (nq/κ)E = χe E. If the electric without the dielectric is E0 then
E = E0 − 4πχe E which gives
E = E0 /ǫ,
ǫ = 1 + 4πχe .
(5)
This result is familiar to us, and ǫ is called dielectric constant. It is worth noting also that the dipole
produced field is −(ǫ − 1)E.
2.2
Dipoles change orientation
In most dielectrics, however, molecules themselves are dipoles. If no electric field is applied the
dipoles are oriented chaotically so that the average polarization is zero. If electric field E is applied
the energy of a dipole changes due to the interaction with the field, Up = −p · E. Since the system
2
Physics 2. Electromagnetism
Lecture 11
prefers to be in the state with minimum energy, the dipoles prefer to be oriented along the electric
field. Thermal motion prevents this orientation, so that the number of dipoles with the dipole
moment p is proportional to exp(−Up /kB T ) ≈ 1 + p · E/kB T for not very large fields. Without going
into details, the average polarization again P = χE, where now χ depends on the temperature. We
arrive at the same result as above.
2.3
Maxwell equations
Let us have a look at the equation
div E = 4πρ
and write the charge density in the right hand side in the form ρ = ρe + ρmic , where the charge
density ρmic described the charges of the molecules (dipoles) while ρe are external charges (that is,
those which we bring into the system and which do not belong to the microscopic charges of the
dielectric). As we already know the microscopic charges produce the field −4πP, that is, we can
write
div(−4πP) = 4πρmic
(6)
Now, subtracting (6) from (2.3) we obtain
div D = 4πρe ,
(7)
where D = E + 4πP = ǫE. Eq. (7) is the Maxwell equation for the electric field in an isotropic
dielectric.
2.4
Anisotropic materials
There are materials, in which dipoles can, for example, more easily rotate into one direction than into
another, which means that the relation between the polarization and the applied electric field can
be more complicated. When electric fields are too strong this relation can even become nonlinear.
However, for moderate fields it can be represented with the help of the dielectric tensor. Namely, we
should write Di = ǫij Ej in addition to (7)
3
3.1
Magnetic properties
Diamagnetism - free charges
Let us first consider free charged particles in the applied magnetic field B. For the particle with a
charge q and mass m we have
mv̇ = mω × v = (q/c)v × B
(8)
3
Physics 2. Electromagnetism
Lecture 11
where we have taken into account that v̇ = ω × v for circular motion, ω being the angular velocity
of rotation. Thus,
ω = −(q/mc)B
(9)
The angular momentum of the rotating particle is L = mr×v = mr×(ω ×r) = mr 2 ω, where r is the
radius of the orbit. the magnetic moment is m = (1/c)(ω/2π)q · πr 2 = (q/2mc)L = −(q 2 r 2 /2mc2 )B,
and is directed against the applied magnetic field. According to the found above, the magnetic field
produced by these charges particles would be
Bm = 4πM = −(2πnq 2 r 2 /mc2 )B
and is directed against the applied magnetic field, thus reducing it. This is called diamagnetism.
3.2
Diamagnetism - bound charges
In the above example the relation between the applied and produced fields was nonlinear, simply
because we assumed that the particles were able to move freely and switching the magnetic field
on changed their trajectories from straight lines to circles, that is, the effect was large. However, if
charged particles were rotating even before the magnetic field is switched on (under influence of some
another force, like, say, electron around a proton), then the influence of the magnetic field would be
weak and we would expect to obtain a linear relation, similar to what we obtained for dielectrics.
Let us assume that a particle with the charge q and mass m is rotating on a circle with the radius r
and velocity v. When the magnetic field is switched on it causes appearance of the electric field, so
that the emf on the circle is (CGS units !)
E · 2πr =
1d
(B · πr 2 )
c dt
Let us assume that the radius r cannot be changed, then
E=
r
Ḃ
2c
If the magnetic field increases from zero to B, the particle velocity changes by
∆v =
Z
(qE/m)dt =
qr
B
2c
The change of the current is ∆I = q∆v/2πr, and the change of the magnetic moment is (CHECK
THE DIRECTION !)
q2 r2
∆m = −
B.
(10)
2mc2
4
Physics 2. Electromagnetism
Lecture 11
Since in the equilibrium the magnetic moments are distributed randomly, it is only this difference
which matters, so that
nq 2 r 2
M = n∆m = − 2 2 B ≡ χb B
(11)
2m c
where χb = −nq 2 r 2 /2m2 c2 < 0 independently of the sign of the charge. Respectively, the induced
magnetic field is 4πM = 4πχb B and the total magnetic field B = B0 + 4πχb B so that B = B0 /(1 −
4πχb ). In diamagnets the magnetic field is weaker relative to what would be without the material.
3.3
Paramagnetism
Paramagnetism is similar to dipole containing dielectrics. In the magnetic field B a magnetic dipole
has the energy Um = −m · B so that the distribution of the dipoles according to their orientation is
determined by the factor exp(m · B/kB T ). Again, without coming into detail, the average magnetic
moment m ∝ B (see calculation in the Appendix) and M = χb B, so that again B = B0 /(1 − 4πχb ).
In this case, however, χb > 0 so that the magnetic field is stronger than it would be in the absence
of the material.
3.4
Ferromagnetism
In ferromagnets a large number of magnetic dipoles spontaneously becomes similarly oriented in
domains, so that the material is in some sense like a paramagnetic but made of a smaller number
but larger dipoles. Since the distribution is determined by exp(m · B/kB T ), the orientation along
the magnetic field is much stronger since the effective magnetic moment m is that of a domain not
a single dipole. Thus, the magnetic field enhancement is much stronger than in paramagnets.
3.5
Maxwell equation
Let us consider the short Maxwell equation (not including displacement current)
rot B =
4π
J
c
(12)
Here J is the total current. We shall write it as follows J = Je + Jmic where Je is the external
current not related to the microscopic features of the medium, while Jmic describes the effects of the
magnetization. From above we conclude that we can write
rot(−4πM) =
4π
Jmic
c
(13)
and subtracting (13) from (12) one finds
rot H =
4π
Je
c
(14)
5
Physics 2. Electromagnetism
Lecture 11
where H = B − 4πM = (1 − 4πχb )B. In this case it is widely accepted to define the magnetic
constant as B = µH and M = χM H.
4
Magnetostatics - current analogy
Let us write down the magnetostatics equations:
div B = 0,
(15)
B = µH,
I
H · dr = 4πI/c,
Z
B · dS = Φ
(16)
(17)
(18)
and compare with the equations for dc currents:
div J = 0,
(19)
J = σE,
I
E · dr = E,
Z
J · dS = I
(20)
(21)
(22)
We see that there is similarity between the behavior of the current density J and magnetic field
B, that is, the equations do not change with the following change of variables: B ↔ J, H ↔ E,
µ ↔ σ, 4πI/c ↔ E, and Φ ↔ I. Now we can understand why magnetic field lines choose to follow
the materials with high µ: it is like currents which prefer materials with lower resistivity.
5
Appendix. Average magnetic moment in paramagnets
The number of dipoles with the magnetic moment m is given by
n = C exp(m · B/kB T )
(23)
where C = const. The averaging is calculated as follows:
hmk i =
Rπ
0
m cos θ exp(mB cos θ/kB T ) sin θdθ
Rπ
exp(mB cos θ/kB T ) sin θdθ
0
(24)
6
Physics 2. Electromagnetism
Lecture 11
where mk = m · B/B and we chose coordinates with ẑ k B. The integrals are calculated as follows.
First, change the integration variable: cos θ = Z:
R1
hmk i = m R−11
Z exp(sZ)dz
−1
exp(sZ)dz
(25)
where s = mB/kB T . Let us define
I0 =
I1 =
Z
1
exp(sZ)dz,
−1
Z 1
(26)
Z exp(sZ)dz =
−1
dI0
dZ
(27)
Integral I0 is easily calculated:
1
(28)
I0 = (es − e−s ).
s
Since we are interested in small s we shall Taylor expand ex = 1 + x + x2 /2 + x3 /6 . . . to obtain
I0 = 2 + s2 /3
(29)
I1 = 2s/3
(30)
and
Substituting all this one has
hmk i = mI1 /I0 =
m2 B
3kB T
(31)
Finally,
M=
nm2
B
3kB T
(32)
If nm = const the magnetization is proportional to the magnetic moment of single dipole m, which
explains why domain structure of ferromagnets is much more efficient than paramagnets.
7