The electric field it can even be shown that under this condition the field Er is necessarily continuously differentiable if the region of integration is bounded1.. 17.1.3 Gauss’s LawGauss’
Trang 1where γ is the gravitational constant, is otherwise similar to Coulomb’s law
in electrostatics
For a distribution of point charges, the electric field E(r) generated by
this ensemble of charges can be calculated according to the principle of su-perposition:
F1←2,3, = q1· E(r1 ) = q1·
k=2,3,
q k(r1 − r k)
4πε0|r1− r k |3 . (17.5) This principle, which also applies for Newton’s gravitational forces, is
often erroneously assumed to be self-evident However for other forces, such as
those generated by nuclear interactions, it does not apply at all Its validity in
electrodynamics is attributable to the fact that Maxwell’s equations are linear
w.r.t the fields E and B, and to the charges and currents which generate
them On the other hand: the equations of chromodynamics, a theory which is formally rather similar to Maxwell’s electrodynamics but describing nuclear
forces, are non-linear, so the principle of superposition is not valid there.
17.1.2 Integral for Calculating the Electric Field
For a continuous distribution of charges one may “smear” the discrete
point-charges, (q k k )ΔV k), and the so-called Riemann sum in equation (17.5) becomes the following integral:
E(r) =
dV r )(r − r )
4πε0|r − r |3 . (17.6) This integral appears to have a singularity∝ |r−r | −2, but this singularity
is only an apparent one, since in spherical coordinates near r =r one has
dV ∝ |r − r|2d(|r − r|)
The electric field
it can even be shown that under this condition the field E(r) is necessarily
continuously differentiable if the region of integration is bounded1 It can then be shown that
This is the first of Maxwell’s equations (16.3), often referred to as Gauss’s law A proof of this law is outlined below (This law is not only valid under static conditions, but also quite generally, i.e., even if all quantities depend explicitly on time.)
1 This is plausible, since for d = 1 the integral of a continuous function is a
con-tinuously differentiable function of the upper integration limit
Trang 217.1.3 Gauss’s Law
Gauss’s law2gives the relation between the electric flux flowing out of a closed surface and the electric charge enclosed inside the surface The proof is subtle and we only give a few essential hints Consider two complementary regions of integration: For the first integration region, “1”, we consider a small sphere of
radius ε (an “ε-sphere”), whose center is situated exactly at the singularity
r = r The second integration region, “2”, is the complement of our
“ε-sphere”, i.e., the outer part (A sketch is recommended.) In this outer region, all differentiations with respect to the components of the vector r 3 can be performed under the integral, and one obtains the exact result 0, because we have for
r = r : (r − r )
|r − r | = 0 (since for example ∂ x
x/r3 = 1/r3− 3x2/r5, such that ∇ · r/r3 = 0) As
a consequence there remains
divE(r) ≡ divE1(r) , where E1(r) ∼ r)ΔV (r − r )
4πε0|r − r |3 , where we assume that the volume ΔV of our ε-sphere,
ΔV = 4πε
3
3 ,
r ) can be considered as
con-stant
We now use the statement, equivalent to the integral theorem of Gauss, that divE1 can be interpreted as a source density, i.e.,
divE1(r) = lim
ΔV →0 (ΔV )
−1
∂ΔV
E1(r) · n(r)d2A
As a consequence, after a short elementary calculation with
R = ε and n = |r − r r − r |
2 Here we should be aware of the different expression: “Gauss’s theorem” is the
divergence or integral theorem, while “Gauss’s law” means the first Maxwell
equation
3
We must differentiate with respect tor, not r .
Trang 3we obtain the result
divE(r) = r)(ΔV )4πR2
This equation,
constitutes the differential form of Gauss’s law The proof can be simplified using the δ-function:
div(r − r )
|r − r |3 ≡ 4πδ(r − r ) , ∀r , including r = r . (17.7)
One can now formally differentiate infinitely often under the integral, and one immediately obtains
divE(r) ≡
dV r )δ( r − r )/ε
Inserting (17.8) into Gauss’s integral theorem, we then arrive at the integral form of Gauss’s law:
∂V
E(r) · n(r)d2A ≡ Q(V )
where Q(V ) is the total amount of charge contained in the enclosed volume4
The theorem can be proved most visually in this integral version, since, due
to the superposition principle, one can assume that one is dealing only with
a single point charge placed at the origin r = 0 The field strength is then
∝ R −2, or more exactly:
E · n = cos ϑ · 4πε0R2 −1
, but d2A = R2dΩ/ cos ϑ Here, ϑ is the angle between the field direction (radial direction) and the
surface normal n (not necessarily radially directed); dΩ is the solid-angle
corresponding to a surface element, i.e.,
dΩ = 4π
These arguments are supported by Fig 17.1
4 However, surface charges, i.e., those located on ∂V , are only counted with a factor 1/2, whereas charges in the interior of V are counted with the full factor 1.
Trang 4Fig 17.1 On the proof of Gauss’s law (= Maxwell I) At the center of an ellipsoid
(here with the axes a = 2 and b = 1) there is a point charge Q, which generates the
electric field E = Q/(4πε0r2)e r, where e r is the radial unit vector Also plotted
are two long lines y ≡ x · 1 and y ≡ x · 0.85, which are meant to illustrate a cone,
through which the field flows outwards The corresponding surface element d(2)A
on the ellipsoid is r2dΩ/ cos ϑ, where r is the distance of the surface from the charge, ϑ the angle between the directions of the surface normal n and the field
direction, and dΩ the solid-angle element On the other hand, the scalar product
n ·E is Q/(4πε0r2
)·cos ϑ Therefore, according to obvious geometrical reasons, the result of an integration over the surface of the ellipsoid is Q/ε0, i.e., Gauss’s law,
as described above In contrast, charges outside the ellipsoid give two compensating
contributions, i.e., then the result is zero
17.1.4 Applications of Gauss’s Law: Calculating the Electric Fields for Cases of Spherical or Cylindrical Symmetry
The integral formulation of Gauss’s law is useful for simplifying the
calcula-tion of fields in the cases of spherical or cylindrical symmetry (see problems
1 and 3 of the exercises in summer 2002, [2])
Firstly, we shall consider spherical symmetry:
x2+ y2+ z2.
As a consequence, the electric field is also spherically symmetrical, i.e.,
E(r) = E(r) · e r , where e r:= r
r
is the radial unit vector
Applying the integral version of Gauss’s law for a sphere of arbitrary
radius r, we easily obtain the following result for the amplitude of the field, E(r), i.e for all 0 ≤ r < ∞:
E(r) = 1
ε0r2
r
Trang 5
A similar result is obtained for cylindrical symmetry, i.e., with
⊥ ) ,5 where
r ⊥:=
x2+ y2.
In this case we obtain
E(r) = E ⊥ (r ⊥)e ⊥ , with r = (x, y, z) , but e ⊥ := (x, y, 0)/r ⊥ ,
and
E ⊥ (r ⊥) = ε0r1
⊥
r ⊥
0
˜⊥d˜r ⊥ r ⊥ ) (17.11)
Using these formulae, calculation of the fields in the interior and exterior
of many hollow or massive bodies with spherical or cylindrical symmetry can
be considerably simplified Externally the field appears as if the total charge
of the sphere or cylinder is at the center of the sphere or on the axis of the cylinder
17.1.5 The Curl of an Electrostatic Field;
The Electrostatic Potential
One knows “by experience” that the curl of an electrostatic field always vanishes (i.e., the electrostatic field is irrotational ) This is in agreement
with the third of Maxwell’s equations or Faraday’s law of induction
curlE = − ∂ B
∂t (see below).
One should keep in mind here that although Maxwell’s equations are based on experimental observations Therefore they should not really be re-garded as universal laws or axioms, which are – as it were – self-evident or valid “before all experience”, although one could conceivably think of them
in that way.6
The experience mentioned above is that by calculating the work done
along a closed path in 3-space,
dr · F (r) = q
dr · E(r) ,
5
We do not consider any dependence on z.
6 The excellent textbook by Sommerfeld derives the whole of electrodynamics from
Maxwell’s equations, from which it starts without any reasoning or justification This is permissible in this case, but it should not be regarded naively as an acceptable modus operandi for theroretical scientists in general
Trang 6no energy can be gained, i.e., one always has
Γ
E(r) · dr = 0 , for any closed path Γ ∈ R3.
In the above we are assuming that the boundary of the region considered
is “connected” (i.e., one considers “one-fold connected regions” without any
“holes”), such that any closed line Γ in the considered region G can be written as the boundary line, Γ = ∂F , of a 2d-manifold F which is completely contained in G With these conditions one obtains from the integral theorem
Γ
E(r) · dr =
F E(r) · n(r)d2A
As a consequence one has
E(r) ≡ 0
However, according to Poincar´e’s lemma, any vector field v(r) which is
irrotational (curl v ≡ 0) in a “one-fold connected open region” G, possesses
in G a potential φ( r) such that
v(r) = −gradφ(r) , i.e., v i(r) = −∂ i φ( r) for i = 1, 2, 3
The potential φ( r) is only determined up to an arbitrary additive
con-stant, similar to the potential energy in mechanics, and the minus sign is mainly a matter of convention (although there are good reasons for it) For any irrotational vector field E(r), and in particular the electric field, this
potential can be calculated using
φ( r) = −
r
r1
where bothr1and the integration path fromr1tor can be arbitrarily chosen.
A potential φ( r) which leads to the standard expression occurring in
electrostatics,
r − r
|r − r |3 , is φ( r) = |r − r1 |
In fact it can easily be shown, with
|r−r | = [(x−x )2+(y −y )2+(z −z )2]1 , that −∂ x
1
|r − r | =
x − x
|r − r |3 For a continuous charge distribution, in accordance with the principle of superposition, one thus obtains the following potential:
φ( r) =
d3r r )
4πε0|r − r | , (17.13)
which is a result that can easily be memorized
Trang 717.1.6 General Curvilinear, Spherical and Cylindrical Coordinates
At this point it makes sense to include a short mathematical section on polar coordinates Hitherto we have always written vectors, e.g., dr, as triples of
three orthogonal Cartesian coordinates We shall now adopt general curvi-linear coordinates u, v, w such that
dr = ∂ r
∂u1du1+
∂ r
∂u2du2+
∂ r
By introducing the unit vectors
e i := ∂r
∂u i
/ | ∂ r
∂u i | (i = 1, 2, 3) one obtains with well-defined functions a i (u1, u2, u3) (for i = 1, 2, 3):
dr =
3
i=1
{a i (u1, u2, u3)du i } e i (u1, u2, u3). (17.15)
Note that the products a i du i have the physical dimension of “length” By
introducing the so-called dual triplet of bi-orthogonal vectors e ∗
j (dual to the triplete i , i.e., for i, j = 1, 2 and 3: e ∗
j ·e i
!
= δ j,i, i.e.,= 1! for i = j, but= 0!
for i = j, implemented by e ∗
1:= e2×e3
e1·[e2×e3 ] etc.), one obtains the short-hand expression
gradφ =
3
i=1
∂φ
a i ∂u i e ∗
Therefore one always obtains the following relation for the total
differen-tial dφ:
dφ = gradφ · dr =
3
i=1
∂φ
∂u i
du i
This formulation of the vectors dr and gradφ does not depend on the
particular choice of curvilinear coordinates In particular, for spherical and cylindrical coordinates we have e ∗
j ≡ e j, i.e., one is dealing with the special
case of locally orthogonal curvilinear coordinates, and the ∗-symbols can be
deleted
a) For spherical polar coordinates we have θ ∈ [0, π] (latitude), ϕ ∈ [0, 2π)
(longitude, meridian, azimuth) and
x = r sin θ cos ϕ , y = r sin θ cos ϕ , z = r cos θ : (17.17)
dr = dre r + r · dθe θ + r · sin θdϕe ϕ , (17.18)
Trang 8gradφ = ∂φ
∂r e r+ ∂φ
r · ∂θ e θ+ ∂φ
r sin θ · ∂ϕ e ϕ , and (see below) (17.20)
∇2φ =1
r
∂2
∂r2(rφ) + ∂
r2sin θ∂θ
sin θ ∂φ
∂θ
2φ
r2sin2θ∂ϕ2 . (17.21) (The first term on the r.h.s of equation (17.21) can also be written as
∂
r2∂r
r 2 ∂φ
∂r
, which is easier generalizable, if one considers
dimensional-ities d ≥ 2, i.e., ∂
r d−1 ∂r
r d −1 ∂φ
∂r
, but is less useful for the comparison
with d = 1, see Part III.)
b) In cylindrical coordinates (or planar polar coordinates)
x = r ⊥ cos ϕ, y = r ⊥ sin ϕ, z = z (i.e., unchanged) : (17.22)
dr = dr ⊥ e r ⊥ + r ⊥ · dϕe ϕ + dz e z , (17.23)
gradφ = ∂φ
∂r ⊥ e r ⊥+ ∂φ
r ⊥ · ∂ϕ e ϕ+∂φ
∇2φ = ∂
r ⊥ ∂r ⊥
r ⊥ ∂r ∂φ
⊥
2φ
r2
⊥ ∂ϕ2
+∂
2φ
In (17.21) and (17.26) we have added the so-called Laplace operator
∇2(= div grad)
These results are particularly important, since they occur in many applica-tions and examples
In fact, with
∇2φ ≡ div gradφ and gradφ = −E
they result from the following very general formula for orthogonal curvilinear coordinates, which can be proved by elementary considerations on source
densities:
divE = 1
a1a2a3
$
a2a3 ∂E1
∂u1 + a3a1
∂E2
∂u2 + a1a2
∂E3
∂u3
%
Capacitors; Capacity; Harmonic Functions; The Dirichlet Problem
Consider two arbitrary metal plates connected to the opposite poles of a cell
or battery The approximate profiles of the equipotential surfaces φ( r) =
constant and the corresponding electrostatic fields
E = −gradφ
Trang 9between the capacitor plates can be readily sketched; but how does the field
vary quantitatively in the space between the two capacitor plates?
This will now be discussed in detail
We recall that the boundary of a conductor is always an equipotential surface (electrons can flow until the potential is equal everywhere) and that the relation
E = −gradφ
implies that the field is perpendicular to the equipotential surfaces
φ( r) = constant
Furthermore in the interior of the conductor all three components of the electric field are zero
In the space S outside or between the conductors there are no charges, so the equation
∇2φ( r) ≡ 0
holds But on the surfaces ∂V iof the two metal plates 1 and 2 the electrostatic
potential φ must have different, but constant values, say
φ |∂V1 ≡ c + U ; φ |∂V2 ≡ c , where U is the potential difference (or “voltage”) provided by the battery.7
A function φ( r) which satisfies Laplace’s equation,
∇2φ = 0 ,
in an open region G (here G = S) is called harmonic.
As a consequence, in the space S between V1 and V2we must find a har-monic function φ( r) which assumes the values
φ |∂V1≡ U , but φ |∂V2≡ 0
(and vanishes, of course, sufficiently quickly as |r| → ∞).
This is a simplification of the somewhat more general Dirichlet problem:
r) one has the task of finding a function Φ(r), which (i)
in the interior of S satisfies the Poisson equation
∇2Φ=! 0, and which (ii) at the boundary of S takes prescribed values:
Φ |∂S = f (!
7 We remind the reader that the potential is only determined relative to some arbitrary reference value Therefore it is always advisable to “ground” the second metal plate, since otherwise the field can be changed by uncontrolled effects, e.g., electrically charged dust particles
Trang 10The Neumann problem, where one prescribes boundary values not for the function φ( r) itself, but for the normal derivative
∂ n φ := gradφ · n
is only slightly different
One can show that a solution of the Dirichlet problem, if it exists, is always unique The proof is based on the linearity of the problem and on another important statement, which is that a harmonic function cannot have a local maximum or local minimum in the interior of an open region.
For example, in the case of a local maximum in three dimensions the equipotential surfaces culminate in a “peak”, and the field vectors, i.e., the negative gradients of the potential, are oriented away from the peak Accord-ing to Gauss’s law there would thus be a positive charge on the peak This would contradict
|S ≡ 0 , i.e., the harmonicity of the potential.
Let us now assume that in S a given Dirichlet problem has two different solutions, φ1 and φ2; then (because of the linearity) the difference
w := φ2 − φ1 would satisfy in S the differential equation
∇2w( r) = 0 ,
with boundary values
w |∂S = 0
Because of the non-existence of a local maximum of a harmonic function in
the interior of S the function w( r) must vanish everywhere in S, i.e.,
φ2 ≡ φ1 ,
in contradiction with the assumption
By applying voltages U and 0 to the capacitor plate V1 and V2,
respec-tively, one induces charges Q1(on ∂V1) and Q2 (on ∂V2) We shall now show that
Q2=−Q1 , even if only V1 is connected to pole 1 of the battery, while V2 is grounded,
but not directly connected to pole 2;
Q2
=
∂V2
ε0 · E · nd2A
is called the induced charge on ∂V (see below)