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HET316 Electromagnetic Waves: Gauss Law 4.1
Maxwells Equations for aStatic Electric Field
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HET316 Electromagnetic Waves: Gauss Law 4.2
Gauss Law
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HET316 Electromagnetic Waves: Gauss Law 4.3
Maxwells Equations for a Static Electric Field
For the next few sections we assume the microscopic picture of
electric fields in which we tacitly assume that the positions of allthe charges are known.
The initial aim is to find a concise way of writing the laws
governing the generation of electric fields by static chargedistributions.
The ultimate aim is to find the complete set of equations governing
electric and magnetic fields.
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4.4
Gauss Law (I).
We shall start with the electrostatic problems already considered.
The crucial observation (which is anything but obvious) is that the
distribution of electrostatic fields is governed by
(1). the properties of electric field lines.
(2). the fact that the field is conservative.
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4.5
Ignoring point (2) for the moment, the relevant properties ofelectric field lines are:
(a). Field lines start on positive charges (or at infinity) and endon negative charges (or at infinity).
(b). Except at points where there are charges or the field falls to
zero, electric field lines do not branch or cross.
(c). The number of field lines starting (or ending) on a point
charge is proportional to the charge of the particle. (This is really
only possible for inverse square law forces.)
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+ -
Field lines start on positive charges
Or at infinity
Field lines end on negative chargesOr at infinity
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4.7
Suppose we imagine a closed surface surrounding a collection ofcharged particles.
The field lines may cross this surface; count a crossing from insideto outside as positive and from outside to inside as negative.
A field line which begins on a charge inside the region and ends on
a charge also inside the region contributes nothing to the numberof lines crossing the surface (since the line cannot split or end
anywhere else and outward crossings must balance inward
crossings since the surface is closed).
Similarly lines which start and end outside the region contribute
nothing to the net number of crossings .
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-+
S
Number of lines crossing a surface (here equal to 12)
remains constant as surface changes
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-+
S
Number of lines crossing a surface (here equal to 12)
remains constant as surface changes
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-+
S
Number of lines crossing a surface (here equal to 12)
remains constant as surface changes
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-+
S
Outgoing lines count as positiveIngoing lines count as negative
Number of lines crossing a surface (here equal to 12)
remains constant as surface changes
+1
+1
+1
+1
+1-1
+1
+1
+1
+1+1
+1 +1
The negative crossing here is
cancelled by a new positive
crossing so the number of lines
crossing the surface remains
unchanged.
+1
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-+
S
If the number of charges inside the region bounded by the surface changesthen the number of field lines crossing the surface does change.
+1
+1
+1
+1
+1-1
-1
-1
-1
-1
The change in number of
lines is proportional to the
change in charge inside
the surface.
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4.13
A little thought will convince that the total number of field lines,e(S), crossing a closed surface, S, must be proportional to the
total charge, Q(S), contained in the region bounded by the surface .
)()( SQSe
This intuitive result (we have not derived it just given a plausiblejustification) is the essence of Gauss Law.
In this form it is an interesting observation, but not much use.
The pay-off comes when we find a way of interpreting the number
of field lines crossing S in terms of the electric field on S .
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4.14
Electric Flux
The quantity we have loosely identified with the number of field
lines crossing S need not really be an integer, all we need is some
quantity, usually called electric flux, which behaves in the same
way.
The simplest approach is to take the total flux leaving a charge q asequal to q.
We shall use the symbol to denote this quantity. For a single
point charge q we then have
q=
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Choose a surface S which is a sphere of radius r, centred on
the charge.
+
r
The surface area of the sphere is
24 rA =
The flux density on this sphere is
clearly uniform over the surfaceand given numerically by
4.15
24 rq
AD ==
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4.16
But this is directly proportional to the strength of the electric fieldon the surface of the sphere.
)(4
)( 020
0 rEr
qrD
==
This suggests that a possible interpretation of the magnitude of the
electric flux density in the vicinity of a point, r, is just the
magnitude of the electric fieldE(r) in the vicinity of the point.
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4.17
Indeed the electric field itself can serve as the electric flux densityvector; we shall define the total electric flux, de crossing a small
plane surface perpendicular to E with area dS as
dSd e E=
dS
E
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4.18
If the area element is not perpendicular to the field direction wecan still calculate the flux across the projection of the area at right
angles to the field and take this as the definition of the elemental
flux.
cosdSd e E=
dS
E
cosdS
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To complete the definition we introduce the elemental area
vector dS which is a vector normal to the element with a
magnitude equal to the area of the element. Then use the
definition of the scalar product
cosSESE dd =
dS
E
cosdS
To give the definition
of the elemental flux
SE dd e =
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The idea of a surface integral is needed to extend the
definition of electric flux to finite surfaces.
Consider a smooth surface S (which may be open or closed)
S
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Divide the surface into N small elements, let dSn be the area
vector describing the n-th small element. For open surfaces
the direction of dSn is defined by the right-hand rule once a
direction has been assigned to the bounding curve.
S
dS1 dS2
dSn
For closed surfacesthe direction is
taken along the
outward normal.
nnn
nn
dSd
ddS
sS
S
=
=
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Assume that the a scalar function,f(r), is sufficiently smooth
that on any given small element it can be well approximated
by its value at the centre of the element (f(rn )). The surface
integral offover S is defined to be the limit of the sum.
S
f(r1 )dS1
f(rn)dSn
f(r2 )dS2=
N
n
nn dSf1
)(r
as the size of each element
goes to zero and the number
of elements goes to infinity.
=
=
N
n
nn
NdA
S
dSffdSn 1
0)(lim r
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For a vector field, f(r), the flux is the surface integral of the
normal component of the field over the surface.
=
=
=
=
N
n
nn
NdA
N
n
nnn
NdA
f
d
dS
n
n
10
10
)(lim
)(lim
Srf
srf
S
=S
f dSf
This is often written in the
symbolic form :
nnnnn ddS Srfsrf = )()(
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The electric flux across a finite surface, S, is then the sum of
the fluxes across all of the small elements into which the
surface is partitioned. Thus
S
= Se dSE
E
dS
If the surface is closed then
it is conventional to indicate
this by drawing a circle on
the integral sign
=S
e dSE
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Gauss Law
We are now in a position to state Gauss law in its usual form.
In fact it can be shown that the total electric flux across anyclosed surface is proportional to the total charge contained
within the region bounded by the closed surface.
4.25
0
)(
sQdS
= SE
This is the integral form of Gauss law in the microscopic
picture.
Gauss law is the first of Maxwells equations.
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It is also common to write this in terms of a second field, D,called the electric displacement field:
4.26
In terms of which Gauss law becomes
)()( 0 rErD =
In the microscopic picture (or in free space) the two fields are
strictly proportional to one another and there is no particularadvantage in using D.
In the macroscopic picture, using smoothed average charge
distributions, the distinction becomes important and both fields
are required.
)(SQdS
= SD
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4.27
Two points need to be emphasised
1. The surface in Gauss Law can be any closed surface.
Its shape is usually chosen for convenience of calculation in a
given problem.
2. By itself Gauss law is not strictly enough to completely
determine the electric field.
If the field has known symmetries, however, then Gauss law may
be enough to determine the distribution of field strengths.
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4.28
Electric Field for a Uniformly Charged Sphere.
The most famous case for which Gauss law can be used to
simplify the field calculation is for radially symmetric volume
charge distributions.
We start with the simplest case;
a spherical volume containing
a uniform charge distribution V.
a
r
v
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4.29
The symmetry of the problem makes it clear that the field strengthcan depend only on the distance from the centre of the sphere
and that the field direction is radial
(ie. points to or away from the centre of the sphere).
a
S r
E(r)
)()( rErrE =
To take advantage of this symmetrychoose the Gaussian surface S to
be a sphere with radius r and the
same centre as the charge
distribution.
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4.30
The calculation is best done in two parts;
(1). Calculate the total electric flux through the Gaussian
surface
(2) Calculate the total charge in the region bounded by the
surface.
a
Sr
E(r)
The surface normal at r is
so
r r
dSd rS =and
dSrEdSrEd )()()( == rrSrE
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4.31
Calculating the total flux across the sphere is now easy, (it doesnot matter if r < a or r > a for this part of the calculation), since r
is constant on the surface of the sphere.
a
Sr
E(r)
r
24)()(
hence
sphereofarea)(
)()()(
rrEd
rE
dSrEdSrEd
S
SSS
=
=
==
SrE
SrE
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4.32
To calculate the charge inside S treat the cases r < a and r > aseparately.
For the r > a case (Gaussian surface outside the surface of thecharged sphere), the total charge inside S is just the total charge,
Q0 , on the sphere:
a
Sr
E(r)
0
3
3
4
SphereChargedofVolume)(
Qa
SQ
V
V
=
=
Gauss Law then becomes
arQ
rrESQ
dS
>== ;4)()(
0
02
0
SE
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4.33
For r < a, the charge inside S is given by
a
S
r
3
0
3
34
SphereGaussianofVolume)(
=
=
=
a
rQ
r
SQ
V
V
Gauss Law then becomes
ara
rQrrE
SQd
S
== ;4)(
)( 3
0
02
0
SE
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4.34
Thus
>
=
arr
ararQrE
;1
;
4)(
2
3
0
0
a r
E(r)
Q0/(40a2)
Note that, outside the sphere
the field is indistinguishable
from the field of a point charge
Q0 at the centre of the sphere.
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4.35
Electric Field for a Uniformly Charged Cylinder.
A Gauss law argument can be easily adapted to the field due to a
uniformly charged cylinder of infinite length.
a
r
E(r)
S
The symmetry in this case suggests
that the field strength can depend
only on the perpendicular distancefrom the axis of the cylinder and
that the direction of the field is
radial in planes perpendicular to
the axis.
)()( rErrE =
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a
r
S
E(r)
S
L
dS= zdS^
dS= rdS^
4.36
The appropriate Gaussian surface to choose is a cylinder with thesame axis as the charged cylinder.
The surface normals on the curved surface are radial, and on the
end surfaces are axial.
On the end surfaces
dSd rS = (On the curved surface)
dSd zS = (On the end surfaces)
0)( == dSrEd zrSE(since r and z are perpendicular)
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a
r
S
E(r)
S
L
dS= zdS^
dS= rdS^
4.37
On the curved surface
Thus the total electric flux acrossthe surface S is equal to the flux
across the curved surface and,
since E(r) is constant on this
surface
dSrEdSrEd )()( == rrSE
rLrE
rE
dSrE
dSrEd
Curved
CurvedS
2)(
surfacecurvedofArea)(
)(
)(
=
=
=
=
SE
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a
r
S
S
L
4.38
The calculation of the total charge inside the surface S is carriedout for two cases:
Gaussian surface outside the charged cylinder, ie. r> a
For which Gauss Law becomes:
LV
V
LLa
SQ
=
=
2
SinsideCylinderChargedofVolume)(
arLrLrE
SQd
L
S
>=
=
;2)(
)(
0
0
SE
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a
r
S
L
S
4.39
Gaussian surface inside the charged cylinder, ie. r< a
For which Gauss Law becomes:
2
2
SinsideVolume)(
=
=
=
a
rL
Lr
SQ
L
V
V
ara
rLrLrE
SQd
L
S
=
=
;2)(
)(
2
0
0
SE
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a
r
S
E(r)=E(r)r
4.40
Thus the magnitude of the field due to a uniformly chargedcylinder is given by :
Note that the field outside thecylinder is the same as the field
due to a charged line at the axis of
the cylinder with the same linear
charge density as the charged
cylinder.
>
= arr
arar
rE
L
;1
;
2)(
2
0
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Electric Field due to an Uniformly Charged, Infinite Slab
Suppose a uniformly charged slab of thickness 2a is parallel to the
XOY plane.
2a
E(z)z
Uniformly charged slab
x
y
z -E(z)z
^
Symmetry suggests that the
electric field strength at a point
(x, y, z) must be independent
of the x and y coordinates and
that the field direction must be
parallel to the z direction:
Symmetry also suggests that
the electric field above theXOY plane points in the
opposite direction to the field
below the plane. zE )(),,( zEzyx =
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The appropriate Gaussian surface is a cuboid with its end faces
parallel to the slab.
2a
2z
L
L
E(z)
Gaussian surface S
Uniformly charged slab
x
y
z
On the upper surface
dSd zS =
On the lower surfacedSd zS =
On the side surfaces dS is
perpendicular to the field.
The total flux crossing S
is thus
2)(2
faceupperofArea)(2
)()(
LzE
zE
dSzEdSzEdLowerUpperS
=
=
+= SE
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4.43
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The total charge contained within S is, for |z| > a
2a
2z
L
L
E(z)
Gaussian surface S
Uniformly charged slab
x
y
z
For which, Gauss Law
becomes:
222
SinsideSlabChargedofVolume)(
LaL
SQ
SV
V
=
=
aLLzE
SQd
S
S
>=
=
z;2)(
)(
0
22
0
SE
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For |z| < a the total charge contained within S is:
2a
2z
Uniformly charged slab
x
y
z
For which, Gauss Law
becomes:
222
SsurfaceGaussianinsideVolume)(
L
a
zLz
SQ
SV
V
==
=
aa
zLLzE
SQd
s
S
=
=
z;2)(
)(
0
22
0
SE
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So that the magnitude of the electric field due to a uniformly
charged slab is
The field outside the slab is
the same as the field due to a
uniformly charged plane with
the same surface chargedensity as the slab.
>
=
a
aazzE S
z;1
z;
2
)(
0
z
Ez(z)
a
-a
V/ (2o)
g
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Electric Fields due to Shells
One of the curious properties of inverse square law forces is that
the field inside a symmetric uniformly charged shell is often zero.
We leave it as an exercise to show that the electric field is zero
(i). inside a uniformly charged spherical shell,
(ii). inside a uniformly charged infinitely long cylindrical shell
(iii). between a pair of equally, uniformly charged, parallel
plates
g
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Electric Fields due to Layered Distributions.
The charge distribution need not be strictly uniform for some of
the results derived above to remain valid.
We leave it as an exercise to show that the external fields for the
spherical, cylindrical and slab charge distributions are unchanged
if the charge distribution depends only on the distance from the
centre for the spherical case, on the perpendicular distance fromthe axis for the cylindrical case, and on the z coordinate alone in
the slab case.