het316 l4 gauss law

5/28/2018 HET316 L4 Gauss Law

1/47

HET316 Electromagnetic Waves: Gauss Law 4.1

Maxwells Equations for aStatic Electric Field


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Gauss Law


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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.


4/47

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.

HET316 Electromagnetic Waves: Gauss Law


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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

4.6HET316 Electromagnetic Waves: Gauss Law


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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





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-+

S





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-+

S

Outgoing lines count as positiveIngoing lines count as negative



+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



12/47

-+

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.



13/47

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 = )()(



24/47

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.



26/47

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


4 34


34/47

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.


4 35


35/47

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 =


4 36


36/47

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)


4 37


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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


4 38HET316 El i W G L


38/47

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


4 39HET316 El t ti W G L


39/47

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




40/47

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




41/47

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 =




42/47

The appropriate Gaussian surface is a cuboid with its end faces

parallel to the slab.

2a

2z

L

L

E(z)

Gaussian surface S


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


4.43



43/47

The total charge contained within S is, for |z| > a

2a

2z

L

L

E(z)

Gaussian surface S


x

y

z

For which, Gauss Law

becomes:

222

SinsideSlabChargedofVolume)(

LaL

SQ

SV

V

=

=

aLLzE

SQd

S

S

>=

=

z;2)(

)(

0

22

0

SE




44/47

For |z| < a the total charge contained within S is:

2a

2z


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




45/47

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



46/47

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



47/47

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.

het316 l4 gauss law

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