2-Gauss
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Gauss’s LawGauss’s Law
Alan MurrayAlan Murray
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Alan Murray – University of EdinburghAlan Murray – University of Edinburgh
Revision : Vector Dot Revision : Vector Dot (Scalar) Product(Scalar) Product
a
ba.b = ab cos a
b
a.b = ab cos(90) = 0
a
b
a.b = ab cos(0) = aba
a.a = aa cos(0) = a²
In Cartesian co-ordinates, a.b = (ax,ay,az).(bx.by,bz) = axbx + ayby + azbz
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Alan Murray – University of EdinburghAlan Murray – University of Edinburgh
Revision : Vector x ScalarRevision : Vector x Scalar
a
In Cartesian co-ordinates, for example, 2a = 2(ax,ay,az) = (2ax,2ay,2az)
2a -2a
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Alan Murray – University of EdinburghAlan Murray – University of Edinburgh
Gauss’s Law : Crude AnalogyGauss’s Law : Crude Analogy
Try to “measure” the rain on a rainy dayTry to “measure” the rain on a rainy day• Method 1 : count the raindrops as they fall, and Method 1 : count the raindrops as they fall, and
add them upadd them up cf Coulomb’s Lawcf Coulomb’s Law
• Method 2 : Hold up an umbrella (a “surface”) Method 2 : Hold up an umbrella (a “surface”) and see how wet it gets.and see how wet it gets.
cf Gauss’s Lawcf Gauss’s Law Method 1 is a “divide –and-conquer” or “microscopic” Method 1 is a “divide –and-conquer” or “microscopic”
approachapproach Method 2 is a more “gross” or “macroscopic” Method 2 is a more “gross” or “macroscopic”
approachapproach They must give the same answer.They must give the same answer.
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Alan Murray – University of EdinburghAlan Murray – University of Edinburgh
Electric Field LinesElectric Field Lines
1C 1C 1C
These are all “correct” as E-field lines are simply cartoonsFor now, adopt a drawing scheme such that 1C = 1 E-line.
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Alan Murray – University of EdinburghAlan Murray – University of Edinburgh
Lines of Electric FieldLines of Electric Field
8C
How many field lines cross out of the circle?
8C 8 lines16C 16 lines
16C32C 32C 32 lines
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Alan Murray – University of EdinburghAlan Murray – University of Edinburgh
Lines of Electric FieldLines of Electric Field
8C
How many field lines cross out of the surface?
8C 8 lines16C 16 lines
16C32C 32C 32 lines
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Alan Murray – University of EdinburghAlan Murray – University of Edinburgh
Gauss’s Law : Cartoon VersionGauss’s Law : Cartoon Version The number of electric field lines The number of electric field lines
leaving a closed surface is equal to leaving a closed surface is equal to the charge enclosed by that surfacethe charge enclosed by that surface
(E-field-lines) (E-field-lines) Charge Enclosed Charge Enclosed
N Coulombs N lines
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Alan Murray – University of EdinburghAlan Murray – University of Edinburgh
Lines of Electric FieldLines of Electric Field
8C
How many field lines cross out of the surface?
8C 0 lines16C 0 lines
16C32C 32C 0 lines
i.e. charge enclosed = 0
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Alan Murray – University of EdinburghAlan Murray – University of Edinburgh
Gauss’s Law Proper (Gauss’s Law Proper ())
(E-lines) (E-lines) proportional toproportional to (Charge Enclosed) (Charge Enclosed) DD = = εεEE DD..dsds = = ((rr)dv)dv
= = ((rr)dxdydz)dxdydz DD..dsds = charge enclosed = charge enclosed ε= εε= ε00 = 8.85 x 10 = 8.85 x 10-12-12 in a vacuum in a vacuum
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Alan Murray – University of EdinburghAlan Murray – University of Edinburgh
Digression/RevisionDigression/RevisionArea IntegralsArea Integrals
This area gets wetter!
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Alan Murray – University of EdinburghAlan Murray – University of Edinburgh
Area Integrals – what’s happening?Area Integrals – what’s happening?
dsds
RainfallRainfall
This area gets wetter!
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Alan Murray – University of EdinburghAlan Murray – University of Edinburgh
Area Integrals – what’s happening?Area Integrals – what’s happening?
ds
ds
Rainfall Rainfall
Clearly, as the areas are the same, the angle between thearea and the rainfall matters …
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Alan Murray – University of EdinburghAlan Murray – University of Edinburgh
Area Integrals – what’s happening?Area Integrals – what’s happening?
dsds
Rainfall, R Rainfall, R
Extreme casesat 180° - maximum rainfallat 90°, no rainfall
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Alan Murray – University of EdinburghAlan Murray – University of Edinburgh
Flux of rain (rainfall) through an Flux of rain (rainfall) through an area area dsds
FluxFluxrainrain = = RR..dsds• ||RR||||dsds||cos(cos())• Rds cos(Rds cos())
FluxFluxrainrain = 0 for = 0 for 90° … cos(cos() = 0) = 0 FluxFluxrainrain = -Rds for = -Rds for 180° … cos(cos() = -1) = -1 Generally, FluxGenerally, Fluxrainrain = = Rds cos(Rds cos())
• -1 < -1 < cos(cos() < +1) < +1
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Alan Murray – University of EdinburghAlan Murray – University of Edinburgh
Area Integrals : Take-home Area Integrals : Take-home messagemessage
Area is a Area is a vectorvector, perpendicular to the , perpendicular to the surfacesurface
Calculating flux of rain, Calculating flux of rain, EE-field or anything -field or anything else thus involves a scalar or “dot” else thus involves a scalar or “dot” product product aa..bb = ab = abcos(cos())
This is what appears in a surface integral This is what appears in a surface integral of the form of the form DD..dsds, or , or RR..dsds, which , which would yield the total rainfall on whatever would yield the total rainfall on whatever surface is being used for integration (here, surface is being used for integration (here, the hills!)the hills!)
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Alan Murray – University of EdinburghAlan Murray – University of Edinburgh
l Coulombs/m
L
Gauss’s law – Worked ExampleGauss’s law – Worked ExampleLong straight “rod” of chargeLong straight “rod” of charge
Construct a “Gaussian Surface” that reflects the symmetryof the charge - cylindrical in this case, then evaluate DD..dsds
ds
ds
E, D
ds
E, D
r
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Alan Murray – University of EdinburghAlan Murray – University of Edinburgh
ds
E, D
r
EvaluateEvaluateDD..dsds
DD..dsds = = DD..dsds curved surfacecurved surface
++DD..dsds flat end facesflat end faces
End faces, End faces, DD & & dsds are perpendicular are perpendicular• DD..dsds on end faces = 0 on end faces = 0
• DD..dsds flat end facesflat end faces = 0 = 0
Flat end faces do not contribute!Flat end faces do not contribute!
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Alan Murray – University of EdinburghAlan Murray – University of Edinburgh
EvaluateEvaluateDD..dsds
DD..dsds = = DD..dsds curved surface onlycurved surface only
l Coulombs/m
L
ds
ds
E, D D & ds parallel,D.ds = |D|ds| = Dds
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Alan Murray – University of EdinburghAlan Murray – University of Edinburgh
EvaluateEvaluateDD..dsds
DD..dsds curved surface onlycurved surface only = = Dds Dds
l Coulombs/m
L
E, DD has the same strengthD(r) everywhere on thissurface.
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Alan Murray – University of EdinburghAlan Murray – University of Edinburgh
EvaluateEvaluateDD..dsds DD..dsds curved surface onlycurved surface only = = dsds = D= Dds = D ds = D area of curved area of curved
surfacesurface = D = D rrLL So 2DSo 2DrrL = charge enclosedL = charge enclosed Charge enclosed?Charge enclosed? Charge/length Charge/length length L = length L = l l LL
l Coulombs/m
rL
22rr
DD
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Alan Murray – University of EdinburghAlan Murray – University of Edinburgh
EvaluateEvaluateDD..dsds DD..dsds = charge enclosed = charge enclosed DrDr LL = = l l LL D(r) = D(r) = ll
rr
DD(r) =(r) =l l âârr
rr
x x
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Alan Murray – University of EdinburghAlan Murray – University of Edinburgh
DiscussionDiscussion
||DD| is proportional to 1/r| is proportional to 1/r• Gets weaker with distanceGets weaker with distance• Intuitively correctIntuitively correct
DD points radially outwards ( points radially outwards (âr) ||DD| is proportional to | is proportional to ll
• More charge density = more fieldMore charge density = more field• Intuitively correctIntuitively correct
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Alan Murray – University of EdinburghAlan Murray – University of Edinburgh
Other forms of charge distribution?Other forms of charge distribution?
Spherical charge Spherical charge distributiondistribution
rr-2-2, r, r-3-3, e , e –r–r … … Choose a spherical Choose a spherical
surface for integrationsurface for integration Then Then DD and and dsds will will
once again be parallel once again be parallel on the surfaceon the surface
Check it out!Check it out!
r
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Alan Murray – University of EdinburghAlan Murray – University of Edinburgh
R
Spherical charge distributionSpherical charge distribution- worked example- worked example
Solid sphere of 4C, Solid sphere of 4C, uniformly uniformly distributeddistributed r>R
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Alan Murray – University of EdinburghAlan Murray – University of Edinburgh
`
Other forms of charge distribution?Other forms of charge distribution?
Sheet of chargeSheet of charge Mirror symmetryMirror symmetry Choose a surface that Choose a surface that
is symmetric about is symmetric about the sheetthe sheet
Then Then DD and and dsds will will once again be parallel once again be parallel or perpendicular on or perpendicular on the surfacesthe surfaces
Check it out!Check it out!
`
D, ds
dsD
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Alan Murray – University of EdinburghAlan Murray – University of Edinburgh
Gauss’s LawGauss’s Law
. Dds enclosed vQ dv
This is Maxwell’s first equation
. 0Bds
As there is not such thing as an isolated “magnetic charge”, no Gaussian surface can ever contain a net “magnetic charge” – they come in pairs (North and South poles).
And we can have Maxwell’s second equation for free!