Gauss’s Law
Alan Murray

Revision : Vector Dot
(Scalar) Product
a.b = ab cos(90) = 0
a
a.b = ab cos θ a
θ b b
a.b = ab cos(0) = ab a.a = aa cos(0) = a²
a a
b
In Cartesian co-ordinates, a.b = (ax,ay,az).(bx.by,bz)
= axbx + ayby + azbz
Alan Murray – University

Revision : Vector x Scalar
2a -2a
a
In Cartesian co-ordinates, for example,
2a = 2(ax,ay,az) = (2ax,2ay,2az)
Alan Murray – University

Gauss’s Law : Crude Analogy
 Try to “measure” the rain on a rainy day
• Method 1 : count the raindrops as they fall,
and add them up
 cf Coulomb’s Law
• Method 2 : Hold up an umbrella (a “surface”)
and see how wet it gets.
 cf Gauss’s Law
 Method 1 is a “divide –and-conquer” or “microscopic”
approach
 Method 2 is a more “gross” or “macroscopic”
approach
 They must give the same answer.
Alan Murray – University

Lines of Electric Field
How many field lines
cross out of the circle?
8C ⇒ 8 lines
16C ⇒ 16 lines
32C
16C
8C 32C ⇒ 32 lines
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Lines of Electric Field
How many field lines
cross out of the surface?
8C ⇒ 8 lines
16C ⇒ 16 lines
32C
16C
8C 32C ⇒ 32 lines
Alan Murray – University

Gauss’s Law : Cartoon Version
 The number of electric field lines
leaving a closed surface is equal to
the charge enclosed by that surface
 Σ(E-field-lines) α Charge Enclosed
N Coulombs ⇒ αN lines
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Lines of Electric Field
How many field lines cross
out of the surface?
8C ⇒ 0 lines
16C ⇒ 0 lines
32C
16C
8C 32C ⇒ 0 lines
i.e. charge
enclosed = 0
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Gauss’s Law Proper ()
 Σ(E-lines) (Charge Enclosed)
proportional to
 ∫∫D.ds = ∫∫∫ρ(r)dv
= ∫∫∫ρ(r)dxdydz
 ∫∫D.ds = charge enclosed
 D = εE
 ε= ε0 = 8.85 x 10-12 in a vacuum
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Digression/Revision
Area Integrals
This area gets
wetter!
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Area Integrals – what’s happening?
Rainfall
Rainfall
ds
ds
This area gets
wetter!
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Area Integrals – what’s happening?
Rainfall Rainfall
ds
ds
Clearly, as the areas are the same, the angle between the
area and the rainfall matters …
Alan Murray – University

Area Integrals – what’s happening?
Rainfall, R Rainfall, R
ds
ds
Extreme cases
at 180° - maximum rainfall
at 90°, no rainfall
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Flux of rain (rainfall) through an
area ds
 Fluxrain = R.ds
• |R|×|ds|×cos(θ)
• Rds cos(θ)
 Fluxrain = 0 for 90° … cos(θ) = 0
 Fluxrain = -Rds for 180° … cos(θ) = -1
 Generally, Fluxrain = Rds cos(θ)
• -1 < cos(θ) < +1
Alan Murray – University

Area Integrals : Take-home
message
 Area is a vector, perpendicular to the
surface
 Calculating flux of rain, E-field or anything
else thus involves a scalar or “dot” product
a.b = abcos(θ)
 This is what appears in a surface integral
of the form ∫∫D.ds, or ∫∫R.ds, which would
yield the total rainfall on whatever surface
is being used for integration (here, the
hills!)
Alan Murray – University

Gauss’s law - Example
Long straight “rod” of charge
Construct a “Gaussian Surface” that reflects the symmetry
of the charge - cylindrical in this case, then evaluate ∫∫D.ds
E, D
ds
E, D
ds
ρ l Coulombs/m
r
ds L
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E, D
Evaluate ∫∫D.ds ds
 ∫∫D.ds = ∫∫ D.ds curved surface
+∫∫ D.ds flat end faces r
 End faces, D & ds are perpendicular
• D.ds on end faces = 0
• ∫∫ D.ds flat end faces = 0
 Flat end faces do not contribute!
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Evaluate ∫∫D.ds
 ∫∫D.ds = ∫∫ D.ds curved surface only
D & ds parallel,
E, D D.ds = |D|×|ds| = Dds
ds
ρ l Coulombs/m
ds L
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Evaluate ∫∫D.ds
 ∫∫ D.ds curved surface only = ∫∫ Dds
D has the same strength
E, D D(r) everywhere on this
surface.
ρ l Coulombs/m
L
Alan Murray – University

Evaluate ∫∫D.ds
 ∫∫ D.ds curved surface only = ∫∫ ds D
 = D∫∫ ds = D × area of curved surface
 = D ×2πrL
 So 2Dπ r L = charge enclosed
 Charge enclosed?
 Charge/length × length L = ρl × L
2πr
ρ l Coulombs/m
L r
Alan Murray – University

Evaluate ∫∫D.ds
 ∫∫ D.ds = charge enclosed
 2πDr× L = ρl × L
 D(r) = ρl
2πr
 D(r) =ρl âr
2πr
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Discussion
 |D| is proportional to 1/r
• Gets weaker with distance
• Intuitively correct
 D points radially outwards (âr)
(
 |D| is proportional to ρl
• More charge density = more field
• Intuitively correct
Alan Murray – University

Other forms of charge distribution?
 Spherical charge
distribution
 ρ α r-2, r-3, e –r …
r
 Choose a spherical
surface for integration
 Then D and ds will
once again be parallel
on the surface
 Check it out!
Alan Murray – University

Other forms of charge distribution?
 Sheet of charge
 Mirror symmetry
 Choose a surface that
is symmetric about
the sheet
 Then D and ds will
once again be parallel `
`
or perpendicular on
the surfaces
 Check it out!
Alan Murray – University

Gauss’s Law
This is Maxwell’s first equation
∫∫ D.ds = Qenclosed = ∫∫∫ ρv dv
And we can have Maxwell’s second equation
for free!
∫∫ B.ds = 0
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).
Alan Murray – University