Physics 2102Physics 2102Lecture 3Lecture 3
GaussGauss’’ Law ILaw IMichael Faraday
1791-1867
Version: 1/22/07
Flux Capacitor (Schematic)
Physics 2102
Jonathan Dowling
What are we going to learn?What are we going to learn?A road mapA road map
• Electric charge Electric force on other electric charges Electric field, and electric potential
• Moving electric charges : current• Electronic circuit components: batteries, resistors, capacitors• Electric currents Magnetic field
Magnetic force on moving charges• Time-varying magnetic field Electric Field• More circuit components: inductors.• Electromagnetic waves light waves• Geometrical Optics (light rays).• Physical optics (light waves)
What? What? —— The Flux! The Flux!STRONGE-Field
WeakE-Field
Number of E-LinesThrough Differential
Area “dA” is aMeasure of Strength
dAθ
AngleMatters Too
Electric Flux: Planar SurfaceElectric Flux: Planar Surface
• Given:– planar surface, area A– uniform field E– E makes angle θ with NORMAL to
plane
• Electric Flux:Φ = E•A = E A cosθ
• Units: Nm2/C• Visualize: “Flow of Wind”
Through “Window”
θ
E
AREA = A=An
normal
Electric Flux: General SurfaceElectric Flux: General Surface• For any general surface: break up into
infinitesimal planar patches
• Electric Flux Φ = ∫E•dA• Surface integral• dA is a vector normal to each patch and
has a magnitude = |dA|=dA• CLOSED surfaces:
– define the vector dA as pointingOUTWARDS
– Inward E gives negative flux Φ – Outward E gives positive flux Φ
E
dA
dA
EArea = dA
Electric Flux: ExampleElectric Flux: Example
• Closed cylinder of length L, radius R• Uniform E parallel to cylinder axis• What is the total electric flux through
surface of cylinder?(a) (2πRL)E(b) 2(πR2)E(c) Zero
Hint!Surface area of sides of cylinder: 2πRLSurface area of top and bottom caps (each): πR2
L
R
E
(πR2)E–(πR2)E=0What goes in — MUST come out!
dA
dA
Electric Flux: ExampleElectric Flux: Example
• Note that E is NORMALto both bottom and top cap
• E is PARALLEL tocurved surface everywhere
• So: Φ = Φ1+ Φ2 + Φ3= πR2E + 0 - πR2E= 0!
• Physical interpretation:total “inflow” = total“outflow”! 3
dA
1dA
2 dA
Electric Flux: ExampleElectric Flux: Example• Spherical surface of radius R=1m; E is RADIALLY
INWARDS and has EQUAL magnitude of 10 N/Ceverywhere on surface
• What is the flux through the spherical surface?
(a) (4/3)πR2 E = −13.33π Nm2/C
(b) 2πR2 E = −20π Nm2/C
(c) 4πR2 E= −40π Nm2/C
What could produce such a field?
What is the flux if the sphere is not centeredon the charge?
q
r
Electric Flux: ExampleElectric Flux: Example
�
dr A = + dA( )ˆ r
�
r E • d
r A = EdAcos(180°) = !EdA
�
r E = !
q
r2
ˆ r
Since r is Constant on the Sphere — RemoveE Outside the Integral!
�
! =r E "d
r A # = $E dA = $
kq
r2
%
& '
(
) * # 4+r2( )
= $q
4+,0
4+( ) = $q /,0
Gauss’ Law:Special Case!
(Outward!)
Surface Area Sphere
(Inward!)
GaussGauss’’ Law: General Case Law: General Case
• Consider any ARBITRARYCLOSED surface S -- NOTE:this does NOT have to be a“real” physical object!
• The TOTAL ELECTRIC FLUXthrough S is proportional to theTOTAL CHARGEENCLOSED!
• The results of a complicatedintegral is a very simpleformula: it avoids longcalculations!
�
!"r E # d
r A =
±q
$0Surface
%
S
(One of Maxwell’s 4 equations!)
ExamplesExamples
! ="#$Surface 0
%
qAdErr
GaussGauss’’ Law: Example Law: ExampleSpherical symmetrySpherical symmetry
• Consider a POINT charge q & pretend that youdon’t know Coulomb’s Law
• Use Gauss’ Law to compute the electric field at adistance r from the charge
• Use symmetry:– draw a spherical surface of radius R centered
around the charge q– E has same magnitude anywhere on surface– E normal to surface
0!
q="
rq E
24|||| rEAE !=="
2204
||r
kq
r
qE ==
!"
GaussGauss’’ Law: Example Law: ExampleCylindrical symmetryCylindrical symmetry
• Charge of 10 C is uniformly spreadover a line of length L = 1 m.
• Use Gauss’ Law to computemagnitude of E at a perpendiculardistance of 1 mm from the center ofthe line.
• Approximate as infinitely longline -- E radiates outwards.
• Choose cylindrical surface ofradius R, length L co-axial withline of charge.
R = 1 mmE = ?
1 m
GaussGauss’’ Law: cylindrical Law: cylindricalsymmetry (cont)symmetry (cont)
RLEAE !2|||| =="
00!
"
!
Lq==#
Rk
RRL
LE
!
"#
!
"#
!2
22||
00
===
• Approximate as infinitely longline -- E radiates outwards.
• Choose cylindrical surface ofradius R, length L co-axial withline of charge.
R = 1 mmE = ?
1 m
Compare with Example!Compare with Example!
!" +
=
2/
2/
2/322 )(
L
L
yxa
dxakE #
if the line is infinitely long (L >> a)…
224
2
Laa
Lk
+
=!
2/
2/
222
L
Laxa
xak
!"#
$%&
'
+
= (
a
k
La
LkEy
!! 22
2==
SummarySummary
• Electric flux: a surface integral (vector calculus!);useful visualization: electric flux lines caught by thenet on the surface.
• Gauss’ law provides a very direct way to computethe electric flux.
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