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
dA
AngleMatters Too
Electric Flux: Planar SurfaceElectric Flux: Planar Surface
Given: planar surface, area A uniform field E E makes angle with NORMAL to
plane
Electric Flux: = EA = E A cos
Units: Nm2/C Visualize: Flow of Wind
Through Window
E
AREA = A=An
normal
Electric Flux: General SurfaceElectric Flux: General Surface For any general surface: break up into
infinitesimal planar patches
Electric Flux = EdA 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 Outward E gives positive flux
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) (2RL)E(b) 2(R2)E(c) Zero
Hint!Surface area of sides of cylinder: 2RLSurface area of top and bottom caps (each): R2
L
R
E
(R2)E(R2)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: = 1+ 2 + 3= R2E + 0 - R2E= 0!
Physical interpretation:total inflow = totaloutflow! 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)R2 E = 13.33 Nm2/C
(b) 2R2 E = 20 Nm2/C
(c) 4R2 E= 40 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( ) r
r E d
r A = EdAcos(180) = !EdA
r E = !
q
r2
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 areal 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
$0Surface
%
S
(One of Maxwells 4 equations!)
ExamplesExamples
! ="#$Surface 0
%
qAdErr
GaussGauss Law: Example Law: ExampleSpherical symmetrySpherical symmetry
Consider a POINT charge q & pretend that youdont know Coulombs 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.
