Transcript Phys 2102 Spring 2002 - Louisiana State University Physics

Physics 2102
Jonathan Dowling
Flux Capacitor (Schematic)
Physics 2102
Lecture 3
Gauss’ Law I
Michael Faraday
1791-1867
Version: 1/22/07
What are we going to learn?
A 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? — The Flux!
STRONG
E-Field
Angle
Matters Too
Weak
E-Field

dA
Number of E-Lines
Through Differential
Area “dA” is a
Measure of Strength
Electric Flux: Planar Surface
• Given:
– planar surface, area A
– uniform field E
– E makes angle  with NORMAL to
plane
• Electric Flux:
F = E•A = E A cos
• Units: Nm2/C
• Visualize: “Flow of Wind”
Through “Window”
E

normal
AREA = A=An
Electric Flux: General Surface
• For any general surface: break up into
infinitesimal planar patches
• Electric Flux F = 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 pointing
OUTWARDS
– Inward E gives negative flux F
– Outward E gives positive flux F
E
dA
E
Area = dA
dA
Electric 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) (2pRL)E
(b) 2(pR2)E
(c) Zero
(pR2)E–(pR2)E=0
What goes in —
MUST come out!
Hint!
Surface area of sides of cylinder: 2pRL
Surface area of top and bottom caps (each): pR2
dA
E
L
dA
R
Electric Flux: Example
dA
• Note that E is NORMAL
to both bottom and top cap
• E is PARALLEL to
curved surface everywhere
• So: F = F1+ F2 + F3
= pR2E + 0 - pR2E
= 0!
• Physical interpretation:
total “inflow” = total
“outflow”!
1
2
3
dA
dA
Electric Flux: Example
•
•
Spherical surface of radius R=1m; E is RADIALLY
INWARDS and has EQUAL magnitude of 10 N/C
everywhere on surface
What is the flux through the spherical surface?
(a) (4/3)pR2 E = -13.33p Nm2/C
(b) 2pR2 E = -20p Nm2/C
(c) 4pR2 E= -40p Nm2/C
What could produce such a field?
What is the flux if the sphere is not centered
on the charge?
Electric Flux: Example
r
q
E = - 2 rˆ
r
(Inward!)
dA = dArˆ
q



(Outward!)
E  dA = EdAcos(180) = -EdA
Since r is Constant on the Sphere — Remove
E Outside the Integral!
 kq 
F =  E  dA = -E  dA = - 2 4 pr 2  Surface Area Sphere
 r 
q
=4 p  = -q /0 Gauss’ Law:
4 p0
Special Case!
Gauss’ Law: General Case
• Consider any ARBITRARY
CLOSED surface S -- NOTE:
this does NOT have to be a
“real” physical object!
• The TOTAL ELECTRIC FLUX
through S is proportional to the
TOTAL CHARGE
ENCLOSED!
• The results of a complicated
integral is a very simple
formula: it avoids long
calculations!

S
F

Surface
E  dA =
q
0
(One of Maxwell’s 4 equations!)
Examples
F

Surface
  q
E  dA =
0
Gauss’ Law: Example
Spherical symmetry
• Consider a POINT charge q & pretend that you
don’t know Coulomb’s Law
• Use Gauss’ Law to compute the electric field at a
distance 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
r
q
E
F =| E | A =| E | 4p r
F=
q
0
q
kq
|E|=
=
2
2
4p 0 r
r
2
Gauss’ Law: Example
Cylindrical symmetry
• Charge of 10 C is uniformly spread
over a line of length L = 1 m.
• Use Gauss’ Law to compute
magnitude of E at a perpendicular
distance of 1 mm from the center of
the line.
• Approximate as infinitely long
line -- E radiates outwards.
• Choose cylindrical surface of
radius R, length L co-axial with
line of charge.
E=?
1m
R = 1 mm
Gauss’ Law: cylindrical
symmetry (cont)
• Approximate as infinitely long
line -- E radiates outwards.
• Choose cylindrical surface of
radius R, length L co-axial with
line of charge.
E=?
1m
F =| E | A =| E | 2pRL
L
F= =
0 0
q
L


|E|=
=
= 2k
2p 0 RL 2p 0 R
R
R = 1 mm
Compare with Example!
L/2
L/2
dx


x
E y = k a 
2
2 3 / 2 = k a  2

2
2
(a  x )
 a x  a  -L / 2
-L / 2
=
2kL
a 4a  L
2
if the line is infinitely long (L >> a)…
2kL
2k
Ey =
=
2
a
a L
2
Summary
• Electric flux: a surface integral (vector calculus!);
useful visualization: electric flux lines caught by the
net on the surface.
• Gauss’ law provides a very direct way to compute
the electric flux.