Transcript PPT - LSU Physics & Astronomy

Physics 2113
Jonathan Dowling
Flux Capacitor (Schematic)
Physics 2113
Lecture: 09 MON 14 SEP
Gauss’ Law II
Carl Friedrich Gauss
1777 – 1855
Gauss’ Law: General Case
• Consider any ARBITRARY
CLOSED surface S -- NOTE: this
does NOT have to be a “real”
physical object!
qins
S
• 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!
(One of Maxwell’s 4 equations!)
Gauss’ Law: ICPP
"+ " = +q
"- " = -q
F1 = +2 + 5 + 7 - 3- 4 - 7 F2 = +3+ 5 +10 - 3- 4 - 6
0
+
F 3 = +2 + 5 + 8 - 5 - 6 - 7
-
23-3 A Charged Isolated Conductor
Learning Objectives
•
•
•
23.17 For a conductor with a cavity
that contains a charged object,
23.14 Apply the relationship
determine the charge on the cavity
between surface charge density
wall and on the external surface.
σ and the area over which the
•
23.18 Explain how Gauss’ law is
charge is uniformly spread.
used to find the electric field
23.15 Identify that if excess
magnitude E near an isolated
charge (positive or negative) is
conducting surface with a uniform
placed on an isolated conductor,
surface charge density σ.
that charge moves to the surface
•
23.19 For a uniformly charged
and none is in the interior.
conducting surface, apply the
23.16 Identify the value of the
relationship between the charge
electric field inside an isolated
density σ and the electric field
conductor.
magnitude E at points near the
conductor, and identify the direction
© 2014 John Wiley &of
Sons,
All vectors.
theInc.
field
•
rights reserved.
Properties of Conductors
Inside a Conductor in Electrostatic
Equilibrium, the Electric Field Is ZERO.
Why?
Because If the Field Is Not Zero, Then
Charges Inside
the Conductor Would Be Moving.
SO: Charges in a Conductor Redistribute
Themselves Wherever They Are Needed
to Make the Field Inside the Conductor
ZERO.
Excess Charges Are Always on the
Surface of the Conductors.
ICPP: Conducting Sphere
• A spherical conducting shell has
an excess charge of +10 C.
• A point charge of –15 C is
located at center of the sphere.
• Use Gauss’ Law to calculate the
charge on inner and outer
surface of spherical shell
R2
+
+
R1
+
–15C
-
(a) Inner: +15 C; outer: 0
(b) Inner: 0; outer: +10 C
(c) Inner: +15 C; outer: –5 C
-
+10 C
S1
+
+
-
S2
"+ " = +5C
"- " = -5C
Hint: E-Field is Zero inside conductor so
Gauss’ Law: Conducting Sphere
• Inside a conductor, E = 0 under
static equilibrium! Otherwise
electrons would keep moving!
• Construct a Gaussian surface
inside the metal as shown. (Does
not have to be spherical!)
–5 C
• Since E = 0 inside the metal, flux
through this surface = 0
• Gauss’ Law says total charge
enclosed = 0
• Charge on inner surface = +15 C
Since TOTAL charge on shell is +10 C,
Charge on outer surface = +10 C - 15 C = -5 C!
+15C
–15C
Faraday’s Cage
• Given a hollow conductor of arbitrary shape.
Suppose an excess charge Q is placed on
this conductor. Suppose the conductor is
placed in an external electric field. How does
the charge distribute itself on outer and inner
surfaces?
(a) Inner: Q/2; outer: Q/2
(b) Inner: 0; outer: Q
(c) Inner: Q; outer: 0
• Choose any arbitrary surface
inside the metal
• Since E = 0, flux = 0
• Hence total charge enclosed = 0
All charge goes on outer surface!
Safe in the
Plane!?
Inside cavity is “shielded”
from all external electric
fields! “Faraday Cage
effect”
Safe in the
Plane!?
Faraday’s Cage: Electric Field
Inside Hollow Conductor is Zero
Safe in the Car!?
E=0
• Choose any arbitrary surface inside
the metal
• Since E = 0, flux = 0
• Hence total charge enclosed = 0
All charge goes on outer surface!
Inside cavity is “shielded”
from all external electric
fields! “Faraday Cage effect”
Field on Conductor Perpendicular to
Surface
We know the field inside the
conductor is zero, and the excess
charges are all on the surface. The
charges produce an electric field
outside the conductor.
On the surface of conductors in
electrostatic equilibrium,
the electric field is always
perpendicular to the surface.
Why?
Because if not, charges on the
surface of the conductors would
move with the electric field.
Charges in Conductors
• Consider a conducting shell, and a negative charge
inside the shell.
• Charges will be “induced” in the conductor to make
the field inside the conductor zero.
• Outside the shell, the field is the same as the field
produced by a charge at the center!
Gauss’ Law: Conducting Plane
• Infinite CONDUCTING plane with
uniform areal charge density s
• E is NORMAL to plane
• Construct Gaussian box as shown.
• Note that E = 0 inside conductor
Applying Gauss' law, we have,
As
e0
= AE
s
Solving for the electric field, we get E =
e0
Gauss’ Law: Conducting ICPP
• Charged conductor of arbitrary shape: no
symmetry; non-uniform charge density
• What is the electric field near the surface
where the local charge density is ?
(a) 0
(b) Zero
+
+
+
+ ++
+
(c) 0
Applying Gauss' law, we have,
+ +
+
+
+
E=0
As
e0
= AE
s
Solving for the electric field, we get E =
e0
THIS IS A
GENERAL
RESULT FOR
CONDUCTORS!
23-4 Applying Gauss’ Law: Cylindrical Symmetry
Learning Objectives
•
•
23.20 Explain how Gauss’ law
is used to derive the electric
field magnitude outside a line
of charge or a cylindrical
surface (such as a plastic rod)
with a uniform linear charge
density λ.
23.21 Apply the relationship
between linear charge density λ
on a cylindrical surface and the
electric field magnitude E at
radial distance r from the
central axis.
•
23.22 Explain how Gauss’ law
can be used to find the electric
field magnitude inside a
cylindrical non-conducting
surface (such as a plastic rod)
with a uniform volume charge
density ρ.
© 2014 John Wiley & Sons, Inc. All
rights reserved.
23-4 Applying Gauss’ Law: Cylindrical Symmetry
Figure shows a section of an infinitely
long cylindrical plastic rod with a uniform
charge density λ. The charge distribution
and the field have cylindrical symmetry.
To find the field at radius r, we enclose a
section of the rod with a concentric
Gaussian cylinder of radius r and height
h.
The net flux through the cylinder from
Gauss’ Law reduces to
yielding
© 2014 John Wiley & Sons, Inc. All
rights reserved.
A Gaussian surface in the form
of a closed cylinder surrounds
a section of a very long,
uniformly charged, cylindrical
plastic rod.
Gauss’ Law: Cylindrical Symmetry
• Charge of q = 10 C is uniformly
spread over a line of length L = 1 m.
E=?
• Use Gauss’ Law to compute
magnitude of E at a perpendicular
distance of 1 mm from the center of
the line.
r = 1 mm
L=1 m
• Approximate as infinitely long
line — E radiates outwards.
• Choose cylindrical surface of
radius r, length L co-axial with
line of charge.
Line of Charge:
 = q/L
Units: [C/m]
Gauss’ Law: Cylindrical Symmetry
• Approximate as infinitely long
line — E radiates outwards.
• Choose cylindrical surface of
radius r, length L co-axial
with line of charge.
F = EA = E2p rL
q lL
F= =
e0
E=?
r = 1 mm
L=1m
l
e0
lL
l
l
E=
=
= 2k
2pe 0 rL 2pe 0 r
r
Acyl = 2p rL
dA || E so cos  = 1
Compare to Finite Line Example From Last Week!
Ey = k l r
L/2
dx
ò (r 2 + x 2 )3/2
- L/2
L/2
é
ù
x
= kl r ê 2 2 2 ú
ë r x + r û- L/2
=
2k l L
• Add the Vectors!
• Horrible Integral!
• Trig Substitution!
r
r 4r 2 + L2
If the Line Is Infinitely Long (L >> r) …
Blah, Blah, Blah…
2k l L
2kl
Ey =
=
2
r
r L
With Gauss’s Law We Got Same
Answer With Two Lines of Algebra!
r
A-Rod