Transcript Lecture04: Gauss`s Law

Physics 121 - Electricity and Magnetism
Lecture 04 - Gauss’ Law
Y&F Chapter 22 Sec. 1 – 5
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Flux Definition (gravitational example)
Gaussian Surfaces
Flux Examples
Flux of an Electric Field
Gauss’ Law
Gauss’ Law Near a Dipole
A Charged, Isolated Conductor
Spherical Symmetry: Conducting Shell with Charge Inside
Cylindrical Symmetry: Infinite Line of Charge
Field near an infinite Non-Conducting Sheet of Charge
Field near an infinite Conducting Sheet of Charge
Conducting and Non-conducting Plate Examples
Proof of Shell Theorem using Gauss Law
Examples
Summary
1
Copyright R. Janow – Spring 2015
FLUID FLUX EXAMPLE: WATER FLOWING ALONG A STREAM
Simplifying Assumptions:
•
•
•
•
constant mass density r incompressible fluid
constant flow velocity parallel to banks
no turbulence (laminar flow)
n̂'
n̂

A 2
Flux measures the (current) flow:
• amount/unit time flowing across an area
• volume flow rate or mass flow rate past point

A 1
2 related fluid flow fields (currents/unit area):
• velocity v represents volume flow/unit area/unit time
• J = mass flow/unit area/unit time

A  n̂A
n̂ are outward unit vectors

 to vector area A


J  rv
Flux = amount of field crossing an area per unit time (field x area)

 

V  L  A 
L  vt
The chunk of

volume flux 

 v  A
fluid moves L

t

t
A


in time t:



r


r


mass
of
solid
chunk
m
V
L
A



v

 

m
L

and
\ mass flux 
r


r



 A v A J A
t
t
Through a closed surface, net flux (fluid flow) = 0
………unless a source or drain is inside surface
Copyright R. Janow – Spring 2015
Flux (symbol F): A vector field magnitude x area
Applies to fluid flow, or gravitational, electric, or magnetic field
Example: Gravitational flux, field ag, take component normal to area dA
“unit normal”
ag
n̂
outward and
perpendicular to
surface dA


dF g  flux of ag through dA

 ag  n̂dA (a scalar)
Dimensions for gravitation flux: N.m
Choose a closed or open surface S and integrate:
- Involves doing “surface integral” of field over S
FS 
 dF 

a
 g  n̂dA
S
S
- Evaluate integrand at all points
on surface S
EXAMPLE : GRAVITATIONAL FLUX THROUGH A CLOSED
IMAGINARY BOX (UNIFORM ACCELERATION FIELD)
• No mass inside the box
• F from each side = 0 since a.n = 0, F from ends cancels
• TOTAL F = 0
• Example could also apply to fluid flow
n̂
n̂
n̂
ag
n̂
Could field still be uniform if a mass (flux source) is in the box?
Could net flux be zero?
Copyright R. Janow – Spring 2015
Gaussian Surfaces are
closed 3D surfaces
• Field lines cross a closed surface:
• An odd number of times
sphere
for charges that are inside S
• An even number of times
for charges outside of S
• Choose surface to match symmetry of the
field or charge distribution where possible
closed
box
cylinder with
end caps
no end caps
not closed
not a GS
The flux of electric field crossing a closed surface equals the net
charge inside the surface (times a constant).
Example: Point charge at center of a spherical “Gaussian surface”
Flux  FE  Field x Surface Area  E x S 
Q
2
x
4

R
 Q /0
2
4 0R
Principle holds for arbitrary charges and closed surfaces?
Flux is a SCALAR, Units: Nm2/C.
Copyright R. Janow – Spring 2015
Electric Flux: Integrate electric field over a surface
Definition: flux dE crossing (vector) area dA
“unit normal”
E
n̂
outward and
perpendicular to
surface dA
Divide up surface S into tiny chunks of dA
each and consider one of them

dA  n̂dA


dFE  flux of E through dA

 E  n̂dA (a scalar)
Flux through surface S (closed or open): Integrate on S
FE 
 dFE
S

  E  n̂dA
To do this: evaluate integrand at
all points on surface S
S
Gauss’ Law: The flux through a closed surface S depends only on
the net enclosed charge, not on the details of S or anything else.
Copyright R. Janow – Spring 2015
F depends on the angle between the field and
chunks of area

F  E  n̂A  EAcos()

A  n̂A
n̂  outward unit vector
F>0
F<0
WHEN E POINTS
FLUX F IS
from inside to outside of surface
E.n > 0 POSITIVE
from outside to inside of surface
E.n < 0 NEGATIVE
tangent to surface
E.n = 0 ZERO
F=0
Copyright R. Janow – Spring 2015
Evaluating flux through closed or open surfaces
Special case: field is constant across pieces of the surface

FE  E  n̂A (a scalar)
Units of FE : Nm2 / C

FE   FE   E  n̂A
SUM F FROM SMALL
CHUNKS OF SURFACE A
small areas
EXAMPLE: Flux through a cube
Assume:
•
•
•
•
small areas
Uniform E field everywhere
E along +x axis
Cube faces normal to axes
Each side has area length2
•
Field lines cut through two
surface areas and are tangent to
the other four surface areas
•
For side 1, F = -E.l 2
•
For side 2, F = +E.l 2
•
•
For the other four sides, F = 0
Therefore, Ftotal = 0
\ FE   Fi  0
i1,6
• What if the cube is oriented obliquely??
• How would flux differ if net charge is inside??
Copyright R. Janow – Spring 2015
Flux of a uniform electric field through a cylinder
• Closed Gaussian surface
• Uniform E means zero
enclosed charge
• Calculate flux directly
• Symmetry axis along E
• Break into areas a, b, c
F tot 


 E  dA  F a  Fb  F c
a ,b ,c

Cap a: E  n̂  - E,

Cap c: E  n̂   E,
cos()  - 1
\ Fa  - E
 dA
 - E Acap
cap a
cos()   1
\ Fc   E
 dA
  E Acap
cap c

 
Area b: E  n̂  0 (E  A) everywhereon b \ Fb  0
\ F tot  0
What if E is not parallel to cylinder axis:
• Geometry is more complicated...but...
• Qinside = 0 so F = 0 still !!
Copyright R. Janow – Spring 2015
Flux of an Electric Field
4-1: Which of the following figures correctly shows a
positive electric flux out of a surface element?
A.I.
B.II.
C.III.
E

D.IV.
I.
II.

E.I and III.
A
A
III.
E
IV.
E

A
E
A

Copyright R. Janow – Spring 2015
Statement of Gauss’ Law
Let Qenc be the NET charge enclosed by a (closed) Gaussian
surface S. The net flux F through the surface is Qenc/0

 Q
F   E  dA  enc
0
Surface
Q enc   0F
See Divergence
Theorem,
for the ambitious
• Does not depend on the shape of the surface, assumed closed.
• Charge outside the surface S can be ignored.
• Surface integral yields 0 if E = 0 everywhere on surface
Example: Derive Coulomb’s Law from Gauss’ Law
Assume a point charge at center, r on spherical Gaussian surface

 
 
E  dA  E  n̂dA is always just EdA
E  dA  E dA

because E and n̂ are always radial, E is constant on S
Q enc
2
F  E  dA  E  4 r 
0
S
r
Qenc
\ E(r ) 
Q enc
4 0 r
2
Coulomb’s Law
Copyright R. Janow – Spring 2015
Shell Theorem: Spherical Shells of Charge
Spherically Symmetric Shell
P Outside
+Q
What are the fields
on Gaussian surfaces
inside and outside?
+Q
P Inside

 Q
F   E  dA  enc
0
S
4-2: What is the flux through the Gaussian surface below?
A.
B.
C.
D.
E.
zero
-6 C./0
-3 C./0
+3 C./0
not enough info

 Q
F   E  dA  enc
0
S
Copyright R. Janow – Spring 2015
Flux through surfaces near a dipole
FLUX POSITIVE
from S1
• Equal positive (+Q) and negative (–Q) charges
• Consider Gaussian (closed) surfaces S1...S4.
•
Surface S1 encloses only the positive
charge. The field is everywhere
outward. Positive flux. F = +Q/0
•
Surface
•
Surface S3 encloses no charges.
•
+
FLUX = 0
S2 encloses only the negative through S4
S1
charge. The field is everywhere
inward. Negative flux. F = -Q/0
Net flux through the surface is zero.
The flux is negative at the upper
part, and positive at the lower part,
but these cancel. F = 0
Surface S4 encloses both charges.
S4
S3
FLUX = 0
through S3
S2
-
Zero net charge enclosed, so equal flux
enters and leaves, zero net flux
through the surface. F = 0
FLUX NEGATIVE
from S2
Copyright R. Janow – Spring 2015
Where does net charge reside on an isolated conductor?
• Place net charge Q initially in the
interior of a conductor
• Charges are free to move, but


can not leak off
F  QE
• At equilibrium E = 0 everywhere
inside a conductor
• Charge flows until E = 0 at every
interior point
E = 0
• Choose Gaussian surface just below surface: E = 0 everywhere on it
• Use Gauss’ Law!


Q
F

Surface
E  dA  0 
enc
0
 Q enc  0
• The only place where un-screened charge can end up is on the outside
surface of the conductor
• E at the surface is everywhere normal to it; if E had a component
parallel to the surface, charges would move to screen it out.
Copyright R. Janow – Spring 2015
Gauss’s Law: net charge on an isolated conductor...
...moves to the outside surface



F
E  dA 
Surface
1: SOLID CONDUCTOR WITH NET CHARGE ON IT
Qenc
0
All net charge on a conductor
moves to the outer surface
E=0
S
2. HOLLOW CONDUCTOR WITH A NET CHARGE, NO CHARGE IN CAVITY
E=0
S’
• Choose Gaussian surface S’ just outside the cavity
• E=0 everywhere on S’, so the surface integral is zero
• Gauss’ Law says:
Q
FS'  0 
enc
0
\ Qenc  0
There is zero net charge on the inner
surface: all net charge is on outer surface
Does this mean that the charge density is zero at each point on the inner surface?
Copyright R. Janow – Spring 2015
Hollow conductor with cavity, charge inside
• Electrically neutral conducting shell, for example
• Arbitrary charge distribution +Q in cavity
• Choose Gaussian surface S completely within
the conductor

 q
F S   E  dA 
S
enc
+
+
-
-
++
+ Q +
+
-
+
+
0
• E = 0 everywhere on S, so FS = 0 and qenc= 0
• Negative charge Qinner is induced on inner surface,
distributed so that E = 0 in the metal, hence…
qenc  0  Q  Qinner
+
+
-
-
-
+
-
-
+
+
+
Gaussian
Surface
Qinner  -Q
• The shell is neutral, so Qouter= -Qinner = +Q appears on outer surface
S
Whatever the inside distribution may be, outside the shell it is shielded.
Qouter is uniform because the shell is spherical, so the field looks like that of a
point charge at the center.
OUTSIDE SPHERICAL SHELL:
• Choose another spherical Gaussian surface S’ outside the spherical shell
• Gauss Law:
F
q'enc
0
Q

 4r 2E(r )
0
E(r ) 
Q
4  0r
2
radially outward
Copyright R. Janow – Spring 2015
Conducting spherical shell with charge inside
4-3: Place a charge inside a cavity in
an isolated neutral conductor.
Which statement is true?
Spherical
cavity
++
+ Q +
+
+
Conducting
sphere
(neutral)
Positive
charge
A. E field is still zero in the cavity.
B. E field is not zero in the cavity,
but it is zero in the conductor.
C. E field is zero outside the
conducting sphere.
D. E field is the same everywhere
as if the conductor were not
there (i.e. radial outward
everywhere).
E. E field is zero in the conductor,
and negative (radially inward)
outside the conducting sphere.

 Q
F   E  dA  enc
0
S
Copyright R. Janow – Spring 2015
Gauss’ Law Example: Find the electric field formula at a
distance r from an infinitely long (thin) line of charge z
lh  qenc
• Cylindrical symmetry around z-axis
• Uniform charge per unit length l
• Every point on the infinite line has the
identical surroundings, so….
• E is radial, by symmetry

A
• E has the same value everywhere on
any concentric cylindrical surface
• Flux through end caps = 0
as E is perpendicular to A
• E on cylinder is parallel to unit vector for A


lh
\ F   E  dA  2rhE 
0
S
E(r ) 
l
2 0r
radially outward
Cylinder area = 2rh
• No integration needed now due to Gauss Law
• Good approximation for finite line of
charge when r << L, far from the ends.
Copyright R. Janow – Spring 2015
Field Lines and Conductors
4-4: The drawing shows cross-sections of three cylinders with
different radii, each with the same total charge. Each cylindrical
gaussian surface has the same radius (again, shown in crosssection). Rank the three according to the electric field at the
gaussian surface, greatest first.
A.
B.
C.
D.
E.

 Q
F   E  dA  enc
0
S
I, II, III
III, II, I
All tie.
II, I, III
II, III, I
I.
II.
III.
Copyright R. Janow – Spring 2015
Use symmetry arguments where applicable
4-5: Outside a sphere of charge the electric field is just like that of a
point charge of the same total charge, located at its center.
Outside of an infinitely long, uniformly charged conducting
cylinder, which statement describes the electric field?
The charge per unit area is s.
A.
B.
C.
D.
E.
Like that of a point charge at the center of the cylinder.
Like a circular ring of charge at its center.
Like an infinite line of charge along the cylinder axis.
Cannot tell from the information given.
The field equals zero

 Q
F   E  dA  enc
0
S
s
Copyright R. Janow – Spring 2015
Use Gauss’ Law to find the Electric field near an infinite
non-conducting sheet of charge
qenc  sA cap
• Cylindrical gaussian surface penetrating sheet
(rectangular OK too)
• Uniform positive charge per unit area s
• E has constant value as chosen point moves
parallel to the surface
• E points radially away from the sheet (both sides),
• E is perpendicular to cylindrical part of gaussian
surface. Flux through it = 0
• On both end caps E is parallel to A
so F = + EAcap on each


qenc sAcap
F   E  dA  2EAcap 

0
0
S
s
\ E
2 0
• Uniform field – independent of distance from sheet
• Same as near field result for charged disk but no integration needed
• Good approximation for finite sheet when r << L, far from edges.
Copyright R. Janow – Spring 2015
Electric field near an infinite conducting sheet of
charge – using Gauss’ Law
qenc  sA cap
• Uniform charge per unit area s on one face
• Second face is immersed in the metal
• Use cylindrical or rectangular Gaussian surface
• End caps just outside and just inside (E = 0)
• E points radially away from sheet outside
otherwise surface current would flow (!!)
• Flux through the cylindrical tube = 0
E normal to surface
• On left cap (inside conductor) E = 0 so F = 0
• On right cap E is parallel to A so F = EAcap
\ F  EAcap
s
r
qenc sA cap


0
0
\ E
s
0
F= 0
• Field is twice that for a non-conducting sheet with same s
• Same enclosed charge, same total flux now “squeezed” out
the right hand cap, not both
• Otherwise like previous result: uniform, no r Copyright
dependence,
R. Janow –etc.
Spring 2015
F=EA
Example: Fields near parallel nonconducting sheets - 1
•
•
•
•
•
Two “large” nonconducting sheets of charge are near each other,
Use infinite sheet results
The charge cannot move, use superposition.
There is no screening, as there would be in a conductor.
s
Each sheet produces a uniform field of magnitude: E 
0
2 0
Oppositely Charged Plates, same |s|
s
-s
+
E0
 +
E0
+

Etot  0
•
•
•
+ 
Etot  2 E0 î
-
-
E0
E0

Etot  0
Left region: Field due to the positively charged sheet is canceled by the field
due to the negatively charged sheet. Etot is zero.
Right region: Same argument. Etot is zero.
Between plates: Fields reinforce. Etot is twice E0 and to the right.
Copyright R. Janow – Spring 2015
Example: Fields near parallel nonconducting sheets - 2
Positively Charged Plates, same s
+
+
+
 +
E0
+
+

Etot  -2 E0 î
•
•
•
s
2 0
s
s
E0
E0 
E0
E0
+
+
Etot  0

Etot  2 E0 î
Now, the fields reinforce to the left and to the right of both plates.
Between plates, the fields cancel.
Signs are reversed for a pair of negatively charged plates
Copyright R. Janow – Spring 2015
Charge on a finite sized conducting plate in isolation
L
S’
S
s1
E
0
or
L
S
• E = 0 inside the conductor
• Charge density s1 becomes the same on
both faces, o/w charges will move to
make E = 0 inside
• Cylindrical Gaussian surfaces S, S’
• S  charges distribute on surfaces
S’
• Same field magnitude on opposite
sides of each conductor, opposite directions
• Same field by replacing each conductor
with 2 charged non-conducting sheets alone:
• same s1
• cancellation inside conductor
• reinforcement outside
Copyright R. Janow – Spring 2015
Proof of the Shell Theorem using Gauss’ Law
(spherical symmetry)
• Hollow shell of charge, net charge Q
• Spherically symmetric surface charge
density s, radius R
• Two spherical Gaussian surfaces:
- S1 is just inside shell, r1 < R
- S2 is outside shell, r2 > R
• qenc means charge enclosed by S1 or S2
OUTSIDE:
(on S2)
R
r2
r1
S1


q
Q
F 2   E  dA  4r22E2  enc 
0
0
S
2
\ E2 
INSIDE:
(on S1)
S2
Q
4 0r22
for r2  R
Shell of charge acts like a
point charge at the center
of the sphere


qenc
2
F1   E  dA  4r1 E1 
0
0
S
1
\ E1  0
for r1  R
Shell of charge creates
zero field inside (anywhere)
At point r inside a spherically symmetric volume charge distribution (radius R):
• only the shells of charge with radii smaller than r contribute as point charges
• shells with radius between r and R produce zero
fieldR. inside.
Copyright
Janow – Spring 2015