Download Electricity and Magnetism Gauss`s Law Electric Potential Energy

Electricity and Magnetism
Gauss’s Law
Electric Potential Energy
Lana Sheridan
De Anza College
Oct 6, 2015
Last time
• motion of charges in electric fields
Overview
• electric flux
• Gauss’s law
• implications of Gauss’s law
• Coulomb’s law from Gauss’s law
• some rules to help solve problems
:
ectric field E of the
Gauss’s
1, we
consideredLaw
the basic idea
. Then we simplified
ndicularGauss’s
components
law relates the electric field across a closed surface (eg. a
sphere) to the amount of net charge enclosed by the surface.
n save far more work
ematician and physi:
ing the fields dE of
nsiders a hypothetical
his Gaussian surface,
es our calculations of
e distribution. For exose the sphere with a
hen, as we discuss in
ct that
sian surface to the
Spherical
Gaussian
surface
?
E
c field on a Gaussian
Fig. 23-1 A spherical Gaussian
As a limited example,
surface. If the electric field vectors
Can from
we quantify
the
“electric field across a boundary”?
ally outward
the
are of uniform magnitude and point
mediately tells us that
radially outward at all surface points,
Flux
Flux is a quantity that makes the idea of the “electric field
through some region” precise.
GAUSS’ LAW
Flux is a flow rate through an area.
stream of vee plane of a
component
e of the loop
between :
v
The area veclane of the
th :
v . (d) The
he area of the
Air flow
v
θ
v
Flux
ImagineLAW
air blowing directly through a square loop of wire of area
GAUSS’
A.
irstream of veAir flow
he plane of a
e component
v
ne of the loop
θ
e between :
v
) The area vecplane of the
ith :
v . (d) The
the area
the of air that passes through in 1 s is V = A × v × (1 s),
Theofvolume
where v is the speed of the air.
(a )
The rate of flow would be V /t = Av .
(b )
v
Flux
Now consider a more general situation: the air does not blow
directly through the loop, but at some angle θ.
If θ = 90◦ , what is the flow rate (flux) through the loop?
Flux
Now consider a more general situation: the air does not blow
directly through the loop, but at some angle θ.
If θ = 90◦ , what is the flow rate (flux) through the loop? Zero!
In that case there is no flow through the loop. The air goes around
the loop.
The flux depends on the angle that the flow makes to the loop /
area.
726
Chapter 24
Gauss’s Law
Flux
The number of field lines that
go through the area A› is the
same as the number that go
through area A.
Normal
A
u
w› u
A›
S
w
,
E
From th
coulomb
lines pe
If the
through
the norm
that the
cross th
lar to th
A 5 ,w.
by w ! 5
areas ar
06
Flux
HALLIDAY REVISED
The area A⊥ = A cos θ.
For other values of θ the flux of air that move through is vA cos θ.
low
v
θ
v
v
(a )
(b )
A
θ
(c )
(d )
We can define flux:
Before we discuss a flux involved in electrostatics, we need to rewrite Eq.
:
1 in terms of vectors. To do this, we first define an area vector A as being a
Φ
=
vA
cos
θ
tor whose magnitude is equal to an area (here the area of the loop) and whose
ection is normal to the plane of the area (Fig. 23-2c). We then rewrite Eq. 23-1
he scalar (or dot) product of the velocity vector :
v of the airstream and the area
Electric Flux
The electric flux, ΦE , through an area A is
ΦE = EA cos θ
where θ is the angle between the electric field vector at the surface
and the normal vector to the surface.
This can be written:
ΦE = E · A
The direction of A is ⊥ to the surface, and the magnitude is the
area of the surface.
Gaussian Surface
Gaussian surface
An imaginary boundary (close surface) drawn around some region
of space in order to study electric charge and field.
The surface can be any shape you like!
It is just a tool for calculating charge or field.
as the scalar (or dot) product of the velocity vector :
v of the airstream and t
:
vector A of the loop:
Electric Flux
:
# $ vA cos & $ :
v ! A,
:
whereau isGaussian
the angle between
v and A
The electric flux ΦE through
surface
is. proportional to
The word “flux” comes from the Latin word meaning “to flow.” That m
the net number of electricmakes
fieldsense
lines
through
that through
surface.
if wepassing
talk about the
flow of air volume
the loop. Howev
:
23-2 can be regarded in a more abstract way. To see this different way, note
can assign a velocity vector to each point in the airstream passing through t
Consider
small
areas ofofallthe
(Fig.
23-2d). Because
the composite
thoseGaussian
vectors is a velocity field, we
terpret Eq. 23-2 as giving the flux of the velocity field through the loop. With
surface
∆A.
terpretation, flux no longer means the actual flow of something through an
rather it means the product of an area and the field across that area.
Gaussian
surface
Total flux through the surface:
1
23-3 Flux of an Electric Field
3
2
∆A
θ
∆A
E
θ
1
Φ<0
Pierce
inward:
negative
flux
E
3
Φ>0
E
2
∆A
Φ=0
Pierce
outward:
positive
flux
Skim: zero flux
To define the flux of an electric
field, consider Fig. 23-3, which shows an ar
X
(asymmetric) Gaussian surface immersed in a nonuniform electric field.
Φ
=
E · (∆A)
E
divide the surface into small squares of area !A, each square being small
to permit us to neglect any curvature and to consider the individual squar
:
flat. We represent each such element of area with an area vector !A, whos
:
nitude is the area !A. Each vector !A is perpendicular to the Gaussian
just the sum of all the flux through each
and directed away from the interior of the surface.
Because
the squares have been taken to be arbitrarily small, the electr
little
area.
:
:
E may be taken as constant over any given square. The vectors !A and
each square then make some angle u with each other. Figure 23-3 sh
enlarged view
surface
of three squares on the Gaussian
and the angle u fo
I
A provisional definition for the flux of the electric field for the G
Formally,
ΦE = E · dA
surface of Fig. 23-3
is
# $ ! % ! !".
:
:
This equation instructs us to visit each square on the Gaussian surface, evalu
:
:
:
:
The word “flux” comes from the Latin word mean
makes sense if we talk about the flow of air volume thro
23-2 can be regarded in a more abstract way. To see this
can assign a velocity vector to each point in the airstrea
(Fig. 23-2d). Because the composite of all those vectors
terpret Eq. 23-2 as giving the flux of the velocity field thr
terpretation, flux no longer means the actual flow of so
• The the
fluxproduct
is positive
rather it means
of an areawhere
and the field acro
Electric Flux through Gaussian Surfaces
Gaussian
surface
1
the field vector point out of
the surface.
23-3 Flux of an Electric Field
3
2
∆A
θ
∆A
E
θ
1
Φ<0
Pierce
inward:
negative
flux
E
3
Φ>0
E
2
∆A
Φ=0
Pierce
outward:
positive
flux
Skim: zero flux
To define the flux of an electric field, consider Fig. 23-3
• TheGaussian
flux is negative
where
(asymmetric)
surface immersed
in a nonun
divide thethe
surface
into
small squares
area !A, each
field
vector
point ofinto
to permit us to neglect any curvature and to consider
the surface.
flat. We represent
each such element of area with an a
:
nitude is the area !A. Each vector !A is perpendicul
andFor
directed
away
from
the
interior
of
the surface.
a closed surface:
Because the squares have been taken to be arbitra
:
E may be taken as constant
I over any given square. T
each square then make some angle u with each ot
Φthree
E · dA
E =squares
enlarged view of
on the Gaussian surfac
A provisional definition for the flux of the elec
surface of Fig. 23-3 is
# $ ! % ! !".
:
:
This equation instructs us to visit each square on the Ga
:
:
:
:
Gauss’s Law
The net flux through a surface is directly proportional to the net
charge enclosed by the surface.
0 ΦE = qenc
We can also write this same thing as an integral:
I
E · dA =
qenc
0
Gauss’s Law
The net flux through a surface is directly proportional to the net
charge enclosed by the surface.
0 Φ = qenc
E
roportional to the net number of
Example: uniform field
z
Let’s draw a Gaussian surface in a uniform field.
Sample Problem
For a cylinder, there is some symmetry: easy to calculate flux!
ugh a closed cylinder, uniform field
form of a
:
tric field E ,
t is the flux
e?
ace by intece.
The
dA
θ
dA
a
θ
Gaussian
surface
E
b
E
E
c
dA
A cylindrical Gaussian surface, closed by end caps, is
immersed in a uniform electric field. The cylinder axis is parallel to
sides
of the cylinder are k to E ⇒ Φ = 0.
the field direction.
Fig. 23-4
ng the flux as
need
consider
theallends.
der cap a,We
the only
' # 0 for
right
cap,to
where
points,
Eq. 23-4,
"
"
2
◦
: 2):
Φ = E (πr
cos(0)
+ E (πr
)=0
E ! dA
# E(cos
0) dA) cos(180
# EA.
c
Finally, for the cylindrical surface, where the angle u is 90° at
Example: uniform field
Sample Problem
The total flux across the boundary is zero!
ugh a closed cylinder, uniform field
form of a
:
tric field E ,
t is the flux
e?
dA
θ
dA
a
θ
Gaussian
surface
E
b
E
E
c
dA
Fig. 23-4 A cylindrical Gaussian surface, closed by end caps, is
ace by inteimmersed in a uniform electric field. The cylinder axis is parallel to
ce.
From Gauss’s Law 0 Φ = qenc , we know:
the field direction.
ng the flux as
der cap a, the
Eq. 23-4,
qenc
=0
right cap, where ' # 0 for all
points,
"
"
:
:
This is always true for
in a uniform electric
E ! any
dA #Gaussian
E(cos 0)surface
dA # EA.
field.
c
(23-5)
Finally, for the cylindrical surface, where the angle u is 90° at
all points,
_c23_605-627v2.qxd
18-11-2009
15:34
Page 612
Gauss’s Law for a Point Charge
HALLIDA
For a point charge, we can imagine a spherical Gaussian surface.
The field will be perpendicular to the surface at every point.
612
CHAPTER 23 GAUSS’ LAW
Φ = Law
4πr Eand Coulomb
23-5 Gauss’
2
Gaussian
surface
Gauss’s
law: law and Coulomb’s law are
Because Gauss’
r
q
E
+
A spherical Gaussian
surface centered on a point charge q.
Fig. 23-8
tion between electric charge and electric
2
0 Φ =
4πrfrom
E =theq other. Here w
able to derive
each
law and some symmetry considerations.
so, Figure 23-8 shows a positive point cha
concentric spherical Gaussian surface of r
1 dA.qBy definition,
kq
differential areas
the are
E=
= 2 outward fr
ular to the
surface
and
directed
2
4π0 r
r
the situation, we know that at any point th
to the as
surface
directed law!
outward from
Same
fromand
Coulomb’s
:
:
between E and dA is zero, we can rewrite E
Question
Imagine a Gaussian surface enclosing a dipole.
What is the net flux through the surface?
Nonconducting sheet of charge
PA R T 3
SS’ LAW: PLANAR SYMMETRY
y
:
E
,
t
e
d
d
e
+
+
+
E+ +
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
617
Again, the sides of the cylinder are k to
E ⇒ Φ = 0.
σ
+
+
We only need to consider the ends.
+
+ Gaussian
+
+
+
+ surface
+
+
+
+
+
A
+E
+
+
+
+
r
+
+
+
+
+
+
+
+
+
+
There is flux only
+
(a) through the
+
+
+
two end faces.
A
E
+
+
+
+
+
+
+
+
+
+
+
(b)
A
E
Fig. 23-15 (a) Perspective view and (b)
side view of a portion of a very large, thin
Φ = E (πr 2 ) cos(0) + E (πr 2 ) cos(0)
= 2πr 2 E
Nonconducting sheet of charge
PA R T 3
SS’ LAW: PLANAR SYMMETRY
y
:
E
,
t
e
d
d
e
+
+
+
E+ +
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
617
Again, the sides of the cylinder are k to
E ⇒ Φ = 0.
σ
+
+
We only need to consider the ends.
+
+ Gaussian
+
+
+
+ surface
+
+
+
+
+
A
+E
+
+
+
+
r
+
+
+
+
+
+
+
+
+
+
There is flux only
+
(a) through the
+
+
+
two end faces.
A
E
+
+
+
+
+
+
+
+
+
+
+
(b)
A
E
Fig. 23-15 (a) Perspective view and (b)
side view of a portion of a very large, thin
Φ = E (πr 2 ) cos(0) + E (πr 2 ) cos(0)
= 2πr 2 E
Then, using Gauss’s law:
0 (2πr 2 E ) = σ(πr 2 )
σ
E =
20
which is what I claimed in the previous
lecture.
w that just outside the
E !Field
s1/"0. Because
the
between
plate.
egative charge having
difference is that now
b to be close to each
uctors, when we bring
ate attracts the excess
onto the inner faces of
on each inner face, the
wice s1. Thus, the elec(23-14)
From Gauss’s
plates
e and toward
thewith
nega-an
uter faces, the electric
– –
+ +
+ +
conducting plates –
–
(a)
(b)
E=0
+
+
+
+
+
+
+
+
+
+
2σ 1
E
–
–
–
–
–
–
–
–
–
–
E=0
(c)
(a) Afind
thin, the
very large
also
field conductbetween
ing plate with excess positive charge. (b) An
airidentical
(or vacuum)
gap
separating
them:
plate with excess negative charge.
(c) The two plates arranged so they are parallel and close.
Fig.we
23-16
Law
can
E=
σ
0
conducting
Some Implications of Gauss’s Law
Rules that make calculating easier!
• If an excess charge is placed on an isolated conductor, that
amount of charge will move entirely to the surface of the
conductor. None of the excess charge will be found within the
body of the conductor.
• A shell of uniform charge attracts or repels a charged particle
that is outside the shell as if all the shell’s charge were
concentrated at the center of the shell.
• If a charged particle is located inside a shell of uniform charge,
there is no electrostatic force on the particle from the shell.
onducting slab placed
Figure 24.16 A conducting
the conductor must be
Some
Implications
Gauss’s
slab inof
an external
electricLaw
field
ere
not
zero,
free
elecS
S
S
E . The charges induced on the
( F 5 qExcess
E ) and Charge
would on a Conductor
two surfaces of the slab produce
ever, would mean that
opposes the
If an excess charge an
is electric
placedfield
on that
an isolated
conductor, that amount
, the existence of elecexternal field, giving a resultant
of charge will movefield
entirely
thethe
surface
of zero to
inside
slab. of the conductor. None
e conductor.
of
the
excess
charge
will
be
found
within
the body of the
re the external field is
conductor.
t the conductor. When
to the left in Figure
Gaussian
surface
n the left surface. The
ve charge on the right
c field inside the conve, the surface charge
magnitude of the interet field of zero inside
equilibrium is on the
instantaneous.
onductor is also zero,
within the conductor.
Figure 24.17 A conductor of
argue with the concept
arbitrary shape. The broken line
E
=
0
inside
the
conductor,
so the Gaussian surface shown cannot
5.6.
represents a gaussian surface
enclose
a net charge.
a conductor
in electrothat can be just inside the conduc-
Charges and Conductors
Excess charge sits on the outside surface of a conductor.
Close to the surface, the electric field lines are perpendicular to the
surface.
1
Figure from OpenStax College Physics.
Some Implications of Gauss’s Law
Uniform Shell of Charge
• A shell of uniform charge attracts or repels a charged particle
that is outside the shell as if all the shell’s charge were
concentrated at the center of the shell.
• If a charged particle is located inside a shell of uniform charge,
there is no electrostatic force on the particle from the shell.
positive
a large, spherical gaussian
is drawnof
concentric
Uniformsurface
Sphere
Charge
eld at a
with the sphere.
r
For points inside the sphere,
a spherical gaussian surface
smaller than the sphere is
drawn.
Gaussian
r
a
rs from
sphere
electric
a
Q
Section
eld due
Gaussian
ield for
sphere
by intea
b
demonChapterConsider a uniform insulating sphere of charge, radius a, charge
Figure 24.10 (Example 24.3) A uniformly charged insulating
ρ, total
charge
d usingdensity
sphere
of radius
a andQ.
total charge Q. In diagrams such as this one,
the dotted line represents the intersection of the gaussian surface
with the plane of the page.
ed uni-How does the electric field strength change with distance from the
ributioncenter?
ss’s law to find the electric field.
surface is drawn concentric
with the sphere.
Uniform Sphere of Charge
he electric field at a
For points inside the sphere,
a spherical gaussian surface
smaller than the sphere is
drawn.
r
roblem differs from
s’s law. The electric
iscussed in Section
e electric field due
found the field for
Chapter 23 by intehis example demoncussions in Chapter
electric field using
a
r
Q
a
a
I
Gaussian
sphere
Gaussian
sphere
b
qenc
· dA
=
Figure 24.10 (ExampleE24.3)
A uniformly charged insulating
0
sphere of radius a and total charge Q. In diagrams
such as this one,
the dotted line represents the intersection of the gaussian surface
Inside sphere (for r < a):
1 4 3
2
4πr E =
Q
4πr E =
πr ρ
0
0 3
symmetry, let’s choose a spherical gaussian surface of radius r, concentric with the
S
S
Q
For this choice, condition
on the surface and E ? d A 5 E ρr
dA.
E (2)=is satisfied everywhere
E =
2
continued
4π0 r
30
ke Q
ke Qr
E =
=
2
r
a3
Outside spherewith
(for
r >ofa):
the plane
the page.
is distributed unicharge distribution
1 field.
2 find the electric
n apply Gauss’s law to
Q
Q / 3 pa
r 5 ke 3 r 1 for r , a 2
3 1 1/4pk
e2
Uniform
Sphere
of aCharge
(2) E 5
art (A). It shows that
m that would exist at
he sphere. That is, if
physically impossible.
hed from inside the
he electric field from
a value from the out-
a
E
kQ
E! e 3 r
a
E!
a
keQ
r2
r
Outside the sphere,Figure
the electric
field is the
same as for a point
24.11 (Example
24.3)
plot of E versus
a uniformly
charge, strength Q,A located
at ther for
center
of the sphere.
charged insulating sphere. The
electric
field inside
thein
sphere
Inside the sphere, field
varies
linearly
the distance from
(r , a) varies linearly with r. The
center and all charge
outside
the
distance
r cancels out!
field outside the sphere (r . a) is
the
Question 3.
Fig. 23-21
8 Figure 23-25 shows four solid spheres, each with charge
Q uniformly distributed through its volume. (a) Rank the spheres
according to their volume charge density, greatest first. The figure
also shows a point P for each sphere, all at the same distance
from the center of the sphere. (b) Rank the spheres according to
the magnitude of the electric field they produce at point P, greatest first.
Question
Page 621, #8
+q
Thea figure
shows four solid spheres, each with charge Q uniformly
c
b
distributed
through its volume.
d
P
Fig. 23-22
+
+
+
P
P
P
Question 4.
+
+
σ(+)
+
+
(a)
e
(b)
(c)
Fig. 23-25
(d)
Question 8.
Rank
according to their volume charge density,
–
– the
– spheres
–
σ
A small
charged
ball shows
lies withina the
hollowPof for
a metallic
greatest
first. 9The
figure
also
point
each sphere, all
spherical shell of radius R. For three situations, the net charges on
Fig. 23-23 Question 5.
at the same distance
the center
the ball and from
shell, respectively,
are (1) of
"4q,the
0; (2) sphere.
$6q, "10q; (3)
$)
4s
s
5s
–
–
–
(–)
Separation
"16q, $12q. Rank the situations according to the charge on (a)
inner surface of the shell and (b) the outer surface, most posi(A) a, b, c, d the
tive first.
d
Rank the situations of Question 9 according to the magnitude
(B) 4dd, c, b, a 10
of the electric field (a) halfway through the shell and (b) at a point
9d
2R from the center of the shell, greatest first.
(C) a and b, c, d
* View
(D) a,All
b, c and d
1
Halliday,
Resnik, Walker
olutions
Here
**
Question 3.
Fig. 23-21
8 Figure 23-25 shows four solid spheres, each with charge
Q uniformly distributed through its volume. (a) Rank the spheres
according to their volume charge density, greatest first. The figure
also shows a point P for each sphere, all at the same distance
from the center of the sphere. (b) Rank the spheres according to
the magnitude of the electric field they produce at point P, greatest first.
Question
Page 621, #8
+q
Thea figure
shows four solid spheres, each with charge Q uniformly
c
b
distributed
through its volume.
d
P
Fig. 23-22
+
+
+
P
P
P
Question 4.
+
+
σ(+)
+
+
(a)
e
(b)
(c)
Fig. 23-25
(d)
Question 8.
Rank
according to their volume charge density,
–
– the
– spheres
–
σ
A small
charged
ball shows
lies withina the
hollowPof for
a metallic
greatest
first. 9The
figure
also
point
each sphere, all
spherical shell of radius R. For three situations, the net charges on
Fig. 23-23 Question 5.
at the same distance
the center
the ball and from
shell, respectively,
are (1) of
"4q,the
0; (2) sphere.
$6q, "10q; (3)
$)
4s
s
5s
–
–
–
(–)
Separation
"16q, $12q. Rank the situations according to the charge on (a)
surface of the shell and (b) the outer surface, most posi(A) a, b, c, d the
←inner
tive
first.
d
Rank the situations of Question 9 according to the magnitude
(B) 4dd, c, b, a 10
of the electric field (a) halfway through the shell and (b) at a point
9d
2R from the center of the shell, greatest first.
(C) a and b, c, d
* View
(D) a,All
b, c and d
1
Halliday,
Resnik, Walker
olutions
Here
**
Question 3.
Fig. 23-21
8 Figure 23-25 shows four solid spheres, each with charge
Q uniformly distributed through its volume. (a) Rank the spheres
according to their volume charge density, greatest first. The figure
also shows a point P for each sphere, all at the same distance
from the center of the sphere. (b) Rank the spheres according to
the magnitude of the electric field they produce at point P, greatest first.
Question
Page 621, #8
+q
The figure shows four solid spheres, each with charge Q uniformly
a
c
b
distributed
through its volume.
d
P
Fig. 23-22
+
+
+
+
+
σ(+)
+
+
(a)
e
–
–
P
P
P
Question 4.
(b)
(c)
Fig. 23-25
(d)
Question 8.
–
– the
– spheres
–
Rank
according to the magnitude of the electric field
σ
9 A small charged ball lies within the hollow of a metallic
they produce at
point P, greatest first.
spherical shell of radius R. For three situations, the net charges on
–
(–)
Fig. 23-23
$)
4s
s
5s
Question 5.
the ball and shell, respectively, are (1) "4q, 0; (2) $6q, "10q; (3)
$12q. Rank the situations according to the charge on (a)
(A)
a, b, c, d "16q,
Separation
the inner surface of the shell and (b) the outer surface, most positive first.
(B) dd, c, b, a 10
4d
9d
Rank the situations of Question 9 according to the magnitude
of the electric field (a) halfway through the shell and (b) at a point
2R from the center of the shell, greatest first.
(C) a and b, c, d
(D) a,All
b, c and d
* View
1
Halliday, Resnik, Walker
Question 3.
Fig. 23-21
8 Figure 23-25 shows four solid spheres, each with charge
Q uniformly distributed through its volume. (a) Rank the spheres
according to their volume charge density, greatest first. The figure
also shows a point P for each sphere, all at the same distance
from the center of the sphere. (b) Rank the spheres according to
the magnitude of the electric field they produce at point P, greatest first.
Question
Page 621, #8
+q
The figure shows four solid spheres, each with charge Q uniformly
a
c
b
distributed
through its volume.
d
P
Fig. 23-22
+
+
+
P
+
+
σ(+)
+
+
(a)
e
–
–
P
P
Question 4.
(b)
(c)
Fig. 23-25
(d)
Question 8.
–
– the
– spheres
–
Rank
according to the magnitude of the electric field
σ
9 A small charged ball lies within the hollow of a metallic
they produce at
point P, greatest first.
spherical shell of radius R. For three situations, the net charges on
–
(–)
Fig. 23-23
$)
4s
s
5s
Question 5.
the ball and shell, respectively, are (1) "4q, 0; (2) $6q, "10q; (3)
$12q. Rank the situations according to the charge on (a)
(A)
a, b, c, d "16q,
Separation
the inner surface of the shell and (b) the outer surface, most positive first.
(B) dd, c, b, a 10
4d
9d
Rank the situations of Question 9 according to the magnitude
of the electric field (a) halfway through the shell and (b) at a point
2R from the center of the shell, greatest first.
(C) a and b, c, d
←
(D) a,All
b, c and d
* View
1
Halliday, Resnik, Walker
Some Implications of Gauss’s Law23-7 APPLYING GAUSS’ L
Faraday Ice Pail
charge of &5.0
outer wall. The
_
_
gested by Fig. 2
_ + +
_
+
uniform becau
+
R
_
_
skewed distribu
+
+
+
not produce an
_+
_
R/2
ution of charge
+
+_
_ +
tive charges rep
+ _
_+
_
The field li
approximately
the shell and t
(b )
(a )
shell the patte
A
charge
placed
inside
a
conducting
shell
appears
on
the
outside
skew ofofthe po
Fig. 23-11 (a) A negative point charge is located within a
the
conductor.
the pattern is th
spherical metal shell that is electrically neutral. (b) As a result,
and the shell w
positive charge is nonuniformly distributed on the inner wall
= 0and
foran
the
Gaussian
shown.)
of the(E
shell,
equal
amountsurface
of negative
charge is unimatter where in
Gaussian
surface
_
+
Summary
• Electric flux
• Gauss’s law idea
• Gauss’s law implications
• Coulomb’s law from Gauss’s law
• problem solving tricks
• electric potential energy
Homework Halliday, Resnick, Walker:
• Ch 23, onward from page 622. Problems: 1, 57, 63, 65, 69
• Look ahead at Chapter 24, Electric potential.