Transcript 幻灯片 1 - SJTU

Chapter 17 Gauss’ Law
Main Points of Chapter 17
• Electric flux
• Gauss’ law
• Using Gauss’ law to determine electric fields
17-1 What Does Gauss’ Law Do?
Consider the electric field lines associated with
a point charge Q
How many field lines
pass through the tissue
forming a sphere？
All the lines (12).
If the sphere is displaced ?
All the lines (12).
Suppose the tissue is some shape
other than spherical, but still
surrounds the charge.
All the field lines still go through.
Now, imagine the paper is crinkled
and overlaps itself; how shall we
deal with the lines that pierce the
tissue three times?
Notice that they go out twice and in once – if we
subtract the “ins” from the “outs” we are left with
one line going out, which is consistent with the
other situations.
Like how many hairs passing through your scalp.
In the electric field, the number of electric field lines
through a curved surface A is defined as electric flux.
The number of field lines penetrating a surface is
proportional to the surface area, the orientation,
and the field strength.
1. Electric flux for uniform electric field
We can define the electric flux
through an infinitesimal area:
En

if
then
is normal to the surface
A finite plane uniform electric field
2. General definition of electric flux

dA
3. Electric flux through a closed surface
The direction of
unclosed surface: choose one side
Outward: positive
Closed surface:
Inward:
negative

E
The electric flux through a closed surface is the sum
of the electric field lines penetrating the surface.
If the electric flux is positive, there are net electric
field lines “out of” the closed surface.
If the electric flux is negative , there are net electric
field lines “into” the closed surface.
The closed surface does not have to be made of real
matter. The imaginary surface is a Gaussian surface
 Act Imagine a cube of side a positioned in a region
of constant electric field as shown below. Which of
the following statements about the net electric flux
Fe through the surface of this cube is true?
(a) FE = 0
(b) FE  2a2
(c) FE  6a2
a
a
 Act Consider 2 spheres (of radius R and 2R)
drawn around a single charge as shown.
– Which of the following statements about the
net electric flux through the 2 surfaces (F2R
and FR) is true?
(a) FR < F2R
(b) FR = F2R
(c) FR > F2R
•Look at the lines going out through
each circle -- each circle has the
same number of lines.
Example: Calculate the electric flux of a constant
electric field through a hemispherical surface of radius
R whose circular base is perpendicular to the direction
of the field.
Solution
Consider an infinitesimal 
strip at latitude 
E
dl

n
r
R

r  R cos

Alternative Solution
Consider a closed surface that
consists of the hemisphere and
the planar circle

E
A1
A2
R
Act A nonuniform electric field given by
What
is the electric flux through the surface of the Gaussian
cube shown in figure below ?
 Act a point charge q is placed at the center of a sphere
of radius of R. Calculate the electric flux through a
circular plane of radius r as shown below.
R
q
r
S S’
the electric flux of the closed sphere
the electric flux of any closed surface t surrounding the q?

E
17-2 Gauss’ Law
1. for a point charge + q
•Suppose a point charge q is at the
center of a sphere，the electric
flux of a spherical surface

dA
q
r
But the result would be the same if the surface was
not spherical, or if the charge was anywhere inside it!
•Suppose a point charge q is out
of an arbitrary closed surface
+q
For the field of point charge, the total electric flux
through a closed surface is only related to the
total (net) electric charge inside the surface.
2 for multiple point charges
Fe of an arbitrary closed surface
qj
qn
qi
q2
q1
3 for continuous charge distributions
We can quickly generalize this to any surface and
any charge distribution;
Gauss’ law
The total electric flux through a closed surface is
equal to the total (net) electric charge inside the
surface, divided by  0 .
Gauss’s law provides a relationship between the
electric field on a closed surface and the charge
distribution within that surface.
Discussion
Electric field Lines leave (+) charges
with source
Electric field Lines return to (-) charges with sink
without source or sink
Electric field Lines never discontinue in empty space
is the electric field at each point on the surface
produced by the total electric charge，Fe
is connected to the algebraic sum of charges
enclosed by the surface.
 ACT If a closed surface surrounds a dipole, the net
flux through the surface is zero.
TRUE
 ACT If the net flux through a closed surface is zero,
then there can be no charge or charges within that
surface.
FALSE
 Act A full Gaussian surface encloses two of the four
positively charged particles. a) Which of the particles
contribute to the electric field at point P on the surface.
b) Determine the electric flux through the surface.
c) If I change the position of q4 outside the surface,
Does Ep change? How about the electric flux ?
q4
+
q3
+
a) The four particles
+ q1
P
+ q2
c) Ep changes
doesn’t change
 Act Consider a point charge q placed in the center
of a cube. Determine the electric flux through each face
of the cube .
q
If the charge is placed at a corner of the cube, what is
the electric flux through surface S
Put the charge in the middle
of a larger cube
q
S
 Example There is an electric field near Earth’s
surface of about 100N/C that points vertically down.
Assume this field is constant around Earth and that
it is due to charge evenly spread on Earth’s surface.
What is the total charge on the Earth?
Solution
choose a concentric sphere of
radius R for Gaussian surface
the total charge
4. Coulomb’s Law and Gauss’ Law
Coulomb’s Law + Superposition of electric field
Gauss’ Law
Now we use Gauss’ Law to get the distribution
of electric field for a point charge.
Draw a concentric spherical Gaussian surface of
radius r
q
Although E varies radially with distance from q,
it has the same value everywhere on the spherical surfa
Gauss’ law and Coulomb’s law , although
expressed in different forms, are equivalent ways
of describing the relation between electric charge
and electric field in static situation.
Gauss’ law is more general in that it is always
valid. It is one of the fundamental laws of
electromagnetism.
17-3 Using Gauss’ Law to determine Electric Fields
Gauss’s law is valid for any distribution of charges
and for any closed surface. However, we can only
calculate the field for several highly symmetric
distributions of charge.
The steps of calculating the magnitude of the electric
field using Gauss’ law
(1) Identify the symmetry of the charge distribution
and the electric field it produces.
*(2) Choose a Gaussian surface that is matched to the
symmetry – that is, the electric field is either parallel
to the surface or constant and perpendicular to it.
(3) Calculate the algebraic sum of the charge enclosed
by the Gaussian surface.
(4) Find the electric field using Gauss’s law.
Example: Determine the electric field of an infinite
uniformly charged line. The charge density per unit
length:+
Solution
Solution
we use as our Gaussian surface
a coaxial cylinder surface with
height l

E

dA
r

E
l
E
r
• Could we use Gauss’ law to find the field of
a finite line of charge?
• What is the electric field of an infinite uniformly
charged thin cylindrical shell?
• What is the electric field of an infinite uniformly
charged cylinder?
Act Two long thin cylindrical shells of radius r1
and r2 are oriented coaxially. The charge density
are
. What is the electric field ?
Solution
Example: Determine the electric field both inside
and outside a thin uniformly charged spherical shell
of radius R that has a total charge Q.
+
Solution
+
+ P
take a concentric sphere of
radius r for Gaussian surface
rr
R
+
O
+
+
From Gauss’ law
E
E
E 0
• outside the sphere ( r > R )
1
r2
R
O
r
• inside the sphere( r < R )
• on surface ( r = R ) E= ？
ACT A uniform spherical shell of charge of radius R
surrounds a point charge at its center. The point
charge has value Q and the shell has total charge -Q.
The electric field at a distance R/2 from the center
________
A) does not depend on the charge of the spherical
shell.
B) is directed inward if Q>0.
C) is zero.
D) is half of what it would be if only the point charge
were present.
E) cannot be determined
Act: A nonconducting spherical shell of radius R is
uniformly charged with total charge of Q. We put
point charges q1and q2 inside and outside spherical
shell respectively . Determine the electric forces on the
point charges.
• inside the sphere( r < R )
O
q1
r1
• outside the sphere ( r > R )
right
Two shells law
q2
r2
Example: Find the electric field outside and inside a solid,
nonconducting sphere of radius R that contains a total
charge Q uniformly distributed throughout its volume.
Solution take a concentric sphere of
radius r for Gaussian surface
r
+
+r +
+ + +
+
R +
+
• outside the sphere ( r > R )
E
• inside the sphere ( r < R )
O
1
E 2
r
R
r
ACT Figure below shows four 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.
a, b, c, d
a and b tie, c, d
Example: A uniformly charged nonconducting sphere
has volume charge density r. Material is removed
from the sphere leaving a spherical cavity. Show that
the electric field through the cavity is uniform and is
given by
where a is the distance of the centers
of the two spheres
Proof: Replace the sphere-with-cavity with two
uniform spheres of equal positive and
negative charge densities.

Consider a point P in the cavity
r1 
Pr
The e-Field produced by +r sphere at p
2
+r
-r
The e-Field produced by -r sphere at p
 ACT A thin nonconducting uniformly charged
spherical shell of radius R has a total charge of Q . A
small circular plug is removed from the surface. (a)
What is the magnitude and direction of the electric
field at the center of the hole. (b) The plug is put back
in the hole. Using the result of part (a), calculate the
force acting on the plug.
Solution (a) Replace the spherical shellwith-hole with a perfect
spherical shell and a circle with
equal negative surface charge
densities.
R
Q
(b)
dq
R
Q
The “electrostatic pressure”( force per unit area)
tending to expand the sphere
Example: Find the electric field outside an infinite,
nonconducting plane of charge with uniform charge
density s
Solution To take advantage of the symmetry
properties, we use as our Gaussian surface a
cylinder with its axis perpendicular to the
sheet of charge, with ends of
area S. The charged sheet passes
through the middle of the cylinder’s 
n
length, so the cylinder ends are

equidistant from the sheet.
E

n

n

E
Ex
O
x
 ACT Nowhere inside (between the planes) an odd
number of parallel planes of the same nonzero charge
density is the electric field zero.
TRUE
 Act Consider two infinite parallel charged plates
with surface charge density of s1and s2respectively.
What is the electric field in the three regions.
I
II
III
x
Example: Find the electric field of a slab of nonconducting
material forming an infinite plane .It has thickness d and carries a
uniform positive charge density r
r
Solution The Gaussian surface
is shown in the figure
Outside the slab
S
d
x
Inside the slab
Ex
S
O
d
x
*Find the distribution of charge according to the
electric field （Differential form of Gauss’ Law ）
Gauss’ law
We now apply Gauss' divergence theorem
If a closed surface S encloses a volume V then the
surface integral of any vector over S is equal to the
volume integral of the divergence of over V
As this is true for any closed surface the result
requires that at every point in space
This is Gauss' law for electric fields in differential
form. It is the first of Maxwell's equations.
The formulas for the divergence of a vector
Cartesian coordinates
Spherical coordinates
Cylindrical coordinates
 Example The electric field of an atom is
Find the distribution of charge of this atom
Solution
Atom is neutral
Alternative Solution
The charge density should be r (r).
We take a concentric spheres for
Gaussian surface
r+dr
r
Summary of Chapter 17
• Electric flux due to field intersecting a surface S:
• Gauss’ law relates flux through a closed surface to
charge enclosed:
• Can use Gauss’ law to find electric field in situations
with a high degree of symmetry