Title: Chapter 17 Gauss
1Chapter 17 Gauss Law
2Main Points of Chapter 17
- Electric flux
- Gauss law
- Using Gauss law to determine electric fields
317-1 What Does Gauss Law Do?
Consider the electric field lines associated with
a point charge Q
All the lines (12).
All the lines (12).
4Suppose the tissue is some shape other than
spherical, but still surrounds the charge. All
the field lines still go through.
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.
5Like 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.
61. Electric flux for uniform electric field
En
We can define the electric flux through an
infinitesimal area
if
then
A finite plane uniform electric field
72. General definition of electric flux
3. Electric flux through a closed surface
The direction of
unclosed surface choose one side
Outward positive
Closed surface
Inward negative
8The 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
9(No Transcript)
10- 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
a
11- 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?
- Look at the lines going out through
- each circle -- each circle has the
- same number of lines.
12 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 q
13Alternative Solution
Consider a closed surface that consists of the
hemisphere and the planar circle
A1
R
14- Act A nonuniform electric field given by
What is the electric flux through the surface
of the Gaussian cube shown in figure below ?
15- 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
S
S
the electric flux of the closed sphere
the electric flux of any closed surface t
surrounding the q?
1617-2 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
But the result would be the same if the surface
was not spherical, or if the charge was anywhere
inside it!
17- 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.
182 for multiple point charges
qj
?e of an arbitrary closed surface
qn
193 for continuous charge distributions
We can quickly generalize this to any surface and
any charge distribution
Gauss law
Gausss law provides a relationship between the
electric field on a closed surface and the charge
distribution within that surface.
20Discussion
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
21- 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
22- 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 ?
a) The four particles
P
c) Ep changes
23- Act Consider a point charge q placed in the
center - of a cube. Determine the electric flux through
each face - of the cube .
q
Put the charge in the middle of a larger cube
24- Example There is an electric field near Earths
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
Earths surface. What is the total charge on the
Earth?
Solution
choose a concentric sphere of radius R for
Gaussian surface
the total charge
254. Coulombs Law and Gauss Law
Coulombs Law Superposition of electric field
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
Although E varies radially with distance from q,
it has the same value everywhere on the
spherical surface
26Gauss law and Coulombs 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.
2717-3 Using Gauss Law to determine Electric Fields
Gausss 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.
28The steps of calculating the magnitude of the
electric field using Gauss law
- 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 Gausss law.
29- Example Determine the electric field of an
infinite - uniformly charged line. The charge density
per unit - length?
Solution
30Solution
we use as our Gaussian surface a coaxial
cylinder surface with height l
E
r
31- 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?
32Solution
33- 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
take a concentric sphere of radius r for
Gaussian surface
r
r
R
O
From Gauss law
- outside the sphere ( r gt R )
R
- inside the sphere( r lt R )
34- 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 Qgt0.
- C) is zero.
- D) is half of what it would be if only the point
charge were present. - E) cannot be determined
35- 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.
q1
q2
- inside the sphere( r lt R )
O
r1
r2
- outside the sphere ( r gt R )
right
Two shells law
36- 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
r
take a concentric sphere of radius r for
Gaussian surface
r
R
- outside the sphere ( r gt R )
E
- inside the sphere ( r lt R )
R
O
r
37- 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
38Proof
Replace the sphere-with-cavity with two uniform
spheres of equal positive and negative charge
densities.
Consider a point P in the cavity
The e-Field produced by r sphere at p
r
-r
The e-Field produced by -r sphere at p
39- 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 shell-with-hole with a
perfect spherical shell and a circle with equal
negative surface charge densities.
R
Q
40(b)
R
Q
The electrostatic pressure( force per unit
area) tending to expand the sphere
41- 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
cylinders length, so the cylinder ends are
equidistant from the sheet.
42- ACT Nowhere inside (between the planes) an odd
number of parallel planes of the same nonzero
charge density is the electric field zero.
TRUE
43- Act Consider two infinite parallel charged
plates - with surface charge density of s1and
s2respectively. - What is the electric field in the three
regions.
III
I
II
x
44- 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
The Gaussian surface is shown in
the figure
Solution
Outside the slab
S
Inside the slab
S
45Find the distribution of charge according to the
electric field (Differential form of Gauss
Law )
Gauss law
We now apply Gauss' divergence theorem
46As 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.
47The formulas for the divergence of a vector
Cartesian coordinates
Spherical coordinates
Cylindrical coordinates
48- Example The electric field of an atom is
Find the distribution of charge of this atom
Solution
Atom is neutral
49Alternative Solution
The charge density should be r (r).
We take a concentric spheres for Gaussian surface
50Summary 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