Download Chapter 3 GAUSS` LAW

Chapter 3 GAUSS’ LAW
• Introduction
• Flux of vector field
• Solid angle
• Gauss’s Law
• Symmetry
• Spherical symmetry
• Cylindrical symmetry
• Plane symmetry
Figure 1 E field at point p due to a charge distribution
(0  0  0 )
• Superposition of symmetric geometries
• Motion of point charges in electric fields
FLUX OF VECTOR FIELD
• Summary
INTRODUCTION
Coulomb’s Law and the Law of Superposition were
introduced in addition to the concept of electric field
defined as:
→
−
−
→
F
E=
(3.1)

The electric field is a vector field just like the gravitational field. It can be expressed in terms of cartesian
coordinates as:
− b
→
E =  + b
 + b

(3.2)
Last lecture focussed on the calculation of the electric
field using Coulomb’s law plus Superposition.
Coulomb’s Law + Superposition
Discrete charges:

−−→ X
E =
=1
1 
c
2 r0
40 0
Vector fields, such as an electric field, magnetic field,
fluid flow, etc, have both a magnitude and direction
that depend on the position. Vector fields can be characterized using two important concepts, namely, flux
and circulation. For example, the rate of loss of water out of a draining bath tub can be related to the
litres/sec flowing down the drain and the circulation
or angular momentum carried away by the flowing water. The concepts of flux and circulation make it possible to express Coulomb’s Law in a form that is more
general.
Flux, Φ is the normal component of the vector
→
−
field, F through any surface. It is given by the integral over the surface.
(3.3)
Continuous charges:
In cartesian coordinates the E field at the point
() is written as an integral over the charge distribution at 0 (0  0  0 ) :
Z
−−−−→
(0  0  0 )0  0  0
1
E() =
(3.4)
rd
0 
40 
20 
This lecture will focus on exploiting Gauss’s law
to calculate electric fields. Gauss’s Law relates the
net flux of the electric field over a closed surface to
the net charge enclosed by the closed surface. This
is of sufficient import to justify a recap of what was
discussed last lecture.
Φ=
Z
 
− −
→
→
F · S
(3.5)
where the scalar product, also called dot product, was
→ −
−
→
defined in P113, namely F ·  S =  cos  where
→
−
 is the angle between the vector F and the normal
−
→
to the surface element S Note that the flux is the
normal component to the surface of the vector field, it
excludes the component along the surface; the surface
component figures in the definition of circulation to be
discussed next lecture. The most interesting case is
when the surface is closed, this is called a Gaussian
surface. In this case the net flux outwards is written
as:
Φ=
I
 
where the symbol
I
− −
→
→
F · S
(3.6)
means a closed surface. By de−
→
finition, for a closed surface the surface vector S is
assumed to point outward, that is Φ is taken outwards.
17
Figure 2 Solid angle subtended by an area element dS
and analog to the angle subtended by a line element dl.
Point enclosed by a closed surface The integral for a closed surface surrounding a point is
SOLID ANGLE
The concept of solid angle should be familiar to you.
It is a measure of the apparent size of a surface area as
seen when viewing from some location. It is analogous
to the two-dimensional problem of the angle subtended
by a line when viewed from a given location. It is
easiest to understand solid angle by comparison with
the concept of angle.
In two dimensions, the infinitesimal angle ∆ subtended by a short line element at a distance r is given
by the projection of the line element  on a concentric circle divided by the radius r of the circle. That

is, ∆ =  sin
The unit of angle is the radian.

The analogous relation for three dimensions gives
the definition of solid angle. Consider a surface area
−
→
element S at a radius r. Then the solid angle ∆Ω is
defined as:
∆Ω =
−
→
S · b
r
 cos 
=
2
2
(3.7)
where  is the angle between the normal to the surface element and unit vector b
r. The solid angle is the
apparent area of the surface element projected onto a
concentric sphere, divided by the square of the radius
of the sphere. The unit of solid angle is the steradian
which is dimensionless. Thus the net solid angle subtended by some surface is given by
Z
Ω =
Z
−
→
b
r · S
2
(3.8)
The sun and the moon happen to subtend almost
identical solid angles at the earth even though there
is a factor of 200 difference in size which is cancelled
by a similar ratio of the distances from the earth. The
solid angle of a closed surface is an important special
case that will be used frequently.
18
Figure 3 Solid angle for a closed surface, a) enclosing
the point, and b) not enclosing the point.
Ω=
I
−
→
b
r · S
= 4
2
(3.9)
since the surface area of a sphere is 42 then a complete sphere must subtend a total solid angle of 4
−
→
steradians. The vector S is always taken to point outwards for a closed surface. Note that any shaped closed
surface subtends 4 steradians relative to a point inside the closed surface, that is, at all angles the closed
surface completely encloses the point independent of
whether the enclosing surface is a sphere, a cube, or
some arbitrary shape.
Point external to closed surface For points
lying outside the closed surface, the net solid angle is
zero. This can be seen by dividing the closed surface
into two halves having the same perimeter as seen from
the point  illustrated in the figure. Both halves subtend the same magnitude solid angle. However, since
−
→
the vector S is always taken to point outwards, then
the cosine has the opposite sign for the two halves and
thus the integral of the solid angle over the external
closed surface is zero since the contributions from the
two halves cancel.
FLUX OF ELECTRIC FIELD FOR AN
ARBITRARY GAUSSIAN SURFACE
ENCLOSING A POINT CHARGE
Since the electric field for a point charge is given by
−
→
E=
1 
b
r
40 2
(3.10)
For an arbitrary closed surface enclosing the point
charge  the net flux is
I
→ −
−
→
Φ =
E · S
→!
I Ã −
b

r · S
=
(3.11)
40
2
But the above discussion of the solid angle gave that
−
→
Z
Z
b
r · S
(3.12)
Ω =
2
This is just the term in brackets, that is
I

Φ=
Ω
40
(3.13)
Also from above we have that for a closed surface enclosing the point charge the total solid angle, in steradians, is given by
−
→
I
I
b
r · S
Ω =
= 4 steradians
(3.14)
2
This is true for any shaped closed surface surrounding
the point. Thus this gives that the net flux of the
electric field due to an enclosed point charge is given
by
I
I
→ −
−
→


Φ=
E · S =
(3.15)
Ω =
40
0
where the 4 factors cancel.
The above argument shows that the net flux out of
any shaped Gaussian surface enclosing a point charge
q is independent of the shape of the Gaussian surface
or the exact location of the point charge within the
enclosed volume. The assumptions made in obtaining
this result are:
a) Point charge
b) Coulomb’s law.
This is a much more powerful statement of Gauss’
Law in that the net flux is independant of the shape of
the closed surface or the location of the enclosed point
charge.
The above proof is consistent with Faraday’s description of the electric field in terms of electric field
lines; that is, the number of field lines intersecting a
concentric sphere is constant independant of the radius of the sphere. Since the number of field lines is
conserved then it can be seen from figure 4 that the
net flux of electric field lines is the same for any closed
arbitrary closed surface enclosing a given point charge
which is what has just been proved mathematically.
GAUSS’ LAW
The above proof that for a point charge the net flux
out of a closed surface enclosing a point charge is independent of the shape of the Gaussian surface, now
Figure 4 A single point charge surrounded by a concentric closed spherical surface and an irregular closed
surface.
can be extended by invoking the Principle of Superposition. Superposition can be used to integrate over the
charge distribution throughout the volume enclosed by
the closed Gaussian surface resulting in the final form
of Gauss’s Law. The Principle of Superposition allows
extension to arbitrary charge distributions. Consider 
charges within the enclosed surface. Each charge produces an  field which add to produce the net electric
field. That is:
−−→ −
→ −
→ −
→
E = E1 + E2 + E3 + 
(3.16)
Now since the flux Φ is just a number, that is, it is a
scalar:
Φ
I
=
 
−−→ −
→
E · S
 
−
→ −
→
E1 · S +
I
=
= Φ1 + Φ2 + Φ3 + 
1
2
3
=
+
+
+ 
0
0 0
I
 
−
→ −
→
E2 · S + 
(3.17)
Thus one has that for an arbitrary distribution of
 charges, the net flux is:
Φ =
I
 
−−→ −
→
E · S =
à 
!
X 

=1 0
(3.18)

Finally, one can write the sum over an arbitrary
continuous charge density distribution  by taking
 =  for the infinitessimal volume   to get the
final form of Gauss’s Law.
Gauss’s Law:
Φ =
I
−−→ −
→
1
E · S =
0
 
Z

(3.19)

19
Figure 5 Closed Gaussian surface enclosing the positive
charge (A), the negative charge (B) and both charges
(C). The net flux out of these surfaces is positive for A,
negative for B, and zero for C.
This relates the net flux out of a closed Gaussian surface to the total charge lying within the enclosed volume.
Note that the assumptions used to derive this are:
• Coulomb’s Law
• Principle of Superposition
Note that the two crucial aspects of Coulomb’s law
that lead to Gauss’ law are that the electric field for a
point charge is:
• exactly proportional to
1
2 ,
• the field is radial.
Gauss’s law is a restatement of Coulomb’s law in a
less transparent but more useful form.
Gauss’ Law actually is one of Maxwell’s four laws
of electromagnetism. Gauss’ Law is completely equivalent to Coulomb’s law for electrostatics or for slowly
moving charges. However, Gauss’ law is more general and applies to electric fields arising from rapidly
moving and accelerating charges. Really one should
derived Coulomb’s Law from Gauss’s Law.
It is interesting to apply Gauss’ Law to the case of
the electric dipole. As seen in figure 5, if the Gaussian
surface encloses only the positive charge then the net
flux out is positive, if the Gaussian surface encloses
only the negative charge then the net flux out is negative, that is the flux is flowing into the surface. If
the Gaussian surface encloses both charges then the
net flux out is zero. However, this does not mean that
there is not electric field. Actually one has flux flowing
inward in some locations and outwards at others, such
that the net total is zero as seen in Figure 5.
20
Figure 6 Spherical shell of charge density 
SYMMETRY
Symmetry is a powerful concept in physics that can
simplify solution of complicated problems. One can
discuss symmetry of systems under rotation, reflection
and time reversal. For example, crystal lattices have
certain spatial symmetries. Coulomb’s law implies a
spherically-symmetric spatial symmetry for the electric field around a point charge. Such very general
symmetry principles are especially powerful when combined with Gauss’ law for calculating electric fields.
For certain charge distributions in physics, one can
use symmetry to identify a particular Gaussian surface
upon which the electric field is uniform and perpendicular to the surface. When such symmetries occur, it is
easy to evaluate the flux integral for the special surface
where the electric field is uniform and perpendicular.
This allows use of Gauss’ Law in order to determine
the electric field for this surface. Three symmetries
will be considered; spherical, cylindrical, and plane.
SPHERICAL SYMMETRY
A) Electric field for a uniformly charged spherical shell
Consider a uniform spherical surface of radius R and
surface charge density  The net charge Q = 42 
From spherical symmetry one can see that the electric
field must be radial and its magnitude can depend only
on r, not angle. Therefore select a spherical Gaussian
surface concentric with the charged shell and apply
Gauss’ law.
Outside the spherical shell:  ≥ 
Gauss’ law gives
Φ=
I

− −
→
→ 
E · S =
0
(3.20)
Figure 7 Concentric Gaussian surface outside a solid
uniform spherical charge distribution.
Figure 8 Concentric spherical Gaussian surface inside a
uniform solid sphere of charge.
Since  is constant and normal to the spherical gaussian
surface of radius r, the surface integral equals 42 .
Thus one obtains
42 =

0
(3.21)
That is:
−
→
E=

b
r
40 2
≥
(3.22)
Note that the E field for a spherical shell of charge of
radius R is the same as for a point charge Q. In fact,
one gets Coulomb’s law by letting R go to zero.
Inside the spherical shell  ≤ 
Since there is no charge within a concentric spherical Gaussian surface inside the charged shell, then the
net flux equals zero from Gauss’ law. From symmetry
therefore we have that the electric field is zero inside
the spherical charged shell.
B) Electric field outside a uniformly charged
solid sphere,  ≥ 
Let the charged sphere, of radius , have a volume
charge density  The net charge  = 43 3 . Again,
from symmetry of figure 7 one can deduce that the
electric field is radial and only depend on radius .
For exactly the same argument given for the spherical shell, the external field is given by:
−
→
E=

b
r
40 2
≥
(3.23)
C) Electric field inside a uniformly charged solid
sphere,  ≤ 
From the arguments for the spherical shell it is obvious that only the charge within the sphere of radius
r contributes to the electric field, the spherical shell
Figure 9
of charge in the region between  and  does not contribute to . Thus one has only to consider the charge
within the sphere of radius , that is, 43 3 . Thus
−
→
E=
4
3
3  
b
r
40 2
≤
(3.24)
Since  = 43 3  then this can be rewritten as:
−
→
E=
1 −
→
r
40 3
≤
(3.25)
Thus the  field is zero at the origin, is proportional to  for 0    , and then falls off as 12
like for a point charge outside of the charged sphere as
shown in figure 9.
In the 1900’s there were two possible models for the
atom. Thomson’s model of the atom was of electrons
distributed in a uniform positive charge distribution
of the size of the atom, that is 10−10  The Rutherford
21
Figure 11 Gaussian surface for infinite line charge 
C/m.
Figure 10 The electric field in the Thomson and
Rutherford models of the atom.
atom envisioned electrons orbiting around a small positive nucleus. These models predict a different E field
at radii  10−10 m as shown in figure 10. In 1911, two
of Rutherford’s students, Geiger and Marsden scattered alpha particles off gold atoms and proved that
the electric field due to the nucleus continued to fall
off as 12 down to the size of the nucleus of 10−15 
confirming the Rutherford model of the atom.
The gravitational field of the earth has the same
radial form as the case of a uniform spherical charge.
The only difference between the gravitational and electrostatic cases is that 10 is replaced by −4 and the
mass density replaces the charge density  This leads
to the conclusion that inside the earth, the radial dependence of the gravitational field is proportional to ,
while outside the earth it has a 12 dependence. Inside
the earth Hooke’s Law applies since the restoring force
is proportional to the displacement . If you dropped
a body into a hole through the center of the earth then
the body would execute simple harmonic motion with
a period of 1.4 hours, that is, the length of this lecture.
Figure 12 Concentric Gaussian cylinder inside a cylindrical shell of charge
Φ=
I
→
− −
→

E · S = 2r  =
0

(3.26)
This gives that the E field is:
−
→
E=

b
r
2 
(3.27)
This is the same answer as obtained previously by
integration of Coulomb’s law.
B) Thin cylindrical shell of charge of radius 
CYLINDRICAL SYMMETRY
A) Electric Field for an infinite line charge
Coulomb’s law was used to derive the electric field due
to a line of charge density  Coulombs/m. This problem has cylindrical symmetry, that is the electric field
must radiate normal to the line charge.
Note that a concentric cylinder of length L and
cylindrical radius r perpendicular to the line charge,
shown in figure 11, contains a charge L. The E field
must be uniform and perpendicular to the cylinder and
parallel to the ends of the cylinder. Thus the flux
through the ends of the cylinder is zero. Using Gauss’s
law we have the net flux through the cylinder is
22
For    then the field is identical to the answer
above.
For    then there is no charge inside the concentric cylindrical Gaussian surface shown in figure 12,
thus the E field is zero.
C) Uniform cylinder of charge of radius 
For    the answer is as above for the infinite line
2
charge. However, for    then only the fraction 
2
of the linear charge density is inside the cylindrical
gaussian surface. This causes the the E field to depend
linearly of  just like the case of the uniform sphere of
charge. i.e. for   
−
→
E=

r̄
2 2
(3.28)
Figure 13 Concentric charged cylindrical shells.
Figure 14 Infinite plane sheet of charge density 
D) Two concentric equal and oppositely charged
cylindrical shells;
I
→
→ −
−

E · S = 2∆ = ∆
(3.31)
Consider that the inner cylindrical shell, of radius a,

 
has a linear charge density of - C/m while the outer
cylindrical shell, of radius b, has a linear charge density That is:
of + C/m as shown in figure 13.
a)   

(3.32)
⊥ =
Obviously E = 0 inside the inner cylindrical shell
2
since there is no charge enclosed by a concentric Gaussian
Note that this is independent of the distance from
surface lying inside with   
the
infinite sheet of charge.
b)     
Between the concentric cylinders, Gauss’s law gives
for a concentric cylindrical Gaussian surface of radius
r perpendicular to the symmetry axis and length l;
SUPERPOSITION OF SYMMETRIC
2 =
−

(3.29)
Thus the electric field between the shells is:
−
→
E =−

b
r
2 
(3.30)
where b is perpendicular to the symmetry axis. Note
that the electric field points inwards since the inner
charge density is negative.
c)   
The radial electric field outside of both cylindrical
shells is zero since the net charge enclosed by a cylindrical Gaussian surface with    is zero if the two
cylinders have equal and opposite line charge densities.
PLANE SYMMETRY
Consider a infinite plane sheet of areal charge density
 C/m2 . By symmetry the electric field must be perpendicular to the sheet charge and equal on both sides.
Consider a small cylindrical Gaussian pill box enclosing an area ∆ of the charged surface shown in Figure
14.
The net flux flowing out of each of the two flat
surface of the pillbox is ∆ and zero flux from the
cylindrical sides. Thus Gauss’s law gives:
GEOMETRIES
Gauss’s law is very powerful for deriving the electric
field for those few case where a symmetry exists such
that a Gaussian surface can be used for which you
know the relative magnitude and direction of the E
field. Unfortunately, for most cases there is no symmetry and thus one must use the brute-force method
and evaluate the integral:
−
→
E=
1
40
Z

b
r
2
(3.33)
However, there are cases where superposition of
symmetric systems can be used to calculate the fields
due to non-symmetric systems. For example, consider
the system shown in figure 15 where a cylindrical hole
of radius 0 is drilled parallel to the axis of a uniform
cylindrical charge distribution of radius  and a distance  between the axes of the cylinders. This can be
solved by considering the superposition of the electric
fields due to the uniform positive cylindrical charge
distribution of radius  plus that from the negative
cylindrical distribution of radius 0 displaced by  that
cancels the charge distribution inside cylindrical hole.
23
Figure 15 Uniform cylindrical charge distribution of
radius  having a parallel cylindrical hole, of radius 0 
drilled a distance  from the axis.
Figure 16 An electron moving transverse to an electric
field
MOTION OF POINT CHARGES IN
ELECTRIC FIELDS
Having used Coulomb’s law, or Gauss’s law, to compute the electric field distribution, then it is possible to
use Newton’s Laws of motion to calculate the motion
of charged objects in this electric field. The acceleration, , of a charged object of mass,  and charge,
 in an electric field, , is given by Newton’s Law
→
−
→
−
→
F =  E = −
a  That is:
→
−
−
→
a =
E

(3.34)
For uniform electric field  one has the simple case of
motion with constant acceleration which you discussed
extensively in P113. It is useful to consider a simple
case to illustrate the application of Newton’s law of
motion to a problem in electrostatics.
Example: Deflection of moving charged particles.
Consider the case encountered in a cathode ray tube
used in a television or computer CRT display. An electron, of mass  and charge , initially at velocity v0
in the + direction, travels perpendicular to a uniform
non-zero  field between 0 ≤  ≤ 1 as shown in figure 16. The deflected electron then travels a further
distance 1 ≤  ≤ 1 + 2 in a field-free region before
hitting the screen. Calculate the deflection when the
electron hits the screen.
Assuming that the E field points in the − direction, then, since the electron charge is − the accel→
eration vector −
a equals:
→
−
b =  bj
a = bi + bj +  k

(3.35)
(3.36)
The charged particle travels a time  in the region of
the  field given by:
 =
1
0
(3.37)
The particle’s deflection when it exits the E field at
time  is:
1  2b
−
→
r =  bi +
(3.38)
 j
2 
The particle now continues in a straight line in the
field-free region at an angle  given by:
tan  =

()(1 0 )
1
=
=

0
02
(3.39)
Thus the net deflection in the bj direction is:
 = 1 +2 = 1 +2 tan  =
 21
( +1 2 ) (3.40)
02 2
Note the important fact that the deflection is proportional to , that is, the deflection of the electron beam
in your television screen reproduces the magnitude of
the electric field derived from the radio signal.
Inserting typical values, where 1 = 3, 2 =
12 0 = 3106 , and = 103 C, then:

Assume that the electron enters the E field at  = 0
and  = 0. Because the initial velocity is only in the
 direction and the  acceleration is zero, then the
velocity vector at time t is:
24
−
→
b = 0bi +  bj
v = bi + bj +  k

=
1
(16 10−19 )(103 )
( (003)2 + (003)(012))
(911 10−31 )(3 106 )2 2
= 80 × 10−2 
Of this 80m deflection, 1 = 09 and 2 =
71. That is, most of the deflection comes from the
angular deflection of the beam.
(3.41)
SUMMARY
Two methods have been presented for calculating the
electric field.
Coulomb’s Law:
Discrete charges:

−−→ X
E =
=1
1 
c
2 r0
40 0
(3.42)
Continuous charges:
In cartesian coordinates the E field at the point
() is written as an integral over the charge distribution at 0 (0  0  0 ) :
Z
−−−−→
(0  0  0 )0  0  0
1
E() =
(3.43)
rd
0 
40 
20 
Gauss’s Law
Φ =
I
−−→ −
→
1
E · S =

0
 
Z

(3.44)

Gauss’s law combined with symmetry was shown to be
a simple method for calculating electric fields for those
systems having a symmetry. In particular, spherical,
cylindrical and planar symmetry were considered.
Once the electric field has been computed, then it
is possible to predict the motion of charged particles in
this field using Newton’s laws of motion. The motion
of an electron in an electric field was discussed.
Reading assignment: Giancoli, Chapter 22
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