Transcript Gauss* Law

Chapter 23
Gauss’ Law
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
23-1 Electric Flux
Learning Objectives
23.03 Identify that an area vector
for a flat surface is a vector that
23.01 Identify that Gauss’ law
is perpendicular to the surface
relates the electric field at
and that has a magnitude equal
points on a closed surface
to the area of the surface.
(real or imaginary, said to be a
Gaussian surface) to the net
23.04 Identify that any surface
charge enclosed by that
can be divided into area
surface.
elements (patch elements) that
are each small enough and flat
23.02 Identify that the amount of
enough for an area vector dA
electric field piercing a surface
to be assigned to it, with the
(not skimming along parallel to
vector perpendicular to the
the surface) is the electric flux
element and having a
Φ through the surface.
magnitude equal to the area of
the element.
© 2014 John Wiley & Sons, Inc. All rights reserved.
23-1 Electric Flux
Learning Objectives (Contd.)
23.05 Calculate the flux Φ
23.07 Calculate the net flux ϕ
through a surface by
through a closed surface,
integrating the dot product of
algebraic sign included, by
the electric field vector E and
integrating the dot product of
the area vector dA (for patch
the electric field vector E and
elements) over the surface, in
the area vector dA (for patch
magnitude- angle notation and
elements) over the full surface.
unit-vector notation.
23.08 Determine whether a
23.06 For a closed surface,
closed surface can be broken
explain the algebraic signs
up into parts (such as the sides
associated with inward flux
of a cube) to simplify the
and outward flux.
integration that yields the net
flux through the surface.
© 2014 John Wiley & Sons, Inc. All rights reserved.
23-1 Electric Flux
Electric field vectors and field
lines pierce an imaginary,
spherical Gaussian surface
that encloses a particle with
charge +Q.
Now the enclosed particle
has charge +2Q.
© 2014 John Wiley & Sons, Inc. All rights reserved.
Can you tell what the
enclosed charge is now?
Answer: -0.5Q
Flux of a Vector. Consider an airstream of velocity v that is aimed at a loop
of area A . The velocity vector v is at angle  with respect to the
loop normal nˆ. The product   vA cos  is known as the flux. In this
example the flux is equal to the volume flow rate through the loop (thus the
name flux).
Note 1 :  depends on  . It is maximum and equal to vA for   0 (v perpendicular
to the loop plane). It is minimum and equal to zero for   90 (v parallel
to the loop plane).
Note 2 : vA cos   v  A. The vector A is parallel to the loop normal and has
magnitude equal to A.
n̂
n̂
(23-2)
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Flux of the Electric Field.
Consider the closed surface shown in the figure.
In the vicinity of the surface assume that we have
a known electric field E. The flux  of the
electric field through the surface is defined as follows:
1. Divide the surface into small "elements" of area A.
2. For each element calculate the term E  A  EA cos  .
3. Form the sum    E  A.
n̂
n̂
4. Take the limit of the sum as the area A  0.
The limit of the sum becomes the integral:

 E  dA
Flux SI unit: N  m 2 / C
Note 1 : The circle on the integral sign indicates that
n̂

 E  dA
the surface is closed. When we apply Gauss'
law the surface is known as "Gaussian."
Note 2 :  is proportional to the net number of
electric field lines that pass through the surface.
(23-3)
© 2014 John Wiley & Sons, Inc. All rights reserved.
Gauss' Law
Gauss' law can be formulated as follows:
The flux of E through any closed surface   0  net charge qenc enclosed by the surface.
In equation form:  0   qenc
Equivalently:
ε0  E  dA  q enc
Note 1 : Gauss' law holds for any closed surface.
Usually one particular surface makes the problem of
determining the electric field very simple.
Note 2 : When calculating the net charge inside a closed
n̂ surface we take into account the algebraic sign of each charge.
n̂
Note 3 : When applying Gauss' law for a closed surface
we ignore the charges outside the surface no matter how
large they are.
n̂
 0  qenc
ε0  E  dA  q enc
Example :
Surface S1 :  0 1   q,
Surface S2 :  0  2   q
Surface S3 :  0  3  0, Surface S4 :  0  4   q  q  0
Note : We refer to S1 , S 2 , S3 , S 4 as "Gaussian surfaces."
(23-4)
© 2014 John Wiley & Sons, Inc. All rights reserved.
23-1 Electric Flux
The area vector dA for an area element (patch element) on a surface is a vector
that is perpendicular to the element and has a magnitude equal to the area dA of the
element.
The electric flux dϕ through a patch element with area vector dA is given by a dot
product:
(a) An electric field vector pierces a
small square patch on a flat
surface.
(b) Only the x component actually
pierces the patch; the y component
skims across it.
(c) The area vector of the patch is
perpendicular to the patch, with a
magnitude equal to the patch’s
area.
© 2014 John Wiley & Sons, Inc. All rights reserved.
23-1 Electric Flux
Now we can find the total flux by integrating the
dot product over the full surface.
The total flux through a surface is given by
The net flux through a closed surface (which is
used in Gauss’ law) is given by
where the integration is carried out over the
entire surface.
© 2014 John Wiley & Sons, Inc. All rights reserved.
23-1 Electric Flux
Flux through a closed cylinder, uniform field
© 2014 John Wiley & Sons, Inc. All rights reserved.
23-2 Gauss’ Law
Learning Objectives
23.09 Apply Gauss’ law to relate
net flux through the closed
the net flux ϕ through a closed
surface.
surface to the net enclosed
23.12 Derive the expression for
charge qenc.
the magnitude of the electric
23.10 Identify how the algebraic
field of a charged particle by
sign of the net enclosed
using Gauss’ law.
charge corresponds to the
23.13 Identify that for a charged
direction (inward or outward)
particle or uniformly charged
of the net flux through a
sphere, Gauss’ law is applied
Gaussian surface.
with a Gaussian surface that is
23.11 Identify that charge
a concentric sphere.
outside a Gaussian surface
makes no contribution to the
© 2014 John Wiley & Sons, Inc. All rights reserved.
23-2 Gauss’ Law
Gauss’ law relates the net flux ϕ of an electric field
through a closed surface (a Gaussian surface) to
the net charge qenc that is enclosed by that
surface. It tells us that
we can also write Gauss’ law as
Two charges, equal in magnitude but opposite in sign, and
the field lines that represent their net electric field. Four
Gaussian surfaces are shown in cross section.
© 2014 John Wiley & Sons, Inc. All rights reserved.
23-2 Gauss’ Law
Surface S1.The electric field is outward for all
points on this surface. Thus, the flux of the electric
field through this surface is positive, and so is the
net charge within the surface, as Gauss’ law
requires
Surface S2.The electric field is inward for all
points on this surface. Thus, the flux of the electric
field through this surface is negative and so is the
enclosed charge, as Gauss’ law requires.
Two charges, equal in magnitude but opposite in sign, and
the field lines that represent their net electric field. Four
Gaussian surfaces are shown in cross section.
© 2014 John Wiley & Sons, Inc. All rights reserved.
23-2 Gauss’ Law
Surface S3.This surface encloses no charge, and
thus qenc = 0. Gauss’ law requires that the net flux
of the electric field through this surface be zero.
That is reasonable because all the field lines pass
entirely through the surface, entering it at the top
and leaving at the bottom.
Surface S4.This surface encloses no net charge,
because the enclosed positive and negative
charges have equal magnitudes. Gauss’ law
requires that the net flux of the electric field
through this surface be zero. That is reasonable
because there are as many field lines leaving
surface S4 as entering it.
Two charges, equal in magnitude but opposite in sign, and
the field lines that represent their net electric field. Four
Gaussian surfaces are shown in cross section.
© 2014 John Wiley & Sons, Inc. All rights reserved.
Coulomb's law  Gauss' law
dA
n̂
Gauss' Law and Coulomb's Law
Gauss' law and Coulomb's law are different ways
of describing the relation between electric charge
and electric field in static cases. One can derive
Coulomb's law from Gauss' law and vice versa.
Here we will derive Coulomb's law from Gauss' law.
Consider a point charge q. We will use Gauss' law
to determine the electric field E generated at a point
P at a distance r from q. We choose a Gaussian surface
that is a sphere of radius r and has its center at q.
We divide the Gaussian surface into elements of area dA. The flux for each element is:
d   EdA cos 0  EdA
Total flux  
2
EdA

E
dA

E
4

r




From Gauss' law we have:  0   qenc  q  4 r 2 0 E  q  E 
q
4 r 2 0
This is the same answer we got in Chapter 22 using Coulomb's law.
© 2014 John Wiley & Sons, Inc. All rights reserved.
23-3 A Charged Isolated Conductor
Learning Objectives
23.14 Apply the relationship
between surface charge
density σ and the area over
which the charge is uniformly
spread.
23.17 For a conductor with a cavity
that contains a charged object,
determine the charge on the cavity
wall and on the external surface.
23.18 Explain how Gauss’ law is used
to find the electric field magnitude E
near an isolated conducting surface
23.15 Identify that if excess
with a uniform surface charge
charge (positive or negative) is
density σ.
placed on an isolated
conductor, that charge moves 23.19 For a uniformly charged
conducting surface, apply the
to the surface and none is in
relationship between the charge
the interior.
density σ and the electric field
23.16 Identify the value of the
magnitude E at points near the
electric field inside an isolated
conductor, and identify the direction
conductor.
of the field vectors.
© 2014 John Wiley & Sons, Inc. All rights reserved.
23-3 A Charged Isolated Conductor
(a) Perspective view
(b) Side view of a tiny portion of a large, isolated
conductor with excess positive charge on its
surface. A (closed) cylindrical Gaussian surface,
embedded perpendicularly in the conductor,
encloses some of the charge. Electric field lines
pierce the external end cap of the cylinder, but
not the internal end cap. The external end cap
has area A and area vector A.
© 2014 John Wiley & Sons, Inc. All rights reserved.
The Electric Field Inside a Conductor
We shall prove that the electric field inside a conductor vanishes.
e
F
v
E
Consider the conductor shown in the figure to the left. It is an
experimental fact that such an object contains negatively charged
electrons, which are free to move inside the conductor. Let's
assume for a moment that the electric field is not equal to zero.
In such a case a nonvanishing force F  eE is exerted by the
field on each electron. This force would result in a nonzero
velocity v , and the moving electrons would constitute an electric
current. We will see in subsequent chapters that electric currents
manifest themselves in a variety of ways:
(a) They heat the conductor.
(b) They generate magnetic fields around the conductor.
No such effects have ever been observed, thus the original
assumption that there exists a nonzero electric field inside
the conductor. We conclude that :
The electrostatic electric field E inside a conductor is equal to zero.
© 2014 John Wiley & Sons, Inc. All rights reserved.
A Charged Isolated Conductor
Consider the conductor shown in the figure that has
a total charge q. In this section we will ask the question:
Where is this charge located? To answer the question
we will apply Gauss' law to the Gaussian surface shown
in the figure, which is located just below the conductor
surface. Inside the conductor the electric field E  0.
Thus  
 E  A  0
(eq. 1).
From Gauss's law we have:  
qenc
0
(eq. 2).
If we compare eq. 1 with eq. 2 we get qenc = 0 .
Thus no charge exists inside the conductor. Yet we know that the conductor
has a nonzero charge q. Where is this charge located? There is only one place
for it to be: On the surface of the conductor.
No electrostatic charges can exist inside a conductor.
All charges reside on the conductor surface.
© 2014 John Wiley & Sons, Inc. All rights reserved.
(23-7)
An Isolated Charged Conductor with a Cavity
Consider the conductor shown in the figure that has
a total charge q. This conductor differs from the one
shown on the previous page in one aspect: It has a
cavity. We ask the question: Can charges reside
on the walls of the cavity?
As before, Gauss's law provides the answer.
We will apply Gauss' law to the Gaussian surface shown
in the figure, which is located just below the conductor
surface. Inside the conductor the electric field E  0.
Thus  
 E  A  0
(eq. 1).
From Gauss's law we have:  
qenc
0
(eq. 2).
If we compare eq. 1 with eq. 2 we get qenc = 0.
Conclusion :
There is no charge on the cavity walls. All the excess charge q
remains on the outer surface of the conductor.
© 2014 John Wiley & Sons, Inc. All rights reserved.
(23-
n̂2
n̂3 S3
S2
n̂1
S1
The Electric Field Outside a Charged Conductor
The electric field inside a conductor is zero. This is
not the case for the electric field outside. The
electric field vector E is perpendicular to the conductor
surface. If it were not, then E would have a component
parallel E to the conductor surface.
Since charges are free to move in the conductor, E would cause the free
electrons to move, which is a contradiction to the assumption that we
have stationary charges. We will apply Gauss' law using the cylindrical closed
surface shown in the figure. The surface is further divided into three sections
S1 , S 2 , and S3 as shown in the figure. The net flux   1   2   3 .
1  EA cos 0  EA
 2  EA cos 90  0
 3  0 (because the electric field inside the conductor is zero).   EA 
E
qenc 1
.
A 0
The ratio  
qenc
0

qenc

is known as surface charge density  E  .
A
0
© 2014 John Wiley & Sons, Inc. All rights reserved.
Symmetry. We say that an object is symmetric under a particular mathematical
operation (e.g., rotation, translation, …) if to an observer the object looks the same
before and after the operation.
Note: Symmetry is a primitive notion and as such is very powerful.
Rotational symmetry
Featureless
sphere
Observer
Example of Spherical Symmetry
Consider a featureless beach ball that
can be rotated about a vertical axis that
passes through its center. The observer
closes his eyes and we rotate the
sphere. When the observer opens his
eyes, he cannot tell whether the sphere
has been rotated or not. We conclude
that the sphere has rotational symmetry
about the rotation axis.
Rotation axis
(23-10)
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A Second Example of Rotational Symmetry
Rotational symmetry
Consider a featureless cylinder that can rotate
about its central axis as shown in the figure.
The observer closes his eyes and we rotate the
cylinder. When he opens his eyes, he cannot
tell whether the cylinder has been rotated or not.
We conclude that the cylinder has rotational
symmetry about the rotation axis.
Featureless
cylinder
Rotation axis
Observer
(23-11)
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Example of Translational Symmetry:
Consider an infinite featureless plane.
An observer takes a trip on a magic
carpet that flies above the plane. The
observer closes his eyes and we move
the carpet around. When he opens his
eyes the observer cannot tell whether
he has moved or not. We conclude that
the plane has translational symmetry.
Translational
symmetry
Observer
Magic carpet
Infinite featureless
plane
(23-12)
© 2014 John Wiley & Sons, Inc. All rights reserved.
Recipe for Applying Gauss’ Law
1. Make a sketch of the charge distribution.
2. Identify the symmetry of the distribution and its effect
on the electric field.
3. Gauss’ law is true for any closed surface S. Choose
one that makes the calculation of the flux  as easy
as possible.
4. Use Gauss’ law to determine the electric field vector:

qenc
0
(23-13)
© 2014 John Wiley & Sons, Inc. All rights reserved.
23-4 Applying Gauss’ Law: Cylindrical Symmetry
Learning Objectives
23.20 Explain how Gauss’ law is
used to derive the electric field
magnitude outside a line of
charge or a cylindrical surface
(such as a plastic rod) with a
uniform linear charge density
λ.
23.22 Explain how Gauss’ law
can be used to find the electric
field magnitude inside a
cylindrical non-conducting
surface (such as a plastic rod)
with a uniform volume charge
density ρ.
23.21 Apply the relationship
between linear charge density
λ on a cylindrical surface and
the electric field magnitude E
at radial distance r from the
central axis.
© 2014 John Wiley & Sons, Inc. All rights reserved.
23-4 Applying Gauss’ Law: Cylindrical Symmetry
Figure shows a section of an infinitely long cylindrical
plastic rod with a uniform charge density λ. The
charge distribution and the field have cylindrical
symmetry. To find the field at radius r, we enclose a
section of the rod with a concentric Gaussian cylinder
of radius r and height h.
The net flux through the cylinder from Gauss’ Law
reduces to
yielding
A Gaussian surface in the
form of a closed cylinder
surrounds a section of a very
long, uniformly charged,
cylindrical plastic rod.
© 2014 John Wiley & Sons, Inc. All rights reserved.
E

2 0 r
n̂1
S1
S2
n̂2
S3
Electric Field Generated by a Long, Uniformly Charged Rod
Consider the long rod shown in the figure. It is uniformly
charged with linear charge density . Using symmetry
arguments we can show that the electric field vector points
radially outward and has the same magnitude for points at
the same distance r from the rod. We use a Gaussian surface
S that has the same symmetry. It is a cylinder of radius r
and height h whose axis coincides with the charged rod.
n̂3
We divide S into three sections: Top flat section S1 , middle curved section S 2 ,
and bottom flat section S3 . The net flux through S is   1   2   3 . Fluxes 1 and  2
vanish because the electric field is at right angles with the normal to the surface:
q
h
 3  2 rhE cos 0  2 rhE    2 rhE. From Gauss's law we have:   enc 
.
0
If we compare these two equations we get: 2 rhE 
h

 E
.
0
2 0 r
© 2014 John Wiley & Sons, Inc. All rights reserved.
0
(23-14)
23-5 Applying Gauss’ Law: Planar Symmetry
Learning Objectives
23.23 Apply Gauss’ law to derive
the electric field magnitude E
near a large, flat, nonconducting surface with a
uniform surface charge
density σ.
23.24 For points near a large,
flat non-conducting surface
with a uniform charge density
σ, apply the relationship
between the charge density
and the electric field
magnitude E and also specify
the direction of the field.
23.25 For points near two large,
flat, parallel, conducting
surfaces with a uniform charge
density σ, apply the relationship
between the charge density and
the electric field magnitude E
and also specify the direction of
the field.
© 2014 John Wiley & Sons, Inc. All rights reserved.
23-5 Applying Gauss’ Law: Planar Symmetry
Non-conducting Sheet
Figure (a-b) shows a portion of a thin, infinite, nonconducting sheet with a uniform (positive) surface
charge density σ. A sheet of thin plastic wrap,
uniformly charged on one side, can serve as a
simple model. Here,
Is simply EdA and thus Gauss’ Law,
becomes
where σA is the charge enclosed by the Gaussian
surface. This gives
© 2014 John Wiley & Sons, Inc. All rights reserved.
E

2 0
Electric Field Generated by a Thin, Infinite,
Nonconducting Uniformly Charged Sheet
We assume that the sheet has a positive charge of
surface density  . From symmetry, the electric field
vector E is perpendicular to the sheet and has a
constant magnitude. Furthermore, it points away from
the sheet. We choose a cylindrical Gaussian surface S
with the caps of area A on either side of the sheet as
shown in the figure. We divide S into three sections:
S1 is the cap to the right of the sheet, S 2 is the curved
surface of the cylinder, and S3 is the cap to the left of the
sheet. The net flux through S is   1   2   3 .
n̂2
S3
n̂3
1   2  EA cos 0  EA.  3  0
S2
n̂1
S1
( = 90)
A
   2 EA. From Gauss's law we have:  

0
0
A

 2 EA 
 E
.
0
2 0
(23-15)
qenc
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Ei 
S
A
S'
The electric field generated by two parallel conducting infinite planes is charged with
surface densities  1 and - 1. In figs. a and b we show the two plates isolated so that
one does not influence the charge distribution of the other. The charge spreads out
equally on both faces of each sheet. When the two plates are moved close to each
other as shown in fig. c, then the charges on one plate attract those on the other. As
a result the charges move on the inner faces of each plate. To find the field Ei between
the plates we apply Gauss' law for the cylindrical surface S , which has caps of area A.
q
2 A
2
The net flux   Ei A  enc  1  Ei  1 . To find the field E0 outside
0
0
the plates we apply Gauss' law for the cylindrical surface S', which has caps of area A.
The net flux   E0 A 
qenc
0

1  1
 0  E0  0.
 0 John Wiley & Sons, Inc. All rights reserved.
© 2014
0
E0  0
A'
0
2 1
(23-16)
23-5 Applying Gauss’ Law: Planar Symmetry
Two conducting Plates
Figure (a) shows a cross section of a thin,
infinite conducting plate with excess
positive charge. Figure (b) shows an
identical plate with excess negative charge
having the same magnitude of surface
charge density σ1.
Suppose we arrange for the plates of Figs.
a and b to be close to each other and
parallel (c). Since the plates are
conductors, when we bring them into this
arrangement, the excess charge on one
plate attracts the excess charge
on the other plate, and all the excess charge moves onto the inner faces of the
plates as in Fig.c. With twice as much charge now on each inner face, the electric
field at any point between the plates has the magnitude
© 2014 John Wiley & Sons, Inc. All rights reserved.
23-6 Applying Gauss’ Law: Spherical Symmetry
Learning Objectives
23.26 Identify that a shell of
uniform charge attracts or
repels a charged particle that
is outside the shell as if all the
shell’s charge is concentrated
at the center of the shell.
charge, apply the relationship
between the electric field
magnitude E, the charge q on the
shell, and the distance r from the
shell’s center.
23.29 Identify the magnitude of the
23.27 Identify that if a charged
electric field for points enclosed by
particle is enclosed by a shell
a spherical shell with uniform
of uniform charge, there is no
charge.
electrostatic force on the
23.30 For a uniform spherical charge
particle from the shell.
distribution (a uniform ball of
23.28 For a point outside a
charge), determine the magnitude
spherical shell with uniform
and direction of the electric field at
interior and exterior points.
© 2014 John Wiley & Sons, Inc. All rights reserved.
23-6 Applying Gauss’ Law: Spherical Symmetry
In the figure, applying Gauss’ law to surface
S2, for which r ≥ R, we would find that
And, applying Gauss’ law to surface S1, for
which r < R,
A thin, uniformly charged, spherical
shell with total charge q, in cross
section. Two Gaussian surfaces S1
and S2 are also shown in cross
section. Surface S2 encloses the
shell, and S1 encloses only the
empty interior of the shell.
© 2014 John Wiley & Sons, Inc. All rights reserved.
23-6 Applying Gauss’ Law: Spherical Symmetry
Inside a sphere with a uniform volume charge density, the
field is radial and has the magnitude
where q is the total charge, R is the sphere’s radius, and r is
the radial distance from the center of the sphere to the point
of measurement as shown in figure.
A concentric spherical Gaussian surface
with r > R is shown in (a). A similar
Gaussian surface with r < R is shown in (b).
© 2014 John Wiley & Sons, Inc. All rights reserved.
n̂2 E0
The Electric Field Generated by a Spherical Shell
of Charge q and Radius R
Inside the shell : Consider a Gaussian surface S1
that is a sphere with radius r  R and whose
center coincides with that of the charged shell.
q
The electric field flux   4 r 2 Ei  enc  0.
0
Thus Ei  0.
Outside the shell : Consider a Gaussian surface S 2
n̂1
Ei
that is a sphere with radius r  R and whose
center coincides with that of the charged shell.
Ei  0
E0 
q
4 0 r 2
(23-17)
The electric field flux   4 r 2 E0 
Thus E0 
q
4 0 r 2
qenc
0

q
0
.
.
Note : Outside the shell the electric field is the
same as if all the charge of the shell were
concentrated at the shell center.
© 2014 John Wiley & Sons, Inc. All rights reserved.
n̂1
Electric Field Generated by a Uniformly Charged Sphere
Eo
of Radius R and Charge q
Outside the sphere : Consider a Gaussian surface S1
that is a sphere with radius r  R and whose
center coincides with that of the charged shell.
S1
The electric field flux   4 r 2 E0  qenc /  0  q /  0
Thus E0 
q
4 0 r
2
.
Inside the sphere : Consider a Gaussian surface S 2
n̂2
Ei
that is a sphere with radius r  R and whose
center coincides with that of the charged shell.
q
The electric field flux   4 r 2 Ei  enc .
0
S2
qenc
4 / r 3
R3
R3 q
2

q  3 q  4 r Ei  3
3
4 / R
r
r 0
 q 
Thus Ei  
r.
3 
 4 0 R 
© 2014 John Wiley & Sons, Inc. All rights reserved.
(23-18)
n̂1
E0
Electric Field Generated by a Uniformly Charged Sphere
of Radius R and Charge q
Summary :
 q 
Ei  
r
3 
4

R
0


S1
E
q
4 0 R 3
n̂2
Ei
O
E0 
S2
R
r
q
4 0 r 2
(23-19)
© 2014 John Wiley & Sons, Inc. All rights reserved.
23 Summary
Gauss’ Law
• Infinite non-conducting sheet
• Gauss’ law is
Eq. 23-13
Eq. 23-6
• the net flux of the electric field
through the surface:
• Outside a spherical shell of charge
Eq. 23-15
Eq. 23-6
Applications of Gauss’ Law
• Inside a uniform spherical shell
• surface of a charged conductor
Eq. 23-11
Eq. 23-16
• Inside a uniform sphere of charge
• Within the surface E=0.
• line of charge
Eq. 23-20
Eq. 23-12
© 2014 John Wiley & Sons, Inc. All rights reserved.