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Gauss’ Law
Chapter 23
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.
Can you tell what the
enclosed charge is now?
Answer: -0.5Q
Gauss’ Law
Can we find a simplified way to perform electric-field calculations?
Yes, we take advantage of a fundamental relationship between electric charge and electric
field: Gauss’ law
So far we considered problems like this:
Given a charge distribution, what is the resulting E-field? (Adding vectors)
Now, we will consider the reverse: given an E-field, what is the underlying charge
distribution?
Let’s assume we have a
closed container, e.g., a
sphere of an imaginary
material that doesn’t
interact with an electric
field
We detect an electric
field
outside the Gauss
surface
and can conclude
There must be a
charge inside
We can even be quantitative about the charge inside the Gauss sphere
The same number of field
lines on the surface point
into the Gauss sphere as
there are field lines
pointing
out
We say in this case
the net electric flux is zero
What is Flux?
(from Latin fluxus meaning flow):
Flux is the scalar quantity, F, which results from a surface integration over a vector field.
In the case of electric flux, FE, the vector field is the electric -field and the surface is the
surface of the Gauss volume
Let’s consider the velocity vector field of a fluid flowing in a pipe
The flux Fv of the velocity
vector field through the
surface of area A is
F v   v dA  v  dA  vA
A
A
F v  vA
Let’s interpret
A
v
dx
Fv 
A
dt
Volume element of fluid which flows through surface A in time dt
F v  vA 
dV
volume flow rate through A
dt
What if we tilt A?
Extreme case: Flux through A is
zero
If we tilt area by an angle ϕ
is a vector with the properties:
Pointing ^ to the surface
If the surface is curved, the orientation of changes on the surface and we divide the
surface into an infinite number of patches, each with area and sum the flux through each
patch over the entire area:
Electric flux is defined in complete analogy
The electric flux of a point charge inside a Gauss sphere
Spherical Gauss surface
(Note that the surface is closed)
We are going to calculate:
indicates that we integrate over a closed surface
The symmetry of the problem makes the
integration simple:
is perpendicular to the surface and its magnitude is the same at
each point of the surface
Concept Check
Considering the result obtained from calculating the electric flux of a point charge
through a Gauss sphere. Do you expect that the flux depends on the radius of the
sphere?
1. Yes, the larger the radius the more flux lines will penetrate through the surface
2. No, the flux is independent of the radius
3. I have to calculate again for a different radius
1
Q
1 Q
Q
2
FE 
A

4

r

4 0 r 2
4 0 r 2
0
result does NOT depend on the radius of the Gauss sphere
More generally the result of the integral does not even depend on the specific
form of the surface, ONLY on the amount of charge enclosed by the surface.
Our considerations suggest:
Flux through any surface enclosing the charge Q is given by Q/ε0
Let’s summarize findings into the general form of Gauss’s law
FE   E  d A 
qenc
0
The total electric flux through a closed surface is equal to the
total (net) electric charge inside the surface divided by ε0
E 
q
0
E 
q
0
E  0
E  0
Electric Flux
The differential 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:
dF  E  d A
d F   E cos   dA
(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.
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.
Sample Problem 23.01
Sample Problem 23.02
Sample Problem 23.02
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
 0 F  qenc
[Gauss' law]
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.
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.
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.
Gauss’ Law and Coulomb’s Law
 0  E  d A   0  EdA  qenc
 0 E  dA  q
 0 E 4r 2   q
E
1
q
4 0 r 2
Sample Problem 23.03
Sample Problem 23.03
A Charged Isolated Conductor
External Electric Field

E
0
[conducting surface]
(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.
Sample Problem 23.05
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
F  EA cos   E 2rh  cos0   E 2rh 
reduces to
qenc

h
 0 E 2rh   h
yielding
E
Gauss’ Law
 0 F  qenc

2 0 r
[linear charge density]
[line of charge]
A Gaussian surface in the
form of a closed cylinder
surrounds a section of a very
long, uniformly charged,
cylindrical plastic rod.
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,
 0  E  d A  qenc
becomes
 0 EA  EA  A
where σA is the charge enclosed by the Gaussian
surface. This gives

E
[sheet of charge]
2 0
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 .
Applying Gauss’ Law: Planar Symmetry
Two conducting Plates
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
E
2 1
0


0
Applying Gauss’ Law: Spherical Symmetry
In the figure, applying Gauss’ law to surface
S2, for which r ≥ R, we would find that
E
1
q
4 0 r 2
[spherical shell, field at r  R]
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.
E  0 [spherical shell, field at r  R]
Applying Gauss’ Law: Spherical Symmetry
Inside a sphere with a uniform volume charge density, the
field is radial and has the magnitude
 q 
E 
r [uniform charge, field at r  R]
3 
 4 0 R 
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).
Summary
Gauss’ Law
• Infinite non-conducting sheet
• Gauss’ law is
Eq. 23-6
• the net flux of the electric field
through the surface:
Eq. 23-13
• Outside a spherical shell of charge
Eq. 23-15
Eq. 23-6
Applications of Gauss’ Law
• surface of a charged conductor
Eq. 23-11
• Within the surface E=0.
• line of charge
• Inside a uniform spherical shell
Eq. 23-16
• Inside a uniform sphere of charge
Eq. 23-20
Eq. 23-12