§2. 3 ELECTRIC FLUX , GAUSS’S LAW
1. Electric Field Lines
A convenient specialized pictorial representation for
visualizing electric field patterns is created by
drawing lines which are called electric field lines.
The electric field lines are related to the electric
field in any region of space in the following manner:
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(1) The tangent direction at every point on an electric field line is
just the direction of the field intensity at that point or the direction
of the force on the positive point charge at that point.
 (2) The electric field lines are denser in the place where the field
intensity is stronger, and the electric field lines are sparser in the
place where the field intensity is weaker.
(3) The electric field lines start on positive charges and terminate
on negative charges , and never intersected each other. It is never
interrupted in region without charge; this is called the continuity of
electric field line.
(4) Keep in mind: electric field lines do not actually exist.
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+
+
+

For a positive point charge, the lines are directed radially
outward.
For a negative point charge, the lines are directed radially
inward.
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The electric field lines for two charges of equal
magnitude and opposite sign (an electric dipole)
+
NOTE: the number of lines leaving the positive charge
equals the number terminating at the negative charge.
The electric field lines for two positive point charges.
+
+
The electric field lines for a point charge +2q and a
second point charge –q.
+
2q
q
++ ++ + + + + +
2. Electric Flux e
Flux Amount
Ring Amount



A  dl
Si
L
 
A

d
s


Ei

dS
S
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
E
S
8
S

nE
S

n

E
1. Uniform electric field


E  n
2 .Uniform electric field


En =
 
 e  ES cos  E  S

S
 e  ES

E
3. Nonuniform electric field, arbitrary

surface


ndS
d e  E  dS


 e   E  dS
S
UNIT：Vm
 
e   E  dS＝ E cos ds

 
Eds  
Where
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A closed surface is defined as one that completely
divides space into an inside region and outside region,
so that movement cannot take place from one region to
the other without penetrating the surface.
For a closed surface, usually define the
normal line at every point on the surface points
out of the closed surface
A closed
surface
A open surface
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n
n

 =0
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 >0
12
 <0

Attention


If the surfaces is closed,    E  dA
S

According to the convention, d A  0 , outward the

closed surface; d A  0,inward the closed surface.
If the surface is unclosed, the positive
normal direction can be free chosen.
 is a scalar. Its SI unit: N·m2/C.
Example
There is a cube surface of edge length a in the uniform
 

electric field E  E0 (i  j )（ E0 is a constant）as shown
in figure. Find the electric flux of every plane and the
cube surface.
Z
Y
X
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3. GAUSS’S LAW
Gauss worked in a wide variety of
fields in both mathematics and
physics including number theory,
analysis, differential geometry,
geodesy, magnetism, astronomy and
optics. His work has had an
immense influence in many areas.
Sometimes known as "the prince of
mathematicians" and "greatest
mathematician since antiquity"
Johann Carl Friedrich Gauss
1777 - 1855
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(1) GAUSS’S
LAW
The net electric flux e of an arbitrary closed surface
in the vacuum is equal to the net charge inside the
surface divided by  0 .
 
 e   E  dS
S
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

16
1
0
1
0
Q
i
 dQ
 
 e   E  dS
S



1
0
1
0
Q
i
 dQ
S： The closed surface, i.e. gaussian surface. It is an
imaginary surface and need not coincide with any real physical
surface.
 
E  d S The close surface integral is over all gaussian surface.
S

E is the total electric field at any point on the surface due to all
charges.

ds
q
Surface element. Its orientation is perpendicular to
the surface and points outward from the inside region.
i
The algebra sum of charges in the closed surface.
S
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It give a simple way to calculate the distribution of
electric field for a given charge distribution with
sufficient symmetry.
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(2) Proving
A spherical gaussian surface of radius r
surrounding a point charge q which is at the
centre of the sphere.

 dS
E
The electric field is normal to the surface
and constant in magnitude everywhere on
the surface.
q
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19
+
r
A spherical gaussian surface of
radius r surrounding a point charge q
which is not at the centre of the sphere.
q
+
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O
r
An arbitrary gaussian surface
surrounding a point charge q.
The net electric flux
through each surface is
the same.
S2
S1
+
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There are many charges
inside the guassian surface.
q3
+
q1
S
+
+
q2
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+
A point charge located outside a closed surface. The
number of lines entering the surface equals the
number leaving the surface.
Zero flux
is not
zero field.
+q
The net electric flux through a closed surface that surrounds no
net charge is zero.
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Conclusion
The net flux e through any closed surface
surrounding the point charge q is given by
 
 e   E  dS
S
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

1
0
1
0
Q
i
 dQ
24
q
0
A system of charges
Continuous
distribution of charges
(3) Physical Meaning
The positive charge is the source of the electrostatic field.
Gauss’s law is valid for the electric field of any system of
charges or continuous distribution of charge.
Guass’s law can be used to evaluate the electric field for
charge distributions that have spherical, cylindrical , or plane
symmetry. The technique is useful only in situations where the
degree of symmetry is high.
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Quick Quiz
Why are the electrostatic field lines never interrupted in
region without charge?
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Quick Quiz
Find the flux through the square.
a
Q
a
2
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Consider whether the following statements are truth.
 Electric flux of the Gauss surface is related with charges in
Gauss surface and is not related with charges out of Gauss
surface.
The field intensity at a point on the Gauss surface is
related with charges in the Gauss surface and is not related
with charges out of the Gauss surface.
If the electric flux of a Gauss surface equals zero, there
must be not charge in the Gauss surface.
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If the electric flux of a Gauss surface equals zero, then
field intensity at every point on the Gauss surface is zero.
 Gauss theorem is tenable only to the electrostatic field
whose distribution is symmetrical in space.
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(4) Application of Gauss’s Law to Symmetric Charge
Distribution
The gaussian surface should always be chosen to take
advantage of the symmetry of the charge distribution ,
so that we can remove E from the integral and solve it.

 
 e   E  dS
S
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
30
1
0
1
0
Q
i
 dQ
The gaussian surface had better satisfies one or more of the
following conditions:
The value of the electric field is constant.

 

 E and dA are parallel. E  dA  EdA


 
 E and dA are perpendicular.
E  dA  0

 E is equal to zero everywhere on the surface.
Note: The surface integral in Guass’s law is taken over
the entire gaussian surface.
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The three symmetries:
Spherical symmetry
Cylindrical symmetry
Plane symmetry
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Problem-solving strategy
Analysis the symmetry of the field intensity distribution .
Select appropriate gaussian surface.
Select appropriate coordinates, apply Gauss’s law.
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(1) A Spherically Symmetric Charge Distribution
Example 1. Electric quantity Q distributes uniformly on a
spherical surface of center O and radius R. Find the field
intensity.
P
O
Analysis the symmetry of the
field intensity distribution
R
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Spherical symmetry
P
O
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The magnitude of the electric field is
constant everywhere on the concentric
spherical surface, and the field is
normal to the surface at each point.
R
35
Select appropriate gaussian surface.
0
 
 e   E  dS 
Q
S
0
r  R
r  R 
 E  dS  E  4r 2
S
E
q
4π 0 R 2
E
q
8π 0 R 2
O
R
r
36
0 r  R
Q
r  R 
2
4 0 r
Note
For a uniformly charged spherical surface, the
field in the region external to the spherical
surface is equivalent to that of a point charge at
the center of the sphere surface.
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Quick Quiz
There are two concentric charged spherical shells of radius
R1 and R2 . Charge quantities distribute uniformly.
R1， Q1
r  R1 ,
E
R1  r  R 2 ,
R2  r ,
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E
E
38
R2， Q2
Example 2.
An insulating solid sphere of radius r has a uniform
volume charge density ρ and carries a total positive
charge Q. Calculate the electric field intensity.
Solution
o
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r
Because the charge distribution
is spherically symmetric, we
select a spherical gaussian
surface of radius r, concentric
with the sphere.
39
Qr 3
Q  3
R
 
 e   E  dS 
S
0
Q
0
R
r
3
Qr
Q3
0R
r  R 
r
 R
 E  dS  E  4r 2
S
E
E
r
0
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R
40
Qr
4 0 R 3
Q
4 0 r 2
r  R 
r  R 
The electric field inside the sphere ( r  R ) varies
linearly with r.
The electric field outside the sphere ( r  R )is the
same as that of a point charge Q located at r = 0.
E
r
0
The expressions of field
intensity match when r = a.
R
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(2) A Cylindrically Symmetric Charge Distribution
an infinite uniform charged straight line
an infinite uniform charged cylinder
… … … …
Example 1
A section of an infinitely long cylindrical plastic rod with a
uniform +. Let us find an expression for the magnitude of

the Eat a distance r from the axis of the rod.
Infinite length
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Analysis the symmetry of the field intensity
distribution
dl1
dE1
P
dE
O
dl２
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dE2
43
Cylindrical Symmetry
The magnitude of the electric
field is constant everywhere on
the coaxial cylindrical surface,
and the field is normal to the
surface at each point.
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2. Select appropriate gaussian surface: a cylindrical
gaussian surface of radius r and length l that is coaxial
with the line charge.
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
dS
r

E

dS 
l
E

dS
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
dS
 
l
 e   E  dS 
S

0
E  dS   0 E  dS
？
curved surface
ends

E 
2  0 r
46
The field intensity of an infinite uniform
charged straight line :

E
2 0 r
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Example 2
The charge density of an infinite uniform charged
cylindrical surface of radius R is  along the
direction of axis , find the electric field intensity.
r
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
E
Quick Quiz
The charge density of an infinite uniform
solid charged cylinder of radius R is
,

find the electric field intensity.
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r R

2
E 2  rl 

r
l
2
 0 R
rR
l
E 2 rl 
0
E 
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r
2
2 0 R
r R

2 0 r
r R
50
3. Planar Symmetry
• Infinite Charged Plane or Sheet.
+
+
+
+
+
+
+
+
+
+
Planar Symmetry: for
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those points whose distance to
the plane is identical, the field
intensity is equal in magnitude
and perpendicular to the plane.
51
Example 3
Find the electric field due to a nonconducting, infinite
plane with uniform surface charge density  .
Research in the Symmetry.
Select Gaussian surface.
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+
+
+
+
+
+
+
+
+
+
Select a cylindrical gaussian surface penetrating an
infinite sheet of charge.


Each end of the gaussian surface, E  dS
Curved surface,

E
+
+
+
+
+
+
+
+
+
+
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

E  dS
   S
   E  d S
dS e S
0
E
dS
 2SE
53

E 
2 0
Conclusion : the field is uniform everywhere.
+
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Example 4
As shown in figure, there are two same parallel
sheets each with surface charge density  1 and  2
respectively .The distance between planes is smaller
far than the size of them. Find the electric field
density at points (a) to the left of the two sheets.(b)in
the between , and (c) to the right of the two sheets.
1
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+
++
++
++
++
++
+
2
55
+
++
++
++
++
++
+
Solution
1

E1

E2
A
2
+
+
++  ++
++ E ++
1 +
++
+
++  ++
+ + E2 + +
+ B +
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1
2
E1 
E2 
2 0
2 0

E1 
1  2 
EA  
i
2 0


1  2 
E2
EB 
i
2 0
Cx

1  2 
EC 
i
2 0
56
＋
+
+
+
+
+
+
－







when 1 = - 2
E A  EC  0
EB


0
Parallel Plate
Capacitor
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