From Chapter 23 – Coulomb’s Law
We found the fields in the vicinity of continuous charge
distributions by integration:
+
+
+
+
Line of Charge:
2k
E
x
Are you interested in learning an easier way?
R
r
x
dE
Charged Plate:

E
2 o
Electric flux
• The field is uniform
• The plane is
perpendicular to the
field
E  E A
Flux for a uniform electric field
passing through an arbitrary plane
n̂
• The field is uniform
• The plane is not
perpendicular to the
field
 E  E A   E A cos 
A  Anˆ
E  E A
Rules for drawing field lines
• The electric field, E , is tangent to the field lines.
• The number of lines leaving/entering a charge is
proportional to the charge.
• The number of lines passing through a unit area normal
to the lines is proportional to the strength of the field in
that region.
# of electric field lines
E
Area 
• Field lines must begin on positive charges (or from
infinity) and end on negative charges (or at infinity).
The test charge is positive by convention.
• No two field lines can cross.
N  E A  E
General flux definition
• The field is not
uniform
• The surface is not
perpendicular to the
field
E
E

n̂
A
 
 E  E nˆ A  E A cos 
A
N


 E   E i nˆ i A i
i 1
If the surface is made up of a
mosaic of N little surfaces
ˆ
 E   E ndA
  E dA
Electric Flux, General
• In the more general
case, look at a small
area element
E  Ei Ai cos θi  Ei  Ai
• In general, this
becomes
 E  lim
Ai 0
 E  A
i
i


surface
E  dA
Electric Flux Calculations
• The surface integral means the integral
must be evaluated over the surface in
question
• In general, the value of the flux will
depend both on the field pattern and on
the surface
• The units of electric flux will be N.m2/C2
Electric Flux, Closed Surface
• The vectors ΔAi
point in different
directions
– At each point, they
are perpendicular to
the surface
– By convention, they
point outward
Closed surfaces
E 
ˆ
  E dA
 E ndA
n̂ or A point in the direction outward from the closed surface
+Q +
–
– 3Q
+Q
Negative Flux
Zero Flux
Flux from a point charge through
a closed sphere
E 
ˆ
  E dA
 E ndA
E nˆ  E nˆ  E
kq
E  2 rˆ  E  constant
r
E 
ˆ
 E  dA  E A
 E ndA
kq
q
2
 E  2 4r  4kq 
r
o
sphere
Gauss’s Law
Gauss asserts that the proceeding calculation for the flux
from a point charge is true for any charge distribution!!!
E 
 E dA  4kq
enclosed
q enclosed

o
This is true so long as Q is the charge
enclosed by the surface of integration.
A few questions
+Q +
–
– 3Q
q enclosed
 E   E dA 
o
• If the electric field is zero for all points on the surface, is
the electric flux through the surface zero?
• If the electric flux is zero, must the electric field vanish for
all points on the surface?
• If the electric flux is zero for a closed surface, can there be
charges inside the surface?
• What is the flux through the surface shown? Why?
Example P24.1
An electric field with a magnitude of 3.50 kN/C
is applied along the x axis. Calculate the
electric flux through a rectangular plane 0.350
m wide and 0.700 m long assuming that
(a) the plane is parallel to the yz plane;
(b) the plane is parallel to the xy plane;
(c) the plane contains the y axis, and its normal
makes an angle of 40.0° with the x axis.


E  EA cos  3.50  103  0.350  0.700 cos0  858 N  m 2 C
  90.0


E  0
E  3.50  103  0.350  0.700 cos40.0  657 N  m
2
C
Example P24.9
The following charges are located inside a submarine:
5.00 μC, –9.00 μC, 27.0 μC, and –84.0 μC.
(a) Calculate the net electric flux through the hull of the
submarine.
(b) Is the number of electric field lines leaving the
submarine greater than, equal to, or less than the
number entering it?
E 
qin  5.00 C  9.00 C  27.0 C  84.0 C 
6



6.
89

10
N m
0
8.85  1012 C 2 N  m 2
2
C2
Example P23.17
A point charge Q = 5.00 μC is located at the center of a
cube of edge L = 0.100 m. In addition, six other identical
point charges having q = –1.00 μC are positioned
symmetrically around Q as shown in Figure P24.17.
Determine the electric flux through one face of the cube.
 E one face 
Q 6q
6 0

 5.00  6.00  106 C  N  m 2
6  8.85  1012 C 2
 18.8 kN  m
2
C
Applying Gauss’s Law
• Select a surface
– Try to imagine a surface where the electric field is constant
everywhere. This is accomplished if the surface is equidistant from
the charge.
– Try to find a surface such that the electric field and the normal to
the surface are either perpendicular or parallel.
• Determine the charge inside the surface
• If necessary, break the integral up into pieces and sum
the results.
– e.g. For a cylinder,
 dA 

top
dA 

bottom
dA 

side
dA  2  r 2   2rh
h
r
Example – a line of charge
E
r
total charge

total length
L
1. Find the correct closed surface
2. Find the charge inside
that closed surface
q enclosed
 E   E dA 
o
Example – a charged plane
E
+Q
total charge

total area
1. Find the correct closed surface
2. Find the charge inside
that closed surface
Qenclosed
 E   E dA 
o
Example – a solid sphere of charge
+Q uniformly distributed
total charge

total volume
E
Inside the charged sphere:
a
r
ra
1. Find the correct closed surface
2. Find the charge inside
that closed surface
q enclosed
 E   E dA 
o
Example – a solid sphere of charge
+Q uniformly distributed
total charge

total volume
E
Outside the charged sphere:
a
r
ra
1. Find the correct closed surface
2. Find the charge inside
that closed surface
q enclosed
 E   E dA 
o
Example – a conducting conductor
• Insulators, like the previous charged sphere, trap excess
charge so it cannot move.
• Conductors have free electrons not bound to any atom.
The electrons are free to move about within the material. If
excess charge is placed on a conductor, the charge winds
up on the surface of the conductor. Why?
• The electric field inside a conductor is always zero.
• The electric field just outside a conductor is perpendicular
to the conductors surface and has a magnitude, /o
Einside = 0
• Consider a conducting slab in
an external field E
• If the field inside the conductor
were not zero, free electrons in
the conductor would
experience an electrical force
• These electrons would
accelerate
• These electrons would not be
in equilibrium
• Therefore, there cannot be a
field inside the conductor
Einside = 0, cont.
• Before the external field is applied, free
electrons are distributed throughout the
conductor
• When the external field is applied, the
electrons redistribute until the magnitude of
the internal field equals the magnitude of the
external field
• There is a net field of zero inside the
conductor
• This redistribution takes about 10-15s and can
be considered instantaneous
Charge Resides on the
Surface
• Choose a gaussian surface
inside but close to the actual
surface
• The electric field inside is
zero (prop. 1)
• There is no net flux through
the gaussian surface
• Because the gaussian
surface can be as close to
the actual surface as
desired, there can be no
charge inside the surface
Charge Resides on the
Surface, cont
• Since no net charge can be inside the
surface, any net charge must reside on
the surface
• Gauss’s law does not indicate the
distribution of these charges, only that it
must be on the surface of the conductor
Field’s Magnitude and
Direction
• Choose a cylinder as
the gaussian surface
• The field must be
perpendicular to the
surface
– If there were a parallel
component to E,
charges would
experience a force and
accelerate along the
surface and it would
not be in equilibrium
Field’s Magnitude and
Direction, cont.
• The net flux through the gaussian
surface is through only the flat face
outside the conductor
– The field here is perpendicular to the
surface
• Applying Gauss’s law
σA
σ
 E  EA 
and E 
εo
εo
Conductors in Equilibrium,
example
• The field lines are
perpendicular to
both conductors
• There are no field
lines inside the
cylinder
Example – a solid conducting sphere of
charge surrounded by a conducting shell
+2Q on inner sphere
-Q on outer shell
E
c
b
r
a
Find the Electric Field:
ra
arb
r c
br c
1. Find the correct closed surface
2. Find the charge inside
that closed surface
q enclosed
 E   E dA 
o
Example P24.31
Consider a thin spherical shell of radius
14.0 cm with a total charge of 32.0
μC distributed uniformly on its
surface. Find the electric field
(a) 10.0 cm and
(b) 20.0 cm from the center of the
charge distribution.
k Q  8.99  10  32.0  10 
E

 7.19 M N
6
9
e
2
r
 0.200
2
C
Example P24.35
A uniformly charged, straight filament 7.00 m in
length has a total positive charge of 2.00 μC. An
uncharged cardboard cylinder 2.00 cm in length
and 10.0 cm in radius surrounds the filament at its
center, with the filament as the axis of the
cylinder. Using reasonable approximations, find
(a) the electric field at the surface of the cylinder
and
(b) the total electric flux through the cylinder.

9
2ke 2 8.99  10 N  m
E

r
2


C 2  2.00  106 C 7.00 m 


0.100 m
E  EA cos  E  2 r  cos0


E  51.4 kN C ,radially outw ard
E  5.14  104 N C 2  0.100 m   0.020 0 m   1.00  646 N  m 2 C
Example P24.43
A square plate of copper with 50.0-cm sides
has no net charge and is placed in a region
of uniform electric field of 80.0 kN/C
directed perpendicularly to the plate. Find
(a) the charge density of each face of the plate
and
(b) the total charge on each face.



  8.00  104 8.85  1012  7.08  107 C m 2


Q   A  7.08  107  0.500 C
2
Q  1.77  107 C  177 nC
Example P24.47
A long, straight wire is surrounded by a hollow
metal cylinder whose axis coincides with that of
the wire. The wire has a charge per unit length of λ,
and the cylinder has a net charge per unit length of
2λ. From this information, use Gauss’s law to find
(a) the charge per unit length on the inner and outer
surfaces of the cylinder and (b) the electric field
outside the cylinder, a distance r from the axis.
0    qin
qin
 
E
2ke  3  6ke
3


radially outw ard
r
r
2 0 r
3