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Lecture 5-1
Electric flux
To state Gauss’s Law in a quantitative form,
we first need to define Electric Flux.
# of field lines N
= density of field lines
x “area”
where “area” = A2 x cos 

Lecture 5-2
Why are we interested in electric flux?
 E is closely related to the charge(s) which cause it.
Consider Point charge Q
If we now turn to our previous
discussion and use the analogy to
the number of field lines, then the
flux should be the same even when
the surface is deformed. Thus should
only depend on Q enclosed.
Lecture 5-3
Gauss’s Law: Quantitative Statement
The net electric flux through any closed surface equals the
net charge enclosed by that surface divided by 0.
 E ndA  
E

Qenclosed
0
How do we use this equation??
The above equation is TRUE always but it doesn’t
look easy to use.
BUT - It is very useful in finding E when the physical situation
exhibits a lot of SYMMETRY.
Lecture 5-4
READING QUIZ 1
Consider a spherical conducting shell with an outer radius b and inner
radius a. The shell is insulated from its surroundings and has a net charge
of +Q. A point charge –q is inserted into the shell without touching the
shell and placed at r = a/2. Which statement is correct ?
A| The net charge on the outer surface does not change.
B| A charge –q is induced on the inner surface of the shell.
C| When the point charge is moved from r = a/2 to r = 0 the charge
distribution on the outer surface does not change.
D| The net charge on the outside of the shell is Q+q
Lecture 5-5
Infinitely long uniformly charged line
 E   (end caps)  ( side)
 ( side)
 E  2 rh
Gauss’s Law:
h
E 
0

2k 
E

2 r 0
r
Same result but much less work!
E
h
r
Lecture 5-6
Uniformly charged thin, infinite sheet

Gauss’s Law!
 E   (end caps)  ( side)
 (end caps)
 2  EA
Qenc
0
A

0

E
2 0
If E=0 on one side, then what
is E between the plates?
A
h
Lecture 5-7
Thin sheet of any charge distribution
tiny disk
 E  ( E ' E disk  )
 ( E ' E disk  )
 E disk   E disk 
    

 
n
 2 0  2 0  

 n
0
E L  E ' E disk 
Just to left of disk
E R  E ' E disk 
Just to right of disk

E n 
0
Lecture 5-8
Physics 241 –Warm-up quiz
The magnitude of the electric field from a single
charged plane of charge density  and infinite area
is E = 0). What is the magnitude of the field
at point A,B and C ? Note that the planes are
infinitely large.


A). E0 in A and C but E= 0 to
the left in B
(B). E= 0 to the left in A and C
but E=0 in B.
(C). E= 0 to the left in A and C
but E=  0 in B.
D). E0 in A and C but E= 0 to
the right in B
A
B
C
Lecture 5-9
Charges and fields of a conductor
• In electrostatic equilibrium, free
charges inside a conductor do not move.
Thus, E = 0 everywhere in the interior
of a conductor.
• Since E = 0 inside, there are no net
charges anywhere in the interior. Net
charges can only be on the surface(s).
 
0
The electric field must be
perpendicular to the surface just
outside a conductor, since, otherwise,
there would be currents flowing along
the surface.
Lecture 5-10
DOCCAM 2
Lecture 5-11
Lecture 5-12
Electrostatic Shielding (Faraday Cage)
• E=0 in empty cavity in a good conductor. Why?
If there are E field lines, then moving a charge from one
end to the other against the E line through the cavity must
require some work W.
On the other hand since E is perpendicular to surface,
W=0 if you move the charge along the surface instead. But
the work should be independent of the path used.
=> Contradiction!
X
Lecture 5-13
DOCCAM 2
Lecture 5-14
Lecture 5-15
Electric Field of a Conductor: Examples
Hollow region
Net charge Q ( > 0)
on the conductor
Net charge
–q < 0
Hollow region
Field ?
Net charge
induced = +q
+ +
+
+
+
+
Net charge (Q-q) on
the outside surface
Lecture 5-16
Cavity in a spherical conductor
q
Total
induced
charge -q
• With +q in the cavity of a
conductor, a net charge –q is
induced on the inside surface of
the cavity.
• If +q is at the center of the spherical
cavity, E is radial by symmetry.
Uniform charge distributions.
• As long as the E contribution
due to all the interior charges,
measured at the outer surface is
zero, then moving the +q charge
does not affect the exterior
surface charge distribution.
Electrostatic shielding
Lecture 5-17
Electrostatic Shielding (Continued)
If you move charge q in the cavity, the
exterior electric fields and the exterior
charge distribution are not affected.
q
Conducting shell electrostatically
shields its exterior from changes
on the inside.
Q’’
Add Q’
+ +
+
+
+
+
If you now add charge Q’ to the
conductor and/or Q’’ on the outside
of the conductor, the interior
electric fields do not change.
Conducting shell electrostatically
shields its interior from changes
on the outside, too.
Lecture 5-18
DOCCAM 2
Lecture 5-19
Lecture 5-20
©2008 by W.H. Freeman and Company
Lecture 5-21
Two conducting plates
In isolation, the electric field near a
charged conducting plate is given by
1
E
0
• properties of conductors
• and, superposition
What happens if these two plates are
pulled toward each other?
Lecture 5-22
Two conducting plates continued
1
0
1
0
2 1   
 
0  0 
• The excess charge on one plate attracts the
excess charge on the other plate.
• The surface charge density vanishes on
the outer faces but is doubled on the inner
faces.
• The magnitude of the electric field is twice
as much as that near an isolated plate and
is zero everywhere else.
Case (c) is not a simple superposition of cases
(a) and (b), because the electric charges are
re-distributed on both plates when they are
brought near each other.
Lecture 5-23
Physics 241 – 10:30 Quiz 2
The magnitude of the electric field from a single
charged plane of charge density  and infinite area
is E = 0). What is the magnitude of the field
at point A on the right side of the two infinitely
large planes of uniformly charged?
a)
b)
c)
d)
e)
   0) to the right
3   0) to the left
  0 to the right
  0) to the left
3   0) to the right
e = 1.610-19 C
0 = 8.8510-12 C2/(Nm2)


A
Lecture 5-24
Physics 241 – 11:30 Quiz 2
The magnitude of the electric field from a single
charged plane of charge density  and infinite area
is E = 0). What is the magnitude of the field
at point A between the two infinitely large planes
of uniformly charged?
a)
b)
c)
d)
e)
   0) to the right
3   0) to the left
  0 to the right
  0) to the left
  0 to the left
e = 1.610-19 C
0 = 8.8510-12 C2/(Nm2)


A
Lecture 5-25
Physics 241 –Quiz 3c
The magnitude of the electric field from a single
charged plane of charge density  and infinite area
is E = 0). What is the magnitude of the field
at point A on the left side of the two infinitely
large planes of uniformly charged?
a)
b)
c)
d)
e)
   0) to the right
3   0) to the left
  0 to the right
  0) to the left
3   0) to the right
e = 1.610-19 C
0 = 8.8510-12 C2/(Nm2)

A
