Download 24-1 A New Look at Coulomb`s Law

24 Gauss’ law
24-1 A New Look at Coulomb’s Law
Gauss’ law relates the electric fields
at points on a (closed) Gaussian
surface and the net charge enclosed
by that surface.
24-2 Flux
(a)The rate Φis equal to v·A
(b)
  v cos A


(c)   vAcos  v  A
(d)A velocity field.Flux means the
product of an area and the field
across that area.
24-3 Flux of an Electric Field
A provisional definition
for the flux of the electric
field for the Gaussian
surface is
 
   E  A
Electric flux through a Gaussian surface
 
   E  dA
The electric flux Φ through a Gaussian
surface is proportional to the net number
of electric field lines passing through that
surface.
Sample Problem 24-1
What is the flux Φ of
The electric field through
This closed surface?
Step one:
 
 
 
 
   E  A   E  dA   E  dA   E  dA
Step two:
 
0
E

d
A

E
cos
180
dA   E  dA   EA


a

b
c

a


 E  dA 
c
 E cos 0dA  EA
 
0
 E  dA   E cos 90 dA  0


b
Step three:
  EA  0  EA  0
Sample Problem 24-2
What is the electric flux
through the right the face,
the left face,and the top
face?
Right face:

dA  dAiˆ

 
 
 r   E  dA   3.0 xiˆ  4.0 ˆj  dAiˆ
 3.0 xdA  3.0 3.0dA  9.0 dA
 36 N  m / C
2
Left face:
 l  12 N  m / C
Top face:
 t   3.0 xiˆ  4.0 ˆj  dAˆj
2

 
 16 N  m / C
2
24-4 Gauss’ Law
Gauss’ law and Coulomb’s law, although
expressed in different forms, are equivalent
ways of describing relation between charge
and electric field in static situations. Gauss’s
law is:
 0  qenc
or
 
 0  E  dA  qenc
Surface S1
The electric field is outward
for all point on this surface.
Surface S2
The electric field is inward
for all point on this surface.
Surface S3
This surface encloses no charge,and
thus qenc=0
Surface S4
This surface encloses no net charge,
because the enclosed positive and
negative charges have equal magnitudes.
Sample Problem 24-3
What is the net electric
flux through the surface
if Q1=q4=+3.1nC,
q2=q5=-5.9nC,
and q3=-3.1nC?

qenc
0

q1  q2  q3
0
 670 N  m / C
2
24-5 Gauss’ Law and Coulomb’s
Law
Gauss’ law as:
 
 0  E  dA   0  EdA  qenc
 0 E  dA  q
 0 E 4r   q
2
1
q
E
40 r 2
Coulomb’s
law
Gauss’ law is equivalent to Coulomb’s law.
24-6 A Charged Isolated
Conductor
If an excess charge is placed
on an isolated conductor,that
amount of charge will move
entirely to the surface of the
conductor .None of the
excess charge will be found
within the body of the
conductor.
An Isolated Conductor with a Cavity
There is no net charge on the cavity walls.
The Conductor Removed
The electric field is set up by the charges and
not by the conductor.The conductor simply
provides an initial pathway for the charges to
take up their position.
The External Electric Field
Conducting surface:

E
0
Sample Problem 24-4
Key idea
The electric flux through the Gaussian surface
must also be zero.The net charge enclosed by
the Gaussian surface must be zero.With a point
charge of -5.0μC within the shell,a charge of
+5.0 μC must lie on the inner wall of the shell.
Can you think of another key idea?
24-7 Applying Gauss’
law:Cylindrical Symmetry
  EA cos 
 E (2rh ) cos 0  E (2rh )
 0  qenc
 0 E 2rh   h

E
20 r
The electric field at any point due to an infinite
line of charge with uniform linear charge
density λis perpendicular to the line of charge
and has magnitude

E
20 r
Where r is the perpendicular distance from
the line of charge to the point.
Sample Problem 24-5
If air molecules break down (ionize) in an
electric field exceeding 3×106N/C,what is the
column?
Key idea
The surface of the column of charge must
 be at
The radius r where the magnitude of E is
3 ×106N/C,because air molecules within that
Radius ionize while those farther out do not.

r
 6m
20 E
Can you think of another key idea?
24-8 Applying Gauss’ law:Planar
Symmetry
nonconducting sheet
The electric field due to an
infinite nonconducting sheet
with uniform surface charge
density σis perpendicular to
the plane of the sheet and has
magnitude

E
2 0
Two Conducting Plates:
E
2 1
0


0
Sample Problem 24-6
Step one:
  
5
E  
 3.84 10 N / C
2 0
  
5
E  
 2.43 10 N / C
2 0
Step two:
El  E   E   1.4 10 N / C
5
Eb  E   E   6.310 N / C
5
24-9 Applying Gauss’
law:Spherical Symmetry
A shell of uniform charge attracts or
repels a charged particle that is outside
the shell as if all the shell’s charge were
concentrated at the center of the shell.
A shell of uniform charge exerts no
electrostatic force on a charged particle
that is located inside the shell.
Spherical shell,field at r ≥R
1
q
E
2
40 r
Spherical shell,field at r ＜R
E 0
Spherical distribution,field at r ≥R
'
1
q
E
2
40 r
Uniform charge,field at r ≤R
E(
q
40 R
3
)r