Transcript Lecture #5 01/25/05

Announcements
Careers in Physics Event:
Dr. Jeffrey Phillips from EPRI
will discuss working in the field of energy production.
11 AM today Olin Lounge
Seminar:
Dr. Jeffrey Phillips, Electric Power Research Institute
Integrated Coal Gasification Combined Cycle Power Plants :
A Cleaner Coal-to-Electricity Option
4:00 PM Olin 101 Thursday
Pick up the handout
Announcements II
Office Hrs:
None after class today; email for an appointment today
Semester Quiz I is on 02/03/05.
Lectures 1-5; Ch 3,22-24
One {letter-sized} page of notes allowed
Hints: Review {and master} HW, and reading quizzes
Study lecture notes, and do optional problems
for extra practice
Electric Flux and Gauss’ Law
 E   E  dA   E cos dA
E
En

•The electric flux through a closed surface is given by
Gauss’ Law
•Usually you can pick your surface so that the integration doesn’t
need to be done
•We must distinguish between field and flux!
E 
 E  dA  4 k q
e
Special optional Readings
For those of you who are interested, “Div, Grad, Curl and all that”
in Ch 2 has a more quantitative analysis of Gauss’ law.
The handout is from the beginning of that chapter and has a more
quantitative treatment of unit normals and surface integrals than
we did in class. You will not need to know this for the course, but
it might aid your understanding.
Gauss’s Law
charge q
charge 2q
charge -q
charge -2q
Which other charge(s) do we have to include in a gaussian surface so as to contain
the +2q charge and have flux equal to:
Zero ?
+3q/e0?
A) -2q
C) q
-2q/e0 ?
B) -q
D) impossible
IMPOSSIBLE
Today’s Reading Quiz
charge q
charge 2q
charge -q
charge -2q
Consider Gauss's law:
Which of the following is true?
E must be the electric field due to the enclosed charge
If a charge is placed outside the surface, then it cannot affect E on the surface
On the surface E is everywhere parallel to dA
If q = 0 then E = 0 everywhere on the Gaussian surface
If the charge inside consists of an electric dipole, then the integral is zero.
Example
A) q/eo
B) -q/eo
C) 0
Figure 24-29.
•What is the flux through the first surface?
•What is the flux through the second surface?
•What is the flux through the third surface?
•What is the flux through the fourthsurface?
•What is the flux through the fifth surface?
D) 2q/eo
Applying Gauss’s Law
•Can be used to determine total flux through a surface in simple
cases
•Must have a great deal of symmetry to use easily
Charge in an infinite triangular channel
What is flux out of one side?
charge q
E 
q
e0
q
S 
3e 0
  S 1   S 2   S 3   E1   E 2  3 S  0
Applying Gauss’s Law
L
R
r
•Infinite cylinder radius R charge density 
•What is the electric field inside and outside the cylinder?
•Draw a cylinder with the desired radius inside the
cylindrical charge
•Electric Field will point directly out from the axis
qin
V

  AE  2 rLE 
e0
e0
r
E
2e 0
r L

e0
2
Applying Gauss’s Law
L
R
r
•Infinite cylinder radius R charge density 
•What is the electric field inside and outside the cylinder?
•Draw a cylinder with the desired radius outside the
cylindrical charge
•Electric Field will point directly out from the center
  AE  2 rLE 
R2 
E
2e 0 r
qin
e0

V
e0
 R L

e0
2
Applying Gauss’s Law
r
E
3e 0
Sphere volume:
V = 4a3/3
R
r
Sphere area:
A = 4a2
•Sphere radius R charge density . What is E-field inside?
•Draw a Gaussian surface inside the sphere of radius r
What is the magnitude of the
electric field inside the
sphere at radius r?
A) R3/3e0r2
B) r2/3e0R
C) R/3e0
D) r/3e0
  AE  4 r E 
2

V
e0
4 r 

3e 0
3
qin
e0
Conductors in Equilbirum
•A conductor has charges that can move freely
•In equilibrium the charges are not moving
•Therefore, there are no electric fields in a conductor in
equilibrium
F  qE  ma
=0
qin  e 0   e 0  E  dA
=0
•The interior of a conductor never has any charge in it
•Charge on a conductor is always on the surface
Electric Fields near Conductors
•No electric field inside the conductor
•Electric field outside cannot be tangential – must be perpendicular
•Add a gaussian pillbox that penetrates the surface
Area A
  0  0  AE 
Surface charge 
qin
e0

A

E  nˆ
e0
e0
•Electric field points directly out from (or in to) conductor
Conductors shield charges
No net charge
•What is electric field outside
the spherical conductor?
Charge q
•Draw a Gaussian surface
•No electric field – no charge
•Inner charge is hidden – except
Charge -q
Charge +q
E
•Charge +q on outside to
compensate
•Charge distributed uniformly
qrˆ
4e 0 r
2
Conductors shield charges
Charge q
Charge -q
Charge +q
•Charge is induced on the
inner and outer surfaces of the
conducting shell, which hides
the interior charge
•How would the situations
change if the charge inside
were on a conductor?
Example
Suppose that the radius of
the central wire is 26 µm,
the radius of the cylinder
1.3 cm, and the length of
the tube 16 cm. If the
electric field at the
cylinder's inner wall is 2.7
x 104 N/C
•What is the charge enclosed on the wire?
•We know all the physical dimensions!
•We know how the field lines point out from a wire
•We know the field on the interior of the cylinder
Example
Suppose that the radius of
the central wire is 26 µm,
the radius of the cylinder
1.3 cm, and the length of
the tube 16 cm. If the
electric field at the
cylinder's inner wall is 2.7
x 104 N/C
•EA=q/eo=E(2rl)+0+0
•What if we know the charge density, l, and not the
field?
•EA=q/eo=E(2rl)+0+0=ll/eo
Practice Problem I
A cube with 1.40 m edges is oriented as shown in the figure
Suppose there is a charge situated in the middle of
the cube.
• What is the magnitude of the flux through the whole
cube?
• What is the magnitude of the flux through any one
side?
A) q/eo
B) q/4eo
C) 0
D) q/6eo
Practice Problem II
A cube with 1.40 m edges is oriented as shown in the figure
Suppose the cube sits in a uniform electric field of 10i ?
• What is the magnitude of the flux through the whole
cube?
• What is the magnitude of the flux through the top
side?
• How many sides have nonzero flux?
A) q/eo
B) q/4eo
C) 0
D) q/6eo
A) 2 D) 1
B) 4
C) 0