Transcript Lectures6and7

Electric Flux
•To find the flux you need to see how much is going through
perpendicular to the surface
E
•Multiply by cos 
En

E = EnA= EA cos 
•For a non-constant electric field,
or a curvy surface, you have to
integrate over the surface
 E   E  dA   E cos dA
•Usually you can pick your surface so that the integration doesn’t
need to be done given a constant field.
Quiz: Flux and field
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) 2q/eoD) 1
B) 4q/4eo
C) 0
D) q/6eo
Electric Flux
•What is electric flux
E
through surface
surrounding a charge q?
R
charge q
ke q
 4 R E  4 R 2  4 ke q
R
 E  4 ke q
2
2
 E  2 ke q  2 ke q
 4 ke q
Answer is
always 4keq
Gauss’s Law
•Flux out of an enclosed region
depends only on total charge inside
E 
qin
e0
charge q
A positive charge q is set down outside a sphere. Qualitatively, what
is the total electric flux out of the sphere as a consequence?
A) Positive
B) Negative
C) Zero
D) It is impossible to tell from the given information
Gauss’s Law
charge q’’
charge q’
charge q
E 
 E  dA  4 k q  4 k q '  4 k  q  q '
e
e
 4 ke q ''
 E   E  dA  4 ke qin 
e
qin
e0
Quiz: Drawing gaussian surfaces
charge q
charge 2q
charge -q
charge -2q
How do we draw surfaces to contain the +2q charge and have flux?:
Zero ?
+3q/e0?
-2q/e0 ?
Quiz: Flux
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 fourth surface?
•What is the flux through the fifth surface?
D) 2q/eo
Quizzes: Flux II
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
Gauss’ Law and Coulumb’s
•Suppose we had
measured the flux as:
 E  4 ke q
•From Gauss’ law:
 E  4 R E  4 ke q
2
E
R
charge q
Keq
R2
So Gauss’ law
implies
Coulomb’s law
•What if we lived in a
Universe with a different
number of physical
dimensions?
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 a long 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
•No flux through end surfaces
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
•No flux through endcaps
qin
V


  AE  2 rLE
2
R  e0
e0
E
2e 0 r
 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