### ppt

```Physics 2112
Unit 3: Electric Flux and Field Lines
Today’s Concepts:
A) Electric Flux
B) Field Lines
Gauss’ Law
Unit 3, Slide 1
First….a comment
Electric Field Lines
What the point of all this flux stuff?
To create Guass’ Law so we can
find the E field without doing an
integral
Guass’ Law
+Q
Always true, sometimes not useful
Unit 3, Slide 2
Electric Field Lines
Field Lines
Electric FieldElectric
is a “vector
field”
“Density” represents magnitude
Unit 3, Slide 3
Electric Field Lines in 3D
Electric Field Lines
Point Charge:
3D Density  area of sphere (1/R2)
Unit 3, Slide 4
Electric Field Lines for multiple charges
Electric Field Lines
+
-
Just add E-Field vectors at every point
Unit 3, Slide 5
CheckPoint: Field Lines - Two Point Charges 1
Compare the magnitudes of the two charges
o |Q1| > |Q2|
o |Q1| = |Q2|
o |Q1| < |Q2|
o There is not enough information to determine the relative magnitudes of
the charges.
Unit 3, Slide 6
CheckPoint: Field Lines - Two Point Charges 1
Compare the magnitudes of the two charges
A. |Q1| > |Q2|
B. |Q1| = |Q2|
C. |Q1| < |Q2|
D. There is not enough information to determine the relative magnitudes of
the charges.
Unit 3, Slide 7
CheckPoint: Field Lines - Two Point Charges 3
Compare the magnitudes of the electric fields at points A and B.
A. |EA| > |EB|
B. |EA| = |EB|
C. |EA| < |EB|
D. There is not enough information to determine the relative magnitudes of
the fields at A and B.
Unit 3, Slide 8
Flux
E
Q
A
Same number of field lines but
different areas.
Flux  Amount of electric field passing through an
surface.  Count field lines.
F = |E|*|A|*cosQ.
Unit 3, Slide 9
Electric Flux “Counts Field Lines”
DATA:
WITH
OWN
“I’m very confused bySAMPLE
the general
concepts
of fluxREPLACE
through surface
areas.
 
F S   E  dA
S
Flux through
surface S


Integral of E  d A
on surface S
Unit 3, Slide 10
Example 3.1: Flux through a plate
|E|=12N/C
Q
A
30o
A 3cm X 3cm copper plate is placed in a constant
horizontal electric field with a strength of 12N/C at an
angle of 30o to the horizontal.
What is the electric flux through the plate?
Unit 3, Slide 11
CheckPoint: Flux from Uniformly Charged Rod
An infinitely long charged rod has uniform charge density of l, and passes
through a cylinder (gray). The cylinder in case 2 has twice the radius and half
the length compared to the cylinder in case 1.
Compare the magnitude of the
flux through the surface of the
cylinder in both cases.
A. Φ1 = 2 Φ2
B. Φ1 = Φ2
C. Φ1 = 1/2 Φ2
D. None of these
TAKE s TO BE RADIUS !
Unit 3, Slide 12
 Matters:
Direction
E

E

E

dA

dA

dA

E

E
FS 

dA

E

For
 a closed surface,
d A points outward

dA

E

E


E  dA  0
S
Unit 3, Slide 13
 Matters:
Direction
E

E

E

dA

dA

dA

E

E
FS 

dA

E

For
 a closed surface,
d A points outward

dA

E

E


E  dA  0
S
Unit 3, Slide 14
Gauss’ Law for simple point charge
Gauss Law


FS 

E  dA 
closed
surface
FS 

1
Q
4  o R
2
4 R
2
 Q

E  d A  enclosed
closed
surface
0
Unit 3, Slide 15
CheckPoint Results: Flux Point Charge (Sphere) 1
A positive charge (blue) is contained inside a spherical shell (black).
How does the electric flux through the two surface elements, dΦA and dΦB change
when the charge is moved from position 1 to position 2?
o dΦA increases and dΦB decreases
o dΦA decreases and dΦB increases
o Both dΦA and dΦB do not change
Unit 3, Slide 16
A positive
charge (blue) is
contained
insidePoint
a sphericalCharge
shell (black).(Sphere)
CheckPoint:
Flux
from
2
How does the flux ΦE through the entire surface change when the charge is
moved from position 1 to position 2?
A. ΦE increases
B. ΦE decreases
C. ΦE does not change
Unit 3, Slide 17
Think of it this way:
+Q
+Q
1
2
The total flux is the same in both cases (just the total number of lines)
The flux through the right hemisphere is smaller for case 2.
Unit 3, Slide 18

 Gauss
Q enclosed
Things to notice
Law
F
E  dA 
S

closed
surface
0
If Qenclosed is the same, the flux has to be
the same, which means that the
integral must yield the same result for
any surface.
Unit 3, Slide 19
“Gausses law and what concepts are most important that we know .”
Q enclosed Gauss Law
Things to notice
 E  dA 
0
In cases of high symmetry it may be possible to bring E outside the
integral. In these cases we can solve Gauss Law for E
E  d A  EA 
E 
Q enclosed
0
Q enclosed
A 0
So - if we can figure out Qenclosed and the area of the
surface A, then we know E!
This is the topic of the next lecture.
Unit 3, Slide 20
Question
-Q
r2
r1
A solid conducting sphere
with net charge –Q and a