Download Electric Flux and Field Lines

Physics 2112
Unit 3: Electric Flux and Field Lines
Today’s Concepts:
A) Electric Flux
B) Field Lines
Gauss’ Law
Unit 3, Slide 1
Unit 3, Slide 2
Your Comments
• This would be extremely confusing if I hadn't had calc 3 yet. As it is, the
concept of flux through object is very familiar but some of the terminology
and formulas is not.
• Felt fairly confident in this material. Would just like examples shown using
the formulas.
• I would like to see more calculations done in class like we see with
homework.
• I am a little bit confuse
• Application of Gauss's Law how to determine E from flux if it's not a surface
• is epsilon just a rewritten version of the k constant, or is it separate?
• This one wasn't too bad. Definitely still go over flux.
• While I admired the narrator's enthusiasm for the conclusions drawn in the
prelectures, could you explain why Gauss' Law is so important? It seems like
such a simple formula for being a major basis for much of electromagnetism.
Sure, it makes painful integrals easier, but is there more to it other than that?
• Easier to follow than the last one, cool concept although I still need to
memorize every of the new names and where they are in the pictures.
• The concepts are not difficult when you take the time to read and understand
how it all works. More practice problems certainly wouldn't hurt.
Electricity & Magnetism Lecture 3, Slide 3
k in terms of fundamental constants

Note
k 
Eline
1
4 o

2k


r 2 o
r
Unit 2, Slide 4
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 6
Electric Field Lines
Field Lines
Electric FieldElectric
is a “vector
field”
“Density” represents magnitude
Unit 3, Slide 7
Electric Field Lines in 3D
Electric Field Lines
Point Charge:
Direction is radial
3D Density  area of sphere (1/R2)
Unit 3, Slide 8
Electric Field Lines for multiple charges
Electric Field Lines
+
-
Just add E-Field vectors at every point
Unit 3, Slide 9
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 10
CheckPoint: Field Lines - Two Point Charges 2
What do we know about the signs of the charges from looking at the picture?
A. Q1 and Q2 have the same sign
B. Q1 and Q2 have opposite signs
C. There is not enough information in the picture to determine the relative
signs of the charges
Unit 3, Slide 11
CheckPoint: Field Lines - Two Point Charges 2
What do we know about the signs of
the charges from looking at the
picture?
A.
B.
C.
Q1 and Q2 have the same sign
Q1 and Q2 have opposite signs
There is not enough
information in the picture to
determine the relative signs of
the charges
B) They are opposite because the lines that are leaving one of the charge
go toward the second charge, even though we dont know which one is
positive or negative
C) There needs to be arrows.
Unit 3, Slide 12
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 13
Question
+q
E
A positive charge is placed at rest in an electric field
represented by arrows above. The charge will:
A. Move with a constant acceleration to the right
B. Move with a constant acceleration to the left
C. Move with a constant velocity to the right
D. Move with a constant velocity to the left
E. None of the above
Unit 3, Slide 14
Question
A negative charge is placed at rest in an electric field
represented by arrows above. The charge will:
-q
E
A. Move with a constant acceleration to the right
B. Move with a constant acceleration to the left
C. Move with a constant velocity to the right
D. Move with a constant velocity to the left
E. None of the above
Unit 3, Slide 15
Question
+
- +
-
(a)
(b)
- +
-
(c)
(d)
+
An electrically neutral
dipole is placed in an
external field. In which
situation(s) is the net force
on the dipole zero?
A. (a)
B. (c)
C. (d)
D. (a) and (c)
E. (c) and (d)
Unit 3, Slide 16
Question
+
- +
-
(a)
(b)
- +
-
(c)
An electrically neutral
dipole is placed in an
external field. In which
situation(s) is the net
torque on the dipole zero?
+
(d)
A. (a)
B. (c)
C. (d)
D. (a) and (c)
E. (c) and (d)
Unit 3, Slide 17
Question (Point Charges)
“Telling the difference between positive and negative charges while looking at field lines. Does field
line density from a certain charge give information about the sign of the charge?”
-q
+2q
What charges are inside the red circle?
-Q
-Q
-Q
-2Q
-2Q
+Q
+Q
+2Q
+Q
A
B
C
D
-Q
E
Unit 3, Slide 18
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 20
“I’m very confused by the general concepts of flux through surface areas. please help”
 
F S   E  dA
S
Flux through
surface S
 
Integral of E  dA
on surface S
Unit 3, Slide 21
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 22
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 23
CheckPoint Results: Flux Unif. Charged Rod
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 !
A) Flux is the amount of field lines passing through a surface. In case 2, the rod
only has half of the length as in case 1. With infinitely charged rods, the lines point
directly outwards in all directions of the rod.
B) the charge enclosed by 1 is double 2 but E dot dA is also double so they are
equal
C) There are no electric field lines running through the planes of the cylinders due
to the perpendicular nature of them. They are only released perpendicular to the
cylinder and thus with a larger length, more lines pass through cylinder 1. Width is
irrelevant.
Unit 3, Slide 24
Positive Net Flux
 Matters:
Direction
E

E

E

dA

dA

dA

E

E

dA

E
For
 a closed surface,
dA points outward

dA

E

E
 
F S  E  dA  0

S
Unit 3, Slide 26
 Matters:
Direction
E

E

E

dA

dA

dA

E

E

dA

E
For
 a closed surface,
dA points outward

dA

E

E
 
F S  E  dA  0

S
Unit 3, Slide 27
Gauss’ Law for simple point charge
Gauss Law
 
1
Q
2
F S   E  dA 
4

R
2
4 o R
closed
surface
  Qenclosed
F S   E  dA 
closed
surface
0
Unit 3, Slide 30
CheckPoint Results: Flux Point Charge (Sphere) 1
A positive
charge (blue) is
contained
insidePoint
a sphericalCharge
shell (black).(Sphere)
CheckPoint:
Flux
from
1
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?
A. dΦA increases and dΦB decreases
B. dΦA decreases and dΦB increases
C. Both dΦA and dΦB do not change
Electricity & Magnetism Lecture 3, Slide 31
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?
A. dΦA increases and dΦB decreases
B. dΦA decreases and dΦB increases
C. Both dΦA and dΦB do not change
A) The electric field lines are dense close to the parent charge, so
moving the charge closer to dA will increase the flux through it, will
decreasing the flux through dB.
B) Because the electric field will have a greater magnitude at
dA, the flux at dA will increase because the charge is closer to
dA.
C) The flux does not change when the charge moves. The flux is not
affected when the charge moves. .
Unit 3, Slide 32
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 35
 Q Law
 Gauss
Things to notice
about
F
E  dA  enclosed
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 36
“Gausses law and what concepts are most important that we know .”
  about
QenclosedGauss 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 
Qenclosed
0
Qenclosed
E
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 37
Question
+q
r
+q
2r
A point charge is surrounded by a sphere of radius r. If
the sphere’s radius is then changed to equal 2r, how
will the total electric flux through the sphere change?
A. It will increase by a factor of r
B. It will increase by a factor of r2
C. It will decrease by a factor of 1/r
D. It will decrease by a factor of 1/r2
E. It will stay the same
Unit 3, Slide 38
Question
-Q
r2
r1
A solid conducting sphere
with net charge –Q and a
radius r1. What is the
electric field within the
sphere some distance r2
from the center? (r2<r1)
A. -kQ/r12
B. -kQ/r22
C. -kQ(r2/r1)2
D. Zero
E. None of the above
Unit 3, Slide 39