Transcript Presentation 01

Workshop: Using Visualization
in Teaching Introductory E&M
AAPT National Summer Meeting, Edmonton, Alberta,
Canada.
Organizers: John Belcher, Peter Dourmashkin,
Carolann Koleci, Sahana Murthy
P01 - 1
MIT Class:
Particle Interactions:
Coulomb’s Law
P01 - 2
Gravitational Vector Field
P01 - 3
Example Of Vector Field:
Gravitation
Gravitational Force:
Mm
Fg  G 2 rˆ
r
Gravitational Field:
GMm / r 2
M
g

rˆ  G 2 rˆ
m
m
r
Fg
M : Mass of Earth
P01 - 4
Example Of Vector Field:
Gravitation
Gravitational Field:
M
g  G 2 rˆ
r
Fg  mg
Created by M
Felt by m
rˆ :
r:
unit vector from M to m
vector from M to m

r
rˆ 
r
M
 g  G 3 r
r
M : Mass of Earth
USE THIS FORM!
P01 - 5
The Superposition Principle
Net force/field is vector sum of forces/fields
Example:
F13 / g13
1
F3  F13  F23
In general:
N
F j   Fij
i1
2
P01 - 6
In Class Problem
Find the gravitational
field g at point P
Bonus: Where would
you put another mass
m to make the field g
become 0 at P?
M
Use g  G 3 r
r
NOTE: Solutions will be posted within two days of class
P01 - 7
From Gravitational to
Electric Fields
P01 - 8
Electric Charge (~Mass)
Two types of electric charge: positive and negative
Unit of charge is the coulomb [C]
Charge of electron (negative) or proton (positive) is
19
e,
e  1.602 10 C
Charge is quantized
Q   Ne
Charge is conserved
n  p  e 

e  e  


P01 - 9
Electric Force (~Gravity)
The electric force between charges q1 and q2 is
(a) repulsive if charges have same signs
(b) attractive if charges have opposite signs
Like charges repel and opposites attract !!
P01 - 10
Coulomb's Law
Coulomb’s Law: Force
on q2 due to interaction
between q1 and q2
q1q2
F12  ke 2 rˆ
r
ke 
rˆ :
r:
1
4 0
 8.9875 109 N m 2 /C2
unit vector from q1 to q2
vector from q1 to q2

r
rˆ 
r
q1q2
 F12  ke 3 r
r
P01 - 11
Coulomb's Law: Example
q3 = 3 C
a=1m
q1 = 6 C

F32  ?

r32
r32 
q2 = 3 C
r  1m

1
2
ˆi 

3
2
r
F32  ke q3q2 3  9 109 N m 2 C2  3C  3C 
r
81109 ˆ

i  3ˆj
2



ˆj m
1
2
 ˆi  3ˆj m
1m 
3
N
P01 - 12
The Superposition Principle
Many Charges Present:
Net force on any charge is vector sum of forces
from other individual charges
Example:
F3  F13  F23
In general:
N
F j   Fij
i1
P01 - 13
Electric Field (~g)
The electric field at a point P due to a charge
q is the force acting on a test charge q0 at
that point P, divided by the charge q0 :
Eq ( P) 
For a point charge q:
Fqq0
q0
q
Eq ( P)  ke 2 rˆ
r
Units: N/C, also Volts/meter
P01 - 14
Superposition Principle
The electric field due to a collection of N point
charges is the vector sum of the individual electric
fields due to each charge
N
Etotal  E1  E2  .....   Ei
i1
P01 - 15
Gravitational & Electric Fields
SOURCE:
Mass Ms
Charge qs (±)
CREATE:
Ms
g  G 2 rˆ
r
qs
E  ke 2 rˆ
r
Fg  mg
FE  qE
FEEL:
This is easiest way to picture field
P01 - 16
PRS Question:
Electric Field
P01 - 17
PRS: Electric Field
:20
Two opposite charges are placed on a line as shown
below. The charge on the right is three times larger
than the charge on the left. Other than at infinity,
where is the electric field zero?
0%
0%
0%
0%
0%
0%
1.
2.
3.
4.
5.
6.
Between the two charges
To the right of the charge on the right
To the left of the charge on the left
The electric field is nowhere zero
Not enough info – need to know which is positive
I don’t know
P01 - 18
PRS Answer: Electric Field
Answer: 3. To the left of the charge on the left
Between: field goes from source to sink.
On right: field dominated by qR (bigger & closer).
On left: because qL is weaker, its “push” left will
somewhere be balanced by qR’s “pull” right
P01 - 19
Electric Field Lines
1. Join end-to-end infinitesimal vectors representing E…the curve
that results is an electric field line (also known as line of force).
2. By construction then, the direction of the E field at any given
point is tangent to the field line crossing that point.
3. Field lines point away from positive charges and terminate on
negative charges.
4. Field lines never cross each other.
5. The strength of the field is encoded in the density of the field
lines.
P01 - 20
PRS Questions:
Electric Field
P01 - 21
PRS: Force
The force between
the two charges is:
50%
25%
25%
0%
1.
2.
3.
4.
Attractive
Repulsive
Can’t tell without more information
I don’t know
P01 - 22
PRS Answer: Force
The force between
the two charges is:
2) Repulsive
One way to tell is to notice that they both must
be sources (or sinks). Hence, as like particles
repel, the force is repulsive.
You can also see this as tension in the field lines
P01 - 23
PRS: Field Lines
Electric field lines show:
1. Directions of forces that exist in space at all
times.
2. Directions in which charges on those lines will
67%
accelerate.
3. Paths that charges will follow.
4. More than one of the above.
33%
5. I don’t know.
Remember: Don’t pick up until you are ready to answer
0%
5
4
3
0%
2
1
0%
P01 - 24
PRS Answer: Field Lines
Answer: 2. Directions charges accelerate.
NOTE: This is different than flow lines (3).
Particles do NOT move along field lines.
P01 - 25
In-Class Problem
P
ĵ
s
q
d
î
q
Consider two point charges of equal magnitude but
opposite signs, separated by a distance d. Point P
lies along the perpendicular bisector of the line
joining the charges, a distance s above that line.
What is the E field at P?
P01 - 26
Two PRS Questions:
E Field of Finite # of
Point Charges
P01 - 27
PRS: Equal Charges
1. E 
3. E 
Electric field at P is:
1
2
3
4
5
2ke qs
 2 d 
s  4 


2
3/ 2
ˆj
2. E  
3/ 2
ˆj
4. E  
2ke qd
 2 d2 
s  4 


2ke qd
 2 d2 
s  4 


ˆi
3/ 2
2ke qs
 2 d2 
s  4 


3/ 2
ˆi
75%
25%
3
2
5. I Don't Know
1
0%
0%
0%
5
2.
3.
4.
5.
4
1.
P01 - 28
PRS Answer: Equal Charges
Electric field at P is:
1. E 
2ke qs
 2 d 
s  4 


2
3/ 2
There are a several ways to see this. For example,
consider d0. Then,
2q ˆ
E  ke 2 j
s
which is what we want (sitting above a point
charge with charge 2q)
ˆj
P01 - 29
PRS: 5 Equal Charges
equal positive charges q sit at the vertices of a
1.Six
regular hexagon with sides of length R. We remove
the bottom charge. The electric field at the center of
the hexagon (point P) is:
1
2.
2
3.
3
4.
4
5.
5
6.
6
100%
0%
0%
6
0%
5
0%
4
0%
3
ˆj
2kq ˆ
2. E   2 j
R
kq ˆ
4. E   2 j
R
6. I Don't Know
2
ˆj
1
2
3
4
5
6
1
2kq
1. E  2
R
kq
3. E  2
R
5. E  0
1.
2.
3.
4.
5.
6.
P01 - 30
PRS Answer: 5 Equal Charges
kq ˆ
Answer : 4. E   2 j
R
• E fields of the side pairs cancel (symmetry)
• E at center due only to top charge (R away)
• Field points downward
Alternatively:
• “Added negative charge” at bottom
• R away, pulls field down
P01 - 31
Charging
P01 - 32
How Do You Get Charged?
• Friction
• Transfer (touching)
• Induction
+q
-
Neutral
+
+
+
+
P01 - 33
Demonstrations:
Instruments for
Charging
P01 - 34
Electric Dipoles
A Special Charge Distribution
P01 - 35
Electric Dipole
Two equal but opposite charges +q and –q,
separated by a distance 2a
Dipole Moment
q
p
2a
-q
p
p  charge×displacement
 q×2aˆj  2qaˆj
points from negative to positive charge
P01 - 36
Why Dipoles?
Nature Likes To Make Dipoles!
Animation
P01 - 37
Dipoles make Fields
P01 - 38
Electric Field Created by Dipole
Thou shalt use
components!
rˆ
r
x ˆ y ˆ
 3  3 i 3 j
2
r
r
r
r
 x x 
E x  ke q  3  3 
 r r 


x
x

 ke q 

  x 2  ( y  a)2  3/ 2  x 2  ( y  a)2  3/ 2 


 



 y y 
ya
ya



E y  ke q  3  3   k e q
2
2 3/ 2
2
2 3/ 2 

r
r


 x  ( y  a) 
 
 
  x  ( y  a) 

P01 - 39
PRS Question:
Dipole Fall-Off
P01 - 40
PRS: Dipole Field
As you move to large distances r away from
a dipole, the electric field will fall-off as:
100%
0%
0%
0%
1.
2.
3.
4.
1/r2, just like a point charge
More rapidly than 1/r2
More slowly than 1/r2
I Don’t Know
P01 - 41
PRS Answer: Dipole Field
Answer: 2) More rapidly than 1/r2
We know this must be a case by thinking
about what a dipole looks like from a large
distance. To first order, it isn’t there (net
charge is 0), so the E-Field must decrease
faster than if there were a point charge there.
P01 - 42
Point Dipole Approximation
Take the limit r  a
Finite Dipole
You can show…
3p
Ex 
sin  cos 
3
4 0 r
Ey 
Point Dipole
p
4 0 r
 3cos   1
2
3
P01 - 43
Shockwave for Dipole
Dipole Visualization
P01 - 44
Dipoles feel Fields
P01 - 45
Demonstration:
Dipole in Field
P01 - 46
Dipole in Uniform Field
E  Eˆi
p  2qa(cos ˆi  sin  ˆj)
Total Net Force:
Fnet  F  F  qE  (q)E  0
Torque on Dipole:
τ  r F  pE
  rF sin( )   2a  qE  sin( )  pE sin( )
p tends to align with the electric field
P01 - 47
Torque on Dipole
Total Field (dipole + background)
shows torque:
Animation
• Field lines transmit tension
• Connection between dipole field and
constant field “pulls” dipole into alignment
P01 - 48
PRS Question:
Dipole in Non-Uniform Field
P01 - 49
PRS: Dipole in Non-Uniform Field
E
A dipole sits in a non-uniform
electric field E
Due to the electric field this dipole will
feel:
67%
33%
0%
0%
1.
2.
3.
4.
force but no torque
no force but a torque
both a force and a torque
neither a force nor a torque
P01 - 50
PRS Answer: Non-Uniform Field
E
Answer: 3. both force and torque
Because the field is non-uniform, the forces on
the two equal but opposite point charges do
not cancel.
As always, the dipole wants to rotate to align
with the field – there is a torque on the dipole
as well
P01 - 51
Continuous Charge
Distributions
P01 - 52
Continuous Charge Distributions
Break distribution into parts:
V
Q    qi   dq
i
V
E field at P due to q
q
dq
 E  ke 2 rˆ  d E  ke 2 rˆ
r
r
Superposition:
E P  ?

E   E  dE
P01 - 53
Continuous Sources: Charge Density
R
dQ   dV
Volume  V   R 2 L
L
w
Area  A  wL
L
Length  L
L
Q

V
dQ   dA
Q

A
dQ   dL
Q

L
P01 - 54
Examples of Continuous Sources:
Line of charge
dQ   dL
Length  L
Q
L

L
Link to
applet
P01 - 55
Examples of Continuous Sources:
Line of charge
dQ   dL
Length  L
Q
L

L
Link to
applet
P01 - 56
Examples of Continuous Sources:
Ring of Charge
Q

dQ   dL
2 R
Link to
applet
P01 - 57
Examples of Continuous Sources:
Ring of Charge
Q

dQ   dL
2 R
Link to
applet
P01 - 58
Example: Ring of Charge
P on axis of ring of charge, x from center
Radius a, charge density .
Find E at P
P01 - 59
Ring of Charge
1) Think about it
E  0 Symmetry!
Mental Picture…
2) Define Variables
dq   dl    a d 
r  a x
2
2
P01 - 60
Ring of Charge
3) Write Equation
rˆ
r
dE  ke dq 2  ke dq 3
r
r
dq   a d
r  a2  x2
x
dEx  ke dq 3
r
P01 - 61
Ring of Charge
dq   a d
4) Integrate
r  a2  x2
x
Ex   dEx   ke dq 3
r
x
 ke 3  dq
r
Very special case: everything except dq is constant
 dq  
2
0
2
a d  a  d    a2
0
Q
P01 - 62
Ring of Charge
5) Clean Up
x
E x  ke Q 3
r
E x  ke Q
E  ke Q
x
a
2
x
2

3/ 2
x
a
2
x
2

3/ 2
ˆi
6) Check Limit a  0
E x  ke Q
x
x 
2
3/ 2
ke Q
 2
x
P01 - 63
In-Class: Line of Charge
ĵ
r̂
P
î
s

L
2

L
2
Point P lies on perpendicular bisector of uniformly
charged line of length L, a distance s away. The
charge on the line is Q. What is E at P?
P01 - 64
Hint: Line of Charge
ĵ
r̂

î
P
r  s 2  x 2
s


L
2
x
dq   dx 
dx 

L
2
Typically give the integration variable (x’) a “primed”
variable name. ALSO: Difficult integral (trig. sub.)
P01 - 65
E Field from Line of Charge
Q
ˆj
E  ke
2
2
1/ 2
s( s  L / 4)
Limits:
Qˆ
lim E  ke 2 j
s  L
s
Qˆ
ˆ
j  2ke j
lim E  2ke
s  L
Ls
s
Point charge
Infinite charged line
P01 - 66
In-Class: Uniformly Charged Disk
( x  0)
P on axis of disk of charge, x from center
Radius R, charge density .
Find E at P
P01 - 67
Disk: Two Important Limits
Edisk

 
x

1
2
2
2 o 
x

R



1/ 2

 ˆi


Limits:
lim Edisk
x  R
Qˆ

i
2
4 o x
***
lim Edisk
x  R
1
 ˆ

i
2 o
Point charge
Infinite charged plane
P01 - 68
Scaling: E for Plane is Constant
1)
2)
3)
4)
Dipole:
E falls off like 1/r3
Point charge: E falls off like 1/r2
Line of charge: E falls off like 1/r
Plane of charge: E constant
P01 - 69