#### Transcript Lecture01: Introduction, Vectors, Scalar and Vector Fields

```Physics 121: Electricity and Magnetism
Introduction
Syllabus, rules, assignments, exams, etc.


Text: Young & Friedman, University Physics
Homework & Tutorial System: Mastering Physics
Course Content:
•
•
•
•
•
•
5 Weeks: Stationary charges –
– Forces, fields, Electric flux, Gauss’ Law,
– Potential, potential energy, capacitance
2 Weeks: Moving charges –
– Currents, resistance, circuits containing resistance and capacitance,
– Kirchoff’s Laws, multi-loop circuits, RC circuits
2 Weeks: Magnetic fields (static fields due to moving charges)
– Magnetic force on moving charges,
– Magnetic fields caused by currents (Biot-Savart’s and Ampere’e
Laws)
2 Weeks: Induction & Inductance
– Changing magnetic flux (field) produces currents (Faraday’s Law)
– Inductance, LR Circuits
2 – 3 Weeks: AC (LCR) circuits,
– Eelectromagnetic oscillations, Resonance
– Impedance, Phasors
Not covered:
– Maxwell’s Equations - unity of electromagnetism
– Electromagnetic Waves – light, radio, gamma rays,etc
1
Copyright R. Janow – Spring 2016
Physics 121 - Electricity and Magnetism
Lecture 01 - Vectors and Fields
Review of Vectors :
• Components in 2D & 3D. Addition & subtraction
• Scalar multiplication, Dot product, Vector product
Field concepts:
• Scalar and vector fields in math & physics
• How to visualize fields: contours & field lines
• “Action at a distance” fields – gravitation and electro-magnetics.
• Force, acceleration fields, potential energy, gravitational potential
• Flux and Gauss’s Law for gravitational field: a surface integral of
gravitational field
More math:
• Calculating fields using superposition and simple integrals
• Path integral/line integral
• Spherical coordinates – definition
• Example: Finding the Surface Area of a Sphere
• Example: field due to an infinite sheet of mass
2
Copyright R. Janow – Spring 2016
Definition: Right-Handed Coordinate Systems
•
•
•
We always use right-handed
coordinate systems.
In three-dimensions the righthand rule determines which
way the positive axes point.
Curl the fingers of your
RIGHT HAND so they go from
x to y. Your thumb will point
in the positive z direction.
z
y
x
This course uses several right hand rules related to this one!
Copyright R. Janow – Spring 2016
Cartesian & Spherical Polar Coordinates – 3D
+z
Cartesian

r  (x, y, z)

r  x î  yĵ  zk̂

r
z

Polar, 3D r  (r, q, f)
rz  r cos(q)
r  (x 2  y 2  z2 )1 / 2
q
q  " colatitude" , in [0,  ] radians
f  " azimuth" , in [0,2  ] radians
+y
x
z  r cos(q)
r2  x2  y 2  r 2 sin2 (q)
x  r cos(f)  r sin(q) cos(f)
y  r sin(f)  r sin(q) sin(f)
Copyright R. Janow – Spring 2016

| r |  rxy  r sin( q)
f
90o
90o
Polar to Cartesian
P
90o
y
+x
Cartesian to Polar
q  cos1 (z / r )
f  tan-1(y / x)
r  (x 2  y 2  z2 )1/ 2
Surface Integral Example: Show that the surface area of
a sphere A= 4R2 by integrating over the sphere’s surface
Find dA – an area segment on the surface of the sphere,
then integrate on angles f (azimuth) and q (co-latitude).
z
dA  dl  dh

dA  r̂ dA
Where:
• dl is a curved length segment of the circle around
the z-axis (along a constant latitude line)
• dh is a segment along the q direction (along a
constant longitude line)
dl  r sin(q) df
q
r
f
y
dh  r dq
x
q  [0,  ]
Angle range for a full sphere:

A   r̂.dA 
surface
f  [0, 2 ]
Factors into 2 simple 1 dimensional integrations
2

0
0
 
2

0
0
r sin(q) dqdf  r {  df} {  sin(q) dq }
2
2


 2r 2  sin(q) dq  2r 2 [  cos(q)] 0  2r 2 ()[ 1  1]
0
 A  4 r 2
Copyright R. Janow – Spring 2016
Right Handed Coordinate Systems
1-1: Which of these coordinate systems are right-handed?
A.
B.
C.
D.
E.
I and II.
II and III.
I, II, and III.
I and IV.
IV only.
y
y
I.
II.
x
z
z
x
x
x
III.
IV.
y
z
z
y
Ans: D
Copyright R. Janow – Spring 2016
There are 3 Kinds of Vector Multiplication
Multiplication of a vector by a scalar:

sA  sA x î  sA y ĵ

A

sA
vector times scalar  vector whose length is multiplied by the scalar
Dot product (or Scalar product or Inner product):

B
f

A
- vector times vector  scalar
- projection of A on B or B on A
- commutative
 
 
AoB  ABcos( )  BoA  A xBx  A yB y  A zBz
unit vectors measure
perpendicularity:
Copyright R. Janow – Spring 2016
î  ĵ  0,
ĵ  k̂  0, î  k̂  0
î  î  1,
ĵ  ĵ  1, k̂  k̂  1
Vector multiplication, continued
Cross product (or Vector product or Outer product):
- Vector times vector  another vector perpendicular to the
plane of A and B
- Draw A & B tail to tail: right hand rule shows direction of C
  
 
C  A  B  - B  A (not commutative)


magnitude : C  ABsin( f)
B
A



where
f
is
the
smaller
angle
from
A
to
B
f
C
- If A and B are parallel or the same, A x B = 0
- If A and B are perpendicular, A x B = AB (max)
  
 
 
distributiv e rule : A (B
Algebra:
  C)   A  B  A  C
associative rules : sA
  (sB)
 B
  (sA)  B  A
( A  B)  C  A  (B  C)
î  ĵ  k̂, ĵ  k̂  î , î  k̂  - ĵ
Unit vector
representation: î  î  0, ĵ  ĵ  0, k̂  k̂  0
 
A  B  (Ax î  Ay ĵ  A zk̂)  (B x î  By ĵ  Bzk̂)
 (A yBz - AzBy ) î  (A zBx - AxBz )ĵ  (A xB y - AyBx )k̂


  
 

Applications:   r  F
L  r p
F  qE 
Copyright R. Janow – Spring 2016
i
j
k
 
qv  B
What’s a “Field” - Mathematical View
•
•
•
•
A FIELD assigns a value to every point in space (2D, 3D, 4D,….)
It may have nice mathematical properties, like other functions:
E.g. superposition, continuity, smooth variation, multiplication,..
A scalar field f maps a vector into a scalar:
f: R3->R1, f: R2->R1 …
• A scalar quantity is assigned to every point in 3D space
ISOBARS
• Temperature, barometric pressure, potential energy
EQUIPOTENTIALS
•
A vector field g maps a vector into a vector:
g: R3->R3, g: R2->R2
• A 2D/3D vector is assigned to every point in 2D/3D space
• Wind velocity, water velocity (flow), acceleration
• Taxing to the imagination, involved to calculate
Example: map of the velocity
of westerly winds flowing
past mountains
Pick single
altitudes and make
slices to create
maps
Copyright R. Janow – Spring 2016
…
FIELD LINES
“FIELD LINES” (streamlines) show wind direction
Line spacing shows speed: dense  fast
Set scale by choosing how many lines to draw
Lines begin & end only on sources or sinks
Scalar field examples
A scalar field assigns a simple number as
the field value at every point in “space”.
•
•
•
Altitude map shows heights of
points on a mountain as function
of x-y position.
All points on a contours have the
same altitude
•
Temperature map portrays ground-level
temperature as function of x-y position
Maps R2 -> R1
Contours far
apart
Contours
closely
spaced
•
Contours
Grade (or slope) is related to
the horizontal spacing of
contours (vector field)
flatter
steeper
Copyright R. Janow – Spring 2016
Side View
Vector Fields
•
The value of a vector field at every point in
space is a vector – with magnitude and direction
•
A vector field (e.g.,gravitational force) can be
generated by taking the gradient of a scalar
field (e.g.,potential energy).
•
Gradient field lines are perpendicular to the
contours (e.g., lines of constant potential
energy)
•
The steeper the gradient (e.g., rate of change
of gravitational potential energy) the larger the
field magnitude is.
DIRECTION
•
Gradient vectors point along the direction of steepest
descent, which is also perpendicular to the contours.
•
Imagine rain flowing down a mountain. The vectors are
also “streamlines.” Water running down the mountain
Copyright R. Janow – Spring 2016
Side View
Height contours portray constant potential energy U = mgh.
Motion perpendicular to a contour at a point is along the gradient.
•
Slope and grade mean the same
thing. A 15% grade is a slope of
15
lim h / x  dh / dx  0.15
x  0
•
100
Gradient is measured along the path.
For the case at right it would be:
lim h / l  dh / dl   15 /101.1  0.148
x  0
•
15%
Gravitational force component along
path l is the gradient of potential
energy
l
F  dU / dl  d (mgh) / dl  mg dh / dl
h
dh / dl  sin(q)
F  mg sin(q)
•
•
q
x
The GRADIENT of height (or gravitational potential energy) is also a field representing
steepness (or force)
Are the GRADIENTS of scalar fields also scalar fields or are they vector fields?
Copyright R. Janow – Spring 2016
Another scalar field – atmospheric pressure
Isobars: lines
of constant
pressure
How do the isobars affect air motion?
What are the black arrows showing?
Copyright R. Janow – Spring 2016
A related vector field: wind velocity
Wind speed and
direction depend on
Copyright R. Janow – Spring 2016
“Action at a Distance” forces are also called fields
• Place a test mass, test charge, or test current at some test point in a field
• It feels a force due to the presence of remote sources of the field.
• The sources “alter space” at every possible test point.
• The forces (vectors) at a test point due to multiple sources add up
via superposition (individual field vectors add & form the net field).
Field Type
Source
Acts on
Definition
Strength
(dimensions)
mass
another
mass
Force per
unit mass at
test point
ag = F g / m
electrostatic
charge
another
charge
Force per
unit charge
at test point
E=F/q
magnetic
electric
current
.length
another
current
.length
Force per unit
current.length
B ~ F/qv or
F/iL
gravitational
Copyright R. Janow – Spring 2016
Summary: Visualizing Physical Fields
Could be:
• 2 hills,
• 2 charges
• 2 masses
Scalar field: lines of constant field magnitude
•
•
•
Altitude / topography – contour map
Pressure – isobars, temperature – isotherms
Potential energy (gravity, electric)
Vector field: field lines show a gradient
•
•
•
•
Direction shown by TANGENT to field line
Magnitude proportional to line density inversely to distance between lines
Lines start and end on sources and sinks of field
(highs and lows)
Forces are fields, with direction related to
gravitational, electric, or magnetic field
Mass or
negative
charge
Magnetic field
around a wire
carrying current
Summary: Scalar and vector fields in mechanics and E&M:
TYPE
MECHANICS (GRAVITY)
ELECTROSTATICS (CHARGE)
FORCE
(vector)
Gravitational Force = GMm / r2
Coulomb Force = kqQ / r2
SCALAR
FIELDS
Gravitational Potential Energy
Gravitational Potential
(PE / UNIT MASS)
Electric Potential Energy
Electric Potential (volts)
(PE / UNIT CHARGE)
Magnetic P. E. (due to a current)
E = Force / Unit Charge
= “Electric Field”
B = Force / Unit Current x Length
= “Magnetic Field”
ag = Force / Unit Mass
VECTOR
= “Gravitational Field”
FIELDS
= Acceleration
of Gravity “g”
– Spring 2016
MAGNETOSTATICS
(CURRENT)
Magnetic Force = q v X B
Gravitational field of a point mass M
Point mass  Spherical symmtery
The gravitational field at a point is the acceleration of gravity g
(including direction) felt by a test mass at that point
• Move test mass m around to map direction
& strength of force
• Field g = force/unit test mass
• Lines show direction of g is radially inward
(means gravity is always attractive)
• g is large where lines are close together
surfaces of
constant
field & PE
inward force
on test mass m
gA
gB
rb
• From Newton force law:

GM
g   2 r̂
r
2
(Newtons/k g or m/s )
rA
M
gB
• Field lines END on masses (sources)
Where do gravitational field lines BEGIN?
• Gravitation is always attractive, lines BEGIN at r = infinity
Why inverse-square laws? Why not inverse cube, say?
The same ideas apply to electric fields
Copyright R. Janow – Spring 2016
gA
Superposition of fields (gravitational)
•
•
•
“Action-at-a-distance”: gravitational field permeates all of space with force/unit mass.
“Field lines” show the direction and strength of the field – move a “test mass” around to map it.
Field cannot be seen or touched and only affects the masses other than the one that created it.
•
What if we have several masses? Superposition—just vector sum the
individual fields.
M
•
M
M
M
The NET force vectors show the direction and strength of the NET field.
The same ideas apply to electric fields
Copyright R. Janow – Spring 2016
An important idea called Flux (symbol F):
Basically a vector field magnitude x area
- fluid volume or mass flow
- gravitational - electric - magnetic
Definition: differential amount of flux dFg of field ag crossing vector area dA
ag
“unit normal”
n̂
outward and
perpendicular to
surface dA


dF g  flux of a g through dA

 a g  n̂dA (a scalar)
“Phi”
Flux through a closed or open surface S: calculate “surface integral” of field over S
Evaluate integrand at all points on
surface S
FS 
 dF 

a
 g  n̂dA
S
S
EXAMPLE : FLUX THROUGH A CLOSED, EMPTY,
RECTANGULAR BOX IN A UNIFORM g FIELD
• zero mass inside
• F from each side = 0 since a.n = 0, F from ends cancels
• TOTAL F = 0
• Example could also apply to fluid flow
n̂
n̂
n̂
ag
What if a mass (flux source) is in the box?
Can field be uniform? Can net flux be zero.
Copyright R. Janow – Spring 2016
n̂
FLUID FLUX EXAMPLE: WATER FLOWING ALONG A STREAM
Self
Study
Assume:
•
•
•
•
constant mass density r 
incompressible fluid
constant flow velocity parallel to banks
no turbulence (laminar flow)
n̂'
n̂

A 2
Flux measures the flow (current):
• flow means amount/unit time across area
• either rate of volume flow past a point …or…
rate of mass flow past point

A 1
Two related fluid flow fields (currents/unit area) appear:
• velocity v represents volume flow/unit area/unit time
• J = mass flow/unit area/unit time

A  n̂A
n̂ is the outward unit vector
 to vector area A


J  rv
Flux = amount of field crossing an area per unit time (field x area)

 

V  L  A 
L  vt
The chunk of

volume flux 

 v  A
fluid moves L

t

t
A


in time t:



r


r


mass
of
solid
chunk
m
V
L
A



v

 

m
L

and
 mass flux 
r


r



 A v A J A
t
t
Continuity: Net flux (fluid flow) through a closed surface = 0
………unless a source or drain is inside
Copyright R. Janow – Spring 2016
Gauss’ Law for gravitational field:
The flux through a closed surface S depends only on the enclosed mass (source of
field), not on the details of S or anything else
Example: spherically symmetric mass distribution, radial gravitational field

GM
Field: g   2 r̂
 r 
2
(Newtons/k g or m/s )

 
GM
d(flux )  f ield . dA  g . dA   2 r̂.dA
r
Find total flux through closed surface A


 
GM
GM
F A   g.dA    2 r̂.dA   2  r̂.dA
A
A
A
rA
r
inward acceleration
of test mass m
Spherical
surface of
constant
field & PE
gA
Integral for surface area of sphere
 
GM
 F A   g.dA   2 x 4 rA2   4GM
A
rA
rA
M
Flux depends only on the enclosed mass
(same flux for any closed surface enclosing M)
FLUX measures the strength of a field source that
is inside a closed surface - “GAUSS’ LAW”
Copyright R. Janow – Spring 2016
gA
Shell Theorem follows from Gauss’s Law
1. The force (field) on a test particle OUTSIDE a uniform SPHERICAL shell
of mass is the same as that due to a point mass concentrated at the
shell’s mass center (use Gauss’ Law & symmetry or see section 13.6)
m
r
r
m
x
x
Same for a solid sphere (e.g., Earth, Sun) via nested shells
m
r
r
x
x
+
r
x
+
2. For a test mass INSIDE a uniform SPHERICAL shell of mass m,
the shell’s gravitational force (field) is zero
m
x x
• Obvious by symmetry for center point
• Elsewhere, integrate over sphere (painful) or apply Gauss’
Law & Symmetry to a concentric sphere inside the shell
3. Inside a solid sphere of mass combine above. Force on a test mass
INSIDE depends only on mass closer to the CM than the test mass.
x
• Example: On surface, measure acceleration
distance
r
from center
• Halfway to center,
Copyright R. Janow – Spring 2016
ag = g/2
g
a
4
3
Vsphere  r 3
Superposition Example: Calculate the field (gravitational) due to
two point masses at a special point
Find the field at point P on x-axis due to
two identical mass chunks m at +/- y0
• Superposition says add fields created at P
by each mass chunk (as vectors!!)
• Same distances r0 to P for both masses
y
m
r0
+y0
r02  x 02  y02
q
+x0
• Same angles with x-axis
cos(q)  x 0 / r0
•
Gm
x 02  y02
q
P
ag
-y0
r0
• Same magnitude ag for each field vector
ag 
ag
(from Newtons law of gravitatio n)
m
y components of fields at P cancel, x-components reinforce each other
a tot  a x  
2Gm cos(q)
r02
 
2Gm x 0
r03
where
• Result simplified because problem had a lot of symmetry
Copyright R. Janow – Spring 2016
r03  [ x 02  y02 ]3 / 2
Direction: negative x
x
Example: Calculate gravitational field due to an infinitely long line of
uniformly distributed mass. Find the field at point P on x-axis
to y  
• Integrate over the source of field holding P fixed
•
Add differential amounts of field created at P by
differential point mass chunks at y (vectors!!)
dm = ldy
•
Include mass from y = – infinity to y =+ infinity
• For symmetrically located point mass chunks dm:
r
y
•
y-components of fields cancel,
•
x-components of fields reinforce
• Mass per unit length l is uniform, find dm in terms of q:
-y

a x  da x where
Gdm
da

da
cos(
q
)


cos(q)
-
x
g
r2

dm  ldy  lx[1  tan2 (q)] dq
 da x  
Glx[1  tan2 (q)]
cos(q) dq  
x 2 [1  tan2 (q)]
• Integrate over q from –/2 to +/2
 ax  
Gl
cos(q) dq
x
Gl   / 2
Gl
cos(
q
)
d
q


2
x   / 2
x
Field of an infinite line falls off as 1/x not 1/x2
Copyright R. Janow – Spring 2016
x
CYLINDRICAL
SYMMETRY
dag
q
q
P
x
l = mass/unit length
to y
 
y  x tan(q)
dy
dtan(q)
x
 x [1  tan2 (q)]
dq
dq
r 2  x 2  y2  x 2 [1  tan2 (q)]
 /2
 cos(q)dq  2
- /2
Example of a “line integral” (path integral)- Work done on
a mass traversing a gravitational field
Self Study
How much work is done on a test mass
 as it traverses a particular path


dW  dU  F  ds  mag  ds
through a field?
B
B

 U    F  ds
evaluate along path
test mass
A
Gravitational field is conservative, meaning U is
independent of path chosen
A
 

F

d
s

F

d
s


B
A
for any path between A & B
B
 
U   F  ds  0 for any closed path that is chosen
S
circulation,or path integral
EXAMPLE
uniform field
U= - mgh
Copyright R. Janow – Spring 2016
U= + mgh
Vectors in 3 dimensions
•
Cartesion unit vector representation:
•
Spherical polar coordinate representation:
1 magnitude and 2 directions
Rene Descartes
1596 - 1650

a  a x î  a y ĵ  a zk̂

a  (a, q, f)
•
z
Conversion into x, y, z components
a x  a sin q cos f
a y  a sin q sin f
a z  a cos q
•

a
az
Conversion from x, y, z components
q
a  a 2x  a 2y  a 2z
1
ax
q  cos a z / a
f  tan  1 a y / a x
x
Copyright R. Janow – Spring 2016
ay
f
a sin(q)
y
Gravitational field due to an infinite sheet of mass
Simple 2 dimensional
Copyright R. Janow – Spring 2016
integration
Self Study
Constant - does not depend
on distance from plane!
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