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60 lectures: 3rd period, Monday to Friday
12 weekly problem sets
12 Tuesday afternoon tutorials and laboratories
2 class tests
1 final examination (November)
Class tutors: Maciej Stankiewicz and Michael Malahe
Use them!
Also check the course website regularly for resources
PHY1004W Second Semester 2008
These lecture notes are not a substitute for
… check for significant errata … files on website
My expectations of you …
1. … that if you come to lectures, then you will engage with what
is happening
2. … that you read M&I daily (before and after lectures)
3. … that you do what I ask you to do
4. … that you will not copy another student’s work, but work
together, where appropriate.
(Collaboration becomes copying when both parties are not
gaining positive learning from the activity.)
5. … spend enough time at home working on what you need to …
… what you can expect from me …
1. … the best course that I can deliver
2. … a reasonable and appropriate homework load.
3. … no mercy in the face of plagiarism
4. … an open door policy
Real world (phenomena)
Physical model (shared, contextual)
idealization
Physical theories (shared, acontextual)
particularization
When making sense of the
ideas in this course, it’s
useful to think about both the
nature of physics and how
you learn physics yourself ...
• Draw one or more pictures which show all the important objects, their
motion and any interactions.
• Now ask “What is being asked?” “Do I need to calculate something?”
• Think about what physics concepts and principles you think will be
useful in solving the problem and when they will be most useful.
• Construct a mental image of the problem situation - do your friends
have the same image?
• Specify a convenient system to use - circle this on your picture.
• Identify any idealisations and constraints present in the situation -
write them down!
• Specify any approximations or simplifications which you think will
make the problem solution easier, but will not affect the result
significantly.
An approach to solving physics problems
Step 1. Think carefully about the problem situation and draw a
picture of what is going on (Pictorial Representation).
• Draw a coordinate axis (or axes) onto your picture (decide where to put the
origin and on the direction of the axes).
• Translate your pictures into one or more diagrams (with axes) which only
gives the essential information for a mathematical solution.
• If you are using kinematic concepts, draw a motion diagram specifying the
object’s velocity and acceleration at definite positions and times.
• If interactions or statics are important, draw idealised, free body and force
diagrams.
• When using conservation principles, draw “initial” and “final” diagrams to
show how the system changes.
• For optics problems draw a ray diagram.
• For circuit problems, a circuit diagram will be useful.
• Define a symbol for every important physics variable in your diagram and
write down what information you know (e.g. T1 = 30 N).
• Identify your target variable? (“What unknown must I calculate?”).
Step 2. Describe the physics (Physics Representation).
• Only now think about what mathematical expressions relate the physics
variables from your diagrams.
• Using these mathematical expressions, construct specific algebraic
equations which describe the specific situation above.
• Think about how these equations can be combined to find your target
variable.
• Begin with an equation that contains the target variable.
• Identify any unknowns in that equation
• Find equations which contain these unknowns
• Do not solve equations numerically at this time.
• Check your equations for sufficiency... You have a solution if your plan has
as many independent equations are there are unknowns. If not, determine
other equations or check the plan to see if it is likely that a variable will
cancel from your equations.
• Plan the best order in which to solve the equations for the desired variable.
Step 3. Represent the problem mathematically and plan a solution
(Mathematical Representation).
• Do the algebra in the order given by your outline.
• When you are done you should have a single equation with your target variable.
• Substitute the values (numbers with units) into this final equation.
• Make sure units are consistent so that they will cancel properly.
• Calculate the numerical result for the target variable.
Step 4. Execute the plan
Step 5. Evaluate your solution
• Do vector quantities have both magnitude and direction ?
• Does the sign of your answer make sense ?
• Can someone else follow your solution ? Is it clear ?
• Is the result reasonable and within your experience ?
• Do the units make sense ?
Have you answered the question ?
1 2 3 4 5In-class “voting” questions
Bring your “12345” sheets along with you to class every day
… and have them ready !
A practice question:
I am really pleased to be back in PHY1004W because:
1. All vacation long I dreamed of physics
2. I missed the smell of this lecture theatre
3. Physics is my best course
4. I am a masochist
5. None of the above.
1 2 3 4 5
Which textbook do you have?
1.
2.
3. None, and I don’t think I need one.
4. None, but I am planning to get one.
5. None, but I share with a friend.
11
The story so far….
Right handed coordinate system:
Unit vectors kji ˆ ˆ ˆ1 ˆ ˆ ˆ kji
, ,
ˆ ˆ ˆ( )
ˆ ˆ ˆ( , , )
( , , )
( , , )
( ) ( , , )
( )
d
x y z x y z
x y z x y z
x x y y z z
x y z
x x y y z z
x y z
x x y y z z
A ,A ,A A A A
B B B B B B
A B A B A B
A A A
A B A B A B
c cA cA cA
A B A B A B
A i j k
B i j k
A B
A
A B A B
A
A B
A B B A2 A
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ1 0
A A
i i j j k k i j j k k i
z
y
x
k ˆ j
i
Vector algebra
12
( ) 0
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ 0 ; ;
A B B A A A
i i j j k k i j k j k i k i j
where and G G A G B
ˆ ˆ ˆ ( ) + ( ) + ( ) y z z y z x x z x y y xA B A B A B A B A B A BA B i j k
easy to remember:always
ˆ ˆ ˆ
x y z
x y z
A A A
B B B
i j k
In polar form in 2D:
and
where is the angle between tails of and .
cosABBA
kBA ˆ sinAB
B
A
The spherical polar coordinate system
2 2 2
ˆ ˆ ˆ
cos sin
sin sin
cos
cos
tan
x y z
x
y
z
x y z
z
y
x
A A A
A A
A A
A A
A A A A
A
A
A
A
A i j k
Spherical coordinates: A, θ, :
z
y
x
k ˆ j
i
Ax
Az
Ay
Aθ
The cylindrical polar coordinate system
2 2
ˆ ˆ ˆ
cos
sin
tan
x y z
x
y
z
x y
y
x
z
A A A
A
A
A z
A A
A
A
z A
A i j k
Cylindrical coordinates: , θ, z :
z
y
x
k ˆ j
i
Ax
Az
Ay
Aθ
z
ˆ ˆ ˆ( ) ( ) ( ) ( )
( )( ) ( )ˆ ˆ ˆ ( )
( ) ( )( ) ( )
( ) ( )( ) ( ) ( ) ( )
( ) ( )( ) ( ) ( )
x y z
yx z
t A t A t A t
dA tdA t dA tdt
dt dt dt dt
d d t d tt t
dt dt dt
d dc t d tc t t t c t
dt dt dt
d d t d tt t t
dt dt
A i j k
A i j k
A BA B
AA A
B AA B A ( )
( ) ( )( ) ( ) ( ) ( )
tdt
d d t d tt t t t
dt dt dt
B
B AA B A B
Differentiation of vector functions
Also:
then
If
If position m
then the instantaneous velocity m s-1
and the instantaneous acceleration m s-2
( ) ( ) ( )ˆ ˆ ˆ( )dx t dy t dz t
tdt dt dt
v i j k
( )( )
d tt
dt
va
( )( )
d tt
dt
rv
( )( ) ( )ˆ ˆ ˆ( )yx z
dv tdv t dv tt
dt dt dta i j k
ˆ ˆ ˆ( ) ( ) ( ) ( )t x t y t z tr i j k
Example of the time derivatives of a vector function
M&I
Chapter 13
Electric Field
Scalar and vector fields
A scalar or vector field is a distribution of a scalar or vector
quantity on a specified surface or throughout a specified
region of space such that there is a unique scalar or vector
associated with each position.
Fields may be time independent, e.g.
or time dependent
( , , )T x y z
( , , , )T x y z t
Examples of scalar fields:
• Temperature, or
• Potential, or
Examples of vector fields:
• Electric field
• Velocity
( )T r( , , )T x y z
( , , )V x y z ( )V r
( )E r
( )v r
Scalar fields
A scalar field can be represented by specifying a finite
number of scalar values at strategic positions in the region of
interest.
It is also possible to draw contour curves - continuous curves
joining points where the scalar values are the same.
In 3D space these contours are surfaces. Such representations
are always incomplete, since an infinite number of contours
or surfaces should really be drawn.
A third way of representing a scalar field is by a mathematical
function.
Scalar fields in 2D ....
... and 3D ...
A vector field is a vector function of position.
Vector fields may be represented
visually by field lines which
are everywhere parallel to the
local value of the vector function.
These lines are sometimes called
“lines of force” in mechanics and
“stream lines” in fluid mechanics.
A vector field may also be represented by lines which are
everywhere a tangent to the vectors. Although we lose track of the
lengths of the vectors, we can keep track of the strength of the field
by drawing lines far apart where the field is weak, and close where
it is strong.
Vector fields
Vector fields may also be represented mathematically, often
using differential equations.
We adopt the convention that the
number of lines per unit area at
right angles to the lines is
proportional to the field strength.
The electric field around a point charge q may be written as
where and
In Cartesian form
and
2
0
ˆ( ) 4
qE r r
r
r
rrr
ˆ 0
2 2 2 2 2 2 3 22 2 20 0
ˆ ˆ ˆ ˆˆ ˆ( ) ( ) ( )
4 ( ) 4 ( )
q x y z q x y z
x y z x y zx y z
i j k i j kE r
kjir ˆˆˆ zyx
Example of a vector field function of position: the electric field
2 2 2
ˆ ˆ ˆˆ
x y z
x y z
r i j kr
r
VPython scripts used in class can be found in the EM section of the
PHY1004W web site.
Also look at these PhET simulations from the University of Colorado:
Digital resources
Electric fieldElectric field
hockey
Charges
and fieldsTravoltage
A proton is at location < 0, 3, −2 > m.
An electron is at location <−1, 0, −6 > m.
What is the relative position vector from the proton
to the electron?
1. < −1, 3, −8 > m
2. < −1, −3, −4 > m
3. < 1, 3, 4 > m
4. < 1, −3, 8 > m
5. < 1, 0, 6 > m
1 2 3 4 5
Electrostatics
Thales of Miletus (640-548 BC) … basic phenomena of
charging on intimate contact (friction)
William Gilbert, 1574-1603, physician to Queen Elizabeth …
amber, rubbed with cloth or fur, acquires the property of
attracting small bodies … the amber has become “electrified.”
Dufay, 1733, originated (?) the “two fluid” theory of
electrification, calling the two sorts of electricity
“vitreous” (on glass) and “resinous” (on amber)
Benjamin Franklin (1747) introduced the terms “positive” and
“negative”
The Electrostatic Force
• Can be either attractive or repulsive (gravity only attractive)
• Can act through empty space
• Very much stronger than gravitational force
• From experiments we find that the electrostatic force decreases
with distance r as , and it is also proportional to the
product of the amount of charge on each of the charges:
• Electrostatic forces are mutual forces of attraction
i.e. they obey Newton’s 3rd law.
21
r
1 2F q q
Coulomb’s Law
Coulomb (1785) put a quantitative basis to
the observations that charged particles
attracted or repelled one another.
... measured the forces using a tensional
balance and found that ...
1 221 2
12
q q
rF 1 2
21 2
12
q qk
rF Coulomb’s Lawor
where: is the force on due to21F 2q 1q
9 2 -2
0
1 constant (from experiment) 9 10 N m C
4πεk
-12 2 -1 -2
0ε permittivity of free space = 8.85 10 C N m
1 2
2
0 12
1ˆ
4
q q
r21 12
F r
Where is the unit vector (magnitude = 1) which
indicates the direction along which the force is acting
i.e. from q1 to q2.
12r
12 ˆr12 12
r r
i.e. but ˆ ˆ
12 21
12 21 12 21
F F
F F r r
Coulomb’s Law
So
Note:
1q12
r
21F
2qˆ21
r
12F
12r
+
+
+
Silicon atoms
Particle Mass Charge
electron 9.11 10-31 kg e = 1.60 10-19 C
positron 9.11 10-31 kg +e = 1.60 10-19 C
proton 1.67 10-27 kg +e = 1.60 10-19 C
antiproton 1.67 10-27 kg e = 1.60 10-19 C
muon 1.88 10-28 kg +e ( +) or e ( )
pion 2.48 10-28 kg +e (π+) or e (π )
Some charged particles
Definition of electric field:
... where is the electric field at the location of charge q2
The electric field
… a region in which a charge experiences a (mechanical) force is
called an electric field.
… we assume that a charge creates a field of influence around it.
Any other charge present in that region will experience a force.
This force is described by Coulomb’s Law.
2q2 1
F = E
1E
2q 2F
1E
1N CUnits:
( , , , )x y z tE E
M&I
13.3
The electric field around a point charge q may be written as
where and
In Cartesian form
and
The electric field of a point charge
2
0
ˆ( ) 4
qE r r
r
r
rrr
ˆ 0
2 2 2 2 2 2 3 22 2 20 0
ˆ ˆ ˆ ˆˆ ˆ( ) ( ) ( )
4 ( ) 4 ( )
q x y z q x y z
x y z x y zx y z
i j k i j kE r
kjir ˆˆˆ zyx
2 2 2
ˆ ˆ ˆˆ
x y z
x y z
r i j kr
r
This vector function is the sum of a 3 component vector:
where and are the three scalar components of
2 2 2 3 2
0
2 2 2 3 2
0
2 2 2 3 2
0
( , , ) 4 ( )
( , , ) 4 ( )
( , , ) 4 ( )
x
y
z
qxE x y z
x y z
qyE x y z
x y z
qzE x y z
x y z
).( , rE
zyx EEE
The electric field …2
2 2 2 3 2
0
ˆ ˆ ˆ( )( )
4 ( )
q x y z
x y z
i j kE r
A note on vector notation
In these notes, a vector in 3D such as
will be written
2
0
ˆ( ) 4
qE r r
r
2 2 2 3 2
0
ˆ ˆ ˆ( )( )
4 ( )
q x y z
x y z
i j kE r
The Matter and Interactions textbooks use a notation which is
similar to the notation use in VPython syntax:
2 2 2 3 2
0
, ,( )
4 ( )
q x y z
x y zE r
You should be comfortable with both!
E_pointcharge_drag.py
At location x there is an electric field in the direction shown
below, due to nearby charged particles.
If an electron were placed at location x, what would be the
direction of the force on the electron?
Ex
x
1
2
3
4
5 zero
1 2 3 4 5
What is the direction of the electric field at the different positions
below? Your arrows should be of the appropriate relative length.
Electric field lines
A
BC
D
E
F
G
+
E
A charge as a projectile
A positive charge q of mass m initially moving at constant
velocity, enters and leaves a region where there is constant,
downward electric field.
0E 0E
(a)Draw the trajectory of the charge as it moves through each of
the three regions.
(b) Write down an expression for the acceleration of the charge in
each of the three regions.
(c) If the mass of the charge is doubled, then what will its
trajectory look like?
The net electric field at a location is the vector sum of the
individual electric fields contributed by all charged particles
located elsewhere.
The electric field contributed by a charged particle is
unaffected by the presence of other charged particles.
1 2netE E E
The superposition principle
1q
2q
1E2E
netE
M&I
13.4
The superposition principle: Example
The negative charge below has twice the magnitude of each positive
charge. Use graphical vector addition to estimate the direction and
relative magnitude of the electric field at each position.
++ _A
B
C
D
E
The superposition principle: an important worked example
A small object with charge Q1 = 6 nC is located at the origin.
A second small object with charge Q2 = 5 nC is located at
m. What us the net electric field at a location
m due to Q1 and Q2 ?
0.05,0.08,0
0.04,0.08,0
1Q
2Q1E
2E
netE
i
j
1 0.04,0.08,0 0,0,0r
0.04,0.08,0 m
11
2 2 21
0.04,0.08,0ˆ
( 0.04) (0.08) (0)
rr
r
0.447,0.894,0 m
9 9
11 12 2
0 1
3 3
(9 10 )(6 10 )ˆ 0.447,0.894,0
(0.0894)4
3.02 10 ,6.04 10 ,0
QE r
r
N C-1
2 0.04,0.08,0 0.05,0.08,0r 0.09,0,0 m
22
2 2 22
0.09,0,0ˆ
( 0.09) (0) (0)
rr
r 1,0,0 m
9 9
22 22 2
0 2
3
(9 10 )( 5 10 )ˆ 1,0,0
(0.09)4
5.56 10 ,0,0
QE r
r
N C-1
The superposition principle: an important worked example cont
3 3 3
net 1 2 3.02 10 ,6.04 10 ,0 5.56 10 ,0,0E E E
3 32.54 10 ,6.04 10 ,0 N C-1
… or we can write a short VPython programme …
Note that norm(A) gives the unit vector of
and mag (A) gives the magnitude of
from visual import *
Q1 = sphere(pos=(0, 0, 0), radius=.3e-2, color = (0,0,1), charge = 6e-9)
Q2 = sphere(pos=(0.05, 0.08, 0), radius=.3e-2, color = (1,0,0),
charge = -5e-9)
location = vector(-0.04, 0.08, 0)
k = 9e9
r1 = location - Q1.pos
E1 = k*Q1.charge*(r1/(r1.x**2+r1.y**2+r1.z**2)**0.5)/mag(r1)**2
r2 = location - Q2.pos
E2 = k*Q2.charge*norm(r2)/mag(r2)**2
Enet = E1 + E2
print Enet
Escale = 3e-6
Earrow = arrow(pos=location, color=(1,.6,0), axis=Enet*Escale,
shaftwidth = .5e-2)
A
A
A
A
compare
The electric field of a dipole
The electric field of a dipole
Along the x-axis:
2 21 10 2 2
1 2 ˆ4
qsx
x s x si
q
s
q E E
xE
x
i
1 12 2
2 22 21 11 10 02 22 2
ˆ ˆ1 1
4 4x
x s x sq q
x s x sx s x s
i iE E E
Along the y-axis:
12
ˆ ˆs yr i j
s
E
E
yE
r
yr
j
i
12
ˆ ˆs yr i j
Then
2 212
s yr2 21
2s yr
1 12 2
2 22 22 21 12 21 10 02 22 2
ˆ ˆ ˆ ˆ1 1
4 4
y
q s y q s y
s y s ys y s y
E E E
i j i j
322 210
2
1 ˆ4
qs
s y
i
Dipole field …2
Far along the x-axis:
If x >>s , then2 2 2 21 1
2 2x s x s x r
and3
0
1 2 ˆ4
x
qs
rE i
xE
x
i
Far along the y- and z-axis: 3
0
1 ˆ( )4
y z
qs
rE E i
Dipole field …3
Interaction between a point charge and a dipole
q
s
q
dipoleE
d s
i+QF
dipole 3
0
1 2 ˆ( )4
qsQ Q
dF E i
Hence force on dipole due to Q = ˆ( )F i
dipoleE
i+Q
on - on +F F F
pointE
on +F on -F
The dipole moment
Write p = qs for a dipole with in a direction from q to q
q
s
q
EE
p
p
p
See movie … oscillation of an electric field in an external
electric field … and try the challenge problem.
Electric field of a uniformly charged sphere of radius R
... see later ...
Q
r
+
+
+
+
+
++
++
+
+
+
2
0
1ˆ
4sphere
Q
rE r
for r > R 0sphereE
for r < R
r
A uniformly charged sphere acts
like a point charge, at locations
outside the sphere.
The dipole and charged ball … worked example in M&I
qs
q
b
Q
R
C
a s
3 2 2 2 20 0
ˆ ˆ1 2 1ˆ 4 4
net dipole ball
qs Q b a
a b a b a
E E E
i jj
j
i
M&I
13.5Choice of system
Consider the following when making sense of things:
Split the Universe up according to:
… the charges that are the source of the field
… the charge that is affected by the field
... and there is the issue of retardation ...
Why bother with a “field”?
... knowing the field at a location means that we know the
force acting on any charge q placed at that location...
... no matter how that field was produced.
q
( , , , )x y z tE E
E
Take q away
E
For how long will
you still detect ?E
A real example (e e+ annihilation): e + e+ → +
... so is the electric field real, or only a construct?
M&I
13.6
M&I
Chapter 14
Matter and
Electric Fields
M&I
14.1
Charged particles
… “net charge” of an object …sum of all the
charges of all its constituent particles …
… conservation of charge …
the net charge of a system and its surroundings
cannot changee+ + e → γ + γ
M&I
14.2Electric interactions between charged particles
Do it yourself … experiments with U and L tapes
M&I
14.3Interaction of charges and neutral matter
The electron cloud around the
nucleus of an atom is described by
a probability distribution :
… bring another charge close to atom …
… the electron cloud will be distorted by
the electric field …
… average location of the electron no
longer at the centre where the nucleus is
located … the atom is “polarized”
+
… can be represented simply: +
E
Such polarized atoms are “induced” dipoles … return back to
original state when external electric field is removed.
Write: p E
where is the dipole moment, is the external field, and
the constant α is the “polarizability”
… which is characteristic of the particular material (measured)
p E
A neutral atom and a point charge
+1E rq1
r
+1E
q1
r
+
s
q2 q2 2E
1F2F
Charge polarizes
the atom …
2 1p E
… which makes electric
field at q12E
2
1 1 12 3 3 3 2 5
0 0 0 0 0
2 21 2 1 1 2 1 1 4 4 4 4 4
E q qpE
r r r r r
2 22
1 11 1 2 1 25 5
0 0
2 21 1 = ˆ ˆ
4 4
q qq q
r rF E r r F
M&I
14.4Conductors and insulators
Conductors: contain mobile charges that can
move through the material.
Insulators: have no mobile charges
Insulators can be polarized:Eapplied
+ +
+ +
+ +
+ +
+ +… so can conductors …
… such as ionic solutions …
Ionic solutions
++
+
++
Eapplied
+
+
+
+
+
Eapplied
Epolarization
Enet
+
+
++
+
Eapplied
Epolarization
Enet= 0
+ + + + + + + + + ++ + + + + + + + +
+ + + + + + + + + ++ + + + + + + + +
Model of a metal
Metal: atoms arranged in regular 3D geometric lattice,
most electrons tightly bound, one or two outer electrons per atom
free to move within the metal (“sea” of electrons) … but are not
easily removed from the metal.
+ + + + + + + + + ++ + + + + + + + +
+ + + + + + + + + ++ + + + + + + + +
+ + + + + + + + + ++ + + + + + + + +
+ + + + + + + + + ++ + + + + + + + +
+ + + + + + + + + ++ + + + + + + + +
+ + + + + + + + + ++ + + + + + + + +
Epolarization
+
+
+
+
+
+
EappliedEpolarization
static equilibrium
Eapplied
conductor insulator
mobile charges yes no
polarization entire sea of mobile individual atoms or
charges move molecules polarize
static equilibrium Enet = 0 inside Enet nonzero inside
location of only on surface anywhere on or
excess charge inside material
distribution of spread out over located in patches
excess charge entire surface
A negatively charged ion is located to the left of a
neutral molecule. Which diagram correctly shows
the polarization of the neutral molecule?
1 2 3 4 5
A point charge is brought near a neutral molecule.
(There is nothing else nearby).
Is it possible for the point charge and the neutral molecule to
repel each other?
1. Yes. The molecule can polarize so that it repels
the point charge.
2. No. The molecule can only polarize in a way
that will attract the point charge.
1 2 3 4 5
In a region of space there is an electric field upward (in the +y
direction), due to charges not shown in the diagram. A neutral
copper block is placed in the region.
Which diagram best describes the charge distribution on the block?
1 2 3 4 5
A negatively charged iron block is placed in a region where
there is an electric field downward (in the –y direction) due
to charges not shown.
Which diagram best describes the charge distribution in
and/or on the iron block?
1 2 3 4 5
M&I
14.5Charging and discharging
An object is charged when its net charge is non-zero…
… and may be discharged by contact … or grounding
An object may be charged by induction …
Try it yourself …
1. 2.+
+
++
++
++3. 4. 5.
++
++
+
++
+
A and B are identical metal blocks.
What is the final charge of block B?
1. +6 nC
2. +3 nC
3. 0 nC
4. 3 nC
5. 6 nC
1 2 3 4 5
What happens?
1. protons move from A to B
2. positrons move from A to B
3. electrons move from B to A
4. both protons and electrons move
5. no charges move
1 2 3 4 5
You neutralize a positively charged tape by running your
finger across it.
What happens?
1. electrons move from skin to tape
2. Cl- ions move from skin to tape
3. protons move from tape to skin
4. + ions move from tape to skin
5. no charges move
1 2 3 4 5
Two aluminum blocks, A and B, are initially neutral. They have
insulating handles, which are not shown. This sequence occurs:
At a time after t4, what is the net charge of A?
1. positive 2. negative 3. neutral
1 2 3 4 5
Example problem:
What force does the charged ball exert on the neutral wire?
R
Q
r
L
… the ball polarizes the wire
… which becomes a dipole with +q and q on either ends…
wire 3
0
1 2
4
qLE
r
on ball on wireF F
on ball ball wireF Q E
wireE
2 2
0 0
1 1 2 0
4 42wire sphere
q Q
rLE E
2
8
Q Lq
r
Then
22 2 3
on ball 3 3 5
0 0 0
1 2 2 1
4 4 8 4 4
qL Q L L Q LF Q
r r r r
Example problem …2
Inside the wire
at static equilibrium:
net 0E
Putting in some sensible numbers:2 3
-9 -3
9 2 -2 -11
on ball 5
10 C 4 10 m9 10 N m C 1.4 10 N
4 0.1 mF
M&I
14.7Sparks in air
... air is an excellent insulator,
consisting mainly of neutral N2 and O2
… during a spark, these molecules are
ionized … N2+ and O2
+
How can electric charge move through air?
Take two charged balls, closely located, but not touching
More charge here
(polarized)
-- -
-
--
- -
- +
+
++
+
+
+
+
+
+
-- -
-
--
- -
- +
+
++
+
+
+
+
+
+
-
Join 2 balls with wire, and free
electrons move onto positive ball
For a 1 m long wire, there are about 1023 free electrons.
The balls are charged about 10-9 coulombs (1010 e)
So in a fraction 1010/1023 (=10 13) of the 1 m long wire are
enough electrons to neutralize the positive ball
… i.e. the electron sea shifts about 10-13 m!
What happens in the case when there is only air between the balls?
? Electrons jump between the balls ?
… how far (“mean free path”) does an electron travel in air
before colliding with a gas molecule ?
Mean free path d of electrons in air
A
de
No. of molecules in cylinder 1
No. of molecules/m3 volume of cylinder 1
At STP, one mole
of air occupies
0.0224 m3
2310 26 10
(1.5 10 ) 10.0224
d
… giving d 5 × 10-7 m
? Positive ions and electrons move in ionized air ?
If the oxygen and nitrogen molecules in air become ionized (how?)
…. Then we have a gas of charged particles (a conductor)
-- -
-
--
- -
- +
+
++
+
+
+
+
+
++
+
+
+-
-
-
-
No particle moves further than one mean free path
(no electrons move between the balls)
what happens here?
E
The spark!
… electrons drift towards the positive ball and positive ions drift
(more slowly) to the negative ball.
As the electrons from the air move onto the positively charged
ball, the electric field between the balls decreases slightly.
… electrons also move off the negative ball to neutralize
positively charged molecules.
The spark only lasts a short time, unless the charge on the balls
is replenished, since the excess charge on the balls will not be
sufficient to maintain an electric field large enough to keep the
air ionized.
A photon of light is given off when a free
electron re-combines with a positive ion
… the energy of the photon is equal to the
difference between the high energy unbound
state and the lower energy bound state.
+-
? How does the air become ionized ?
Need E = 3 106 N C-1 (experimentally determined) to
maintain the air in an ionized state
What is the electric field between the atomic “core”
(nucleus + inner electrons) and a single outer electron …
-199 2 -2 11 -1
22-10
0
1 1.6 10 C 9 10 N m C 1.4 10 N C4 10 m
eE
r
… which is much larger than the experimentally observed
E = 3 106 N C-1
… so if it’s not having a strong electric field, what ionizes the air …?
… fast moving charged particles knock electrons out of atoms…
muons from cosmic rays, particles from radioactive sources, …
Once there is a single free electron, which is then accelerated in
an electric field, a “chain reaction” can start, as 2, 4, 8, …
electrons can be knocked out of molecules …
… and the air becomes ionized.
… need about to knock one electron from a molecule 2.4 10-18 J
2-19
9 2 -2
-100
1.6 10 C1 ( )( ) 9 10 N m C4 10 m
e eU
r
Then 18
critical 2.4 10 JeE d-18
6 -1
critical -19 -7
2.4 10 J30 10 N C
1.6 10 C 5 10 mE
…close enough?
1 2
3
Drift speed of free electrons in a spark
2 1812
2.4 10 Jmv
-18
6 -1
-31
2 2.4 10 J2 10 m s
9 10 kgv
The magnitude of the electric field is not uniform between
the two charged metal balls, and is largest near the balls …
Why is this the case?
How can a region of ionization propagate though space?
General approach:
1. Think about the distribution ... draw it! ... are there
any symmetries?
2. Cut up the distribution into pieces and consider the
electric field from a single piece.
3. Write down an expression for the electric field from
that one piece.
4. Repeat for all other pieces and sum (integrate) over
the entire distribution.
5. Check your result.
M&I
Chapter 15
Electric Field of Distributed Charges
The electric field of an uniformly charged rod
... with total length L and total charge Q
Magnitude of : r22
0r x y y
Then
0
22
0
ˆ ˆˆ
x y y
r x y y
i jrr
Magnitude of : E 22 20 0 0
1 1
4 4
Q QE
r x y y
E
yr
j
i
y
0
0y
x
Q
Q Q
y L 0ˆ ˆx y yr i j
M&I
15.2
A uniformly charged rod … 2
22 20 0 0
1 1
4 4
Q QE
r x y y
Then 0
22 220 0 0
ˆ ˆ1
4
x y yQ
x y y x y y
i jE
3 32 22 22 2
0 00 0
1 1
4 4x
x Q Q xE y
Lx y y x y y
3 32 2
0 0
2 22 20 0
0 0
1 1
4 4y
y y Q y yQE y
Lx y y x y y
0zE
1E
2E
1 2E = E E
... now sum up the contributions of all the pieces
The y-components of all the sum to zeroE
322 2
0
1
4x x
Q xE E y
L x y
As 0y 2
32
2
2 20
1 1
4
L
L
x
QE x dy
L x y
The x-components:
and set y0 = 0
A uniformly charged rod …3
A uniformly charged rod …4
2
32
2
2 20
1 1
4
L
L
x
QE x dy
L x y
Evaluate the integral ... try it yourself ... or look it up ...
2
2
2 2 20
1
4
L
L
x
Q yE x
L x x y
220
1
4 2x
QE
x x L
220
1
4 2
QE
r r L
Write or22
0
1 ˆ
4 2
Q
r r LE r
A uniformly charged rod …5
220
1 ˆ
4 2
Q
r r LE r
Check the result ... units? ... direction?
Special cases and :r L L r2
0
1 ˆ
4
Q
rE r
Another special case: r L
Then 2 22 2 2 2r r L r L r L
0
21 ˆ
4
Q L
rE rand
from visual import *
scene.x = scene.y = 0
scene.height = 800
scene.width = 600
kel = 9e9
Q = 1e-8
N = 50.
L = 1.0
dl = L/N
Escale = 3e-5
rod = []
for y in arange (-(L/2.)+(dl/2.), (L/2.), dl):
a = sphere(pos=(0,y,0), color=color.red, radius=0.01, q=Q/N)
rod.append(a)
obs = []
dy = L/4.
r = 0.05
for y in arange (-(L/2.), (L/2.)+dy, dy):
for theta in arange(0,2*pi,(2*pi/6.)):
pt = vector(r*cos(theta), y, r*sin(theta))
obs.append(pt)
for pt in obs:
E = vector(0,0,0)
ar = arrow(pos=pt, color=(1,.5,0), axis=Escale*E, shaftwidth=0.01)
for source in rod:
r = pt - source.pos
E = E + norm(r)*kel*source.q/mag(r)**2
ar.axis = Escale*E
if pt.y == 0:
print '%e' %mag(E)
Erod.py
Electric field along the axis of a uniformly charged thin ring
E
j
i
Q
2
Q
Q
... with radius R and total charge Q
k
Magnitude of : r2 2r R z
ˆ ˆ ˆ ˆˆ ˆ0 0 cos sin 0
ˆ ˆ ˆ cos sin
z R R
R R z
r i j k i j k
i j kr
Magnitude of : E 2 2 2
0 0
1 1
4 4
Q QE
r R z
z
M&I
15.4
A uniformly charged thin ring …2
2 2
0
1
4
QE
R z
Then2 2 2 2
0
ˆ ˆ ˆ1 cos sin ˆ
4 2
Q R R zE
R z R z
i j kE r
2Q Q
From thinking about the symmetry, = 0x yE E
and322 2
0
1
4 2z
Q zE
R z
3 32 2
2
2 2 2 20 0
0
1 1
4 2 4z
Q z QzE d
R z R zThen
A uniformly charged thin ring …3
322 2
0
1
4
QzE
R zAlong the axis of the ring:
Special cases:
0E
Another special case: z R
Then 3 32 22 2 2 3R z z z
3
0
2
0
1
4
1
4
QzE
z
Q
z
and
Exact centre of the ring, z = 0:
(point charge)
zE
z
Ering.py
Electric field at a few other positions:
Electric field along the axis of a uniformly charged disc
zE
j
i
R
2
area of ring 2
area of disc
Q
Q R
... with radius R and total charge Q
k Magnitude of : zE
32
2 2 2 2 2 20 0
1 1
4 4z
Q z Q zE
z z z
z
r
M&I
15.5
Again, only is nonzerozE
uniformly charged disc …2
322 2
0
1
4z
Q zE
z
2
2Q
Q R
3 32 2
2
2 2 2 20 0
2
1 1ˆ ˆ ˆ 4 4
z z
Q zQ z R
Ez z
E k k k
32
22 2
0
1
2z
Q zE
R z
3 122
2 22 22 2
0 0 0
1 1 1
2 2
R
z
Q Q zE z d
R R R zz
uniformly charged disc …3
Write
0
12
Q A zE
R
2A R
then
Special case: z R
and
122 2
0
12
Q A zE
R z
If is extremely small, then /z R 1 1z
R
0
2
Q AE
… which is true near any large uniformly charged plate
Edisk.py
Edisk_add_rings.py
A charged disk viewed edge on:
The capacitor
…consider two uniformly charged metal disks,
of area A, a close distance s apart, carrying
charges Q and Q
s
+
+
+
+
+
Q Q
+
+
1 2 3x x x
k
0 z s
At location 2:
What are the directions of the electric fields at
locations 1, 2 and 3?
2Ε Ε Ε
2
0 0
0 0
( )1 1
2 2
2 1
Q A z Q A s zE
R R
Q A s Q A
R
2
0
ˆ( ) Q A
E k
M&I
15.6
The capacitor …2
At location 3 (“fringe field):3Ε Ε Ε
3
0 0
0
1 12 2
2
Q A z s Q A zE
R R
Q A s
R
3 1
0
ˆ( ) 2
Q A s
RE k E
+
+
+
+
+
sQ Q
+
+
1 2 3x x x
k
0 s z
3 0
2
0
2
2
Q A s
E R s
Q AE R… very small, if s R
Electric field of a uniformly charged spherical shell
r
+
+
+
+
+
+
+
++
+
+
+
i 2Every 0point on sphere
1ˆ
4sphere
Q
rE E r
For r > R:
i
Every point on sphere
0sphereE E
For r < R:
r
... with radius R and total charge Q
12
3
4
5
6
M&I
15.7
Esphere_outside_rings.py
Esphere_rings.py
1 2 3 4 5
A negatively charged hollow plastic sphere is near a negatively
charged plastic rod. What is the direction of the net electric field at
location P, inside the sphere?
1
4
3
2
5 zero magnitude
1 2 3 4 5
You stand at location A, a distance d from the origin, and hold a small
charged ball. You find that the electric force on the ball is 0.008 N.
You move to location B, a distance 2d from the origin,
and find the electric force on the ball to be 0.004 N.
What object located at the origin might be the source of the field?
0. A point charge
1. A dipole
2. A uniformly charged rod
3. A uniformly charged ring
4. A uniformly charged disk
5. A capacitor
1 2 3 4 5
You stand at location A, a distance d from the origin, and hold a small
charged ball. You find that the electric force on the ball is
0.08 N. You move to location B, a distance 2d from the origin,
and find the electric force on the ball to be 0.01 N.
What object located at the origin might be the source of the field?
0. A point charge
1. A dipole
2. A uniformly charged rod
3. A uniformly charged ring
4. A uniformly charged disk
5. A capacitor
1 2 3 4 5
You stand at location A, a distance d from the origin, and hold a small
charged ball. You find that the electric force on the ball is 0.009 N.
You move to location B, a distance 2d from the origin,
and find the electric force on the ball to be 0.00899 N.
What object located at the origin might be the source of the field?
0. A point charge
1. A dipole
2. A uniformly charged rod
3. A uniformly charged ring
4. A uniformly charged disk
5. A capacitor
... with radius R and total charge Q
2
0
1ˆ
4sphere
Q
rE r
For r > R:
For r < R:
... with radius R and total charge Q
(think of a series of concentric spherical
shells, all uniformly charged)
r E
Contribution to at r due to all
concentric spherical shells
between r and R is zero
Contribution to at r due to all
concentric spherical shells
between 0 and R is
E
E
2
0
1
4
Q
r
Electric field of a uniformly charged solid sphereM&I
15.8
343
343
volume of inner shells
volume of sphere
Q r
Q R
Therefore, inside the sphere:
343
2 2 3 3430 0 0
1 1 1
4 4 4
rQ Q QrE
r r R R
( )E r
rR
Electric field of a uniformly charged solid sphere …2
Try it yourself …
M&I
15.9The hollow 3/4 cylinder
What is the direction of the electric field due to the two charged
rods at each of the positions shown?
_ _ _ _ _ _ _ _ _
1. 2. 3. 4. 5. zero
1 2 3 4 5
A
+ + + + + + + + +
+ + + + + + + + +
+ + + + + + + + +
B
C
DE
F
G
H
What is the direction of the electric field at the centre of the ring
in each case?
_ __
_
_
_
_
_
_
_
_
_ _
__
_
_ __
_
__
_
_
_
+++
+
+
+
+_
_
_
_
_
_
__
++
+
++
+
1. 2. 3. 4. 5. zero
1 2 3 4 5
A B C
What is the direction of the electric field due to the charged ring
at the position shown in each case?
_ __
_
_
_
_
_
_
_
_
_ _
__
_
_ __
_
__
_
_
_
+++
+
+
+
+_
_
_
_
_
_
__
++
+
++
+
1. 2. 3. 4. 5. zero
1 2 3 4 5
A B C
