Download Lesson # 11 – Electric Fields

Lesson # 6 – Properties of Electric Charge and Coulomb’s Law
Pre-Lesson Notes
Unit: Fields
Date:
Concept: Properties of Electric Charge (textbook section 7.1)
Definitions:
Law of electric charges
Law of conservation of charge
Insulator
Conductor
Grounding
Charging by induction
Charging by contact
Net Charge:
Explanation:
Lesson # 7 – Coulomb’s Law
Pre Lesson Notes:
Unit: Fields
Date:
Concept: Coulomb’s Law (textbook section 7.2)
Definitions:
Electric force
Coulomb’s Law
Coulomb’s constant
Comparing Coulomb’s Law and Universal Gravitation
The superposition principle
Equations:
In 1785 Charles Coulomb used a torsion balance to measure the force that one charged sphere exerts
on another charged sphere to find how the force between two electrically charged objects depends on
the magnitudes of the two charged objects and on their separation. Coulomb could not measure the
absolute magnitude of the electric charge on the metal spheres. However, he could divide the charge in
half by touching a charged metal sphere with a third identical uncharged metal sphere.
Part A: Coulomb’s Law
Two point-like objects with a net charge exert a force upon one another along a line from the centre of
one object to the centre of the other. The magnitude of the electric force that object 1, with electric
charge q1, exerts on object 2, with electric charge q2, when they are separated by a center-to-center
distance r is given by the expression below. Note that this is also the equal magnitude electric force that
object 2, with electric charge q2, exerts on object 1, with electric charge q1:
k q1 q 2
Fq1 on q2  Fq 2 on q1 
r2
where k = 9.0 x 109 Nm2 /C2. We assume that the objects are much smaller than their separation r (i.e.
point-like objects).
1. The diagrams show two charged objects and their separation. Rank the size of the force that the left
object exerts on the right object from the strongest to the weakest force. Explain how you made the
ranking.
A
B
C
Ranking:
Why:
+2q
+3q
+2q
d
d
A
C
+q
-q
d
d
d
+q
+q
+q
+q
½d
B
d
Ranking:
Why:
3d
A
-q
C
+q
+q
+q
+q
+q
B
+q
Ranking:
Why:
d
A
+q
d
B
-q
+q
+q
C
+2q
+q
2d
Ranking:
Why:
+3q
+q
1/3d
2. For each of the cases above, compare the force that the left object exerts on the right object to the
force that the right object exerts on the left object. Explain how you know.
Basic Electric Force Sketches and Calculations
In all of the following examples, the charges are shown in Coulombs and each vertical
dotted line, drawn for your reference, is 1 m from the next. Sketch each force, as well as
the total force (bold or in a different colour) on each charge. Don’t forget that k = 9 x
109. The 1st is done for you (& a bit of the others too):
e.g. #1
F =k∙1∙1/12 = k
1
+1
+1
FNet(C) = k + k/4
+1
F2=k∙1∙1/22 = k/4
FNet (B)=0
= 1.125 x 1010 N [R]
e.g. #2
+3
+1
+3
FNet(C) = 3k + 9k/4
= 4.725 x 1010 N [R]
e.g. #3
+2
+2
+2
-3
+1
-3
FNet(C) = 4.5 x 1010N [R]
e.g. #4
FNet(A) = 6.75 x 109 N [R]
e.g. #5
+4
-2
+4
FNet(B) = 0 N
e.g. #6
+1
-3
-2
+4
Lesson # 8 – Pith Ball Lab
Charge Splitting
1. Imagine you had two containers, one with VA = 400 mL of water in it, and one empty, VB = 0. Now if you
could connect them at the bottom so the water was free to flow, how much would be in each container
once they “equalize”?
2. Create an equation to show how to calculate the final volume, V’, in each container.
3. Now imagine you have three containers this time: VA = 400 mL, VB = 0, VC = 100 mL. A and B are connected
together to equalize. Then B and C are connected together to equalize. Use diagrams to prove that final
volumes are 200 mL, 150 mL, and 150 mL. Notice that the total must be the same before and after.
4. Gravity is the force that causes these liquids to equalize. Electrostatic repulsion works in a very similar
way to equalize charges when objects are connected to allow charges to flow. Thus, if a -10 nC charged
metal sphere is touched to a neutral metal sphere, the electrons which repel each other, will spread out
as much as they can. Thus, each sphere will split the total, ending up with -5 nC each so that the
electrons can spread out as much as possible.
5. Try these practice questions:
a) qa = -6 C, qb = -2 C, qc = -5 C. If A is touched to B and then B is touched to C, what will the final charge
on each be?
b) qa = -6 C, qb = +2 C, qc = -6 C. If A is touched to B and then B is touched to C, what will the final
charge on each be? (As a first step, you can still just add qa and qb and divide by 2 to get the
distributed average charge.)
c) qa = -8 C, qb = +6 C, qc = 0 C, qd = +3 C. If A is touched to B and then B is touched to C, and then C is
touched to D, what will the final charge on each be?
d) qa = -8 C, qb = +6 C, qc = 0 C, qd = +3 C. If D is touched to C and then C is touched to B, and then B is
touched to A, what will the final charge on each be?
6. A sphere with a -4µC charge is touched to an identical but uncharged sphere.
a) If they are then placed 1.0 cm away from each other, what force will each feel?
b) What mass would they need to be so that all forces would cancel?
7. Consider these four charges: qa = -8 nC, qb = +6 nC, qc = 0 C, qd = +3 nC.
a) A, B, and C are laid out in order, 5.0 cm apart. Sketch the situation and the forces each would feel.
b) Calculate the total electrostatic force on each of A, B, and C.
c) D is touched to C and then C is touched to B, and then B is touched to A. What will the final charge
on each be?
d) A and B are put beside each other again. What force will they feel this time?
(#6 answers : 360 N
23 tonnes)
Quick Lab: Measuring the Electrostatic Force
Procedure:
- Get a retort stand, a test tube clamp, a straw, and two strung pith balls.
- Weigh each pith ball. Hold the string in such a way that you are not measuring the weight of the string,
nor are you lifting the ball. Try to get two pith balls of almost exactly the same mass. Write it here: m =
____ g = _____kg
- Attach the test tube clamp to the retort stand and clamp the straw into it. If the straw isn’t already split on
the unfastened end, check the other end or get a pair of scissors if necessary and cut a ½ cm slit into the free
end of the straw. Slide the pith ball strings into this slit so that they hang freely beside each other. Your
teacher may have an apparatus set up as a model for you.
- Slide the strings up or down through the straw slit to make sure the pith balls are lined up perfectly side by
side. Also try to ensure that the strings converge to a sharp point at the top where they’re clamped to the
straw; i.e., ensure there is no gap between the strings there are the top.
- Measure the length of the string from the straw to the center of either pith ball.
- Write it here: L = ______ cm = ________ m
- Touch your fingers to the pith balls to discharge them (they should be side to side, touching each other).
- Rub the ebonite rod with fur. Slowly bring it near the pith balls.
1.What reaction do they first have to the approaching rod?
-
Allow one or both of the balls to touch the rod.
2. Now describe the reaction of the pith ball(s) to the rod.
3. Explain this reaction.
-
Touch the pith balls to discharge them again (they should be side to side).
Recharge the ebonite rod and quickly tap it on both pith balls simultaneously so they both get charged.
Remove the rod.
4. Sketch what the apparatus looks like now, with the pith balls repelling each other.
-
Somehow carefully measure the distance between the pith balls from their centers. Write it here:
-
r = ____ cm = _____ m
Carefully measure the angle between the strings, or use trig to calc. it. Write it here:  = ______.
-
Touch the pith balls to discharge them again (they should be side to side).
Recharge the ebonite rod, more than before and quickly tap it on both pith balls simultaneously so they
both get charged. Remove the rod. Aim to get a much larger “r” than last time.
Carefully measure the distance between the pith balls from their centers.
-
Write it here: r = ____ cm = _____ m
Carefully measure the angle between the strings. Write it here:  = ________.
-
5. a) What 3 forces are acting on each ball?
b) Draw a F.B.D. for the left pith ball.
6. Since the pith balls are not accelerating, it must be true for each one that the net Force = _______ N.
7. Use the FBD and trig to write a relationship for the horizontal forces:
8.
Use the FBD and trig to write a relationship for the vertical forces:
9.
Draw a triangle with the two sides being the strings and the base being the distance between the balls.
Divide this triangle down the center to create two right angled triangles sitting back to back. Label all
the values you measured during the lab on this diagram.
* for the first pith ball trial (not so far apart) answer these questions:
10. Show a calculation for Fg.
11. Show a calculation for T (force of tension).
12. Show a calculation for Fq.
13. Calculate q in coulombs.
14. If every electron has a charge of 1.6 x 10-19 C, how many electrons are in each pith ball?
* for the second pith ball trial (when they were really far apart) answer these questions:
15. Redo #10-14 for the second pith ball trial.
16. What would it mean if one pith ball was higher than the other (if they were lop-sided)?
17. State some sources of error for this experiment.
Lesson #9 – Multiple Forces in Multiple Directions!
Part A: Gravity vs. Electricity – Fight!
Represent and Reason. Imagine two point-like charged objects of mass m1 and m2 that have electric
charges q1 and q2, respectively. Complete the table below that compares their electric and gravitational
interactions.
Electric
Gravitational
What property of
the objects
determines
whether they
participate in the
interaction?
What is the
direction of the
force between
the interacting
objects?
Write an
expression for
the magnitude
of the force that
one object
exerts on the
other.
It is an attractive
force.
How does the
magnitude of the
force depend on
properties of
the objects?
How does the
magnitude of
the force
depend on the
distance
between the
objects?
It is directly
proportional to
the masses m1
and m2.
The electric
charge
determines
whether they will
interact.
Example:
1. Calculate the force of repulsion between 2 electrons which are separated by 1m. (Remember,
the charge on one electron is -1.6x10-19 C).
Answer: F on left electron is 2.3x10-28 N [L]
2. Compare your answer to the gravitational force felt between them at separation.
(me = 9.11 x 10-31 kg)
Answer: F on left electron 5.54x10-71 N [R]
Part B – Complete Multiple Charges sketching handout!
Lesson #10 – The Electric Field
Pre-Lesson Notes
Unit: Fields
Date:
Concept: Electric Field (textbook section 7.3)
Definitions:
Electric field
Electric Field Lines
Electric Dipoles
Uniform Electric Fields
Earth’s Electric Field
Electric Fields in Nature
Equations:
One of the reasons why electrostatic forces seem so mysterious is that they have an effect at a distance,
without any objects being in contact. We are familiar with one other force that also has this property.
One way of explaining how gravity can have an effect on objects far away is with the idea of a field. We
say that the earth has a gravitational field that extends throughout space and that the earth is the
source of the field. The earth acts on distant objects “through” its gravitational field.
A: In Search of the Electric Field
For this investigation you will need an electroscope, an ebonite rod, an acetate strip, and a piece of
aluminum foil. An electroscope can be used as an electric field detector. If there is a fairly strong electric
field at the location of the metal sphere atop the electroscope, the metal leaves inside will repel each
other.
1. Observe. Charge the rod and move the electroscope in the region of space around the rod. Always
be careful with the glass electroscope. Don’t let the rod and sphere touch! Describe what you
observe.
2. Reason. Based on your observations, where is the electric field around the rod strongest? What
happens to its strength further away? What is the source of this field?
The sphere of the electroscope acts as a test particle in the rod’s electric field. The properties of a test
particle are such that it does not affect the field of the source (its charge and size are small).
3. Observe. Place the electroscope on a table and bring the charged rod nearby. Place the piece of
paper between the rod and sphere – don’t let them touch! Observe. Next, place the aluminum foil
between the rod and sphere – don’t let them touch! Observe. What effect do different materials
have on an electric field?
4. Apply. How would you protect sensitive electronics from unwanted electric fields?
5. Predict and Test. What would happen if there is aluminum foil around your cell phone? How many
bars do you think you will get? Try it out!
B: Picturing the Electric Field
1. Represent and Reason. For each situation pictured below, represent with arrows the gravitational
force or the electric force that a test particle would experience due to the sources at the points
shown. In the study of electricity we will always choose all test particles (test charges) to have a
positive charge.
The planet Earth
●
An object with a large negative
charge
An object with a large positive
charge
●
●
Earth
●
●
●
●
●
●
+
●
-
●
●
A force field (a field representing forces) is a way of representing the force vectors due to a source at
every point in space. This would require many, many vectors so instead we draw field lines. To do this
we use a set of rules for electric fields:
(1) Electric field lines follow the “path” of the vectors. The vector is always tangent to the field line.
(2) E-field lines start on positively charged objects and end on negatively charged objects.
(3) The magnitude of the field at a point is represented by the density or concentration of the lines
near that point.
(4) A corollary to this idea is that the number of lines leaving or terminating on a charged object is
proportional to the magnitude of its electric charge.
2. Represent and Reason. The table below gives four examples of point-like charged objects. Study
the given examples and draw E-field vectors and E-field lines for each source.
(a) a charge of +q
(b) a charge of +2q
Electric Field Lines
Electric Field Vectors
●
3.
●
●
● ● ●
● + ●
●
●
●
●
●
●
●
●
●
●
●
●
●
●
● ● ●
● + ●
●
●
●
●
(c) a charge of -q
●
●
●
(d) a charge of -2q
● ● ●
● - ●
● ●
●
●
●
●
●
●
●
●
●
●
● ● ●
● - ●
●
●
●
●
●
●
●
●
+
Apply. A region of space has an electric field as shown to the right. Note that we do not know
the location of the source of this field. A particle with a negative charge is placed at point A and
then at point B. Draw the force vectors at positions A and B. Explain how to decide the direction
of the force on the particle at that position in space. Explain in terms of the field line density
how you determined the lengths of the vectors.

A
B
Lesson # 11 – Electric Fields
A: Problem Solving
To help deepen our understanding of fields, please read the following 2 lessons online. You can use a
computer in the classroom or your tablet.
http://www.physicsclassroom.com/class/estatics/Lesson-4/Electric-Field-Intensity
http://www.physicsclassroom.com/class/estatics/Lesson-4/Electric-Field-Lines
In the diagram below, beads 1 and 2 carry charges 1.0 nC (nanocoulombs) and 2.0 nC, respectively. P is
just a point in space, not a charge. The electric force exerted on bead 2 by bead 1 is 12 N.
1
2
P
First let’s think about the effect of bead 2 on bead 1.
1. Reason. Find the electric force exerted by bead 2 on bead 1. Yes, you have enough information: use
a basic law of physics!
2. Calculate. Find the electric field due to bead 2 at the location occupied by bead 1.
3. Reason. Is your D#2 answer greater than, less than, or equal to the electric field due to bead 1 at
the location occupied by bead 2? Explain.
4. Reason. Given that beads 1 and 2 feel different fields, it’s reasonable to expect that they also feel
different forces. But they don’t! To reconcile this apparent conflict, explain in intuitive terms how
beads 1 and 2 can end up experiencing the same force even though they feel different fields.
5. Reason. The charge on bead 1 is now tripled. How does that affect:
(a) the electric field due to bead 2 at the location occupied by bead 1? Explain.
(b) the electric field due to bead 1 at the location occupied by bead 2? Explain.
(c) the force felt by bead 2? Explain.
(d) the force felt by bead 1? Explain.
Definition of the Electric Field.
The electric field describes the electrical effects throughout space of charged source objects on other
charged particles. The strength and direction of the electric field only depend on the charges of the
source and the position in space. To measure the electric field we need to use a charged particle. The
value of the electric field at a point in space (for example, 10N/C) means that if we place a particle at
that position, it will experience 10 N of force for each coloumb of its charge.
E 
FE
q
Use the equation to find the magnitude of the electric field and use a sign convention to show the
directions, just like you do with forces.
Electric Field Sketches and Calculations
1. How does the magnitude of the total field at the middle x spot
compare to the total field at the right x spot? (All distances are equal.)
-3
x
+1
x
2. Try again with a new set of charges:
x
-3
x
+3
+2
3. What is the ratio of the total fields at the two marked spots below?
x
-2
x
+1
+1
4. (a) Calculate the electric field midway between the +1 and the -3
charge. (All charges are in nC.)
+4
+1
30 cm
-3
x
30 cm
30 cm
(b) Calc. FQ on a 0.5 C charge placed at the x.
(c) Calc. FQ on a 20 mC charge placed at the x.
Answers: (1) 6:1
(2) 13:19
(3) 59:37
(4) 500 N/C
250 N
10 N
Lesson #12: Non-Collinear Fields
If charges are not in a perfectly straight line, we still need to add up forces on them, or the fields around
them, but now we’ll need to use trigonometry and vector components.
E.g. #1) A - 3.0 nC charge is 5.0 cm above the midpoint between two +4.0 nC charges which are 10 cm
apart, as shown:
a) Calculate the net force on the negative charge.
b) How would you calculate the ratio of the net force on each charge?
c) What is the total field at the midpoint between the two positive charges?
d) Imagine this diagram now as a side view, with the bottom charges sitting on the ground. What new
charge would they need in order to levitate the -3.0 nC charge if it weighs 0.25 g?
-3
+4
+4
E.g. #2: Three +4.0 nC charges are arranged at the corners of a
square.
a) What is the total field at the missing corner?
b) What is the net force on the bottom right charge?
+4
+4
+4
Lesson #13: Fields and Potential Energy
Pre-Lesson Notes
Unit: Fields
Date:
Concept: Potential Difference, Electric Potential, Electric Potential Energy due to Point Charges (section
7.4, 7.5)
Definitions:
Work and Electric Potential Difference
Electric Potential Energy
Electric Potential
Electric Potential Due to a point charge
Electric Potential energy of two point charges
Equations:
Electric Potential:
Electric Potential Energy:
Electric Potential Due to a point charge:
Electric Potential energy of two point charges:
Imagine forcing two repelling positive charges toward each other. This would require work. By doing
work to the charges, you give them potential energy. This situation is very similar to gravitational
potential energy.
Consider that the force on an object near the earth can be given as Fg = m⋅g
Work is F x d, so if we lift the charge up (in line with gravity)
m⋅g x h (where h is the distance or height that we lift it.) Doing work results in giving something
energy; that’s why we then get the equation: Eg = m⋅g⋅h.
So now let’s imagine that we do work against the electrostatic force. This time, F x d would be FQ and
the distance we now call r. So the equation becomes:
(Remember that energy isn’t a vector so never put a “hat” on the symbol E for energy; whereas the
symbol for field 𝜀 should have a vector “hat”: 𝜀⃑ )
Questions:
1. Calculate the electric potential energy stored in an electron and a proton which
are: (a) 0.5 m apart; and (b) 1 m apart.
2. Calculate the work done to move the electron from 0.5 m away to 1 m away.
3. The proton and an electron are now released from rest 1 m apart. Estimate the speed of the electron
when they are 0.5 m apart. (Assume the proton doesn’t move.)
Answers: -4.6 x 10-28
-2.3 x 10-28
-2.3 x 10-28
22.5 m/s
Electrostatic Work and the Conservation of Energy
Although energy (E) is not a vector, it can be (+) or (–), and this has some significance. Consider the
energy in the following two scenarios, and how much work is required to change from the 1st to the 2nd
(all charges are in mC) :
1st
Eq1 =
+2
+4
2m
2nd
You push
+2
1m
+4
You push
Eq2 =
Work = ∆E = E2 – E1
(done to go from
1st scenario to 2nd)
W = Eq2 – Eq1
=
=+
J
(the work needed is (+) so you DO work to make it happen; i.e., you need to INPUT energy.)
Now let’s consider instead what happens when they start close together and are released:
1st
(now remove the barriers)
+2
1m
+4
2nd
+2
2m
+4
Work = ∆E = E2 – E1
(done to go from
1st scenario to 2nd)
W = Eq2 – Eq1
=
=J
(the work needed is (-) so you GET work from them; i.e., an energy OUTPUT is possible!)
These charges will accelerate away from each other. Using the “Law of the Conservation of Energy” we
can calculate their speeds at any place.
Example: What is the speed of each when they are at 2.0 m apart?
Lesson #13: Using the Conservation of Energy for Field Calculations
Imagine two charged particles, such as two electrons, which find themselves close to each other. They
will obviously repel each other, and if they are free to move, they will accelerate away from each other.
Their speed (v) is increasing, so the kinetic energy (Ek) is
.
Their separation (r) is increasing, so the electrostatic energy (Eq) is
.
This kinetic energy must come from somewhere. Indeed, the change in Ek comes from Eq. Recall that
Etotal1 (initial) = Etotal 2 (final)
Notice how the equation for electrostatic energy has 2 q’s in it, but the equation for kinetic energy only
has one m in it. So the two electrons (at first being at rest) will have a total energy of:
Total energy = 1st electron’s kinetic E + 2nd electron’s kinetic E + electrostatic energy between them
(you see, Eq =
k ∙ q1 ∙ q2
, so this one equation contains and takes care of both q1 and q2.)
r
But after accelerating for a little while, the total energy will be:
(and notice here how both m’s are
present, using two expressions; and both q’s are present too, but using only one expression. This way,
both types of energy are represented for both particles.)
Thus, ET1 = ET2 becomes :
Since they’re both electrons, q1 = q2 and m1 = m2 and they will have the same speeds too. So you
can re-arrange the last equation to find that:
Examples:
1. Using the mass and charge of an electron (found in your textbook or on the physics classroom
wall,) calculate the speed two electrons will have if they are released from r = 1.00 mm apart,
when they are:
a. 2.00 mm apart
b. 10 cm apart
c. 1.00 m apart
d. Their maximum possible speed (at r2 = infinity.) Yes, it’s true that they can accelerate
forever and yet reach a finite final speed! More on this later…
2. Try a similar solution backwards: starting from ET1 = ET2 find how close two electrons would need to
initially be so that they will eventually reach speeds of one thousandth the speed of light.
Answers: 1b) 501 m/s
1d) 503.6 m/s
2. r = 2.8 nm
8
O
A “Repulsive” Question
Imagine that two oxygen atoms which gained two valence electrons are repeatedly fired straight at each
other in a particle accelerator, always with the same speed as each other.
a) If they are fired from 1.0 m away, and they get to a minimum separation of 1.0 cm before they
are stopped by their mutual repulsion, what initial velocity must they have had? (Answer: 1.85
m/s)
b) If this time they are fired from really really far away (i.e., EQ is approx. zero), what speed would
they need so that they will end up 1.0 mm away before stopping? (Answer: 5.9 m/s)
c) This time, they are somehow pushed very close together and released. What initial separation
would be necessary so that they will end up going a final maximum speed of 3.0 x 108 m/s?
(Answer: 3.8 x 10-19 m)
16
Lesson #14: Milli-kan…Can you?
Pre-Lesson Notes
Unit: Fields
Date:
Concept: The Millikan Oil Drop Experiment
Definitions:
Fundamental physical constants
Elementary charge
Charge of a proton
Explanation:
Millikan’s experiment
Equations:
Electric Potential due to a point charge:
Electric Potential:
In 1909, Robert Millikan devised the first method for direct measurement of the electric charge of a single
electron. Using the apparatus described below, he discovered that the electric charges on the drops were all
whole-number multiples of a lowest value, the elementary electric charge itself, and thus that electric charge
exists in basic natural units.
Typical Millikan questions: (try them now!)
1. Calculate the charge on a 0.050 g oil drop which is suspended between two plates 6.0 cm apart,
using a potential difference of 24 V. (Answer: 1.2 µC)
2. What voltage would accelerate a 0.00020 g oil drop which has 6000 excess electrons, upwards
at 2.2 m/s2 between two plates 1.0 cm apart? (25 MV)
Millikan Oil Drop Example
An oil droplet with a mass 5.0 x 10-16 kg accelerates at 2.20 m/s2 upwards between two charged metal
plates. The top plate is negative and the positive plate is 3.00 cm below it. A 9.0 V battery is connected
to them.
a) Calculate the charge on the oil droplet.
b) What excess or deficit of electrons would the droplet have?
Lesson # 15 – Magnetic Fields
Pre-Lesson Notes
Unit: Fields
Date:
Concept: Magnets and Electromagnets (section 8.1)
Definitions:
Auroras
Permanent Magnets
Magnetic Field Lines
Earth’s magnetic field
Principle of Electromagnetism
Fields can describe the force of any interaction that has an effect throughout space. Another force you have no
doubt played with is magnetism. We need to develop some new rules and ideas to describe magnetic effects. For
example, with magnetism we won’t use test particles we will use test-compasses!
A: Searching for a Field
You need a bar magnet and a compass.
A compass is a special device whose north end points in the direction of the magnetic field at that
position in space.
1. Observe. Check that your compass works – it should jiggle at the slightest touch. Place the compass
at each point in the region around the bar magnet as shown below. Draw an arrow showing the
magnetic field vector ( B ) at that position in space.












2. Observe and Reason. How does the interaction of the magnet and compass depend on the distance
from the bar magnet and its location around the bar magnet? How can you tell? Modify any of your
field vectors if necessary.
3. Observe. Use the applet: http://falstad.com/vector3dm/. Set the field selection to “solenoid”. Set
the display to “Field Vectors” and then “Field Lines”. It may be easier to see if you choose “Show Y
Slice”. We can’t examine the field inside the bar magnet with a compass, but we can study it using a
solenoid. A solenoid is a coil of wire whose field is very similar to that of a bar magnet. Sketch the
magnetic field lines around and inside the bar magnet. Add arrows to the field lines to match your
field vectors above.
4. Explain. Point out two regions where the field strength is particularly strong and weak. How can you
tell?


B: The Birds and the Bees of Magnetism
You need an alligator wire, a compass and a 9V battery. WARNING: Never clip both ends of the wire to
the battery – it will overheat!
1. Observe. Clip only one end of the wire to the battery. Move the middle of the wire above the
compass. Describe your observations. Is there any evidence for a magnetic interaction between the
compass and the wire?
Reminder: The conventional current flows from the positive terminal to the negative terminal of the
battery. We can imagine positive charges flowing through the wire in a direction opposite to that of the
actual electron flow. We understand that it is only electrons that flow in most electrical circuits, but we
are stuck with using the conventional current for historical reasons.
2. Observe. Place the wire over the top of the compass such that it is parallel to the
needle. Briefly touch the second alligator clip to the other battery terminal – DO
NOT CLIP IT ON! Hold the wire in place only long enough for the needle to
stabilize. Record your observations. Illustrate the direction of the conventional
current flow and the compass needle. Is there any evidence for the existence of a
new magnetic field?
3. Observe. Now hold the compass just above the wire. Repeat your observations.
Magnetic fields (B) are created by moving electric charges. Charges at rest (static charges) do not
create magnetic fields. To illustrate the motion of the charges and the magnetic fields, we need a new
set of symbols. For a vector pointing out of a page draw:  For a vector pointing into a page draw: 
4. Observe and Explain. Use the applet: http://falstad.com/vector3dm/. Set the field
selection to “current line”. Set the display to “Field Vectors”. Select “Show Z
Slice”. Sketch a sample of the field vectors around the wire. Is the current flowing
into or out of the page? Does this representation agree with your compass
observations? Explain.
Field Vectors
5. Observe. Now set the display to “field lines”. Sketch the field lines and the current
in the box to the right. Which way should the arrows on the field lines point? Now draw them!
Field Lines
The Right Hand Rule for Current Flow relates the motion of positive charges (conventional current) and
the magnetic field they create. When the thumb of your right hand points in the direction of the
conventional current, your fingers curl in the direction of the magnetic field lines.
6. Predict Imagine the wire is oriented across our page as shown below. It is difficult to draw the field
lines from this point of view. We can simply show the direction of the field vectors as they pass
through the page surface. Explain whether the field lines will point into or
out of the page in the region around the wire.
7. Test. Switch back to field vectors and choose “Show X Slice”. You may
need to rotate the image to understand what you are seeing. Explain any
differences with your prediction. Draw field vectors illustrating your
observations using  and .
Electromagnetism: Right-Hand Rules Summarized
Note: All right hand rules are based on the assumption that current travels from positive to negative.
For electron flow from negative to positive, simply use the left hand instead!
RHR for a Straight Conductor
Your thumb points in the direction
of the current and your curled
fingers describe the direction of
the produced magnetic field.
RHR for a Current Carrying Loop (or Helix)
Your fingers curl around in the direction of
the looping current and your thumb then
points to the North side of the magnetic
field produced.
RHR for Electromagnetic Force
Your thumb points in the direction
of the current and your fingers
point in the direction of the
magnetic field. Then, the force
on the conductor is directed
outward from the palm of your hand.
RHR for Moving Charge through a Uniform Field
Your thumb points in the direction of the velocity
of positive charge, your fingers point in the direction
of the magnetic field. Then the force on the conductor
is directed outward from the palm of your hand.
Lenz's Law
The magnetic field of an induced current always opposes the change in magnetic field that is causing the
induced current
Lesson #16 – Magnetic Forces on Charges
Pre-Lesson Notes
Unit: Fields
Date:
Concept: Magnetic Force on Moving Charges and on current-carrying conductor (section 8.2, 8.3)
Definitions:
Charge in a magnetic field
Right hand rule for a moving charge in a magnetic field
Equations:
Magnetic force (two equations)
A: Charges at Rest
This is a class demonstration. We want to study the interaction between a magnetic field and a charged
object that is stationary. An electromagnet will create the magnetic field and rubbing a balloon will
create a static electric charge.
1. Observe. For each situation, record your observations in the chart below.
Charged balloon, magnet turned Charged balloon, magnet current Charged balloon, magnet turned
on
reversed
off
2. Reason. Is there any evidence from your observations for the existence of a magnetic interaction
between a charged object at rest and a magnetic field? Justify your answer using your observations.
B: Charges on the Move
http://webphysics.davidson.edu/physlet_resources/bu_semester2/c12_force.html
This is a class demonstration. We want to study the interaction between a magnetic field and moving
charges. An old TV will produce moving charges. A permanent magnet will provide the magnetic field.
1. Observe. The bar magnet is held perpendicular to the beam with the north
pole beside the TV. To a good approximation, the magnetic field points straight
out from the end of the bar magnet. Record your observations. Complete the
diagram showing the velocity of the current, the magnetic field and the force
acting on the charges.
2. Observe. The bar magnet is held perpendicular to the beam with the north pole
above the TV. Record your observations and complete the diagram.
3. Observe. The bar magnet is held parallel to the TV with the north pole pointing
towards you. Record your observations and complete the diagram. Why do the
field lines look so different in this example?
4. Reason. Two conditions must be met in order for magnetic fields to exert a force on a charged
object. From your observations in parts A and B, what are these two conditions?
The force that results from the movement of a charged object in a magnetic field is given by:
Fm  q v B sin  , where q and v are the charge and speed of the object, B is the strength of the magnetic
field the object is moving through and θ is the angle between the magnetic field lines and the velocity.
The direction of the force is determined by the Right Hand Rule for the Magnetic Force on Positive
Charges: The thumb points in the direction of the velocity of the object with a positive charge, the
fingers in the direction of the magnetic field lines and the palm of your hand points in the direction of
the force. If the object has a negative charge, just reverse you answer from the right-hand rule.
5. Reason. Use the equation for the magnetic force to verify your conditions from the previous
question. Explain.
6. Predict. Use the right hand rule to predict how the beam will deflect when the bar
magnet is held perpendicular to the beam with the south pole on the left side of t he
tube. Draw a picture to illustrate your prediction. We will try this out as a class. Move
on for now.
C: The Motion of Charges in a Magnetic Field
Consider an electron traveling with an initial velocity v moving through a
magnetic field that is oriented into the page as shown to the right.
1. Reason. What is the direction of the force acting on the electron? Explain
and draw the force on the diagram.
prediction




v
●












B


2. Reason. A short moment later, where might the electron be located?
Draw it and its new velocity vector. At this later moment, what will the direction of the force be?
Draw this on the diagram. Explain.
3. Predict. Assume the electron never leaves the magnetic field. What will the complete path of the
electron look like? How will the directions of the velocity and force vectors compare? What type of
motion results? Explain.
4. Reason. How much work does the magnetic field do on the electron? How does the electron’s speed
and kinetic energy change? Explain.

D: Bubbles, Tiny Bubbles
The image to the right shows the paths of particles
travelling through a bubble chamber, a chamber of
superheated fluid. A charged particle travelling
through the fluid causes a bubble to form. This
leaves a bubble trail behind the particle. A strong
magnetic field passes through the bubble chamber
and is oriented out of the page. The trails of five
particles are labeled.
A
E
C
D
B
1. Apply. Indicate the charge of the particle that
produced each trail. Explain how you can tell.
2. Apply. Rank the momentum of the particles
that produced each trail. Explain.
3. Apply. Why do you think trails C and D spiral inwards?
Lesson #17: Electromagnetic Disturbances
We have studied the creation of both electric and magnetic fields separately, but now it is time to build
up the complete picture of what happens when current flows through a wire.
A: The Physics of a Wire
Consider a wire connected between the negative and positive terminals of a power supply. A steady
current travels through the wire. The diagram to the right illustrates this situation and shows two sets of
field lines.
1. Represent. Choose a colour to represent the conventional
current. Draw an arrow that represents the flow of
conventional current.
2. Represent and Explain. Choose a colour to represent the
electric field. Explain how you decide which lines are the
electric field lines. Label and colour code these lines. Add
arrows showing the direction of the field lines.
-
+
3. Represent and Explain. Choose a colour to represent the magnetic field. Explain how you decide (as
if you haven’t done question #2) which lines are the magnetic field lines.
B: A Closer Look
Now imagine that we zoom in to a small region of space located above the middle of the wire. The
diagram below illustrates this along with the electric and magnetic field lines. Note that one field points
in or out of the page and is represented with the circles.
1. Represent. Complete the diagram by: (a) labelling and
colour coding the electric and magnetic field lines and (b)
showing the direction of all the field lines and current

2. Reason. How do the directions of the electric and
magnetic field lines compare in this region of space?
wire
3. Explain. A positive test charge  is placed at rest in this region of space. In this investigation, we will
ignore the effects of the magnetic field since they are relatively small. What will happen to the
positive charge which is initially is at rest? Explain and draw vectors representing any forces.
Below is a third diagram showing the same region of space as above. But for this example, we have
switched the polarity of the power supply, reversing the positive and negative terminals shown in the
first diagram.
4. Represent and Explain. Complete the diagram as you did in
question #4. Describe the how the fields have changed due
to the reversed terminals.

5. Explain. What will happen to the positive test charge which
is initially at rest?
wire
C: Shaking Things Up
Now imagine that we continuously vary the polarity of each terminal of the wire from positive to
negative.
6. Reason. Describe the motion of the charges in the wire.
7. Reason. Describe the motion of the positive test charge as the polarity varies.
8. Observe. When the polarity of the wire changes, the fields don’t change everywhere immediately.
They begin to change closest to the wire and then spread outwards through space. Use the
simulation: http://falstad.com/emwave1. Describe what this changing pattern looks like.
9. Observe. Freeze the simulation. Static electric or magnetic fields drop off in a steady way at
positions father from the sources. Is this happening in this situation? Explain.
An alternating current produces a changing pattern of electric and magnetic fields that spreads
outwards from the wire through space. In 1864, James Clerk Maxwell studied these phenomena in detail
and mathematically described the wave-like motion of these oscillating fields. From his calculations he
determined the outwards speed of the rippling fields to be given by:
v2 
4k
o
Where k is Coulomb’s constant (k = 8.99 x 109 Nm2/C2) and o is a similar constant for magnetism (o =
1.257 x 10-6 Tm/A).
10. Predict. Determine the velocity of this wave including units (1A = 1C/s) (1T = 1 kg/Cs). What are the
implications of this result?
Rules for Electromagnetic Waves.
What we have discovered is an electromagnetic wave which works according to these ideas:
1) An EM wave is created by the acceleration of electrical charges (our varying current in this case).
2) Energy from this wave may be absorbed by causing other charges to accelerate (our distant charge).
3) The frequency of the wave’s oscillation and the properties of the object containing the distant
charge determine how much that charge will respond to the wave and how much energy it will
absorb. (a kind of resonance)
Lesson #18: Understanding EM Waves
A genuine, real-life EM wave is very tricky to visualize – it has much more detail than a simple water
wave. To help us picture what takes place when an EM wave travels through space we will use the line
model of an EM wave.
A: The Line Model of an Electromagnetic Wave
To the right is a screen capture from
the simulation we used in our
previous investigation. It represents a
two-dimensional slice in space and
the field values along that slice. To
help us talk about this image we will
R
define the vertical direction as the ydirection and the x- and z-directions
as the horizontal plane. The wire is
positioned
along the y-axis
with its middle
at the origin.
This image
represents the
field values in
the xy-plane.
G
R
G
R
G
y
x
z
+
x
E
1. Represent. Imagine we draw a line along the x-axis starting at the
surface of the wire. Remember that the simulation does not show
the field strength using the arrow lengths. Sketch a graph of the
electric field vs. position along the +x-axis (the white line). Do the
same for the magnetic field. Upwards is positive for the E-field and
outwards is positive for the B-field.
+
x
B
-
We can put these two graphs together and illustrate the field vectors to give us the line model of an
electromagnetic wave. This is a picture of the electric and magnetic field vectors along one line through
space at one moment in time. One extraordinary feature of an EM wave is the fact that the amplitude of
the peaks decreases very, very little allowing EM waves to travel extremely far through space- even
across the entire universe!
2. Reason. Examine
the line model of
an EM wave shown
to the right. Rank
the magnitude of
the electric field at
points A, B, C and
D in space. Explain.
A
E

B
C
D



B
3. Reason. In what direction is the EM-wave propagating (travelling)? Draw an arrow on the diagram
above.
4. Reason. How do the directions of the electric and magnetic fields compare with the direction of
propagation?
5. Reason. An electron is located at point C in the diagram above. What will initially happen to it?
B: Wave, Meet Conductor. Conductor, Meet Wave.
Imagine that the EM wave depicted above in question #2 is a radio wave. It now passes by a long, thin
conducting wire that is oriented along the y-axis.
wire
1. Reason. As the wave propagates (travels) past the wire, would the electric field due to the radio
wave cause the charges in the wire to move? If so, would the charges move in a direction along
the wire? Explain.
2. Reason. As the wave propagates past the wire, would the magnetic field due to the wave cause the
charges in the wire to move in a direction along the length of the wire? Explain.
3. Reason. Imagine a light bulb is connected to two halves of the same wire in the middle. The bulb
and wire are again placed in the path of the previous radio wave and is oriented parallel to the yaxis. Would the bulb ever glow? Explain.
wire
4. Reason. Would it glow if the wire was oriented parallel to the z-axis? Explain.
5. Reason. Would it glow if the wire was oriented parallel to the x-axis? Explain.
An antenna is a conductor designed to capture the energy from an electromagnetic wave. The fields of
the wave cause charges in the antenna to accelerate transferring electromagnetic energy from the wave
into the electrical energy of the moving charges. An electric current is created or induced and this
current can be passed into a circuit providing a radio station signal, a Bluetooth signal, a wireless router
signal, a TV signal, a cellphone signal, … so many possible signals!
6. Predict. In order to best detect the oncoming radio wave (that is, to maximize the induced current in
the circuit), how should the antenna be oriented relative to the wave? Explain.
Lesson #19: Polarization
When a light wave is received by our eyes, we cannot tell which direction the electric field vectors point
in – but some creatures can! Certain materials react differently to electric or magnetic field vectors in
different directions.
A: Through the Polarizing Glass
You will need a pair of polarizing filters. Be careful since some of them are glass!
1. Observe. Look at the room lights through one of the polarizing filters. Describe how the filter affects
what you see. Does rotating the filter have any effect?
2. Observe. Hold a second polarizing filter in front of the first, and look at the room lights again.
Describe how the filter affects the light you see. How does rotating one of the filters with respect to
the other affect what you see?
A light beam is linearly polarized when the electric field
in all parts of the beam oscillates along a single axis.
The direction along which the electric field oscillates is
called the direction of polarization of a light beam. An
unpolarized beam is made up of many rays of light,
each of which have the electric field oscillating in
different, random directions.
The light transmitted by a polarizing filter (or polarizer)
depends upon the relative orientation of the polarizer
and the electric field in the light wave. Every polarizer
has a direction of polarization, which is often marked
by a line on it. The electric field of the transmitted
wave is equal to the component of the electric field of
the incident wave that is parallel to the direction of
polarization of the polarizer. Representations of all
these can be used to draw a polarization diagram.
3. Interpret. Describe what is happening according to the
polarization diagram to the right.
y
E

y
x
z
E

x
z
polarized beam
filter with horizontal
direction of
polarization
unpolarized beam
Reason. Do the room lights produce polarized light? Explain how you can tell from your observations.
Draw a polarization diagram illustrating how the room lights can be completely blocked by the filters.
x
5. Observe. Describe other sources of polarized light from around the classroom or amongst your
belongings.
6. Reason. Describe how you should orient the polarizers with respect to one another so that the light
transmitted through the polarizers would have (1) maximum intensity or (2) minimum intensity.
7. Represent. Draw a polarization diagram for the situation of minimum intensity you described above.
x
When two polarizers are oriented with respect to one another such that the light is at a minimum
intensity, the polarizers are said to be crossed.
8. Reason. Suppose that you had a polarizer with its direction of polarization marked. How could you
use this polarizer to determine the direction of polarization of another unmarked polarizer? Explain
your reasoning.
9. Reason. A beam of polarized light is incident
on a polarizer, as shown in the side view
diagram. The direction of E-field of the light
makes an angle  with respect to the
polarizer’s direction of polarization (see front
view diagram). The amplitude of the electric
Polarizer
Incident
light
E vector

Side View
Front view
field of the incident light is Eo. Write an expression for the amplitude of the transmitted component
of the electric field. Represent this in a polarization diagram.
x
10. Predict. A beam of unpolarized light travels through a pair of crossed filters – no light is transmitted
from the last filter. What will happen when a third filter whose direction of polarization makes a 45o
angle with both filters is placed between two crossed filters? Explain and draw a polarization
diagram. We will test this as a class.