ELECTRIC FIELDS
Physics 30S
Outcomes
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S3P-4-14: Define the electric field qualitatively as the region of space around a charge
where a positive test charge experiences a force.
S3P-4-15: Diagram electric fields using lines of force with respect to a positive test
charge. Include: single point charges (positive and negative), near two like charges, near
two unlike charges, between a single charge and a charged plate, between two
oppositely charged parallel plates
S3P-4-16: Define the electric field quantitatively as a force per unit charge (E = F/q)
and solve problems using the unit field concept (F = qE).
S3P-4-17: Solve problems for the motion of charges between parallel plates where Fnet
= Fe + Fg .
S3P-4-18: Describe a simplified version of Millikan’s experiment for the determination of
the elementary charge (solve for charge when Fe = Fg).
S3P-4-19: Define the elementary charge and convert between elementary charges and
coulombs. Include: q = Ne
Introduction
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Definition: An electric field is the region of space
around a charge where a positive test charge
experiences a force
Electric field strength is measured by force per unit
charge
E = Fe/q
How Do We Know?
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http://www.youtube.com/watch?v=7NvKtPq0Hbc
Comparing Electric and Gravitational
Fields
Gravitational Fields
Electric Fields
Force per unit mass
Force per unit charge
g = Fg/m
E = Fe/q
Only attracts
Can attract or repel
Unit: N/kg
Unit: N/C
Strength depends on size (mass)
of object creating field
Strength depends on size (amount
of charge) of object creating field
Important Physicists
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Charles-Augustin de Coulomb (1784)
 Definition
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of electrostatics (Coulomb’s law)
Michael Faraday (1830s)
 Electromagnetic
induction
 Electric charge per mole of electrons
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Robert A. Millikan (1923)
 Millikan’s
Oil Drop experiment
Units
Elementary Charge: 1 e.c. = 1.6 x 10-19 C
 Coulomb: unit of charge
 1 C = 6.25 x 1018 e.c.
 1 μC = 1.0x106 C
 Field: N/C
 q = n x e.c.
(charge is equal to number of electrons x elementary
charge)
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Positive Test Charge
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Imagine that we could create a small charged
particle whose own charge was too small to affect
the surrounding electric field
Used to map out electric fields
Always positive!
Rules for Drawing Diagrams
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Arrows indicate direction that a positive test
charge would move in
 repulsion from positive, attraction to negative
Number of lines is proportional to electric field
strength
Equipotential lines: lines to show locations where
the field has equal strength
Direction of Electric field at any point is tangent to
the field lines
Drawing Electric Fields – Single point
Charge
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All lines radiating either towards (negative charge)
or away (positive charge) from point
Near Like Charges
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Lines are similar to single point charges, but are
altered in the space between charges
Lines are pushed away from the midpoint
Near Unlike Charges
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Lines are straight between at the midpoint (pulled
tight between the two charges)
Electric dipole: two charges of equal magnitude and opposite sign
Single Charge and Charged Plate
Point charge still radiates, but
lines close to the plate are
pulled towards/away from the
plate
Parallel Plates
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Straight lines between plates
Some rounding at the edges
Interactive Problems
www.explorelearning.com:
 Charge
launcher
 Coulomb force
Homework
1.
2.
3.
Draw the electric field for a dipole (one positive,
one negative).
Draw the electric field for two positive point
charges.
How would the field diagram change for two
negative point charges?
Problems
Suppose we could drop a positive test charge in an
electric field. Draw an arrow in the field diagram to
represent the force on the test charge if:
a) it is just to the right of a positive point charge
b) it is just to the right of a negative point charge
c) it is centered between two equal negative point
charges
d) it is centered between two equal point charges of
opposite sign (dipoles)
Electric Fields – Intro Questions
Example 1
a) A toaster passes -3.00 C of charge through the
heating coils. How many electrons move through the
coils?
b) How many electrons would it take to make up a
charge of –12.4 μC?
a)
q = -3.00 C
1 electron has a charge of -1.6 x 10-19 C
1.88 x 1019 electrons
b)
q = -12.4 x 10-6 C
1 electron has a charge of -1.6 x 10-19 C
7.75 x 1013 electrons
Example 2
A test charge of +5.00 C is placed a distance of
2.00 m to the right of the point charge +q. The
strength of the electric field produced by this
charge is +10.0 N/C. What is the electric force on
the test charge? What is the direction of this force?
E = Fe/q
Fe = Eq
= (+10.0 N/C)(+5.00 C)
Fe = +50.0 N
Direction:
Positive force indicates repulsion.
Test charge will move to the right.
Example 3
A test charge of +5.00 C is placed a distance of
2.00 m to the right of the point charge -q. The
strength of the electric field produced by this
charge is -10.0 N/C. What is the electric force on
the test charge? What is the direction of this force?
E = Fe/q
Fe = Eq
= (-10.0 N/C)(+5.00 C)
Fe = -50.0 N
Direction:
Negative force indicates attraction.
Test charge will move to the left.
Example 1b (Optional Extra)
a) How many electrons would it take to make up a
charge of –1.28 x 10-18 C?
b) A lightning bolt passes -25.0 C of charge in a
strike. How many electrons are in the strike?
a)
q = -1.28 x 10-18 C
1 electron has a charge of -1.6 x 10-19 C
8.0 electrons
b)
q = -25.0 C
1 electron has a charge of -1.6 x 10-19 C
1.6 x 1020 electrons
Example 2b (Optional Extra)
A test charge of +3.00 C is placed a distance of
5.00 m to the right of the point charge +q. The
strength of the electric field produced by the point
charge is +8.00 N/C. What is the electric force on
the test charge? What is the direction of this force?
E = Fe/q
Fe = qE
= (+3.00 C) (+8.00 N/C)
Fe = +24.0 N
Direction:
Positive force indicates repulsion.
Test charge will move to the right.
Example 3b (Optional Extra)
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A test charge of +3.00 C is placed a distance of
5.00 m to the right of the point charge -q. The
strength of the electric field produced by the point
charge is -8.00 N/C. What is the electric force on
the test charge? What is the direction of this force?
E = Fe/q
Fe = qE
= (+3.00 C) (-8.0 N/C)
Fe = -24.0 N
Direction:
Negative force indicates attraction.
Test charge will move to the left.
Example 4
A positive charge on the left of a test charge creates
an electric field of 10.0 N/C. Another positive
charge to the left of a test charge creates an
electric field of 15. 0 N/C.
a) What is the total electric field acting on the test
charge?
b) If a charge of 2.0 C is placed in between the two
charges, what is force acting on the charge?
Example 4 - Solution
E1 = 10.0 N/C, positive test charge will move to the right
E2 = 15.0 N/C, positive test charge will move to the right
Enet = 25.0 N/C, positive test charge will move to the right
Example 4 - Solution
Fnet = F1 + F2
Fnet = qE1 + qE2
Fnet = (2.00 C) (10.0 N/C) - (2.00 C) (15.0 N/C)
Fnet = -10.0 N
Fnet is to the left
Homework
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1. Intro to Electric Fields Homework Handout
 Finish
 #3-6
# 1-2
Questions?
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Handout #1-6
Adding Electric Fields
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Electric Fields must be added using vector addition.
Draw a diagram to represent the direction of the
fields.
Use Pythagoras to find the magnitude of the
resultant field, and then use inverse tangent and
your diagram to determine the direction.
Remember, the direction of the field is the direction
of the force that a positive test charge would
experience!
Adding Electric Fields Handout –
Question 3
The electric field due to Q1 has a magnitude of 220 N/C at point W.
The electric field due to Q2 at point W has a magnitude of 420 N/C.
What is the net electric field (magnitude & direction) at point W?
What is the net electric force (magnitude & direction) experienced
by an electron if placed at point W?
W
Q2
Q
1
E = 470 N/C @ 28° S of W;
F = 7.6 × 10-17 N @ 28° N of E
One Last Extra Example – Electric
Fields Handout 2
A negative charge creates a field of magnitude 100 N/C at x. A positive charge
creates a field of magnitude 75 N/C at x.
a) Find the magnitude and direction of the net electric field at X.
b) If an electron is located at x, find the magnitude and direction of the net
force acting on the electron.
E = 125 N/C, E 37°S
F = 2.0 x 10-17 N, W37°N
Homework
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Adding Electric Fields Handout
#1, 2, 4
Millikan’s Oil Drop Experiment
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Elementary charge: there is a fundamental unit of
charge. All quantities of charge must be a multiple
of this fundamental unit.
Make Fe equal to Fg
Charge floats
Allowed Millikan to solve for charge
The Apparatus
Robert A. Millikan
Nobel Prize in Physics (1923)
Cartoon of Millikan’s Experiment:
http://www.youtube.com/watch?v=91E6KvCvRf0&NR=1
Motion of Charges between Parallel
Plates
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Fnet = Fg+ Fe
If Fnet is zero, then Fg and Fe must be equal!
Millikan Examples – Question 1
A negatively charged droplet of oil (m=2.0 x 10-5
kg) is placed between two oppositely charged
plates (upper plate is positive). The oil drop is
balanced when the electric field is adjusted to 4.2
x10-7 N/C.
a) What is the charge on the oil drop?
b) How many elementary charges are on the oil
drop?
a) q = -470 C
b) There are 2.9 x 1021 electrons on the oil drop
Question 2
An oil drop, whose mass is found to be
4.95 x 10-15 kg, is balanced between two large
horizontal plates with the upper plate positive.
The electric field strength between the plates is
E = 5.10 x 104 N/C.
a. What is the charge on the oil drop, both in
coulombs and in elementary charges?
b. Is it an excess or deficit of electrons?
Question 2 Solution
a) q = -9.51 x 10-19 C
= 5.94 e.c.
b) The electrical force on the oil drop must be
upwards in order to counteract gravity. Thus, the
drop is being attracted by the upper plate and
repelled by the lower plate. Since the upper plate
is positive, the oil drop must be negative. There is
an excess of electrons!
(≈ 6 electrons)
Question 3
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Suppose Millikan was able to isolate an electron so
that it balances between parallel plates. What is
the magnitude and direction of the electric field?
(mass of an electron is 9.11 x 10-31 kg and charge
of an electron is -1.6 x 10-19 C). What is the
magnitude and direction of the electric force acting
on the electron?
E = 5.6 x 10-11 N/C, down
F = 8.9 x 10-30 N, up
Homework
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Millikan: Electric Fields and Big Four Worksheet
 #8
van de Graaff Generator
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A van de Graaff is an apparatus able to hold a
charge on the outer ball. Any excess charge
collects on the outer rim of the ball.
A van de Graaff in Action
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http://www.youtube.com/watch?v=fTHOPe7mGIk
As negative charge collects on the ball, it is transferred to
the confetti paper. The paper pieces acquire negative charge and
repel each other, causing the explosion seen in the video.
Question 1
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A charged paper confetti of mass 1.0 x 10-7 kg is
hovering above a field of -70. N/C. What is the
charge on the confetti? How many electrons make
up this charge?
q = -1.4 X 10-8 C
There are 8.8 x 1010 electrons in the charge.
Question 2
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A pie plate of mass 0.005 kg carries a charge of 9.80 X 10-4 C. The plate hovering above the ball
of the van de Graaff. What is magnitude and
direction of the electric field of the van de Graaff?
What is the magnitude and direction of the force
acting on the pie plate?
E = 50.0 N/C, down
F = 0.0490 N, up
Question 3 - Challenge
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Suppose you have been abducted by mad scientists
and you have no idea where you are. You are
being forced to solve parallel plates questions by
applying Fnet = Fe + Fg. Suppose an electron (mass
is 9.11 x 10-31 kg and charge is -1.6 x 10-19 C) is
hovering in a electric field of magnitude 1.48 x 1010 N/C. Are you still on Earth? Why or why not?
Bonus: What planet could you be on?
No. You are not on Earth because g ≠9.8 m/s2. Since g = 26 m/s2, you could
be on Jupiter.
Homework
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van de Graaff: Electric Fields and Big Four
Worksheet
 #6
Using the Big Four
Just like in the other units, we can apply our
knowledge of the Big Four equations to Electric
Fields!
Example 1
A test charge is placed between charged plates. The
electric field strength is 10.0 N/C. The test charge
has a charge of q = +2.00 C and a mass of 4.00 x
10-4 kg. Assume the positive plate is on top.
a. Find the acceleration of the particle.
b. If the particle is initially at rest, what is the object’s
velocity after 6.0 ms?
a) 5.0 x 104 m/s2
b) Vf = 3.0 x 102 m/s
Example 2
A charge, q1 = -5.00 C, is placed in an electric field
between two charged plates (upper plate is
positive). The electric field strength is E = 4.00 N/C.
The mass of the charged particle is m = 2.00 x 10-4
kg.
a. Determine the magnitude and direction of
the acceleration of the charged particle.
b. If the particle is initially at rest, what will
the displacement be after 8.00 ms?
a) a = 1.00 x 105 m/s2 , up
b) d = 3.20 m
Example 3
A charge, q1 = 600.0 µC, is placed in an electric field between
two charged plates (upper plate is positive). The charge
accelerates until it crashes into the negative plate, located 1.4
mm away. The electric field strength is E = 800.0 N/C. The
mass of the charged particle is m = 2.00 x 10-6 kg.
a. Determine the acceleration of the charged particle.
b. If the particle is initially at rest, what will the final velocity be
when it contacts the plate?
a) a = 2.4 x 105 m/s2
b) v = 26 m/s
Homework
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Electric Fields and the Big Four (Uniform Fields)
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1-5, 7
The Plan
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Day 1: Intro to electric fields and diagrams
 Tesla coil, diagrams
Day 2: Field Problems
 Review & explorelearning.com
 Class questions and homework #1-2
Day 3: More Field Problems / Work Class
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Review questions from last day
Make explicit what negatives mean (negative handout)
Homework: 3-6
Day 4: Adding Electric Fields with Vectors
Day 5: Millikan’s oil drop and questions
Day 6: Van de Graaff problems
Day 7: Big Four
Day 8: In-Class Assignment (optional)
Day 9: Review
Day 10: Test