Unit 1
Lesson 1
Coulomb’s Law and the Electric Field
With this lesson, we will begin our study of a physical phenomenon that appears in
many guises: electricity. We see electrical forces at work in nature not only in the
awesome lightning display of a thunderstorm, but also in the functioning of our smallest
cells, which depends on the balance of electrically charged ions and their movement
through the cell membranes. Human technology has found parallel uses for these
forces: synthetic membranes tested in water-purification studies show promise of
“electrically” removing undesired ions from water, and the electrostatic air cleaner
produces electric fields rivaling those in a thundercloud: a
potential difference
between a thin wire and flat collecting plates ionizes the air, and the “flying” electrons
attach themselves to dust particles, which are then pulled to the collecting plates by
electric forces. Since forces that hold atoms together are ultimately electrical, the study
of electricity is the study of one of nature’s truly grand designs.
As you progress in your study of physics you will see the design unfold further: charges
whose position is constant produce electric fields, charges whose velocity is constant
produce magnetic fields as well as electric fields, and charges that accelerate produce
that special combination of electric and magnetic fields we know as electromagnetic
radiation (radio waves, x-rays, light, microwaves, etc.).
1-1: Electric Forces and Fields
OBJECTIVES: Distinguish between insulators and conductors and describe how each
behaves when subjected to an electric field.
Calculate, for a group of point charges at rest, the resultant force on
one of the charges caused by all of the others, or calculate the total
electric field at some point in space caused by all of the charges.
PREREQUISITES: Adding and subtracting vectors in rectangular and polar form
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Using the concept of a force field to describe the force a particle
would experience at any point in space
Reading Assignment
Young and Freedman, University Physics. Twelfth Edition.
Chapter 21, Sections 1 through 6 (Section 21-7, optional).
Study in your textbook
Commentary
The concept of an electric field is so central to our understanding of forces on charges
particles that it is worthwhile to study these topics together, even though this makes for
a rather long reading assignment for this section. The optional reading relates less
directly to the objectives, but you may find it useful for background.
In the reading assignment, notice that the electric field at a point is defined as the force
that would act on a unit charge placed at that point. You will also see conductors
described as materials in which electric charges are free to move, and insulators as
materials in which charges are not free to move. Let us combine these ideas to gain a
clearer picture of the essential behavior of conductors and insulators. If we establish a
nonzero electric field in a conductor, the conduction electrons (which are free to move)
feel a force equal to their charge times the electric field and thus initially accelerate, and
a current (movement of charge) gets started. If we establish an electric field in an
insulator, on the other hand, there will be essentially no current flow since the charges
are basically not free to move in response to the field. (In an insulator, charges can
move through distances like an atomic radius before they are stopped by forces within
the atom. This movement in response to an electric field sets up “dipoles,” which you
will study in a later lesson.)
The behavior of conductors in an electric field is crucial to many of the applications of
Gauss’s law that you will see in Lesson 2. You must keep in mind that if there is a net
movement of charges in a conductor, the charges are responding to an electric field.
Conversely, if we find that there is no net movement of charges in a conductor, this
indicates that there is zero electric field in that conductor.
The existence of different systems of electromagnetic units is discussed briefly in the
text. We will use the same system of units in this syllabus that is used in the text, but
you should be aware of the other systems since the units associated with them are still
seen in certain contexts. The most basic difference between systems concerns the
units and procedures used to measure electrical charge. In electrostatic (esu) systems
of units, charge is defined from Coulomb’s law- that is, in terms of the attraction or
repulsion of stationary charges separated by a unit distance. This allows us to choose
units for charge (e.g., the statcoulomb) that yield a proportionality factor of 1
2
(dimensionless) in Coulomb’s law. In electromagnetic (emu) systems, charge is defined
in terms of the current flow in parallel conductors that produces a specified magnetic
force between the conductors. This approach, which makes current the fundamental
unit and charge a derived unit, is used in the SI system. It leads, in general, to a
proportionality constant in Coulomb’s law that is neither dimensionless nor equal to 1;
this experimentally determined constant is analogous to
in the universal law of
gravitation. We will express forces in newtons, charges in coulombs, and the
proportionality constant as
, where
is the so-called permittivity of free space.
The approximation
is quite convenient and accurate enough
for our purposes in this course.
You are familiar with the principle of superposition from your study of waves. In its
general form, this principle says that if cause has effect , and cause has cause ,
then
and
taken together will have effect (
).
Within the context of our present study, where charges exert forces on each other, the
principle of superposition definitely does apply and has the following significance: if a
charge feels a force
when only charge is present, and feels a force
when only
charge
both
is present, it will experience a force that is the vector sum of
and
and
when
are present. Since the electric field at the point where a charge
is
located is the ratio of the electrostatic force on this charge to the size of the charge
the same superposition principle also hold for electric fields: if charge
,
by itself causes
an electric field
at some point in space when present by itself, and if charge
an electric field
at this same point in space when only B is present, the electric field
at that point when both
and
are present will be the vector sum
causes
.
It is a relatively straightforward matter to apply this (vector) principle of superposition to
determine the resultant Coulomb’s-law force (or electric field) that is produced by a
given arrangement of point charges. To simplify our results we will consider twodimensional charge distributions, but the equations are easily generalized.
If we ask for the total force on a point charge
point charge
in the presence of point charge
, the answer may be written
,
where
and
is the force on
due to
and
Coulomb’s law, the force is
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(1)
is the force on
due to
. From
directed along the line between
and
.
Therefore, in SI units,
,
where
is the distance between
) directed from
to
and
and
. In order to add
is a unit vector (length
and
in component form, we may
write
.
If we rewrite
in the same form the sum indicated in Eq. (1) becomes
(2)
Notice in Eq. (2) that the same factors appear in both the
and
components of
Study Examples 21.3 and 21.4 of the text in addition to the following examples.
EXAMPLE
1-1:
Calculate the resultant
Coulomb’s law force on the
charge in the figure
at right.
Solution:
To calculate
, we proceed as
follows.
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.
Step 1. Pick axes and . Let
be positive along
and
at
in the plane
of the screen.
Step 2. Sketch all forces ( and
). This gives the signs of the coponents right
away:
,
Step 3. Calculate magnitudes of
.
and
from
:
,
Step 4. Add components to get the resultant:
EXAMPLE 1-2: Calculate the resultant Coulamb’slaw force on the
charge
in the diagram at right, given
.
Solution: Step 1. Choose convenient coordinate axes, as in the diagram below.
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Step 2. Sketch in all forces. This gives component signs and shows that
.
Step 3. Calculate magnitudes from
:
,
.
Step 4. Add components to get the resultant:
.
If a tree falls in the forest when there is no living creature nearby to it, does it make a
sound? An abstraction such as the electric field invites a similar question: Does an
electric field surround a charge even if no second charge is present to feel the resultant
force? For our purposes, the answer to the second question, at least, is a definite yes.
The main point to be made here, though, is that the electric field due to a single point
charge or even a group of charges cannot be everywhere represented by a single
number, or even by a single vector. Both the magnitude and direction of the electric
field depend on position. In Figure 1-1, fields
and
are all different in both
magnitude and direction because the points
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and
are in different locations in
Figure 1-1.
space: the electric field
and
is a function of coordinates like
and
and it changes as
change. This is again illustrated in Figure 1-2 for six points with respect to an
assemblage of 20 charges. The arrows are not rigorously correct, but they show what
is going on generally: the electric field in the region between the lines of charge is fairly
constant, while the electric field outside this region is quite small. We often idealize this
situation to say that the field between the “plates” is constant, and the field outside is
zero. Even in the “ideal” situation, however, the electric field is a vector function of the
coordinates- its value depends on where you are located in space.
Figure 1-2.
Study Examples 21.5 and 21.6 of the text and the following additional example.
Examples 21.10 and 21.11 in the text involve the analysis of the field due to a
continuous charge distribution. These are good examples of the use of calculus, in
particular integration, in this course.
EXAMPLE 1-3: Calculate
, the electric field due to the
charge, at point
in the diagram at right, given
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Solution: Step1. Choose axes as in the figure at right. The
direction of at is the direction in which a positive
charge would move if placed at
of
at
is (
)
the ratio of force
divided by
Step 2. Sketch
. The magnitude
, which is
on a very small test charge
.
as in the figure and calculate
.
Step 3. Calculate the magnitude from
.
Step 4. Calculate
in component form:
.
Practice Exercise 1-1
Write your solutions to the following problems in your notebook.
1. Calculate the resultant Coulomb’s-law force on
the
charge in the diagram at right,
given
.
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2. Calculate the electric field (in rectangular component
form) at point
in the diagram at right, given
Check your answers to Practice Exercises 1-1 with those given in the key at the end of
Unit 1. Correct any errors in your solutions and review Section 1-1 as necessary before
you begin working on the next section.
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1-2: Particle Motion in Electric Fields
OBJECTIVE: Apply the definition of electric field to solve problems involving forces on
a charged particle in an electric field, where
(a) the particle is at rest under the influence of additional forces, like gravity
or tension, or
(b) the particle moves in a constant electric field.
These problems may require you to calculate any of the following quantities:
force, acceleration, time, position, velocity, work, kinetic energy. For vector
quantities you must be able to calculate components, magnitudes, and
directions.
PREREQUISITES: Using Newton’s laws of motion to solve problems involving
rectilinear motion
Stating and applying the work-energy theorem
Solving problems involving planar motion under constant
acceleration
Reading Assignment
Study in your textbook Examples 21.7 and 21.8.
Commentary
In the preceding section, we considered the problem of determining the electric forces
and fields that result from arrangements of point charges. We now turn to the problem
of analyzing the motion of a charged particle in an electric field.
The case of the charged-particle in a uniform electric field is analogous to that of motion
in a uniform gravitational field; it is discussed in the text in Examples 21.7 and 21.8.
The following examples show how this analysis can be extended to other situations
involving constant fields.
EXAMPLE 1-4: A stationary particle whose mass is
and whose charge is
is suspended
by a mass-less string under gravity in the
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presence of an electric field of magnitude
as shown in the
figure. Calculated the angle .
Solution:
The sum of all forces must be zero for the
particle to remain at rest. First, we choose a
coordinate system and draw a free-body
diagram as shown. Then we resolve the
forces (including the electrical force
)
into their
and
components.
Force
x-component
y-component
electrical
)
tension
gravity
;
(1)
;
.
(2)
Dividing (1) by (2) gives
so that
.
EXAMPLE 1-5:
An electron (
and
) with
enters a region of space with uniform electric field
.
(a) How much time will it take for the electron to be stopped by the electric field?
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(b) How far will it have traveled in coming to rest?
(c) How much work is done on the electron in bringing it to rest?
(d) What was the kinetic energy of the electron at the start of the problem?
Solution:
(a)
;
.
(b)
,
.
(c) Since the force is constant and parallel to the displacement, the work done is
.
(d)
.
Note that the work done equals the change in kinetic energy.
EXAMPLE 1-6:
An electron (
) circles a
stationary proton (
) at a distance of
.
What is the electron’s speed?
Solution: The field due to the proton is constant in magnitude at the radius of the
electron’s orbit; thus the electron experiences a constant centripetal force of
.
But for uniform circular motion,
so that
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and
.
By inserting the given values for
and
we obtain
.
Practice Exercise 1-2
Write your solutions to the following problems in your notebook.
1. A particle of mass
and charge
is suspended at rest near the earth’s
surface by a massless string as shown in
the diagram. The fixed charge is
.
Find the mass
.
2. The diagram at right shows an electron
(
)
traveling with velocity
the
in
direction between a pair of charged
“deflection plates”
electric field between
constant and equal to
long.
these
If the
plates is
in the
direction, calculate
(a) the time the electron spends between the deflecting plates;
(b) the acceleration of the electron while it is between the plates;
(c) the electron’s component of velocity when it emerges from the plates;
(d) the angle between the electron’s initial velocity and its velocity upon
emerging;
(e) the amount by which the electron has been deflected in the -direction when
it emerges from the plates.
3. Assuming the electron in problem 2 enters the plate at
equation for
.
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find the
Check your answers with the key and review Section 1-2 as necessary before going on
to the self-check test. To see if you have achieved the objectives in this lesson, try to
solve the problems in Self-Check Test 1 without using any reference materials.
Self-Check Test 1
Write your solutions to the following problems in your notebook.
1. Calculate the resultant Coulomb’s-law force on
the
charge in the diagram at right,
given
2. A stationary
mass has a charge
and is supported vertically by a
massless string, with a massless spring
attached on which a
force is
exerted as shown in the diagram at right.
There is also a
charge located as
shown. Find the charge .
3. What is the main difference between a conductor and a perfect insulator?
Check your solutions with the key and review Lesson 1 as necessary.
Homework Set 1
When you have demonstrated mastery of the content of this lesson, log into Mastering
Physics and work Homework Set 1.
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