I-1
Unit I - CHARGED PARTICLES AND ELECTRIC FIELDS
References:
PHYSICS FOR SCIENTISTS AND ENGINEERS, Serway & Beichner, 5th ed., Ch. 23
FUNDAMENTALS OF PHYSICS, Halliday, Resnick, & Walker, 6th ed., Ch. 22, 23
Unit Objectives
When you have completed Unit I you should be able to:
1. State COULOMB'S LAW, its limitations, and define all terms.
2. Given a collection of point charges, use Coulomb's Law and the principle of
superposition to determine the net force on one of the charges due to the others.
3. Define the concept of ELECTRIC FIELD in terms of force on a test charge.
a. State the SI units of the electric field.
b. Given a diagram on which an electric field is represented by flux lines, determine the
direction of the field at any given point, identify regions where the field is relatively
strong and where it is relatively weak, and identify where positive or negative
charges must be located to produce the given field pattern.
c. Given two or more point charges, find the electric field at a given point in the
vicinity of the charges.
4. Apply COULOMB’S LAW and the concept of ELECTRIC FIELD to analyze the motion of
a particle of specified charge and mass in a uniform electric field, where:
a. the particle is at rest under the influence of additional forces, i.e., gravity, tension,
etc.
b. the particle is in motion in the electric field.
The problems at the end of this unit were chosen to require you to apply concepts from your
earlier studies in MECHANICS such as acceleration, momentum, work, energy, uniform
circular motion, torque, etc.
I-2
Unit I - CHARGED PARTICLES AND ELECTRIC FIELDS
In this opening Unit we will consider the Electric Forces exerted and experienced by charged
objects and develop the idea of the Electric Field. As with the gravitational force, the electric
force is conservative.1 The electric field will be defined in terms of the electric force, hence the
electric field is a conservative field as is the gravitational field.
Coulomb's Law
Suppose we have two point charges, one having a charge Q, the other a
q
charge q, that are separated by a distance R. The magnitude of the electric
force on q (the victim charge experiencing the force) due to Q (the agent
R
charge causing the force on q) is given by COULOMB'S LAW:
Q
where k is a constant.
As you would expect from Newton’s 3rd Law the force on Q due to q is equal in magnitude but
opposite in direction. That is, the magnitude of the force on Q due to q is given by
The directions of these forces can be found by the following convention.
If we wish to find the force on q due to Q, then let’s adopt the convention
that R will be the position vector of the victim (in this case q) relative to
the agent (Q). That is, R is a vector drawn from the agent to the victim.
Recalling the idea of the unit vector from previous work, the vector R̂ Q
(read “R hat”) is a unit vector in the same direction as R. Thus, the
magnitude and direction of the force on q due to Q can be written as
q
R̂
R
where R̂ is a unit vector pointing from Q toward q. Applying the same definitions the force on
Q due to q is given by
where this time
points from q toward Q. This convention results in the forces always being
equal in magnitude but in opposite directions.
The definition of the direction given above is rigorous but a bit cumbersome. Here is an easier,
more straightforward way to determine the direction of the forces between point charges.
Recall the rules regarding charges proposed by our first world famous home-grown American
Physicist, Ben Franklin:
1. There are two types of charges in nature. Let’s call them “+” and “–“
2. Objects having charges with opposite signs attract one another
3. Objects having charges with the same sign repel one another
If you think about it a little, you will realize that if the sign of the charges q & Q are known, rules
#2 & 3 easily give the direction of the force on either charge.
1
You might want to review the definition of a conservative force in your Mechanics notes.
I-3
Unit I - CHARGED PARTICLES AND ELECTRIC FIELDS
Figure I-1 below illustrates how the forces between charges q and Q are oriented when #2 and
#3 above are used to determine the directions of FQ on q and Fq on Q for the three possible
combinations of signs on the charges. Figure I-1(A) illustrates the directions of the forces in
Franklin’s rule #2 where the charges have opposite signs and the forces are attractive. Figures
I-1(B) and (C) illustrate the directions of the forces in Franklin’s rule #3 where the charges
have the same signs and the forces are repulsive.
Charge Units
In classical electromagnetic theory, the electron and proton possesses the smallest possible
charge: –1 and +1, respectively. This unit of charge is called the elementary charge or e.c. In
the SI system of units the unit of charge is the COULOMB (abbreviation: C). By definition:
1 COULOMB = 6.25 x 1025 e.c.
Using this definition, the charge on an electron in units of Coulombs is given by
charge on 1 e- in C
=
25
-1 e.c.
6.25 x 10 e.c.
1C
\ the charge in C on 1 e– = 1.6 x 10-19 C.
Similarly, the charge in C on 1 p+ = +1.6 x 10-19 C.
If the units of all of the quantities which are substituted into Coulomb’s Law in SI units, then
units of the constant k in Coulomb’s Law must be:
2
k=
2
FR
(Force)(Dis tance)
N×m
Þ
Þ
qQ
(Charge units)2
C2
2
In SI units k is:
In many texts Coulomb’s Law is written as
This is no big deal because the constants k and
are equal. \ The numerical value of
I-4
Unit I - CHARGED PARTICLES AND ELECTRIC FIELDS
Summary of Coulomb’s Law Stuff
The magnitude of the force between point charges Q and q:
and where the direction of the force on a particular charge is found by
Franklin’s “Unlike charges repel” and “Like charges attract” rules.
and
1 C = 6.25 ´ 1018 e.c. or 1 e.c. = 1.6 ´ 10-19 C
You should gain an appreciation of the enormity of the electric force between charge objects.
For example, Richard Feynman in his E & M text states that if two people having 1% more
electrons than protons stand at arms length from one another, the magnitude of the force of
electrical repulsion between them would be of sufficient magnitude to lift a weight the size of
the earth!! However, if you think the electric force is large, imagine the size of the nuclear
force that prevents the electric repulsive forces between the protons in the nucleus from
tearing the nucleus apart!
To get a feeling for doing calculations with Coulomb’s Law complete the following:
Suppose we have an electron and a proton 1 meter apart. Calculate magnitude of the
gravitational force between them, then the electric force between them, and compare them
by finding the ratio of the electric force to the gravitational force.
FG =
GmM
=
R2
(
)(
)(
)2
(
=
FE =
kqQ (
=
R2
)
(I-4 #1)
)(
(
)(
)2
=
FE
= ____________________ » 1039
FG
)
For answers to fill-ins
see “In-Text Fill-in
Answers” page I-23-25
(I-4 #2)
¬ right?
This shows that, at least in the case of a proton and an electron, FE is » 1039 times larger than
FG !! This again illustrates how huge the electric force is compared to the gravitational force.
I-5
Unit I - CHARGED PARTICLES AND ELECTRIC FIELDS
An important thing to notice is the similarity between the form of Newton’s Law of Gravitation
FG = GmM / R2 and Coulomb’s Law FE = kqQ / R2 . Both vary directly with the product of a
physical property of the two objects (mass or charge) and inversely with the square of the
distance between them. Since the two equations have a similar form, they have similar
solutions. For example, you may have been asked in your study of Gravitation the following
question.
Two objects of mass m and M separated by a distance R exert a gravitational force on one
another. If both masses are doubled, what change must be made in their separation to
keep the gravitational force the same?
And, of course, from your previous work in gravitation, you know the correct answer is: “Double
their separation.”
Now if in this unit you are asked:
Two point charges q and Q are separated by a distance R exert an electric force on one
another. If both charges are doubled, what change must be made in their separation to
keep the electric force the same?
Since the two equations have similar forms, the answer here is also: “Double their separation.”
Now let’s do a little practicing with this idea.
Given the following point charges shown in Figure I-2 separated by the indicated distances:
+1 C
(A)
+1 C
+1 C
1 cm
(B)
+2 C
(C)
+1 C
1 cm
+1 C
2 cm
Figure I-2
+2 C
(D)
+1 C
2 cm
(E)
+2 C
+2 C
2 cm
Compare the electric forces between the pairs of charges by finding the following ratios.
1.
FA
= __________
FB
3.
FB
= __________
FE
5.
FD
= __________
FC
2.
FA
= __________
FD
4.
FE
= __________
FA
6.
FE
= __________
FC
(I-5 #1)
I-6
Unit I - CHARGED PARTICLES AND ELECTRIC FIELDS
Now the electric force is just another type of force to be considered along with any other type
of push or pull. All of the techniques you used to handle forces in MECHANICS you can also
apply to electric forces. To illustrate this, work through the following problem.
Given three point charges situated as shown in Figure I-3,
where
Q1 = –1 C (remember:  = 10-6)
Q2 = +2 C
Q3 = +3 C.
Determine the magnitude and direction of the net force on
due to Q1 and Q2 . (Caution:
consistent!)
Be sure your units are
Step 1. The force on Q3 due to Q1 is:
F1 on 3 =
k(
)(
(
F1 on 3 =
)
2
)
=
(
)(
(
)(
)
)
, at an angle of
degrees (I-6 #1)
(for now let’s agree to measure angles CCW from the +x axis)
Step 2. The force on Q3 due to Q2 is given by
F2 on 3 =
(
)(
(
F2 on 3 =
)(
)
)
, at an angle of
degrees
(I-6 #2)
Step 3. The NET FORCE on Q3 due to Q1 and Q2 can be found by
(I-6 #3)
Final Answer: Fnet on 3 =
,
At this point try problems 1 through 8 at the back of this unit.
(I-6 #4)
I-7
Unit I - CHARGED PARTICLES AND ELECTRIC FIELDS
Electric Field
We commonly refer to the existence of a GRAVITATIONAL FIELD about the earth. By this we
mean that the earth exerts a gravitational force on objects in the space around the earth. We
can then define a quantity called the GRAVITATIONAL FIELD INTENSITY (GFI) that
describes the force per unit mass exerted by the earth at each point in space about the earth.
For example, a 5 kg mass experiences a gravitational force of about 50 N at the earth’s
surface. Therefore, the force exerted on it per unit mass is 50 N/5 kg = 10 N/kg. Thus we say
that the gravitational field intensity at its location is 10 N/kg. Note that this quantity is a vector
since the force is a vector. So we say that the gravitational field intensity is 10 N/kg, toward the
earth’s center. Similarly, we can find the GFI anywhere else around the earth since we know
that the gravitational force varies as 1/R2. For example, the GFI at 2 earth radii is
; at 3 earth radii:
; etc., etc.
Now the ELECTRIC FIELD INTENSITY (electric field or E-field for short)
is a similar idea. The electric field is the force per unit charge in the
space around a charged object. Suppose we have a point charge q
located at a point P a distance R from another point charge Q. The
magnitude of the electric force on q due to Q is (in terms of the variables Q
given):
(
)(
)(
)
FQ on q =
(
)
q
R
Point P
(I-7 #1)
Now the electric field at the point P, EP , is the electric force exerted by the agent charge Q at P
per unit victim charge. Since in this situation the victim charge is q, then
EP =
FQ on q
=
(
)(
)
, (direction to be defined below)
(I-7 #2)
q
( )
This expression then, is the electric field at point P due to the point charge Q. Notice that the
electric field only depends on the charge causing the field (the agent charge) and the distance
from Q to the point P. The electric field exists around the charge Q whether there are any other
charges in the vicinity or not.
Example: Suppose Q = +1 C and the point P is 1 mm from Q. The magnitude of the electric
field at point P is
(
)(
)
Ü Units??
=
(I-7 #3)
EP =
(
)
Looking at the units of EP (N/C) we see that the value of EP tells us the amount of force that
will be exerted on each coulomb of charge placed at point P. Therefore, if we place a +2 C
charge at point P it will experience a force of magnitude
Fon 2 mC =
(I-7 #4)
For +5 C at P: Fon 5 mC =
(I-7 #5)
For –6 C at P: Fon -6 mC =
(I-7 #6)
I-8
Unit I - CHARGED PARTICLES AND ELECTRIC FIELDS
Generally then, if the electric field EP is known at some point and we place a victim charge q at
that point, the force on q will by given by
Fon q = (_____)(_____)
(I-8 #1)
Notice that in the equation above that Fon q is equal to the product of q and E. Thus, when q is
(+) then Fon q and E are vectors in the same direction and q is (–) then Fon q and E are in
opposite directions. This brings us to a simple way to define the direction of the electric field:
The ELECTRIC FIELD, E , at any point in space is defined as in the direction of
the electric force on a (+) test charge if the charge were placed at the point
Example: Suppose we have
Þ
Q
point P
What is the direction of the E-field due to Q at point P if Q is (+)? To answer this ask yourself:
“If a (+) test charge is placed at point P, what will be direction of the force on the test charge?”
Since charges with like signs repel one another, the force on the test charge is to the right.
Thus, the E-field of charge Q at point P is to the right. Suppose Q is (-). Apply the definition for
the direction of E and show yourself that the direction of the E-field of Q at point P is to the left.
In Figure I-4, answer the following.
A
What is the direction of E due to the point charge –Q at point
A?_________ point B? _________ point C? _________
(I-8 #2)
B
-Q
C
Figure I-4
From these examples you can see that we can generalize that the
direction of the electric field, E, about an isolated (+) point charge is radially outward and about
a (–) point charge E is radially inward. These E-fields are illustrated by sketching “electric field
lines” or “force field lines” about the charge or charges which show the symmetry and direction
of the field about the charge. For example, the electric field about an isolated (+) point charge
looks Figure I-5 (A) below and for a (–) charge like the Figure I-5 (B).
(A)
Figure I-5
(B)
+
-
Note that these diagrams just show the E-field in two dimensions. Realize that the E-field of
the charges are spherically symmetric in three dimensions about the charges.
I-9
Unit I - CHARGED PARTICLES AND ELECTRIC FIELDS
The arrowheads give the general direction of the field. Although the magnitude of the E-field
cannot be obtained from the diagram, we can get an idea of the relative strength of the field at
one point as compared to another. In the diagrams above, since the magnitude of the electric
field of a point charge varies as 1/(distance)2, you know that as we move closer to the charge
the magnitude of E increases. In the diagrams, then, we can generalize by saying that as the
field lines get closer together the magnitude of E increases.
Suppose we have a region where the E–field looks like Figure I-6.
As we move through the region from A
toward D the magnitude of E is greatest at A,
becomes less as we move toward B,
increases again as we move toward C then
decreases between C and D. The magnitude
of the E-field at points B and E are about the
same since the spacing between the lines in
those regions is about the same. The direction of the E-field is always tangent to the flow of
the field line pattern at the point in question. For example, the direction of E at point G is
horizontal to the right, at F roughly 45 up from the right, at A approximately 30 up from the
right, at E perhaps 5 to 10 degrees down from right, etc.
Suppose you are told that a uniform downward E-field exists in the
region shown at the right. Remember, uniform means constant in
magnitude & direction. In the region draw a field diagram that shows the
field is uniform and directed downward.
(I-9 #1)
At this point try problems 9 & 10 at the back of this unit.
Now let’s consider how we would determine the E-field at a point due to a collection of point
charges. If you recall that the E-field is a vector quantity, the solution is simple, merely find the
electric field at the point due to each individual point charge then find the vector sum of all the
E-fields at the point. In other words, suppose we have n point charges. The total electric field
at a chosen point will be given by,
i=n
i=n
i=1
i=1
Etotal = E1 + E2 + E3 + ××× + En = å Ei = k å
Qi
Ri2
R̂ =
1 i=n Qi
å R̂
4pe 0 i=1 Ri2
Example: Given the point charges and a point P
shown in Figure I-7. Find the total E-field at point P.
Solution:
Direction of the E-field at P due to each charge: Imagine placing a (+) test charge at P.
Franklin’s rules (p. I-2) tells us the +1 C charge exerts a force on the test charge to the right
since it repels the test charge. Therefore, applying the definition for the direction of the E–field,
the E–field at P due to the +1 C charge, E+1mC , is also to the right.
I-10
Unit I - CHARGED PARTICLES AND ELECTRIC FIELDS
N×m2 (1´ 10-6 C)
=
, Right = (9 ´ 10
)
, Right
(R1)2
C2 (3 ´ 10-2 m)2
kQ1
E+1mC
9
E+1 mC = 1 x 107 N/C, to the Right
Now you calculate E-3 mC at P due to the –3 C charge.
E-3 mC =(
)
(
(
E-3 mC =
)
, _________ Ü Direction?
)
,
(I-10 #1)
The ETotal at P is given by the vector sum:
Etotal at P = E+1mC + E-3mC = (1 x 107 N/C, Right) + (7.5 x 106 N/C, Left)
=
,
(I-10 #2)
Summary - The Electric Field
The Electric Field is a conservative vector field where
1. the magnitude of E at a given point describes the force that will be exerted
per unit of charge placed at the point.
2. the direction of E at a given point is in the direction of the force on a positive test
charge if placed at the point.
Units of E
in the SI system
(We’ll see in Unit III that the Volt/meter = V/m is also a legitimate SI unit for E)
The magnitude of the E-field of a point charge Q a distance R from Q is given by,
The E–field at a point due to n point charges = The vector sum of the E–fields at the point
due to all n charges.
The force on a point charge q in an electric field E is given by,
where F and E are in the same direction if q is (+) and in the opposite direction if q is (–).
At this point try problems 11 & 12 at the back of this unit.
I-11
Unit I - CHARGED PARTICLES AND ELECTRIC FIELDS
Let’s get used to the mechanics of a charged particle in an E-field by comparing it qualitatively
with the mechanics of a mass in a gravitational field (for brevity call it the “ G-field”).
In Figure I-8, the mass M is located near the surface of the earth (AKA the ground) so the Gfield in the region shown is uniform. If the only force on M is
= Mg, down and the G-field is
uniform, when M is moved to point A, B, C, D or anywhere else in the region shown, FG will
also be (Mg, down). Figure I-8 shows the G-field is uniform by the fact that the lines
representing the G–field are parallel, equally spaced, and all in the same direction,
Figure I-9 shows a positive charge q in an E-field. Here too the field lines are parallel, equally
spaced, and in the same direction. This indicates that the E-field is ____________ (I-11 #1) in
magnitude and direction. By the definition of the direction of E-field we know that the direction
of the electric force, , on the +q charge due to the E-field is directed ____________ (I-11 #2)
and its magnitude given by FE = ______ (I-11 #3). Since E is the same everywhere in the region
shown in Figure I-9, then FE on +q at A, B, C, D or anywhere else in the region is equal to
______, directed ____________ (I-11 #4).
Notice: In a uniform downwardly directed G-field a mass M experiences a constant force (=
Mg) in the same direction as the G-field AND a positive charge q in a uniform downwardly
directed E-field also experiences a constant force (= qE) in the same direction as the E-field.
Hmmm - - - there seems to be a similarity here. In uniform fields, M in a G-field and +q in an E
-field, both feel a constant force in the direction of the fields. Let’s see if the similarity
continues. In all the situations below, let the only force on M or q be due to the G-field or Efield, respectively, and let the charge q have a mass m.
Kinematics & Dynamics (We’re still referring to Figures I-8 & 9)
To keep M at rest or have it move up or down at constant speed, a constant force of
magnitude ______, directed __________ must be applied to M.
(I-11 #5)
To keep +q at rest or have it move it up or down at constant speed, a constant force of
magnitude ______, directed __________ must be applied.
(I-11 #6)
I-12
Unit I - CHARGED PARTICLES AND ELECTRIC FIELDS
If the net force on M is only the force due to the G-field and M is free to move, it will move with
(constant or changing?) acceleration (in the direction or in the opposite direction?) of the Gfield. The magnitude of the acceleration will be given by Fnet = Mg = Ma where a = ______.
(I-12 #1)
If the net force on +q is only the force due to the E-field and +q is free to move, it will move
with (constant or changing?) acceleration (in the direction or in the opposite direction?) of the
E–field. The magnitude of the acceleration will be given by Fnet = qE = ma where a = ______.
(I-12 #2)
Since the acceleration of M in the G-field is constant, the kinematics equations for constant
acceleration and the projectile equations developed in your study of Mechanics can be used to
describe the motion of M in the uniform G-field. In the equations __________ (I-12 #3) should be
substituted for the acceleration.
Since the acceleration of +q in the E-field is constant, the kinematics equations for constant
acceleration and the projectile equations developed in your study of Mechanics can be used to
describe the motion of +q in the uniform E-field. In the equations __________ (I-12 #4) should
be substituted for the acceleration.
Suppose M were tossed into the G-field with some initial velocity at an angle  relative to the
horizontal, its path would be a __________ (shape?) since the force on M and its resulting
acceleration are both __________ (constant? changing?) in magnitude and direction. (I-12 #5)
Suppose +q were tossed into the E-field with some initial velocity at an angle  relative to the
horizontal, its path would be a __________ (shape?) since the force on +q and its resulting
acceleration are both __________ (constant? changing?) in magnitude and direction. (I-12 #6)
At this point complete Worksheet I.1.
Work & Energy (We’re still referring to Figures I-8 & 9)
If M is released in the G-field, it will move (in the same direction or the opposite direction?) of
the field and (gain or lose?) kinetic energy, K, indicating that FG is doing (+ or –?) work on M.
(I-12 #7)
If +q is released in the E-field, it will move (in the same direction or the opposite direction?) of
the field and (gain or lose?) K indicating that FE is doing (+ or –?) work on +q.
(I-12 #8)
If an outside agent (you) were to do + work on M thereby increasing its gravitational potential
energy (UG), M would have to be moved (upward or downward?) or in other words (in the
same direction or the opposite direction?) of the G-field.
(I-12 #9)
If an outside agent (you) were to do + work on +q thereby increasing its electrical potential
energy (UE), +q would have to be moved (upward or downward?) or in other words (in the
same direction or the opposite direction?) of the E-field.
(I-12 #10)
I-13
Unit I - CHARGED PARTICLES AND ELECTRIC FIELDS
If we wanted to move M in such a way that its UG remained constant, M would have to be
moved toward (A, B, C, or D?) in Figure I-8 or generally moved ____________ to the -field.
(I-13 #1)
If we wanted to move +q in such a way that its UE remained constant, +q would have to be
moved toward (A, B, C, or D?) in Figure I-9 or generally moved ____________ to the E-field.
(I-13 #2)
If you responded correctly above you found that the electric potential energy UE, of +q is
increased when it was moved it in a direction opposite to the direction of the E-field. Similarly,
when +q moves in the direction of the E-field its UE decreased. How does the UE of +q change
when +q is moved perpendicular to the E-field?
(I-13 #3)
At this point go back to the previous page and consider how answers (I-11 #1 through 6), (I-12
#1 through 10) and (I-13 #1 through 3) would change if the charge in Figure I-9 were negative.
In the discussion above we used the simplest case of a charged particle in a uniform E-field
but the concepts summarized below apply also for charges in non-uniform E-fields.
Let’s summarize the qualitative ideas developed for moving charges around in an E–field.
Given a charge in an E–field:
When a charge moves in the direction of the E–field:
its UE decreases if the charge is (+)
its UE increases if the charge is (–)
When a charge moves opposite to the direction of the E–field:
its UE increases if the charge is (+)
its UE decreases if the charge is (–)
When the charge moves in a direction perpendicular to the E–field:
the UE of the charge, whether (+) or (–), remains constant
Let’s practice a bit with the ideas developed so far.
Figure I-10 shows a portion of two charged parallel
conducting plates whose dimensions are large compared to
the distance between them. If the region containing points
A, B, C, and D is near the (horizontal) center of the plates
away from the edges, the E-field between the plates is
essentially uniform.
The direction of the uniform E-field in this region is __________
(I-13 #4)
At which of the points A, B, C or D is the magnitude of the E -field the greatest?
_____________
(I-13 #5)
I-14
Unit I - CHARGED PARTICLES AND ELECTRIC FIELDS
If the magnitude of the E -field is E and a charge –Q is placed at point B, it will experience an
electric force, FE = _____, directed ________
(I-14 #1)
The force on a +Q charge placed at point A will be FE = _____, directed ________
(I-14 #2)
If the mass of the –Q particle is twice the mass of the +Q and the charges are free to move,
how will the magnitude of their accelerations compare? __________
(I-14 #3)
If the –Q charge is at point A and is moved horizontally, its UE will (increase, decrease, remain
the same)?
(I-14 #4)
If the +Q charge is placed at point A and is free released, it will move ___________ and (gain,
lose) kinetic energy, K.
(I-14 #5)
At which point will the –Q charge have its greatest UE? Point _______
(I-14 #6)
Calculating Work and Energy
From Mechanics: The WORK done by a force
on an object is defined as the component of
the force on the object parallel to the displacement times its displacement, i.e.,
(If F is constant, where Ds is the displacement of the object)
(If F is a function of the distance, s, along the path from A to B)
(Where W NET is the work done by the net force FNET)
The following 2 problems should refresh your memory a bit regarding Work-Energy
relationships.
1. As shown in Figure I-11 a constant force is exerted
on a 0.5 kg hockey puck initially at rest resulting in a
horizontal displacement of 2 m while the force acts.
a) What is the work done by the force, F?
(
)(
) (
)
o
W = FiDs = _________ _________ cos ____ = ____________
b) If the surface is frictionless,
(I-14 #7)
= _________ (I-14 #8)
\ the WNET done = ___________ (I-15 #9) and the gain in K = ___________ (I-14 #10)
c) If there is a friction force of 2 N to the left,
= _________
(I-14 #11)
in K = __________ (I-14 #12).
d) The speed of the puck after traveling 2 m = __________ (I-14 #13).
resulting in a gain
I-15
Unit I - CHARGED PARTICLES AND ELECTRIC FIELDS
2. Figure I-12 shows the hockey puck attached to a string with
one end fixed at point C so that the puck rotates without friction
in a horizontal circle. The puck rotates with uniform circular
motion about point C. The puck then, experiences a centripetal
force, FC , due to the string directed toward point C.
a) The component of FC in the direction of the puck’s
displacement at any instant is ____________. (I-15 #1)
C
R
Figure I-12
b) The work done by FC on the puck during one revolution is
___________. (I-15 #2)
Recall that expression
on the previous
page is for the more general case where F is a
function of the position along a path that is not
necessarily a straight line. Let’s review its meaning.
As Figure I-13 illustrates, suppose we move an
object along the path shown going from point A to B
along small displacements Ds1, Ds 2, Ds 3,×××, etc. If
the force on the object along each small
displacement is constant, (that is,
is constant
along displacement
,
moving it from A to B is
is constant along displacement
, etc,) then the work done in
or
If we make the magnitude of each
shrink to super teensy weensy, then the path is
approximated by the smooth path shown in Figure I-14 and the expression above can be
written in fancy mathematics as:
, where F is a function of the position along the path.
Before applying this relation, let’s do a quick review of some of the ideas on the previous
pages.
I-16
Unit I - CHARGED PARTICLES AND ELECTRIC FIELDS
Given the situation shown in Figure I-15, suppose we want to hold +q at
rest in the position shown. To accomplish this a force
F = _______, _______ (I-16 #1) must be applied to the charge. If +q is
released and free to move, the charge will move to the (left?, right?,
up?, down?) with constant (speed?, acceleration?) (I-16 #2).
The charge also has electrical potential energy, UE, when it is in the
E-field.
Which way does +q have to be moved to increase its UE? (I-16 #3)
Which way can +q be moved and still keep UE constant? (I-16 #4)
If q were (–), which way would it have to be moved to increase its
?
(I-16 #5)
Now let’s consider how much
will change as the charge is moved in the E-field. If +q is
moved to the left at constant speed (i.e., K = 0), it will be in equilibrium. The force that I (the
agent) must exert to move it must be equal but opposite in direction to the force on the charge
due to E. The force on +q due to E is given by FE = qE, to the right. Thus, the force the agent
must apply when moving +q to the left at constant speed must be F = qE, to the left.
Therefore, the work I do moving the charge to the left between two points A and B is
The E-field is a conservative field like the gravitational field, hence the work done under the
conditions stated is equal to the change in the potential energy,
, of the charge q. That is,
This expression, it turns out, is the general expression for the work done (and the resulting
)
by an external agent on any charge q (+ or –) that is moved along any path in an E-field. Let’s
show that the expression gives qualitatively the correct answers as we move a test charge
around in the E-field of Figure I-15.
Suppose q is (+). If it is moved to the left
expression
= Eds cos(180o) = – Eds and thus the
will be +, meaning that positive work is done resulting in an increase in the
of the charge
(since
is +). The equation then gives the correct sign because we know a + charge
increases its
when moved in the opposite direction to the E-field.
I-17
Unit I - CHARGED PARTICLES AND ELECTRIC FIELDS
Suppose q is (+) or (–). If it is moved perpendicular to the field we know that no work is
required since
= 0. Let’s see what the equation says:
The equation gives the correct result. It says that if the charge is moved perpendicular to the
E-field its
is zero.
Suppose q is (+) and we move it to the right. q then, should lose potential energy (
be –) indicating negative work has been done. The equation says:
should
= (–)
qualitatively correct again!
Go through the same exercises with a (–) charge to show yourself that moving a –q to the left
in Figure I-15 decreases the
of the charge and moving it to the right increases its
.
At this point try problem 13 at the back of this unit.
Q
q
Now consider two point charges Q and q separated by a distance R
R
as shown at the right. Let Q be (+) and fixed and q be movable.
If Q and q have like signs, q would have to moved to the (right?, left?) (I-17 #1). to increase the
UE of q.
If q is moved along a circular path centered about Q, the work done moving q at constant
speed along this path is ________ (I-17 #2) and its UE is ________ (I-17 #3).
The force on q due to Q is given by Coulomb’s Law. That is,
, directed __________
(I-17 #4).
The E-field of Q at distance R from Q is given by
, directed __________
(I-17 #5).
Answer I-17 #1 - #5 for the situation where Q is (–) and q (+).
I-18
Unit I - CHARGED PARTICLES AND ELECTRIC FIELDS
Let’s derive the expression for the work done in moving a point charge q in the
another point charge Q.
-field of
Let both Q and q be (+) and let’s push q
toward Q radially from R1 to R2; i.e., from
its initial position at S = 0 to point P. The
teensy weensy bit of work dW, done
pushing q a teensy weensy distance
to the left toward Q is given by
, where
is the
-field of charge Q.
The work done moving q from S = 0 to S = S1 is
Note that behind the integral sign we have 2 variables: S and R. To evaluate the integral we
must have only one. Referring to the figure above note that if q is moved a distance S toward
Q and the resulting separation between q and Q is R then R = R1 – S. Differentiating we get
dR = – dS.
If we choose to evaluate the integral in terms of R we just substitute dS = – dR and put the
limits of integration in terms of R. Note in the figure above when q is at S = 0, R = R1. When q
is at point P, S = S1 and R = R2. Thus, in terms of R the work done moving q from its original
location to P (or R1 to R2) is given by
Or if we evaluate the integral in terms of S, just substitute R = R1 – S and grind away:
Orgy of algebra –
let x = R1 – S
\ dx = – dS
Limits: when S = 0 then x = R1
When S = S1 then x = R1 – S1
= R1 – (R1 – R2)
= R2
As we would expect, we get the same answer as above!
I-19
Unit I - CHARGED PARTICLES AND ELECTRIC FIELDS
Let’s test the expression above and show that it gives the right qualitative result for all
combinations of Q and q. In the equation, realize that R1 was the initial separation of the
charges and R2 was their final separation.
Case 1 – Q and q have the same sign
We already know that if q is pushed toward Q, (+) work is being done and thus the U
of the system increases. Let’s see if the equation says W is (+).
In the equation:
The stuff in the [ ] is (+) and since R1 > R2 the stuff in the ( ) is (+) therefore W is (+)
and U increases when q is pushed toward Q. Thus the equation agrees with our
prediction that (+) work is done and U increases.
If q is moved away from Q, (–) work is done (energy is transferred OUT of the
system) and U then decreases. In the equation the stuff in the [ ] is (+) but the stuff in
the ( ) is (–) since R2 > R1. Thus the equation agrees with our prediction that (–)
work is done and U decreases.
Case 2 – Q and q have opposite signs
If q is moved toward Q the work done is (+ or –?) (I-19 #1) and thus the U of the system
(increases or decreases?) (I-19 #2). In the equation [ ] is (+ or –?) (I-19 #3) and ( ) is
(+ or –?) (I-19 #4); therefore, the equation says (+ or –?) (I-19 #5) work is done and U
(increases or decreases?) (I-19 #6). Does the equation give the same results as your
prediction?
If q is pulled away from Q the work done is (+ or –?) (I-19 #7) and the U of the system
(increases or decreases?) (I-19 #8). In the equation [ ] is (+ or –?) (I-19 #9) and ( ) is
(+ or –?) (I-19 #10); therefore, the equation says (+ or –?) (I-19 #11) work is done and U
(increases or decreases?) (I-19 #12). Does the equation give the same results as our
prediction?
Suppose Q and q have opposite signs and we move q from A to
B to C as shown in the diagram at the right. The path A to B is
the arc of a circle centered on charge Q. What is the total work
done?
because _____________________________________________ (I-19 #13)
________________________________
(I-19 #14)
Thus
________________________________
(I-19 #15)
I-20
Unit I - CHARGED PARTICLES AND ELECTRIC FIELDS
The process above can be used to find the work done
moving q along any path from A to C. If we want to know the
work done moving q along the curved path shown in the
sketch at the right. We just divide the path into teensy
weensy radial and circular segments centered on Q. The
work done moving q along this path or any path from A to C
will be
Note that
depends only on the end points and not on the path taken. This is, as you will
recall from your Mechanics course, a characteristic of a conservative field.
Generally then, we can write that for point charges Q and q where we change their separation
from R1 to R2 (it doesn’t matter which one is moved) the work done is given by
and since the work done by an outside agent (us) equals the change in electric potential
energy U, then
What if we want to know U at a particular point not the change in potential energy, U? As
with all potential energy expressions we have to ask “Potential energy relative to what?”. In
other words we must state a reference level to measure potential energy from. In your study of
gravitation we adopted a convention that U = 0 when the Fgrav = 0. Since,
,
then Fgrav = 0 when R =
¥ thus Ugrav = 0 when R = ¥ by agreement. This gave us
Since the electric force between point charges varies as 1/R2 also, let’s agree that
when
, that is, when R = ¥ . Thus for point charges Q and q the potential energy of
the system (not of a charge) is equal to the work required to bring one of the charges from ¥
to within a distance R of the other one. The potential energy of the two point charge system
when the charges are separated by a distance R is given by
.
I-21
Unit I - CHARGED PARTICLES AND ELECTRIC FIELDS
But
, therefore the electric potential energy of two point charges, Q and q, separated by
a distance R is given by
Work through the following example.
Given two point charges, Q = –5 C and q = –2 C. Find the work done by bringing either
charge from ¥ to a point 1 cm from the other.
= _____________
(I-21 #1)
The potential energy of this 2 charge system is
U = ____________ = ____________ = ____________
Equation
Substition
(I-21 #2)
Answer
If one of the charges is released when they are 1 cm apart, find the kinetic energy of the
moving charge when they are 2 cm apart. Since energy is conserved:
Assume q was the charge released, we want to know Kq when R = 2 cm. Solve for Kq at 2 cm.
= __________ + __________ + __________ - __________ - __________
= __________
(I-21 #3)
At this point try problem 14 at the back of this unit.
I-22
Unit I - CHARGED PARTICLES AND ELECTRIC FIELDS
Above we’ve shown how to find the potential energy of a
system of two point charges. What if we have more than two
charges? Let’s use the example of three charges then
extend the idea to n charges. The work done bringing Q1
and Q2 from ¥ to a distance of R12 from one another is
to bring Q2 and Q3 together:
to bring Q1 and Q3 together:
The total work done bringing the system of charges into the configuration shown is
Generally then, to find the total potential energy of a system of n charges just add up the
potential energy of each charge relative to each of the other charges.
At this point try problem 15 &16 at the back of this unit.
I-23
Unit I - CHARGED PARTICLES AND ELECTRIC FIELDS
In-Text Fill-In Answers
Page
Fill-In #
Fill-In Answer
1
FG » 1 x10-67 N
2
FE » 2 x 10-28 N
I-5
1
1/2, 2/1, 2/1, 1/1, 2/1, 4/1
I-6
1
2
I-4
I-7
F1 on 3 = 16.9 N, 180o
F2 on 3 = 21.6 N, -37o
3
“finding the vector sum F1 on 3 + F2 on 3 ”
4
Fnet on 3 » 13 N, -88o
1
FQ on q =
kqQ
R2
3
4
5
6
kQ
R2
EP = 9 x 109 N/C
1.8 x 104 N
4.5 x 104 N
5.4 x 104 N
1
Fon q = qE
2
E at A about 45o down from left or directly toward –Q,
E at B to the left or directly toward –Q, & E at C up or directly toward –Q
I-9
1
At least 3 equally spaced, parallel lines pointing down
the page should be drawn. 3 or more equally spaced
shows constant magnitude, parallel lines show
constant direction, and arrows on the lines indicate
the direction of the field.
I-10
1
E-3 mC = 7.5 x 106 N/C, to the Left
2
ETotal at P = 2.5 x 106 N/C, to the Right
2
I-8
EP =
I-24
Unit I - CHARGED PARTICLES AND ELECTRIC FIELDS
In-Text Fill-In Answers
Page
Fill-In #
Fill-In Answer
I-11
1
2
constant or uniform
downward ( FE on a positive charge is in the direction of the E-field)
3
4
5
6
FE = qE
qE, downward
Mg, upward
qE, upward
I-12
1
2
3
4
5
6
7
8
9
10
constant, in the direction of the G-field, a = g
constant, in the direction of the E-field, a = qE/m
aM = g = 9.8 m/s2
aq = qE/m
parabola, constant
parabola, constant
in the direction of the -field, gain K, + work
in the direction of the E-field, gain K, + work
upward, opposite the direction of the -field
upward, opposite the direction of the E-field
I-13
1
2
3
4
5
I-14
1
2
3
4
5
6
7
8
9
10
11
FE = QE, downward
FE = QE, upward
a+ = 2a–
remain the same
upward and gain K
point D
W = 16 joules
= 8 N, to the right
WNET =16 J
 = 16 J
= 6 N, to the right
12
13
 = 12 J
B or D, perpendicular
B or D, perpendicular
UE does not change, it remains constant
up
the E at A, B, C, & D are the SAME since E is
uniform between parallel plates
m/s
I-25
Unit I - CHARGED PARTICLES AND ELECTRIC FIELDS
In-Text Fill-In Answers
Page
Fill-In #
Fill-In Answer
I-15
1
2
zero
Work done in 1 rev. = zero because Fc is always perpendicular to the
displacement of the puck Ds
I-16
1
2
3
4
5
F = qE, to the left
to the right with constant acceleration
to the left (opposite to the direction of the E -field)
up or down (perpendicular to E -field)
to the right (in the direction of the E -field)
I-17
1
2
3
to the left
zero
zero
, to the right
4
5
I-19
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
EQ =
kQ
, radially outward from Q
R2
(–)
decreases
(–)
(+)
(–) work
decreases
(+)
increases
(–)
(–)
(+) work
increases
q is moving perpendicular to the
-field of Q
I-26
Unit I - CHARGED PARTICLES AND ELECTRIC FIELDS
In-Text Fill-In Answers
Page
Fill-In #
I-21
1
2
I-22
1
Fill-In Answer
= 9 Joules
I-27
Unit I - CHARGED PARTICLES AND ELECTRIC FIELDS
End of Unit Problems
1. How far apart should two electrons be if the force each exerts on the other is equal to the
WEIGHT of an electron?
[Ans: 5 m]
2. A charge Q is placed at two opposite corners of a square. A charge q is located at each of
the other two corners.
a) If the net force on either Q is zero, how do Q and q compare? [Ans:
]
b) Could q be chosen so that the net force on every charge is zero? Explain your answer.
[Ans: No]
3. Two charges, q and 4q, are separated by a distance D. What is the sign, magnitude and
position of a third charge, Q, that causes all three charges to be in equilibrium?
[Ans:
and is
from q between the two charges]
4. Charge Q is divided into two parts q and Q – q. What is the relationship between Q and q if
the two parts, placed a given distance apart, are to experience a maximum repulsive force?
(Hint: This is a max/min calculus problem.)
[Ans: q = Q /2]
5. Two equally charged particles, held 3.2 mm apart, are released from rest. Upon being
released, the initial acceleration of the first particle is 7.0 m/s2 and that of the second is
9.0 m/s2. If the mass of the first particle is 6.3 x
kg, find
a) the mass of the second particle
[Ans: 4.9 x
kg]
b) the charge on each particle
[Ans: 7.1 x
C]
6. Two identical balls of mass m are hung from non-conducting
threads of length L as shown in the sketch at the right. They each
carry an identical charge q. Assume that the angle  is so teensy
weensy that the approximation you learned in trig that tan » sin
for small  is true. Using this approximation show that
where x is the separation between the balls.
7. In the Bohr model of atomic hydrogen an electron revolves about a proton in a circular orbit
having a radius of 53 pm.
a) Calculate the centripetal force acting on the electron.
[Ans: 8.2 x
N]
22
b) Calculate its centripetal acceleration.
[Ans: 9 x 10 m/s2]
c) Find the angular momentum of the electron about the nucleus.
[Ans:
]
I-28
Unit I - CHARGED PARTICLES AND ELECTRIC FIELDS
End of Unit Problems
8. A particle of charge Q lies midway between two
identical fixed charges each having charge q and
separated by a distance 2b.
a) What is the net force on Q when it is midway between the two fixed charges?
[Ans: Fnet = 0]
b) Suppose Q is displaced from the midpoint by an amount x as shown in the sketch
above. If Q and q have like signs, show that the net force on Q is given by
c) Show that if x << b, then if the particle is released from its position shown in the sketch,
it will oscillate with SHM with an angular frequency  given by
where m is the mass of the particle.
9.
Given the electric field diagram at the right due to
charges Q and q.
a) Describe the relative magnitudes of the -field
at points A through H.
b) What must be the signs on Q and q to produce
this field?
c) What is the direction of the -field at each
point?
d) If the magnitude of charge Q were doubled, how
would the magnitude and direction of the -field
at points A, D, E, and H change?
10. Assume a water molecule to be a dipolar molecule consisting of two charges, +e and –e,
separated by a distance of 0.04 nm where e is the magnitude of the charge on the
electron. Calculate the net torque on a water molecule about its center due to a 500 N/C
electric field if the molecule is oriented
a) perpendicular to the field.
[Ans: 3.2 x
]
b) parallel to the field.
[Ans: zero]
I-29
Unit I - CHARGED PARTICLES AND ELECTRIC FIELDS
End of Unit Problems
11. Two equal and opposite charges have a magnitude of 0.20 C and are held 15 cm apart.
a) What is the magnitude and direction of the -field at a point midway between the
charges.
[Ans: 0.64 MN/C, toward the (–) charge]
b) What would be the magnitude and direction of the force on an electron if it were placed
midway between the charges?
[Ans: 0.10 pN, toward the (+) charge]
12. Four point charges are at the corners of a square as shown
at the right. If q = 0.10 C and a = 5.0 cm,
a) what is the net force on the +2q charge?
[Ans: 0.18 N,
]
b) what is the -field at the location of the +2q charge if
the +2q charge is removed from its position?
[Ans: 0.90 MN/C,
]
c) what is the -field at the center of the square?
[Ans: 3.1 MN/C, to the right]
13. Two large parallel plates (A and B) have equal but opposite charges
and are separated by a distance of 1.0 meter as shown in the
sketch at the right. The -field between them is uniform with a
magnitude of 200 N/C. A +6 C charge is placed between the plates
and experiences a force to the LEFT due to the -field.
a) Which plate is positive?
b) What is the direction of the
c)
d)
e)
f)
[Ans: Plate B]
-field between the plates?
[Ans: To the left]
What is the magnitude of the force on the charge due to the -field? [Ans: 1.2 mN]
How much work is done by an external agent when moving the charge at constant
speed
i. in the direction of the -field?
[Ans: –0.6 mJ]
ii. In the direction opposite to the direction of the -field?
[Ans: +0.6 mJ]
iii. in a direction perpendicular to the -field?
[Ans: 0 J]
iv. along a path incined at an angle of
relative to the -field? [Ans: –0.52 mJ]
In each case in part (d), find the change in the potential energy of the charge and
indicate whether the potential energy increases, decreases, or remains the same.
[Ans: (i) U = –0.6 mJ, decreases, (ii) U = +0.6 mJ, increases, (iii) U = 0, remains
the same, (iv) U = –0.52 mJ, decreases]
Suppose the charge is placed against the plate B and released from rest. Find its
kinetic energy, K, just before striking plate A.
[Ans: 1.2 mJ]
I-30
Unit I - CHARGED PARTICLES AND ELECTRIC FIELDS
End of Unit Problems
14. a) Two protons, initially an infinite distance apart, are projected toward one another with
equal initial kinetic energies to make a head-on collision. Find their kinetic energies if
their distance of closest approach is to be 1.0 fm.
[Ans: 0.12 pJ]
b) If one of the protons is initially at rest, what initial K must the other proton have so that
their minimum separation is again 1.0 fm?
[Ans: 0.46 pJ]
15. Four charges having charges +q, +q, –2q, and –2q are located at the corners of a square
of side L with the identical charges diagonally across from one another. Find the potential
energy of this system of charges assuming the reference level to be at infinity.
[Ans:
]
16. Point charges obey Coulomb’s law just as planets, etc., obey Newton’s law of gravitation.
Since both laws are inverse-square laws, the energy requirements for the trajectory of a
point charge moving in the vicinity of another very massive point charge are the same as
for a satellite moving about a planet. That is, if
E = K + U < 0 the trajectory is an ellipse
E = K + U = 0 the trajectory is an parabola
E = K + U > 0 the trajectory is an hyperbola
Suppose an electron passes a helium nucleus (charge +2e) at a distance of 2 Angstroms.
What is the minimum kinetic energy for the electron at this position which will permit it to
avoid being bound to the nucleus?
[Ans: 2.3 aJ]