Lesson 3 – Forces and Fields
© Lawrence B. Rees 2007. You may make a single copy of this document for personal use without written permission.
3.0 Introduction
In the first two lessons we discovered that the force between point charges can be
described in terms of threads. If a field particle is at rest, the force on it from a source particle is
given by Eq. (2.2):
r e
r
F=
q f lν
ε0
If the field particle is moving; however, we can use stubs to calculate the relativistic corrections
caused by its motion. The stub can be found from the thread by Eq. (2.7):
r
r
s = rˆh × l
and the magnetic or stub force is just [Eq. (2.13)]:
r r
r
e
Fs =
q f νβ f × s .
ε0
In these equations:
ν
r
F is the (vector) force on the field particle in newtons (N).
g0 is a constant called the “permittivity of free space” or just “epsilon naught.” It
equals 8.85 ×10–12 C2/Nm2.
q is the charge of the field particle in coulombs (C).
rf
l is the (vector) length of threads at the field particle. It has units of meters (m)
is the density of threads at the field particle in units of threads/m3.
r̂h is the unit vector from the source (when the thread is emitted) to the field particle. It is
dimensionless.
r
β s is the ratio of the velocity of the source particle to the speed of light, s = vs / c.
r
β f is the ratio of the velocity of the field particle to the speed of light, f = vf / c.
β
β
The thread model provides a rather simple structure that allows us to view the forces of
moving charges from a fundamental perspective; however, these forces have historically been
defined in terms of a different model that makes use of the concept of fields. In this lesson, we
will see how the thread model and the field model are related and we will learn some basic tools
we will need to work with electric and magnetic fields.
1
3.1 The Electric Field of a Stationary Point Charge
Let us assume that in a region of space there are several source particles at fixed
positions. If we place a small field particle in that region, the field particle will experience a net
force. We will assume that the field particle is much less massive than the source particles, so
that its influence on the motion of the source charges is negligible. Then, by observing the
motion of the field particle, we can determine the force acting on it. That is, we can 1)
experimentally measure the field particle’s position as a function of time, 2) take two derivatives
of position with respect to time to obtain the acceleration, and 3) multiply the acceleration by the
mass of the field particle to obtain the force.
r
∆r 
∆ 
r
r dpr
∆t
r
F=
= ma ⇒ m   = F
dt
∆t
Once we have found the force, we can use the results of Sect. 1.9 to find the electric field:
(3.1 Definition of the Electric Field)
r
r
F = qf E
r
F is the force (a vector) on a field particle in an electric field. Its units are newtons (N).
q f is the charge of the field particle in coulombs (C).
r
E is the electric field in units of newtons/coulomb (N/C) or volts/meter (V/m).
From this we note the following:
• The electric field is a vector quantity. That is, it has both magnitude and direction.
• The electric field will generally be different at every point in space because the force
on a field particle will vary from point to point. A vector quantity defined at every
point in space is called a “vector field.”
• The force on a positive field particle is in the direction of the electric field, but the
force on a negative charge is opposite in direction to the electric field. (If it’s not
obvious, this is true because two vectors that differ by a minus sign are in opposite
directions.)
• The SI units of electric field can either be written as newtons/coulomb (the units we
used in Lesson 1) or volts/meter. Volts/meter is the more common way of expressing
it.
2
We also know that the electric field of a point charge is:
Coulomb’s Law for Electric Fields
(Electric Field of a Point Charge)
r
r F
qs
e r
=
l 0ν 0 =
rˆ
E=
qt ε 0
4πε 0 r 2
(3.2)
where:
q is the charge of the source charge in coulombs (C).
rs
r is the vector from the source charge to a field particle.
r
r
r̂ is the unit vector pointing in the direction of r . Thus, r = rr̂.
Notice that the electric field doesn’t depend on the sign or magnitude of the field particle. In fact,
we really don’t need a field particle at all to define the electric field. The electric field, however,
does depend on the location a field particle would occupy in space, if there were a field particle.
We call any such location a “field point.” Eq. (3.2) then becomes a recipe for finding the electric
r
field at any field point defined by r . For these reasons, we now want to change our definition of
r
r just a little:
r
r is the vector from the source charge to a field point.
In Lesson 1, we already used the name “Coulomb’s law” for the equation of force
between two point charges. Of course, this equation gives us exactly the same information as Eq.
(3.2), as the electric field is just the force of Eq. (1.1) divided by qf. Since we will primarily deal
with fields in this course, we will generally mean the electric field equation, Eq. (3.2), when we
refer to Coulomb’s law.
In Lesson 1, we said that when we use Coulomb’s law, it is often easier to multiply both
the numerator and the denominator r to eliminate the unit vector. That is still true, of course, in
Coulomb’s Equation for the electric field as well.
(3.3 Coulomb’s law, alternate form)
r
E=
r
qs r
4πε 0 r 3
It may seem strange at first to be concerned about electric fields. If we define the electric
field in terms of forces, then why not just deal directly with forces? Part of it is tradition, but part
of it is practical as well. We note that the force on a field particle located at a certain point in
space depends on two different things: 1) the configuration of the source charges and 2) the
charge of the field particle. On the other hand, the electric field depends only on the
configuration of the source charges. A set of source charges, therefore, produces a unique
r
r
electric field but a different "force field" for different field particles. That is, by writing F = q f E ,
we can conveniently separate the contribution of the source charge from the contribution of the
field particle in the expression for the force.
3
For example, in Example 1.4 we found that the electric field at a certain point in space
was
r
E = −64.4 MV / m xˆ − 5.97 MV / m yˆ .
What if we had a field particle with a charge of –3.14 C or +7.32 C ? To find the force on
each of these charges, all we need to do is multiply the electric field by the charge:
(
(
)
)
r
r
FA = q A E = − 3.14 × 10 −6 C (− 64.4 MV / m xˆ − 5.97 MV / m yˆ ) = +202 N xˆ + 18.7 N yˆ
r
r
FB = q B E = + 7.32 × 10 −6 C (− 64.4 MV / m xˆ − 5.97 MV / m yˆ ) = −471 N xˆ − 43.7 N yˆ
Let’s rework the examples of the Sect. 1.8 but solve for the electric field rather than the
force.
Example 3.1. A proton is located at the origin of a coordinate system. Find the electric field (a
vector!) at the point 125 nm x̂ -67.0 nm ŷ .
r
r
As before, the vector r from the source to the field point is r = 125 nm x̂ -67.0 nm ŷ .
We obtain r by Pythagoras’s relationship:
r = xxˆ + yyˆ = x 2 + y 2 = 142 nm
Now, we can just plug numbers into Eq. (1.5).
r
E=
(
)
r
qs r
1.6 × 10 −19 125 × 10 −9 xˆ − 67.0 × 10 −9 yˆ
N / C = (63108 xˆ − 33826 yˆ )N / C
=
−12
−9 3
4πε 0 r 3
4π × 8.85 × 10 142 × 10
(
)
Or, since we only have three significant digits, it’s better to write the answer as:
r
E = (63.1xˆ − 33.8 yˆ )kN / C
Example 1.4. The Electric Field of Multiple Charges
A 3.20 :C charge is located at the point (1.00 cm, 3.00 cm) and a – 2.42 :C charge is
located at the point (–3.00 cm, 2.00 cm). Find the electric field at the point P = (-1.00 cm, 1.00
cm). Then find the force on a 1.24 :C charge located at this point.
4
+3.20 :C
!2.42 :C
P
We need to calculate the electric field from each of the two charges and then add them
together as vectors. We find the vector from each source charge to the field point. The vector on
the left 1 and the vector on the right 2.
r
r1 = 0.02 xˆ − 0.01yˆ
r1 = 0.02 2 + 0.012
r
r2 = −0.02 xˆ − 0.02 yˆ r2 = 0.02 2 + 0.02 2
Now all we need to do is plug the known values into Coulomb’s law (electric field form):
r
r
q1 r1
(− 2.42×10 −6 )(0.02 xˆ − 0.01yˆ ) N / C = −38.9 MN / C xˆ + 19.5 MN / C yˆ
=
E1 =
3
4πε 0 r13
4π × 8.85 × 10 −12 0.0005
r
r
q 2 r2
(3.20 ×10 −6 )(− 0.02 xˆ − 0.02 yˆ ) N / C = −25.4 MN / C xˆ − 25.4 MN / C yˆ
=
E2 =
3
4πε 0 r23
4π × 8.85 × 10 −12 0.0008
r r
r
E = E1 + E 2 = −64.4 MN / C xˆ − 5.97 MN / C yˆ
r
r
F = q f E = (1.24 × 10 −6 C )(− 64.4 MN / C xˆ − 5.97 MN / C yˆ ) = −79.8 N xˆ − 7.4 N yˆ
(
)
(
)
Now, let’s calculate the magnitude and angle of the electric field at point P.
E = 64.4 2 + 5.97 2 MV / m = 64.6 MV / m
tan θ =
Ey
Ex
=
θ = −175°
5
− 5.97 N
− 64.2 N
Things to remember:
r
r
• We define the electric field by the relation F = q f E
• If a field particle is positive, the force on it is in the direction of the electric field. If the field
particle is negative, the force is opposite in direction to the electric field.
r
qs
• The electric field of a point charge is given by Coulomb’s law E =
rˆ
4πε 0 r 2
• We can use Coulomb’s law to calculate the electric fields of several point charges by
evaluating the electric field from each source charge at a given field point and adding the fields
vectorially.
3.2 The Fields of Uniformly-Moving Point Charges
In what we’ve said to this point, we have assumed that the source charges are stationary.
When there are moving source charges, we know that the force on a field particle is modified by
the magnetic or stub force. In Lesson 2 we found that for point charges:
(
)
(
)
r
r
r e
r e
r
r
r
F = ν q f l + ν q f β f × rˆh × l = q f E + q f β f × rˆh × E .
ε0
ε0
The magnetic field of a point charge was taken to be
(3.4)
Magnetic Field of a Point Charge
r 1
r
B = rˆh × E
c
The SI units for magnetic field are tesla (T). One tesla is a huge magnetic field. The
magnetic field of the earth ranges from about 30 T to 60 T. In practice, magnetic fields are
often measured in the non-SI units of gauss. 1 T = 10,000 gauss.
r
r
vf
Finally, we recall that β f =
, so:
c
r
r
r
r
1
F = q f E + q f β f c ×  rˆh × E 

c
r
r
r
= qf E + qf vf × B
Usually, we can remember that the charge and velocity are those of the field particle but
the fields are produced by the source charges. We then drop the subscripts because we’re lazy,
and write the equation in its familiar form:
6
(3.5)
Lorentz Force Equation
r
r
r r
F = qE + qv × B
r
F is the force on a field particle moving in electric and magnetic fields. It is in
newtons (N).
q is the magnitude of the field particle’s charge in coulombs (C).
r
v is the (vector) velocity of the field particle in m/s.
r
E is the electric field of the source charges in volts/meter (V/m) or newtons/coulomb
(N/C).
r
B is the magnetic field of the source charges in tesla (T).
r
qE is called the “electric force.”
r r
qv × B is called the “magnetic force.”
At this point, we have only demonstrated that Eq. (3.5) holds when the source is a single
point charge. We know, of course, that if there are two point charges exerting forces on a field
particle, the total force is just the vector sum of the two forces.
(
)
(
r
r
r
r
r
r r
r r
Ftot = qE1 + qv × rs1 × E1 + qE 2 + qv × rs 2 × E 2
)
r
r
where E1 is the electric field of charge 1, rs1 is the vector from source charge 1 (at the time the
thread is emitted from source 1) to the field particle, and so forth. We see that the total fields are
just
r r
r
E = E1 + E 2
r 1r
r 1r
r
r
r
B = rs1 × E1 + rs 2 × E 2 = B1 + B2
c
c
so that Eq. (3.5) holds for two or any number of source charges as well.
From these equations, we recognize that the magnetic field is a quantity that is derived
from the electric fields of the point charges. In fact, we have seen that the magnetic field comes
from the relativistic velocity correction provided by the stubs. The magnetic field is not
independent of the electric field; it is only the by-product of the relativistic motion correction to
the Coulomb force.
You may have noticed that in SI units the magnetic field is much, much smaller than the
electric field because there is additional factor of the speed of light in the denominator of Eq.
7
(3.4). It might seem that the magnetic field would never contribute significantly to the total force
on a field particle because of this. There are, however, situations where the individual electric
field sums to zero vectorially, but the total magnetic field is not zero. In Lesson 2, we saw that
this was true for a current-carrying wire. It is also true for permanent magnets.
If it seems strange to you that the electric field can be zero while the magnetic field is
nonzero, think about forces and torques on a basketball you’re spinning between your hands.
Each hand exerts a force in the opposite direction, making the net force zero. The formula for
r r r
torque is τ = r × F ; however, we can’t conclude that the total torque is zero because the total
r
r
force is zero. Each force has its own r and its own F . We have to calculate the torque for each
force separately and then add the results together to get the total torque. In the diagram, each
torque is out of the page, so the total torque is out of the page as well. The torque vector points in
the direction of the axis of rotation. The direction of rotation is determined by a right-hand rule.
Put the thumb of your right hand in the direction of the torque, and the ball will turn about this
axis in the direction your fingers point.
r
F
r
r
r
r
r
F
Figure 3.1. Torques and forces on a basketball.
r 1
r
Magnetic fields are related to electric fields by a cross-product as well: B = rˆh × E .
c
Even if the electric field vectors of each source charge sum to zero at a point in space, the
r
r
magnetic field need not be zero. Each magnetic field has its own rh and its own E , and we must
r
take the cross product of each r̂h with each E before we add the magnetic of each source charge.
Equations (3.4) and (3.5) bear out some of the conclusions about magnetic fields we
reached in Sect. 2.7:
The magnetic field is a vector field.
The magnetic force on a positive field particle is in the opposite direction to the force
on a negative field particle.
There is no magnetic force if either the field particle or the source charge is at rest.
r
Note that if the source is at rest, r̂h is parallel to E , so the cross product is zero.
The magnetic force is always perpendicular to both the magnetic field and to the
velocity (because cross products are always perpendicular to each vector in the cross
product).
8
Magnetic fields can never make a particle go faster or slower, they only change the
direction a particle moves.
The last point may not be such an obvious consequence of Eq. (3.5), so let’s see why it’s
true. If a force pushes in the direction an object is moving – so the force is parallel to the velocity
– the force causes the object to speed up. If a force pushes opposite to the direction the object is
moving, the force causes the object to slow down. If a force pushes perpendicular to the object’s
velocity, the object changes direction without speeding up or slowing down. (In vector
terminology, it is only the component of the force parallel or antiparallel to the velocity that
changes an object’s kinetic energy.) Since the magnetic force is always perpendicular to the
velocity, it cannot change the object’s kinetic energy.
Now, let’s think about the real world for a minute. If we have charged objects or
permanent magnets, they don’t create lots of little arrows in space to tell us what the electric and
magnetic fields are. To determine electric and magnetic fields, we have to measure the force on
field particles. By putting stationary field particles at various points in space, we can use Eq.
r
(3.5) with v = 0 to find the electric field. To use Eq. (3.5) to find the magnetic field, we have to
do a little more work. (Note that there is no such operation as “vector division” so we can’t write
r r r
B = F / qv .) First of all, let’s assume that there is no net electric field (that is, each source
charge produces an electric field, but these all sum to zero) and that our field particle is moving
in the +x direction so that v y = v z = 0 . Then:
Fx
= v y Bz − v z B y = 0
q
Fy
Fy
= v z B x − v x B z = −v x B z ⇒ B z = −
qv x
q
Fz
F
= v x B y − v y Bx = v x B y ⇒ B y = z
q
qv x
From this single measurement we found the y and z components of the magnetic field, but we
didn’t get any information about the x component of the magnetic field.
Think About It
If we have both electric and magnetic fields in some region of space and we wished to
determine the fields by measuring the force on field particles that could be either stationary or
moving, what would be the minimum number of measurements we’d need to make? Describe the
measurements.
9
Things to remember:
r 1
r
• The magnetic field of a point charge can be obtained from its electric field: B = rˆh × E
c
• The force on a charge in an electric and magnetic field is given by the Lorentz force law:
r
r
r r
F = qE + qv × B
r
r
• The direction of the magnetic force is given by the cross product of v and B when the field
particle is positively charged and opposite that direction when it is negatively charged. The
direction is always perpendicular to both the velocity and the magnetic field.
• Source charges must be in motion to produce a magnetic field.
• Field particles must be in motion to feel the magnetic force.
• Uniform magnetic fields cannot change the kinetic energy of a charge.
3.3 Electric Potential and Voltage
In Lesson 1, we discussed how potential energy is often a useful concept in describing the
motion of an object moving in static electric fields. As you may recall, we found that potential
energy could not be defined when the source charges are moving, however. If the force on a field
charge is in the x direction, we can obtain the potential energy from the force by the relation:
U ( x) = − ∫ Fdx
If the force is caused by a static electric field, we can rewrite this in terms of the electric field as:
U ( x) = − ∫ q f Edx or
V ( x) ≡
U ( x)
= − ∫ Edx
qf
where q f is the charge of the field particle and V(x) is called the “electric potential” or “voltage.”
When we are talking about charged objects, we usually use the term “electric potential” and we
usually reserve the term “voltage” for circuits; however, they are really the same thing. From this
definition we can conclude that the electric potential of a source measured at a point in space is
just the potential energy on a filed particle located at the point, divided by the charge of the field
particle. This is in exact analogy to the electric field being the force on a field particle divided by
the charge of the field particle. Electric potential is used instead of potential energy partially out
of tradition and partially for convenience, in much the same way as the electric field.
To make this a little more concrete, consider the potential energy of a field particle in the vicinity
of a source charge. From Lesson 1 we found
1 qs q f
U (r ) =
4π ε 0 r
So V (r ) =
qs
4π ε 0 r
10
1
Note that the electric potential depends on the value of r, but it has no dependence on the any
characteristics of the field particle. Note that the electric potential is a scalar function whereas the
electric field is a vector function. This is an important distinction.
If we know the electric field and the electric potential at a point in space, and we also
know the charge of a field particle located at that point, we can deduce the force and potential
energy through the relations;
r
r
F = q f E , and U = q f V .
volts
meter = volts . The
meter
term voltage is related to the fact that we measure electric potential in units of volts.
Since V = − ∫ Edx , we see the units of electric potential are
The key equations that relate electric field to potential are obtained by dividing Eqs. (1.4)
and (1.6) by q:
Electric Potential and Electric Field
V ( x) = − ∫ Edx,
V (r ) = − ∫ Edr
dV (r )
dV ( x)
,
Er = −
dr
dx
or more generally
r
∂V
∂V
r
∂V
zˆ
yˆ −
xˆ −
E = −∇V (r ) = −
∂z
∂y
∂x
Ex = −
(3.6 )
The last equation states that the electric field is minus the “gradient” of the electric potential. The
idea of a gradient may or may not be familiar to you. However, it is very useful to understand
what it means. The gradient is an example of an “operator.” Mathematically, an operator can be
defined as “something which does something to something else.” It is like a recipe or a computer
subroutine that takes in something and spits out something else. Operators include multiplication,
square roots, derivatives, etc. The gradient is an “operator” that takes in a scalar field and returns
a vector field according to a given recipe. Here the gradient takes in the electric potential and
returns the electric field. The gradient is written as L and often pronounced "del."
Now, let’s find the specific recipe the gradient operator uses to return a vector field.
• Let f be any scalar field; that is, f is a function of x, y, and z that yields a scalar for given values
r
of x, y, and z. As a concrete example let us take f (r ) ≡ f ( x, y, z ) = αx 2 y + β sin(kz ) where , ,
and k are constants.
β
α
11
∂f
, read “the partial of f with respect to x,” is the derivative of f
∂x
∂f
= 2αxy ,
with respect to x with all other variables, here y and z, held constant. Thus
∂x
∂f
∂f
= αx 2 , and
= kβ cos(kz ) .
∂y
∂z
∂f
∂f
∂f
• The definition or recipe for the gradient of f is: ∇f ( x, y, z ) =
zˆ
yˆ +
xˆ +
∂x
∂y
∂z
• A partial derivative such as
• This then gives ∇f ( x, y, z ) = 2αxyxˆ + αx 2 yˆ + kβ cos(kz ) zˆ.
Think About It
Now if I give you a function of x, y, and z, you can give me the gradient of that function.
But what does a gradient really mean?
Note that if f only has x dependence, the gradient is just df / dx except it is a vector that
points in the direction of x. In general, ∇f points in the direction that f increases most rapidly. Its
magnitude is the derivative of f along that direction. So, in short, the gradient shows how much f
changes and in what direction it changes. The electric field is minus the gradient of the electric
potential, so the electric field is a measure of how rapidly the electric potential changes and it
points in the direction the electric potential decreases most rapidly. Note that the units of electric
field, V / m, are a reminder of the fact that the electric field tells us by how many volts the
electric potential changes per meter. (We also use N / C as units for the electric field, but this is
identical to V / m.)
Things to remember:
• Electric potential can be defined only for static distributions of charge.
U
• The electric potential of a charge q is defined as V = , so electric potential or voltage is
q
closely related to potential energy.
• We can obtain the electric potential from the electric field by V ( x) = − ∫ Edx, V (r ) = − ∫ Edr .
dV ( x)
dV (r )
, Er = −
.
• We can obtain the electric field from the electric potential by E x = −
dx
dr
r
• More generally E = −∇V . The magnitude of the gradient tells us how rapidly a scalar function
varies in space. The direction of a gradient tells us in which direction the function increases
most rapidly.
12
3.4 Field Vectors, Field Lines, and Field Contours
Now that we’ve defined the electric and magnetic fields, we want a good way of
visualizing the fields. It turns out that there are three particularly useful ways of doing that.
A. Field vectors
In the field vector model, we simply draw an arrow at every point in space. Examples of
two different fields are sketched below. I have drawn just a few of the infinite number of field
vectors.
Figure 3.2. Two fields represented by field vectors.
The problem with these sketches is that it is often hard to visualize the field because there are too
many arrows pointing in different directions.
Note that a field vector sketch is essentially the same thing as a sketch of the threads or
stubs at selected points in space. From the equations in Sect. 3.0, we saw that the electric and
magnetic fields could be given in terms of the threads and stubs by the relation
(3.7)
r
r F
e r
=
E=
lν
q f ε0
r e 1r
sν
B=
ε0 c
As we saw before, up to a factor of e / ε 0 , the electric field vector at some point in space is just
the density of threads at the point multiplied by the thread vector. Similarly, the magnetic field is
e / cε 0 times the density of threads multiplied by the stub vector.
13
B. Field lines
Field lines are a set of curved lines which point along the direction of the field, or more
precisely, field lines are the curves which have field vectors as tangents at each point. You have
probably placed iron filings on a sheet of paper and put a magnet underneath to see the iron
filings line up along the magnetic field lines.
To construct a field line, we begin at some point in space and draw a very short (infinitely
short, in the limit) line segment in the direction of the field. For an electric field, the direction of
the field at a point is given by the direction of the thread or the direction of the force on a small,
positive, stationary test charge. For a magnetic field, the direction of the field at a point is the
direction of a stub at that point. After we draw the first segment, we move to the end of the
segment and find the direction of the field at this new point in space. From there, we draw
another short line segment in the direction of the new field, and so on. To represent the same
fields as those in Fig. 3.2 by field lines, we have the following drawings:
Figure 3.3. Two fields represented by field lines.
Note that this representation of the field is easier to interpret than the drawings of the field
vectors. Algebraically, however, it is harder to describe field lines than field vectors.
14
Figure 3.4 Aligning the threads of a moving charge.
In Sect 2.8 we said that we could visualize threads and stubs better if we aligned them in
an orderly fashion. Since the threads are very small, connecting threads together produces lines
that always point in the direction of the electric field. Connecting stubs gives us lines that always
point in the direction of the magnetic field. Thus the aligned threads and stubs become electric
and magnetic field lines.
A vector must give information about magnitude as well as direction. Field lines clearly
tell us the direction of fields, but to be very useful, they must also convey information about
magnitude. As it turns out, we can determine the magnitude of a field from its field lines as well.
This comes as a direct consequence of Eq. (3.7). From this equation, we see that the electric field
r
r
is proportional to l (and the magnetic field is proportional to s ) multiplied by the density of
threads. (Recall that the density of stubs is the same as the density of threads.) Now let’s align N
of these threads (stubs would work in exactly the same way) as described in Sect. 2.8. We place
the N threads a box whose sides are parallel to the threads and whose top and bottom are surfaces
that are perpendicular to the threads. Rather unimaginatively, we call surfaces that are
perpendicular to field lines “perpendicular surfaces.” The area of each of these surfaces is A. We
arrange the threads so that each head lies on the top surface, as shown in Fig. 3.5.
A
Figure 3.5 Threads aligned in a box.
15
As long as the box is small, it is very nearly a rectangular prism of height and surface
area A. Given this information, we can calculate the magnitude of the electric field using Eq.
(3.7):
E=
e
ε0
lν =
e
ε0
l
N
e N
=
.
Al ε 0 A
Thus, E is proportional to N / A, or since each thread lies on a different field line, we can
generalize this to:
Field ∝
number of field lines piercing a perpendicular surface
area of the surface
That is, whenever we construct field lines by aligning threads or stubs, the field is larger where
the field lines are closer together. Whenever we draw field lines, we will require this condition
to be met.
Think About It
Where is the field strongest in the two frames of Fig. 3.3? In Fig. 3.4, we aligned the
uncorrected threads head to tail, but the stubs are not aligned that way. Will the proximity of
field lines give us information about electric fields or magnetic fields? based on these lines,
where is the field the strongest?
C. Field Contours
Sometimes we can also represent a field graphically with a series of perpendicular
surfaces called a “field contour.” If the field lines are all parallel, a perpendicular surface is flat,
but if the field lines are not parallel, we allow the surface to curve so that it is perpendicular to
every field line.
First, let’s think about how we would construct a single perpendicular surface. Consider a
single field line. Make a small two-dimensional surface that is perpendicular to that field line at
some point. Now go to an adjacent field line and make a second small surface that is
perpendicular to this line and that has an edge joining the first surface. Continuing in this way,
construct a large surface that is everywhere perpendicular to all the field lines. If the number of
field lines is finite, the surface is a bit bumpy with lines connecting small flat sections. however,
as we let the number of field lines become very large, the perpendicular surface becomes
smooth.
A single perpendicular surface provides us with information about the direction of the
field. The direction of the field at any point in space is perpendicular (or “normal”) to the
16
surface. Since there are two directions normal to such a surface, we usually include an arrow to
specify the field direction, as in Fig. 3.6.
Figure 3.6 Two fields represented by field contours. (Field lines are shown in gray.)
Remember that the field contours, shown in black in this figure, are really surfaces in three
dimensions, and not merely lines. For reference, the field lines are shown in gray in the figure.
A single perpendicular surface, as a single field line, gives no information about the
strength of a field. It can only indicate a field’s direction. To be useful, we need to construct a set
of perpendicular surfaces that are closer together where the field is stronger. Such a set is what
we mean by a “field contour.” If we think of the earth’s gravitational field as the vector field, the
field contours are a set of concentric spherical shells. These shells are closer together near the
surface of the earth, as the gravitational field is stronger near the earth.
To be a little more precise, the field strength is given by:
Field ∝
number of contour elements ( perpendicular surfaces ) pierced by a segment of field line
length of the line segment
While we can always produce a set of perpendicular surfaces, we cannot always produce a field
contour. That is, sometimes, there does not exist a set of perpendicular surfaces that satisfies the
field strength formula above. However, we can produce field contours for two important cases:
1) electric field contours for any arrangement of static charges, 2) magnetic field contours for
arrangements of long, parallel, current-carrying wires.
17
Note that field contours carry essentially the same information as the field lines or the
field vectors. Because of geometry, sometimes it is easier to use field lines, and sometimes it is
easier to use field contours.
Things to remember:
• Given field vectors, know how to construct a field line.
• Electric field lines always outward from positive and inward toward negative charges.
• Given field lines, know how to construct perpendicular surfaces.
• For field lines, know that the direction of the field is tangent to the field line at a point.
• For field lines, know the following recipe for the magnitude of the field:
Take a section of one perpendicular surface. Count the number of field lines passing
through the surface. The magnitude of the field is proportional to the number of field
lines piercing the surface section divided by the area of the section. (To be
mathematically accurate, we should say something about drawing an infinite number of
field lines so we don’t have to worry about rounding errors.)
• For field contours, know that the direction of the field is normal to the field contour at a point.
• For field contours, know the following recipe for the magnitude of the field:
Take a segment of one field line. Count the number of field contours pierced by the
field line. The magnitude of the field is proportional to the number of surfaces pierced
by the segment divided by the length of the segment.
3.5 Static Electric Fields and Potentials in Conductors
There are a number of important characteristics of the static electric fields and electric
potentials on conductors that we will use throughout the semester. As you recall, a conductor is a
material in which there are many electrons that can move freely throughout the material with no
energy loss. In practice, there is no such thing as a perfect conductor, but for our applications,
most conductors come very close to being perfect.
Note that in this section, however, we are only going to deal with static distributions of
charge. If we move any charges around, we have to wait for everything to settle down so that the
charges are no longer moving. This is very important to keep in mind!
When we say that electrons can move freely in a conductor, we mean that there are no
frictional forces or attractive force that we must overcome for the conduction electrons to start
moving. That is, if there is any electrical field, no matter how small, inside a conductor, electrons
will move. This means that if we have a static situation – no electrons are moving – there can be
no electric field inside the conductor. Thus, there can be no static electric field inside a
conductor.
If we take a conductor and place electric charges near it, we know those charges will
attract or repel electrons in the conductor. Clearly, there must be spots where the net charge is
not zero on the conductor. On the other hand, if a small region within a conductor has a net
positive charge the electric field in this region would not be zero. Nearby electrons would be
attracted into the positive region until the net charge becomes zero. However, the electrons that
18
move into this region must come from somewhere. They can’t come from within the conductor
and leave a net negative charge somewhere else in the conductor. In the end, the only place there
can be any imbalance of charge is on the surface of the conductor.
Let’s say that a negative charge is on the outside of a conductor, as in Fig. 3.7. The
negative charge repels electrons, so the free electrons in the conductor move as far as they can
from the negative charge. They won’t all move to the most distant spot on the conductor, as they
repel each other. The electrons therefore tend to spread over the part of the conductor’s surface
that is opposite the external charge. Hence the surface of conductor near the external charge has
positive charge and the opposite surface has net negative charge. However, the positive and
negative surface charges produce their own electric field. The electric field they produce exactly
cancels the electric field of the external charge within the conductor.
19
(a)
(b)
(c)
Figure 3.7 A negative charge near a conductor. Fields of (a) the charge alone,
(b) the conductor alone, and (c) both the charge and the conductor.
20
In Fig. 3.7, note how surface charge on the conductor produces a field inside the
conductor that is exactly opposite the external charge. The field produced by the charge on the
outside surface of the conductor goes from positive charges to negative charges, just as it does
inside the conductor. When the fields are combined we see 1) the field near the external charge is
the field of a point charge, the field far from the charge and conductor is the field of a point
charge, 3) the field is perpendicular to the surface of the conductor, and 4) there is no electric
field within the conductor.
Since the electric field is perpendicular to the surface of the conductor, the surface of the
conductor must be a “perpendicular surface” and one element of a field contour. If we put a very
small test charge on the surface of the conductor, it takes no work to move it anywhere on the
surface since the force is always perpendicular to the direction of motion. We can conclude,
therefore, that the surface has constant potential energy and constant electric potential. Such a
surface is called an “equipotential” surface. Note that perpendicular surfaces are always
equipotential surfaces.
Things to remember:
• There can be no static electric field inside a conductor.
• The static electric field on the surface of a conductor must be perpendicular to the conductor’s
surface.
• In the static case, all the charge on a conductor must reside on its surfaces.
• In the static case, the surfaces of a conductor are perpendicular surfaces and equipotential
surfaces.
3.6 Force and Motion in Uniform Fields
There are a large number of applications where charged particles are accelerated in
uniform electric and magnetic fields. We’ll look at a few of these applications in Sect. 3.7, but
first, let’s derive a few rules that will be helpful in analyzing the motion of charged particles in
uniform fields.
We can produce a uniform electric field by placing charges on two parallel metal plates,
as illustrated in Fig. 3.8. (Such an arrangement is called a capacitor. We’ll study capacitors in
detail in Lesson 5.)
21
E
e−
Figure 3.8 An electron in a uniform electric field.
Each charge on one such conducting plate tries to move as far as possible from all the
other charges on the plate. This leads to a very uniform distribution of charge over the plate,
except near the outside edges. As long as we are not close to one of these edges, the electric field
lines point directly from the positive plate to the negative plate. The field lines can’t bend to one
side or the other because the charge distribution is uniform on the plate. Also, the field lines
can’t be closer in one region of the plate than another because the charge distribution is uniform.
We say that “symmetry” requires the field lines to be uniform over the entire plate. Symmetry
arguments are very important in physics and we will make frequent use of them in later lessons.
Now we know that each plate must have the same electric potential or voltage
everywhere on it because it is a conductor. We call the difference in voltage from one plate to the
other )V and the distance between the two plates ) x. From Eq. (3.8), we know that the electric
field between the plates is given by the expression:
∆V
E=−
.
∆x
The minus sign tells us that when the voltage is larger on the right plate (the voltage will larger
where there is more positive charge) the electric field points to the left (away from the positive
charge). Often, we won’t worry about the minus sign as it just tells us the direction of the field.
Thus, if the voltage difference between the plates is 300 V and the separation between the plates
is
1.00 cm, the magnitude of the electric field is 30,000 V / m.
If we place an electron just inside the negative plate, it will experience a force opposite in
direction to the electric field, toward the positive plate. Its change in potential energy is
22
∆U = q∆V = −e∆V .
In the case described above,
U = −e (300 V) = −4.80H10−17 J.
This is obviously a very small number, so physicists have invented a new unit, the
“electron volt” or “eV” (pronounced “ee-vee”) to measure the energy of small particles. This is
the energy an electron gains when it moves through a 1 V potential difference. Thus 1eV = e H
1V = 1.60H10−19 J. So we could write the above energy as
U = −e (300 V) = −300 eV.
You can see by this example how simple electron volts are to use. This is one of the few non-SI
units we will use in this course.
Since energy is conserved, we know that if the electron loses potential energy, it must
gain kinetic energy so that K + U = 0. Hence:
∆K = − ∆U = − q∆V .
(3.8 Kinetic energy and voltage)
If the electron starts at rest, ∆K = K = 12 mv 2 , so:
v=
2
q∆V
m
where
v is the velocity a charge gains in being accelerated from rest through a voltage )V, in
m/s.
q is the charge in coulombs (C).
m is the mass of the charge in kg.
We could go further to describe more general motion of charges in uniform electric
fields; however, all we really need to note is that the force is constant in magnitude and direction.
This is much like the force of gravity on object near the earth’s surface. So the general result is
very much like the motion of projectiles that you studied in your mechanics course. Since we
really won’t need to deal with such motion very often, we’ll just move on to the magnetic force.
As soon as we start talking about the motion of charged particles in magnetic fields, we
find that we are forced into a three-dimensional world. Consider a positive charge moving in a
23
uniform magnetic field. We can draw both the magnetic field vector and the velocity vector in
r
r r
the plane of the screen, but the force F = qv × B is either into the screen or out of the screen.
B
B
v
Figure 3.9 A positively charged particle moving in a uniform magnetic field.
If the charge is moving as in Fig. 3.9, the force is out of the screen. If the charge were negative,
the force would be into the screen.
Think About It
We have made use of both magnetic field lines and magnetic field vectors in Fig. 3.9.
How are they used differently? What information can we deduce from each?
To draw vectors going into the screen or out of the screen on the two-dimensional screen
itself, we need a convention. We denote “into the screen” by crosses, (××××), and “out of the
screen” by circles, (
). These symbols are supposed to remind you of the head of an arrow
coming out of the screen toward you and the fletching of an arrow going into the screen away
from you.
Now let’s consider a uniform magnetic field pointing into the page, as illustrated in Fig.
3.8. A positive charges is drawn with its velocity vector at two different times. Using the Lorentz
force law, determine the direction of the force on each of these charges.
24
v
r
F
r
F
v
v
(a)
(b)
Figure 3.10 A positive charges moving in a magnetic field.
The force on the left charge is upward and the force on the right charge is toward the left.
Now let’s start with the charge in Fig. 3.10 and think about how it moves over a period of
time. At the instant shown in Fig. 3.10 (a), the charge is traveling to the right with a force in the
upward direction. Since the force is perpendicular to the direction of the charge’s motion, the
force will simply deflect the charge upward without changing its speed. As the charge moves, it
is deflected a little upward. But the direction of the magnetic force changes as well, pointing a
little to left so as to remain perpendicular the charge’s velocity. When the charge is moving
upward, as in Fig. 3.10 (b), the force is toward the left, and so on. Since the force is always
perpendicular to the velocity, the force acts very much like the force of earth’s gravity on the
moon or the force of the string on a ball being swung in a circular path. Similarly, the path of the
charge will be circular.
Think About It
If the charge in Fig. 3.10 (b) were moving outward from the screen what force would
there be on it? If the charge had a component of its velocity to the right and a component coming
out of the screen as well, what would happen? Explain in your own words why the most general
motion of a charged particle in a uniform magnetic field is helical.
25
r
Figure 3.11. The path of a positively charged particle moving in uniform magnetic field.
We’d like to be able to find the radius of the path of a charged particle moving in a
uniform magnetic field. We can determine this by noting that for particles moving in a circular
r
mv 2
path, there must be a centripetal force Fc =
(−rˆ) . The − r̂ indicates that the direction of
r
the force is toward the center of the circular path. Physically, this force is provided by the
r
r
r
r r
Lorentz force. Since B and v are perpendicular, Eq. (3.5) gives us FL = qv × B = qvB sin θ (− rˆ)
r
r
r
. The angle between v and B is θ = 90° so FL = qvB (−rˆ) . Equating these expressions, we
have:
mv 2
= qvB
r
mv = qBr
p = qBr
In the last step we have made use of the formula for momentum, p = mv.
26
The first thing we want to find from this equation is an expression for the radius of the proton’s
orbit. This is called the “cyclotron radius.”
(3.9 Cyclotron radius)
r=
p
qB
r is the orbital radius of a charged particle moving in a uniform magnetic field in m.
p is the momentum of the charged particle in kg m/s.
q is the charge of the particle in coulombs (C).
B is the magnetic field in tesla (T).
This expression tells us that the radius of the orbit of a charged particle moving in a magnetic
field is directly proportional to the particle’s momentum and inversely proportional to the
magnetic field. Although we used Newtonian mechanics to derive this expression, it turns out
that it is also valid relativistically.
Things to remember:
• The change in kinetic energy of a particle in an electric field is ∆K = − q∆V .
• The general motion of a charged particle in a uniform magnetic field is a helical path.
• The cyclotron radius is p=qBr. Be able to derive this expression.
3.7 Devices Using Electric and Magnetic Fields
Physicists use a number of important devices that make use of the Lorentz force. We will
look at three of these.
Linear Accelerators
Particle accelerators (“atom smashers” in an outdated vernacular) are devices that
accelerate beams of particles such as protons, electrons, alpha particles, or heavy ions. These
beams are used for a variety of purposes included probing the structure of matter at
submicroscopic levels, determining the composition of materials, implanting ions into solids, and
creating radioisotopes for medical purposes.
27
One of the simplest particle accelerators is the linear accelerator. The basic idea is
simple: if we put a positive charge behind a proton and a negative charge in front of it, the proton
will experience a force in the forward direction. To accomplish this, what is typically done is to
put voltages on sections of a tube through which the beam passes. Inside these
+V
−V
+V
Figure 3.12 A proton being accelerated between sections of a beam tube.
conducting sections, the electric potential remains fairly constant and the electric field is
therefore quite small. However, there is a large electric field between segments with positive and
negative voltage. Since the particles drift without accelerating through the segments of tube, they
are called “drift tubes.” For the beam particles to gain high energy, such an acceleration process
has to occur many times, so there must be a large number of drift tubes along the beamline, as
indicated in Fig. 3.12. If the voltages on each drift tube were fixed, however, the protons would
feel forces alternately in the forward and then in the reverse directions. To keep the proton
accelerating in the same direction, the signs of the voltages in the drift tubes must constantly be
changing so that the section behind the protons is always positive and the section ahead of the
protons is always negative. This is accomplished by using a high-frequency alternating current
source. One more important characteristic of the linear accelerator is that the sections become
longer and longer as the particle moves down the beamline. The frequency of the drift tube
voltage remains constant, but the protons move farther and farther between the cycles as they
accelerate down the tube.
The Cyclotron
One disadvantage to linear accelerators is that the beamlines must often be very long. In the
1930s, Ernest O. Lawrence invented an accelerator that allowed particles to move in a spiral path
while they were being accelerated. This device was called a “cyclotron.” The operation of a
cyclotron is similar to that of a linear accelerator, except that the particles being accelerated are
placed in a strong magnetic field to keep the particles moving in a circular path. The drift tubes
are replaced by “dees,” so called because of their D-shape. A dee is a conductor in the shape of a
tuna can chopped in half. One dee is kept at a positive potential and one at a negative potential,
causing the beam particles to accelerate when they are located between the dees. As with drift
28
Figure 3.13 A proton accelerated in a cyclotron.
tubes, the sign of the voltage on the dees must oscillate to keep the beam particles accelerating.
While the particles are within the dees, they drift in circular paths. The faster the particle goes,
the larger the radius of the circular path becomes.
One important parameter of a cyclotron is the radius that is needed for particles that have
a given kinetic energy. In the precious section, we found in the cyclotron radius to be given by
the expression p = qBr. We can also write this in terms of kinetic energy, but this expression is
not valid for relativistic particles:
1 2 p2
K = mv =
2
2m
p = 2mK = qBr
r=
2mK
qB
One other interesting quantity is the frequency at which we need to change the polarity
(positive or negative, that is) of the voltage on the dees. This period must match the period of the
particle as it travels in its circular path. We know that a charged particle moving in a circular
orbit travels once around the orbit in one period.
29
v=
2π r
T
2π m r
T
2π m r 2π m r 2π m
T=
=
=
p
qBr
qB
p = mv =
Surprisingly, the period is independent of radius or velocity for non-relativistic particles.
(Relativistically, the mass increases as the charge goes faster, however.) Since the orbital period
is the number of seconds per orbit, the frequency – the number of orbits per second – is just the
reciprocal of the period. This frequency is called the “cyclotron frequency.”
f =
(3.10 Cyclotron Frequency)
qB
2π m
f is the orbital frequency of a charged particle moving in a uniform magnetic field in
cycles/second.
m is the mass the charged particle in kg.
q is the charge of the particle in coulombs (C).
B is the magnetic field in tesla (T).
The Mass Spectrometer
Another application of electric and magnetic fields to particle beams is the mass spectrometer. Ions of known mass and charge are accelerated through an electric potential provided
by charged plates. The ions then form a beam which passes through a hole in one plate into a
region of uniform magnetic field oriented perpendicular to the velocity of the beam. The
V
Figure 3.14 Mass spectrometer
30
magnetic field then exerts a force on the particles, causing them to move in a circular path. The
radius of the path depends on the momentum and charge of the particles. Eventually a detector
counts the number of particles which arrive at a particular location in the device.
The Electric Field. Let us assume that ions are introduced with zero velocity to the point
labeled "A" in Figure 3.14. Here they are repelled by the positively charged electric plate and
attracted to the negatively charged plate. As we found in the previous section, the kinetic energy
is: K = q V.
The Magnetic Field. As with the cyclotron, we know
(3.8 Mass spectrometer radius)
r=
p
=
qB
2mK 1 2m∆V
=
qB
B
q
r is the radius of the ion orbit in m.
p is the momentum of the ion in kg m/s.
q is the charge of the ion in coulombs (C).
B is the magnetic field strength in tesla (T).
K is the kinetic energy of the ion in joules (J).
V is the voltage across the accelerating plates in volts (V).
Since the velocity of the ion is very slow compared to the speed of light, it was not necessary for
us to use a relativistic expression for momentum.
Note that the kinetic energy of ions in the mass spectrometer is independent of mass, but
the radius is proportional to the square root of mass. In this way, ions of different atomic weight
can be separated and counted. Such a device can also be used to produce samples of a single
isotope; however, more efficient methods of isotope separation exist if large quantities of a
single isotope are required.
Things to remember:
• Be able to describe how linear accelerators, cyclotrons, and mass spectrometers work.
• Knowing the formula for the cyclotron radius, find the cyclotron frequency.
• Given the charge on an ion, the mass of the ion, the accelerating voltage, and the magnetic
field, be able to find the radius of curvature for an ion in a mass spectrometer.
31