Physics 2B Summer Session
Unit 1 Lecture Notes
Motivation
The Electromagnetic Force (EM) is one of the Four Fundamental Forces
found in nature:
1.
2.
3.
4.
Gravitational Force
Electromagnetic Force
Strong Nuclear Force
Weak Nuclear Force
The last two don’t affect our daily life on earth (see Physics 2D)
The EM is much, much stronger than Gravity
E&M determines the structure of matter, both atomic & molecular
So E&M is an integral part of Chemistry, Biology, Engineering, Etc.
Prolific use of electronic devices is found in all branches of science
Properties of Electric Charge
™
Charge is a fundamental property of matter, like mass
so charge doesn’t exist apart from some kind of particle
™
Charge was first discovered by the Greeks ( ελεκτροη )
™
Charge comes in two flavors – positive & negative
™
Like charges repel – Unlike charges attract
™
Charge is Quantized – electron ( e ) & proton ( e )
™
Charge appears to be balanced in nature
There are an equal number of plus & minus charges
So we get Neutral Atoms and a Neutral Universe
™
Conservation – charge is neither created nor destroyed
−
MKS Unit = Coulomb (C)
+
1 e + = 1.6 ×10−19 C
1 C = 6.25 × 1018 e +
1
First a quick word about units, symbols and the like.
Take a look at the PDF file “MKS Units” on the course website under the
“Handouts” section. It is important to know that this system is closed under
all of the formulas and equations in this course. In other words, your solution
will come out in the correct fundamental units provided that you input all
your data in the other appropriate fundamental units of the system.
For example, you wouldn’t expect the distance formula s = vt to yield an
answer in miles per hour if you put in v = 20 meters/second and
t = 22 minutes . It was fairly easy in mechanics (Physics 2A) to cancel
physical units – remembering, say, that a Joule is equal to a kilogram-meter
per second squared. Here, however, there are many new and colorfullynamed units which can be described in terms of other fundamental units in a
variety of ways. So be careful with units.
Coulomb’s Force Law
Let’s begin with two electric point charges, q1 and q2 separated by a distance
r and calculate the force on q2 by q1. Note that we have drawn a line from q1
to q2 and extended that line with a unit vector which points away from the
source charge q1.
q2
r
q1
r̂
The force is found using
F12 = k
q1q2
rˆ
r2
where the constant
Coulomb Force Equation
N im2
= 9.00 × 10
k=
4πε 0
C
1
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Note the resemblance to the gravitational force equation
F12 = G
m1m2
rˆ
r2
which also
Is an inverse square law with respect to distance,
Depends on the product of a fundamental particle property
and has Force Symmetry, i.e., F12 = −F21
For a numerical exercise, try to follow example 23-1 on page 574.
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Multiple Charges and the Superposition Principle
How do we attack problems where there is more than one source charge?
Fortunately, Nature is logical and accommodating – the forces from separate
charges are themselves separate and can be added together, as a vector sum
Qqi
q
rˆi = kQ∑ 2i rˆi
2
ri
ri
F = ∑ Fi = ∑ k
Example 23-2 page 577
y
F2
F1
F
Q
F2
r
r
F1
F2X
F1X
Q
y
q1
q2
x
-a
a
Consider the figure above. If the information is given without graphics, we first
draw a picture as it always helps us get things straight. For multiple charges, let Q
be the “target” charge. Then, the rule of superposition says we simply add.
But this is a VECTOR SUM !!! So we must add the x and y components
separately. Note from the drawing that the x components are equal and opposite
and thus cancel, saving us time & effort. The y components (vertical) involve the
sine of the angle (or you can just use similar triangles)
F1y =
kQq1 y ˆ
kQq1 y ˆ
=
j
j
3
2
2
2
r12 r1
a +x
(
)
The two charges are identical, so their sum is just twice that, or
F = F1y + F2 y =
(
2 kQqy ˆ
j
3
2
2
2
a +y
)
(where q = q1 = q2 )
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A Few Tips on Problem Solving
The following method may seem extremely elementary and pedantic but believe
me – from personal experience – it definitely works. This is especially true we go
beyond the “toy” problems in these courses and start working in the real world.
The drill below helps to organize our thinking, clarify the problem and lead to its
ultimate solution in spite of the difficulty.
1. Draw a picture or diagram
This clarifies who’s doing what to whom and also makes it possible to set up
a coordinate system where necessary i.e., when vectors are involved. In fact,
we just used it in the example above and you’ll see the value of this again
immediately in the next section.
2. Specify the variables (data)
Write down the given data and single out the unknown, e.g., F = ? This can
be particularly useful when it’s not clear which formula applies. You can
check your reference sheet and find the one equation which contains all the
data and unknown variables.
3. Write down the basic formula(s)
Again, this is often a visual aid to help you check that you’re headed in the
right direction. It also gives you a written record of your work so you can go
back later and proofread your derivation for errors.
4. Derive a working formula
This may seem unnecessary unless you have to put several formulas together
to achieve the final result. However, the benefits are manifold. For example,
you may often find that certain variables or constants cancel, saving you not
only time, but possible errors when entering data in your calculator.
5. Substitute (plug in) the variables
If you’ve followed the steps, you simply “Plug-and-Chug” at this point.
Remember: this method looks superfluous when working simple exercises, but it
becomes extremely powerful as you begin tackling the really nasty problems. I
introduce it here because I believe that, by practicing these steps early on, this
method will become second nature and of tremendous use to you as the course
progresses and, indeed, should serve you well – even as a mental exercise – as your
careers progress.
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More Tips on Problem Solving
There are three ways to tell if you’ve gotten the answer right, even without being
told what it is. In this course “all will be revealed” so you can always check and
correct your work – albeit after an exam. However, you’re being trained to do
independent research, either for yourself or someone else, and that work will be
leading-edge, generally without an “answer section” in the back of some book.
So how can we know? Well, we can’t really, until we test the results
experimentally. However, we can ask ourselves the following questions:
1. Are the units correct?
If you’re solving a Mechanics problem for energy, say, and the units come
out in kg·m/s2 which are the units force (Newtons) then you’ve missed
something along the way and you’d better go back and check your work.
2. Does the solution behave properly as the variables change?
Suppose you’ve calculated the electrostatic force at some distance from a
charge distribution. Does the force decrease with distance? Does it increase
if you increase the charge density in the distribution? If not, you might want
to review your work for errors.
3. Does the solution converge to a known solution in some limit?
For example, any charged body of finite extent must look more and more
like a point charge as you move further away. Look at the limit as the
distance r → ∞ . The expression should converge to that of single point
charge equal to the total charge of the body. If not, then . . . you know.
Example:
Recall the solution to our three-charge problem above
F=
2 kQqy
(
a2 + y2
)
3
2
If we increase either q or Q the force increases as expected. What if y = 0 so that
the Q lies exactly midway between the two positive charges? The only forces are
horizontal and cancel to zero also as expected. Lastly, when Q is far away, y >> a
( )
2
so the denominator reduces to y
F=
3
2
= y 3 and the entire expression collapses to
2kQq ˆ
j as expected (Why?)
y2
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THE ELECTRIC FIELD
For more complicated arrangements or continuous charge distributions, the
calculations can be a chore especially if we have to do them over & over again for
different charges which might pass through the same location, or if we move the
same charge around to different spots.
So we introduce the concept of a field which describes at every point in space the
force exerted by some extended charge distribution (possibly elsewhere) on a unit
point charge placed there, in other words, we calculate the force on any charge
and then divide by the magnitude of that same charge:
E=
F (Q )
Q
Electric Field Defined (Unit = Newtons/Coulomb)
This field must be defined everywhere in space, and is analogous to a velocity
vector field of a river describing the flow of water at all points simultaneously, or
a birds-eye view of merging rush-hour traffic.
EXAMPLE:
Single Point Charge q
E=
EXAMPLE:
F (Q ) 1 ⎛ kqQ ⎞ kq
= ⎜ 2 rˆ ⎟ = 2 rˆ
Q
Q⎝ r
⎠ r
Multiple Point Charges (by superposition)
E = ∑ Ei = k ∑
qi
rˆi
ri 2
Now, once we know the Electric Field everywhere, the force calculation for any
arbitrary point charge Q is easy:
F = QE
Force on charge in Electric Field
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Conductors vs. Insulators
Definition:
A conductor is a material containing charges which are free to move.
In an electric field, a charge feels a force F = qE and, since Newton proclaims
that F = ma , in a perfect conductor a completely unobstructed charge must
accelerate without bound:
ma = qE ⇒ a =
qE
m
Hence, inside a conductor, free charges will move around until the field they
themselves produce by their separation cancels any externally applied field. In
other words,
1. The (static) electric field inside a conductor is always zero.
2. If excess charge is placed on a conductor, whether or not it’s in an
external field, all such charge must reside at the surface, otherwise
there would be a field around any such internal charges, and they would
move until condition (1) is satisfied.
3. The electric field at the surface of a conductor must be perpendicular
to the surface for, if not, free charges at the surface would move until any
parallel component vanishes. The remaining perpendicular field cannot
move charges off the surface.
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CONTINUOUS CHARGE DISTRIBUTIONS
Consider a blob of material which has some non-neutral charge distribution
within it. Divide it up into little pieces and replace the discrete variables q
and E by the differentials dq and dE as shown in the drawing below.
Instead of summing the vector contributions from the discrete charges, we
integrate the differential Electric Field elements over the entire charge
distribution at the point of interest:
E = ∫ dE = ∫
k dq
rˆ
r2
where r is the distance from each dq to the point of interest P.
Note that the vector dE points in the direction of r̂ .
P
dq
r
dE
Now to actually evaluate the integral, we have to convert dq to something
that has location and charge information in it.
Recall from mechanics how we can find the mass of an object by integrating
its mass density (which can vary from point to point) over the entire
volume. We do the same by introducing charge density in the three
dimensionalities:
1-D Line Charge Density
2-D Surface Charge Density
3-D Volume Charge Density
dq
dL
dq
σ=
dA
dq
ρ=
dV
λ=
dL = differential Length
dA = differential Area
dV = differential Volume
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ELECTRIC FIELD LINES
Field lines always start on positive charges and end on negative charges.
In a field line diagram, we draw continuous lines rather than vector arrows, and
can do this by simply connecting the arrows to form “streamlines”. Hence, instead
of vector lengths showing the field strength, it is the density of lines that do so.
EXAMPLES that you should “know”
Configuration
Symmetry
Point, Sphere
Line, Cylinder
Sheet
Dipole
Finite Disk
radial
cylindrical
planar
axial
axial
See pages 601 through 603 for some configurations
It is extremely important that you learn what field lines look like for some of the
most common charge configurations. You’ll need this qualitative picture in your
mind in order to calculate the electric flux (next section) and you’ll need this to
use Gauss’ Law – a much easier way to deal with continuous charge distributions.
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ELECTRIC FLUX
Image you’re holding a tennis racket on a windy day. How much air is
flowing through the web of the net? Well, it depends on three things:
1. The Size of the racket
2. The wind Speed
3. The racket’s Orientation
If the wind direction is normal (perpendicular) to the surface, it goes straight
through while if the racket is held so that the wind vector is parallel to the
surface, nothing will pass through but simply go across.
Recall from mechanics the area vector A whose magnitude varies directly
with size and which points in the direction perpendicular to the surface at
that point. Then for any sufficiently small (nearly flat) differential area
element dA , the flux (wind passing through it) can be defined by the
product of the wind speed vector v , the area element dA and the cosine of
the angle between the vectors. (Why?) But this is just the vector dot product
v • dA
Similarly, we can define the electric flux through any surface (flat or not) by
the number of electric field lines passing through it. We’ll assign the
capital Greek letter Phi to represent the flux and sum up all the differential
elements as an integral and write
Φ E = ∫ E • dA
Definition of Electric Flux
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GAUSS’S LAW
Consider some localized charge which may be either discrete or continuous and
imagine an arbitrarily shaped surface which completely surrounds it like a net
around a school of fish. This is called a Guassian Surface. Electric field lines due
to the enclosed charges may leave and/or enter the surface in some complicated
fashion. However, Gauss’s Law says that the total flux through the surface is
proportional to the total charge enclosed:
Φ E = ∫ E • dA =
QENCLOSED
ε0
Gauss’s Law
Now, in general, we don’t gain much as this is no easier to calculate than before.
But, in some cases, we can cut the problem down to size in no time.
™ The point is to SIMPLIFY the calculation
™ This requires a priori knowledge of the field lines
™ The problem requires symmetry to use Gauss’s Law effectively
™ So we must reduce the expression by specially constructing each
enclosing surface so that we always have:
• E normal to the surface which eliminates the Dot Product
θ = 0 → cos θ = 1 → E • dA = E dA cos θ = E dA
• E uniform over the entire surface collapses integral
∫ E dA = E∫ dA = E × Surface Area
• The entire expression then reduces to
E=
QENCLOSED
ε0 A
Pretty cool, huh?
BEWARE !!! This expression only works when the Gaussian Surface satisfies the
conditions above: E • A = EA and E is constant magnitude over the entire surface.
The following are the three geometries in which Gauss’s Law works immediately.
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Cylindrical Symmetry
For a uniform line charge – a wire or rod – the field lines point radially away
from the axis. We construct a hollow cylinder (picture an exhausted cardboard
toilet-paper roll) coaxially around the line charge which ensures that the field lines
are always perpendicular to the lateral surface of the cylinder. Since the charge
density is constant along the axis, the field can have no skew component and no
flux leaves through the ends. By the same symmetry, the electric field can vary
only with distance r from the axis so the electric field must be constant at the
surface.
L
r
end view
λ
This cylinder is our Gaussian Surface and we give it an arbitrary length L. The
lateral surface area is
A = Circumference × Length = 2π rL
and the total charge enclosed is its length times the linear charge density inside,
QENC = λ L
Hence, the simplified Gauss’s Law immediately gives us
E=
QENC
λL
1 ⎛ 2λ ⎞ 2 k λ
=
=
=
ε 0 A 2πε 0 rL 4πε 0 ⎜⎝ r ⎟⎠
r
where we have substituted the constant k using the identity k ≡
1
.
4πε 0
Note that the arbitrary length L we chose has canceled out of the expression.
Comments
This case applies to any charge distribution with cylindrical symmetry; a line, a
solid cylinder or even a hollow cylinder. At the radial distance r from the axis, the
lateral area used for the field calculation A = 2π rL will always be the same.
However, care must be taken when computing the enclosed charge. If the linear
density λ is not given, you must either derive it to apply the formula or else
compute it yourself from the given data. See the practice exam for examples.
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Spherical symmetry
This applies to any point charge or spherically symmetric charge ensemble. The
electric field points directly outward (or inward, if the net charge is negative) in all
directions isotropically. Hence, our Gaussian Surface should be a sphere whose
radius is equal to the distance from the center of the body to the point of interest.
This takes care of the two requirements that the Electric Field is uniform and the
field lines pass straight through the surface.
The surface area is A = 4π r 2 . Now we need only calculate the charge enclosed.
For a point charge, this is just Q. For a solid sphere, the total charge is
Q = ∫ ρ dV
For a uniform density, this is just the density times the volume, otherwise we have
to integrate. Note that we cannot have an arbitrary density function ρ ( r , θ , φ )
since this would violate the symmetry requirement. Hence only a radial
dependence is permissible ρ = ρ ( r ) .
If we are outside the entire body, we use the body’s total charge. However, if we’re
inside a solid charged non-conducting sphere, then the enclosed charge is only the
charge that lies inside our Gaussian Sphere.
Outside a hollow sphere, we integrate the charge only over the solid part. If we’re
inside the hollow part, no charge is enclosed, so the electric field must be zero.
See textbook for rectilinear symmetry – flat sheets. We’ll do this in class.
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