Transcript Document
adapted from http://www.lab-initio.com/
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Announcements
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Announcements
Exam 1 is Tuesday, February 17, from 5:00-6:00 PM. According
to the Student Academic Regulations “The period from 5:00 –
6:00 PM daily [is] to be designated for common exams. If a
class or other required academic activity is scheduled during
common exam time, the instructor of the class that conflicts
with the common exam will provide accommodations for the
students taking the common exam.”
E-mail me by the end of Friday of next week if for some reason
you have a non-class-related time conflict for Exam 1 (5:006:00 pm, Tuesday, February 17). Include in your email your
recitation section letter and the details of your circumstance.
(That doesn’t mean I accept responsibility for fixing the
conflict.)
Announcements
Exam 1 is Tuesday, February 17, from 5:00-6:00 PM.
The spring Career Fair is also February 17. The exam starts well
after scheduled Career Fair activities have ended.
If you have an interview or MUST attend a Career Fair activity
during your recitation time on February 17, contact your
recitation instructor. In such a case, you may wish to attend a
recitation at a time other than your scheduled time.
It is unlikely that Career Fair activities will conflict with the 5-6
PM exam time, but if there is such a conflict, e-mail me the
details as soon as you find out about it.
Announcements
Homework for tomorrow:
1: 53, 90ab, 91a, 98 (also express the force in unit vector
notation), 100 (reminder: all solutions must begin with
starting equations)
You only need to do part a of problem 91.
You don’t need to be paranoid about using OSE’s, but don’t
start a problem with a “random” equation from your book.
Note the change.
Announcements
If a homework problem asks for the magnitude and direction of
a vector, we are generally satisfied if you provide its
components, or express it in unit vector notation.
You might want to be paranoid and verify beforehand that this
is OK.
Announcements
Labs begin this week.
If you have lab this week, don’t miss it!
3L01 is “odd” 3L02 is “even”
“Odd” labs meet this week (3L01, 3L03, 3L05, etc.). “Even” labs
meet next week.
You will need to purchase a lab manual. Go to the department
office, room 102, to purchase your lab manual! The cost is
$25.00. Do not ask me for a lab manual; I do not have them!
http://campus.mst.edu/physics/courses/24lab/
Announcements
For those who want to use Mastering Physics for practice, I “assigned”
all of the problems from the chapters we will study this semester.
Mastering Physics is OPTIONAL. You will get no course points for using
it. You will not lose course points for not using it.
Instructions for signing on are here:
http://campus.mst.edu/physics/courses/24/CourseInformation/Mastering
_Physics_for_Physics_2135.pdf.
Once you open your Mastering Physics packet, you won’t be able to resell your text.
Mastering Physics is based on the full text. Use the table here to see
which full text chapters correspond to our text:
http://campus.mst.edu/physics/courses/24/CourseInformation/physics_2
135_textbook.htm.
Announcements
There will be occasional errors in the lecture videos. After an
error is discovered, it will be corrected or pointed out with a
“callout” box in the video. It takes me about 15 hours to record
and edit an hour’s worth of lecture video, so “polished”
corrections will likely not be made until Summer 2015.
The Big Picture, Part I
In Lecture 1 you learned Coulomb's Law:
q1q 2
1
F =
,
2
12 4πε 0 r12
r12
+
-
Q1
Q2
Coulomb’s Law quantifies the force between charged particles.
The Big Picture, Part II
In Lecture 1 you also learned about the electric field.
There were two kinds of problems you had to solve…
The Big Picture, Part IIa
1. Given an electric field, calculate the force on a charged
particle.
F
E=
q
F= qE
-
F
E
You may not be given any information about
where this electric field “comes from.”
Or: given the force on a charged particle, determine the electric field that caused the force.
The Big Picture, Part IIb
2. Given one or more charged particles, calculate the electric
field they produce. We’ll focus on this topic today.
q
E=k 2
r
q
E=k 2 rˆ
r
I strongly recommend you
start with this and do x and y
components separately.
Avoid for now.
These are examples of “legal” equations:
F = qE
F
E=
q
F= q E
F
E=
q
F = q E
F
F
q =
E E
E
E
These are INCORRECT: q =
=
F
F
(do you know why?)
q = E/F works *IF* q is positive. It is safer to write |q| = E/F.
Today’s agenda:
The electric field due to a charge distribution.
You must be able to calculate electric field of a continuous distribution of charge.
The Electric Field
Due to a Continuous Charge Distribution
(worked examples)
finite line of charge
general derivation: http://www.youtube.com/watch?v=WmZ3G2DWHlg
ring of charge
disc of charge
infinite sheet of charge
infinite line of charge
semicircle of charge
Instead of talking about electric fields of charge distributions,
let’s work some examples. We’ll start with a “line” of charge.
Example: A rod* of length L has a uniformly distributed total
positive charge Q. Calculate the electric field at a point P
located a distance d below the rod, along an axis through the
left end of the rod and perpendicular to the rod.
Example: A rod* of length L has a uniformly distributed total
negative charge -Q. Calculate the electric field at a point P
located a distance d below the rod, along an axis through the
center of and perpendicular to the rod.
I will work one of the above examples at the board in lecture.
You should try the other for yourself.
*Assume the rod has negligible thickness.
Example: A rod of length L has a uniform charge per unit length
and a total positive charge Q. Calculate the electric field at a
point P along the axis of the rod a distance d from one end.*
P
d
L
To be worked at the board in lecture…
*Assume the rod has negligible thickness.
Example: calculate the electric field due to an infinite line of
positive charge.
There are two approaches to the mathematics of this problem.
One approach is that of example 1.10. See notes here. An
alternative mathematical approach is posted here. The result is
2k
E
20 r
r
This is not an “official”
starting equation!
The above equation is not on your OSE sheet. In general, you
may not use it as a starting equation!
If a homework problem has an infinite line of charge, you would
need to repeat the derivation, unless I give you permission to
use it.
Example: A ring of radius a has a uniform charge per unit
length and a total positive charge Q. Calculate the electric field
at a point P along the axis of the ring at a distance x0 from its
center.
To be worked at the
blackboard in lecture.
P
x
x0
Homework hint: you must provide this derivation in your solution to any
problems about rings of charge (e.g. 1.53 or 1.55, if assigned).
Visualization here (requires Shockwave, which downloads automatically):
http://web.mit.edu/viz/EM/visualizations/electrostatics/calculatingElectricFields/RingIntegration/RingIntegration.htm
y
Ex
a
P
x0
x
x
kx 0 Q
2
0
a
2 3/2
E y Ez 0
E
x0 is positive!
Also “legal” answers:
E
kx 0 Q
2
2
x
a
0
ˆi
3/2
E
kx 0 Q
2
x
0 a
2 3/2
, away from the center
These equations are only valid for P along the positive x-axis!
Awesome Youtube Derivation: http://www.youtube.com/watch?v=80mM3kSTZcE
(he leaves out a factor of a in several steps, but finds it in the end).
What would be different if
Q were negative? If P were
on the negative x-axis?
Example: A disc of radius R has a uniform charge per unit area
. Calculate the electric field at a point P along the central axis
of the disc at a distance x0 from its center.
P
R
x0
x
Example: A disc of radius R has a uniform charge per unit area
. Calculate the electric field at a point P along the central axis
of the disc at a distance x0 from its center.
The disc is made of
concentric rings.
r
P
R
x
x0
2r
dr
The ring has infinitesimal thickness, so you can imagine it as
a rectangular strip.
Imagine taking a ring
and cutting it so you can
lay it out along a line.
The length is 2r, the
thickness is dr, so the
area of a ring at a radius
r is 2rdr.
Caution! In the previous example, the radius of the ring was R. Here the radius of the disc is
R, and the rings it is made of have (variable) radius r.
Example: A disc of radius R has a uniform charge per unit area
. Calculate the electric field at a point P along the central axis
of the disc at a distance x0 from its center.
dq
The charge on each
ring is dq = (2rdr).
Let’s assume is positive
so dq is positive.
r
charge on ring = charge per area ×area
P
R
x0
x
dEring
We previously derived an equation for
the electric field of this ring. We’ll call
it dEring here, because the ring is an
infinitesimal part of the entire disc.
dE ring
kx 0 dq ring
x
2
0
r
2 3/2
Example: A disc of radius R has a uniform charge per unit area
. Calculate the electric field at a point P along the central axis
of the disc at a distance x0 from its center.
Let’s assume is positive
so dq is positive.
dq
dE ring
r
P
x0
R
E disc
dE
disc
ring
disc
kx 0 2rdr
x
x
2
0
r
2 3/2
x
dEring
kx 0 dq ring
2
0
r
2 3/2
kx 0
R
0
x
kx 0 (2rdr )
x
2
0
r
2r dr
2
0
r
2 3/2
2 3/2
Let’s assume is positive
so dq is positive.
dq
r
P
R
x0
x
dEring
E disc kx 0
R
0
x
2r dr
2
0
r
2 3/2
x2 r
0
kx 0
1/ 2
2 1/2
E disc
You know how to integrate
this. The integrand is just
(stuff)-3/2 d(stuff)
R
x0
2k x 0
x 0 x 2 R 2 1/2
0
0
Kind of nasty looking, isn’t it.
P
R
x0
x
Edisc
x0
x0
E x 2k
x 0 x 2 R 2 1/2
0
As usual, there are
several ways to write
the answer.
E y Ez 0
x
x
0
ˆi
E 2k 0
x 0 x 2 R 2 1/2
0
Or you could give the magnitude and direction.
Example: Calculate the electric field at a distance x0 from an
infinite plane sheet with a uniform charge density .
An infinite sheet is “the same as” disc of infinite radius.
Esheet
x
x
0
lim 2k 0
R
x 0 x 2 R 2 1/2
0
1
Take the limit and use k
to get
40
Esheet
20
.
This is the magnitude of E. The direction is
away from a positively-charged sheet, or
towards a negatively-charged sheet.
Example: Calculate the electric field at a distance x0 from an
infinite plane sheet with a uniform charge density .
Esheet
20
.
Interesting...does not depend on distance from the sheet. Does
that make sense?
This is your fourth Official Starting Equation, and the only one
from all of today’s lecture!
I’ve been Really Nice and put this on your starting equation sheet. You don’t have to
derive it for your homework!
Example: calculate the electric field at “center” of semicircular
line of uniformly-distributed positive charge, oriented as shown.
+Q
R
y
You don’t have to follow the steps in the exact
order I present here. Just let the problem tell you
what to. You may do things in a different order;
that’s probably OK.
x
ds
To be worked at the
blackboard in lecture.
d
R
dE
Help!
We covered a lot of material in a brief time.
If you want to explore a slightly different presentation of this at
your leisure, try the MIT Open Courseware site.
http://ocw.mit.edu/courses/physics/8-02sc-physics-ii-electricity-andmagnetism-fall-2010/index.htm E&M main page
http://www.youtube.com/watch?v=OsWDUqJQcpk&feature=relmfu lecture
on electric fields
http://ocw.mit.edu/courses/physics/8-02sc-physics-ii-electricity-andmagnetism-fall-2010/electric-fields/ all course material on electric fields
Homework Hints (may not apply every semester)
Your starting equations so far are:
q1q 2
F k 2
12
r12
F0
E=
q0
q
E=k 2
r
Esheet
20
.
(plus Physics 23 starting equations).
This is a “legal variation” (use it for charge distributions):
dq
dE=k 2 .
r
You can remove the absolute value signs if you know that dq is
always positive.
Homework Hints (may not apply every semester)
Suppose you have to evaluate this integral in your homework…
a
dx
a r x
0
2
Let u=(a+r-x)2. Then du=-dx and u=a+r when x=0, u=r when
x=a.
Or look it up if you have tables that contain it.
Homework Hints (may not apply every semester)
The integrals below are on page 522 of your text.
dx
x
2
a
2
3
2
x dx
x
2
a
2
3
2
Your recitation instructor will supply you with needed integrals.
The above integrals may or may not be needed this semester.
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