Transcript Slide 1

http://www.lab-initio.com (nz183.jpg)
Tickets on sale at the
Havener Center today.
The $4.00 price is
today only.
Every penny of the
proceeds goes to
programs serving
children in our
community.
Today’s lecture is
unofficially
brought to you by
The Museum of Electricity
Announcements
 Exam 1 is Tuesday, February 17, 5:00-6:00 pm.
Exam rooms (on next slide) will be posted on the Physics
2135 web site under “Course Information.”
 Wednesday of this week is deadline to submit the
appropriate memo or e-mail regarding an exam conflict. Follow
this web link for instructions on what to do.
 One of the homework problems for tomorrow is Special
Homework #4. You can find it on the web here.
 Physics 2135 Test Rooms, Spring 2015:
Instructor
Dr. Kurter
Dr. Madison
Dr. Parris
Mr. Upshaw
Dr. Waddill
Sections
F, H
E, G
K, M
A, C, J, L
B, D
Special Accommodations
Exam is from
5:00-6:00 pm!
Room
104 Physics
199 Toomey
125 Butler-Carlton (Civil Eng.)
G-3 Schrenk
G-31 EECH (Electrical Eng.)
Testing Center
Know the exam time!
Find your room ahead of time!
If at 5:00 on test day you are lost, go to 104 Physics and check the exam
room schedule, then go to the appropriate room and take the exam there.
More Announcements
 Exam 1 special arrangements:
Test Center students: you need to make an appointment
with Testing Center. By before now.
I will send you a confirming e-mail by the end of the
Tuesday before the exam. If you do not receive the e-mail,
you are NOT on the Test Center list!
More Announcements
 LEAD Schedule (lead.mst.edu)
Physics Learning Center, 2-4:30 pm and 6-8:30 pm Monday,
Wednesday, rooms 129/130 Physics.
Tutor available, 7-9 pm Monday through Thursday; check
the LEAD schedule.
Other LEAD Tutors available; check the schedule here. (You
can get one-on-one tutoring if you don’t like the group
environment of the Physics Learning Center).
Burns & McDonnell Student Success Center (B&MSSC)
Tutors available; check the schedule here. (Another place to
get one-on-one tutoring).
More Announcements
 If any of tomorrow’s homework problems involve a ring of
charge: there is no starting equation for V on the axis or at the
center of ring of charge. You must derive any equation you use.
 If any of tomorrow’s homework problems involve a
spherically-symmetric charge distribution: do not use the
starting equation for V due to a point charge. You must derive
any equation you use.
Today’s agenda:
Electric potential of a charge distribution.
You must be able to calculate the electric potential for a charge distribution.
Equipotentials.
You must be able to sketch and interpret equipotential plots.
Potential gradient.
You must be able to calculate the electric field if you are given the electric potential.
Potentials and fields near conductors.
You must be able to use what you have learned about electric fields, Gauss’ law, and
electric potential to understand and apply several useful facts about conductors in
electrostatic equilibrium.
Example 1: potential and electric field between two parallel
conducting plates.
Assume V0<V1 (so I have a direction to draw the electric field).
Also assume the plates are large compared to their separation,
so the electric field is constant and perpendicular to the plates.
Also, let the plates be separated
by a distance d.
E
V0
d
V1
V  V1  V0  
plate 1
plate 0
d is dx
Ed
x
V    E dx   E  dx  Ed
d
d
0
0
|V|=Ed
y
E
z
V0
dl
d
V1
The famous “Mr.
Ed equation!*”
V
E
, or V  Ed
d
I’ll discuss in lecture why the
absolute value signs are needed.
*2004, Prof. R. E. Olson.
Important note: the derivation of
V  Ed
did not require rectangular plates, or any plates at all. It works
as long as E is uniform and parallel or antiparallel to d.
If E is uniform but not parallel or antiparallel to d , then use.
V  E  d
Example 2: A rod* of length L located along the x-axis has a
total charge Q uniformly distributed along the rod. Find the
electric potential at a point P along the y-axis a distance d from
the origin.
(*“rod” means “really thin”)
dq
dx
* dV  k
k
r
x 2  d2
y
P
d
r
dq
dx
x
L
dq=dx
Q
x
=Q/L
L
V   dV
0
*What are we assuming when we use this equation?
V
L
0
dx
Q L dx
k
k 
2
2
L 0 x 2  d2
x d
y
A good set of math tables will
have the integral:
P
d
r
dq
dx
x
L
Q
x

dx
x d
2
2

 ln x  x 2  d 2
kQ  L  L2  d 2
V
ln 
L 
d




Include the sign of Q to get the correct sign for V.
What is the direction of V?

Example 3: Find the electric potential due to a uniformly
charged ring of radius R and total charge Q at a point P on the
axis of the ring.
dq
r
R
P
x
Every dq of charge on the
ring is the same distance
from the point P.
x
Q
dq
dq
dV  k
k
r
x2  R2
V
ring
dV  
ring
kdq
x2  R2

k
x2  R2

ring
dq
dq
r
R
P
x
x
Q
Homework hint: you
must derive this equation
in tomorrow’s homework!
V
V
k
x R
2
2
kQ
x2  R2

ring
dq
You must also derive an equation for the
potential at the center of a ring if you
need it for homework! In the next slide I
will show you how easy the derivation is.
Include the sign of Q to get the correct sign for V.
Could you use this expression for V to calculate E? Would you
get the same result as I got in Lecture 3?
Example 4: Find the electric potential at the center of a
uniformly charged ring of radius R and total charge Q.
dq
R
Every dq of charge on the
ring is the same distance
from the point P.
Q
dq
dq
dV  k
k
r
R
kdq k
kQ
V   dV  
  dq 
ring
ring R
R ring
R
Example 4: A disc of radius R has a uniform charge per unit
area  and total charge Q. Calculate V at a point P along the
central axis of the disc at a distance x from its center.
dq
dq
.
Start with dV  k
r
r  x2  r2
r′
P
R
Q
x
x
The disc is made of
concentric rings. The area
of a ring at a radius r′ is
2r′dr′, and the charge dq
on each ring is (2r′dr′),
where =Q/R2.
Each ring is a distance r  x2  r2 from point P.
dq
dq
dV  k
r
r  x  r
2
r′
2
P
R
x
x

k    2 r  dr   
x 2  r 2
Q
This is the (infinitesimal) potential
for an (infinitesimal) ring of radius r′.
On the next slide, just for kicks I’ll replace k by 1/40.
dq
r  x2  r2
r′
P
R
Q
x
x
1
V   dV 
ring
40

V
x 2  r 2
20
R
0


20

2rdr

ring


x 2  r2 20

Q
x R  x 
20 R 2
2
2

R
0

Q

R 2
rdr
x 2  r2
x2  R2  x

dq
r  x2  r2
r′
P
R
x
x
Q
Q
V
20 R 2

x2  R 2  x

Could you use this expression for V to calculate E? Would you
get the same result as I got in Lecture 3?
Example 5: calculate the potential at a point outside a very long
insulating cylinder of radius R and positive uniform linear charge
density .
dq
I would prefer to not start with dV  k
and integrate. Why?
r
Reason #1: a mathematical pain. What do I pick for my charge element
dq? Little cubes? Ugh. Circular rings that I have to integrate from 0 to R
and from - to  along the axis? Long cylindrical shells that I have to
integrate from 0 to R? There must be an easier way.
Reason #2: I suspect that we would eventually find (read your text) that
V= at any finite distance from the cylinder. Not very useful.
Reason #3: we have previously derived a simple expression for E
outside a cylinder of charge. This can easily be integrated to find V.
To be worked at the blackboard in lecture…
Things to note:
λ
R
V r 
ln .
2πε 0 r
V is zero at the surface of the cylinder
and decreases as you go further out. This
makes sense! V decreases as you move
away from positive charges.
λ
R
Vr  VR 
ln
2πε 0 r
If we tried to use V=0 at r= then

Vr  V 
ln   (V is infinite at any finite r).
2πε 0 r
λ
dq
That’s why we can’t start with dV  k .
r
Things to note:
λ
R
Vr  VR 
ln
2πε 0 r
For >0 and r>R, Vr – VR <0.
Our text’s convention is Vab = Va – Vb. This is explained on page
84. Thus VrR = Vr – VR is the potential difference between points
r and R and for r>R, VrR < 0.
In Physics 23, Vba = Va – Vb. I like the Physics 23 notation
because it clearly shows where you start and end. But Vab has
mathematical advantages which we will see in Chapter 4.
See your text for other examples of potentials calculated from
charge distributions, as well as an alternate discussion of the
electric field between charged parallel plates.
Remember: worked examples in the text are “testable.”
Make sure you know what Vab means, and how it relates to
V.
Vif = Vf – Vi so Vif = -Vif
Special Dispensation
For tomorrow’s homework only: you may use the equation for the
electric field of a long straight wire without first proving it:
Eline 

20 r
.
Of course, this is relevant only if a homework problem requires you
to know the electric field of a long straight wire.
You can also use this equation for the electric field outside a long
cylinder that carries charge.
Homework Hint!
Problems like 3.32 and 3.33: you must derive an expression
for the potential outside a long conducting cylinder. See
example 3.10. V is not zero at infinity in this case. Use
f
V   E  d .
i
If 3.32 and 3.33 are not assigned, don’t be disappointed. We
can still get you on this in problems in chapter 4!
Homework Hints!
In energy problems involving potentials, you may know the
potential but not details of the charge distribution that
produced it (or the charge distribution may be complex). In
that case, you don’t want to attempt to calculate potential
q1q 2
energy using U  k
. Instead, use U  q V .
r12
If the electric field is zero everywhere in some region, what
can you say about the potential in that region? Why?
PRACTICAL APPLICATION
For some reason you think practical applications are important.
Well, I found one!
The recipes on the Museum’s web site are sometimes slow to
load, so I have copied a couple here…
Charred Cheese Sandwich
Take two pieces of yummy white bread, and place a piece of good ol'
American cheese in between them.
Place the uncooked sandwich in the broilng & frying unit on your DL-1000
Electro Range (as yet unavailable).
Put on your fire retardant safety suit.
Make sure you have a fully charged fire extinguisher handy!
Set the cooking voltage to 19,000 volts at 3,600 watts, hit the timer switch
and dive for cover!
The recipes on the Museum’s web site are sometimes slow to
load, so I have copied a couple here…
Jacob’s Ladder Kebabs
This recipe will taste best if you can get the freshest ingredients. We use
red & green peppers, and Italian sausage. However feel free to add
mushrooms, onions, or any of your other favorites!
Cut your veggies and sausages into small cubes, and place them on the
optional Kebab-O-Matic cooker attachment on your DL-1000 Electro Range
(as yet unavailable).
Crank up the power and run like heck!
More Cooking with High Voltage
Application: Deep Space Propulsion Systems
Dr. Joshua Rovey, MAE Dept.
Today’s agenda:
Electric potential of a charge distribution.
You must be able to calculate the electric potential for a charge distribution.
Equipotentials.
You must be able to sketch and interpret equipotential plots.
Potential gradient.
You must be able to calculate the electric field if you are given the electric potential.
Potentials and fields near conductors.
You must be able to use what you have learned about electric fields, Gauss’ law, and
electric potential to understand and apply several useful facts about conductors in
electrostatic equilibrium.
Equipotentials
Equipotentials are contour maps of the electric potential.
http://www.omnimap.com/catalog/digital/topo.htm
Equipotential lines are another visualization tool. They
illustrate where the potential is constant. Equipotential lines
are actually projections on a 2-dimensional page of a 3dimensional equipotential surface. (“Just like” the contour
map.)
The electric field must be perpendicular to equipotential lines.
Why?
Otherwise work would be required to move a charge along an
equipotential surface, and it would not be equipotential.
In the static case (charges not moving) the surface of a
conductor is an equipotential surface. Why?
Otherwise charge would flow and it wouldn’t be a static case.
Here are electric field and equipotential lines for a dipole.
I’ll discuss in lecture some
implications this figure has
for charged particle motion.
Equipotential lines are shown in red.
“Toy”
http://www.falstad.com/vector2de/
Today’s agenda:
Electric potential of a charge distribution.
You must be able to calculate the electric potential for a charge distribution.
Equipotentials.
You must be able to sketch and interpret equipotential plots.
Potential gradient.
You must be able to calculate the electric field if you are given the electric potential.
Potentials and fields near conductors.
You must be able to use what you have learned about electric fields, Gauss’ law, and
electric potential to understand and apply several useful facts about conductors in
electrostatic equilibrium.
Potential Gradient
(Determining Electric Field from Potential)
The electric field vector points from higher to lower potentials.
More specifically, E points along shortest distance from a higher
equipotential surface to a lower equipotential surface.
You can use E to calculate V:
b
Vb  Va   E  d .
a
E
You can use the differential version of this equation to calculate
E from a known V:
dV  E  d  E d

dV
E 
d
For spherically symmetric charge distribution:
dV
Er  
dr
In one dimension:
dV
Ex  
dx
In three dimensions:
V
V
V
Ex  
, Ey  
, Ez  
.
x
y
z
V ˆ V ˆ V ˆ
or E  
i 
j 
k   V
x
y
z
dV
E 
d
dV
Er  
dr
V
V
V
Ex  
, Ey  
, Ez  
.
x
y
z
Calculate -dV/d(whatever) including all signs. If the result is
+, E points along the +(whatever) direction. If the result is -,
EE points along the –(whatever) direction.
Example (from a Fall 2006 exam problem): In a region of
space, the electric potential is V(x,y,z) = Axy2 + Bx2 + Cx,
where A = 50 V/m3, B = 100 V/m2, and C = -400 V/m are
constants. Find the electric field at the origin
V
E x (0,0,0)  
   Ay2  2Bx  C
 C
(0,0,0)
x (0,0,0)
V
E y (0,0,0)  
 (2Axy) (0,0,0)  0
y (0,0,0)
V
E z (0,0,0)  
0
z (0,0,0)
V

E(0,0,0)   400  ˆi
m

Today’s agenda:
Electric potential of a charge distribution.
You must be able to calculate the electric potential for a charge distribution.
Equipotentials.
You must be able to sketch and interpret equipotential plots.
Potential gradient.
You must be able to calculate the electric field if you are given the electric potential.
Potentials and fields near conductors.
You must be able to use what you have learned about electric fields, Gauss’ law, and
electric potential to understand and apply several useful facts about conductors in
electrostatic equilibrium.
Potentials and Fields Near Conductors
When there is a net flow of charge inside a conductor, the
physics is generally complex.
When there is no net flow of charge, or no flow at all (the
electrostatic case), then a number of conclusions can be
reached using Gauss’ Law and the concepts of electric fields
and potentials…
Summary of key points (electrostatic case):
The electric field inside a conductor is zero.
Any net charge on the conductor lies on the outer surface.
The potential on the surface of a conductor, and everywhere
inside, is the same.
The electric field just outside a conductor must be
perpendicular to the surface.
Equipotential surfaces just outside the conductor must be
parallel to the conductor’s surface.
Another key point: the charge density on a conductor surface
will vary if the surface is irregular, and surface charge collects at
“sharp points.”
Therefore the electric field is large (and can be huge) near
“sharp points.”
Another Practical Application
To best shock somebody, don’t touch them with your hand;
touch them with your fingertip.
Better yet, hold a small piece of bare wire
in your hand and gently touch them with
that.