Transcript Document

V
Q
4pe0 R
Q
4pe0 r
R
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R
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Announcements
• No lecture September 26th. Everything else goes on as usual
– 3rd workshops September 26-30th
– Preflight #7 due October 1st, 7am
• Review Session: October 5th 3-5pm
– you chose it over office hours! Room TBA
– This time is purposely long to allow students to attend and leave to
accommodate workshops
– I will try to find an alternate time and home for the 3-5pm workshop
that Friday (Brian’s group, B&L 269)
• Homework #2 due September 26th at 9am
• 1st exam is Tuesday, October 9th, 8-9:30am Hubbell Aud.
– Covers material of Lecture 1-6, Workshop 1-3, Homework 1-3
– Question: “Practice exams”?
• I could post an old exam I’ve given, BUT the problem is I don’t have one
that overlaps exactly this material. Let me see what I can do…
– Question: “can we bring in a ‘cheat sheet’ with formulas?”
• That’s a reasonable request. One (1) 8.5”x11” sheet of paper with
whatever you like written on both sides. Don’t abuse it, please.
Your Questions and Comments…
• “Lolrus http://i19.tinypic.com/35dbvxj.jpg”
– No, this is not what Prof. McFarland goes to Japan to study
– Neutrinos are much smaller in size, less massive, perhaps less cute
– Shouldn’t that be “I am the lolrus? Goo goo ga-joob” anyway?
Your Questions and Comments…
• “At the end of a lecture, could you have a slide with all the
equations we learned? It would be a helpful reference.”
– Mmm, is learning in really done terms of equations
– There is a relatively accurate concepts summary at the end
• Prof. McFarland wlil try to make sure he doesn’t miss the “math
equivalent” when relevant
• “what is a partial derivative??? ”
– Hopefuly the “” means you have encountered this in 143/162?
– Short answer: a derivative with respect to one variable while holding
other variables constant
– We will make occasional use of this in a few places in the class
• If it’s causing problems for many of you, please let Prof. McF know
• You have some examples of calculating electric field from potential in this
lecture, workshop, homework. Practice these. Ask a TA, Prof.
• “[on preflight] [Detailed description of your frustrating concern]”
– Sometimes it’s hard to help without a one on one conversation
– Please try e-mail (all you have this week ) or face-to-face
– Prof. McFarland says, “I want to try to help, and it’s OK to talk to me
about concerns over occurrences in workshop, lecture, etc.”
Last time…
• The electrostatic force is the essence
of right-thinking conservatism
– conservation of energy makes
electrical potential a useful concept
• If we know the electric field E,
•Potential due to point charges
•Physically, DV is what counts
•The place where V=0 is “arbitrary” (at infinity)
Reading Assignment: Chapter 23.5-8
examples: 23.10,12
2
Lecture 5, Concept 2
• A single charge ( Q = -1mC) is fixed at
the origin. Define point A at x = + 5m
and point B at x = +2m.
B
´
What did
you say on
Wednesday?
(2) VAB = 0
´
-1mC
– What is the sign of the potential difference
between A and B? (VAB  VB - VA )
(1) VAB < 0
A
(3) VAB > 0
Maybe it’s a hard problem!
Or, maybe one or two of you
may have been tempted to
not answer carefully on your
way out the door… 
x
Lecture 5, Concept 2
• A single charge ( Q = -1mC) is fixed at
the origin. Define point A at x = + 5m
and point B at x = +2m.
B
´
– What is the sign of the potential difference
between A and B? (VAB  VB - VA )
(1) VAB < 0
(2) VAB = 0
A
´
-1mC
(3) VAB > 0
•The simplest way to get the sign of the potential difference is to
imagine placing a positive charge at point A and determining whether
positive or negative work would be done in moving the charge to point B.
•A positive charge at A would be attracted to the -1mC charge;
therefore NEGATIVE work would be done to move the charge from
A to B.
•You can also determine the sign directly from the definition:
Since
, VAB <0 !!
x
Today’s Lecture…
• Equipotentials and conductors
• E from V
• Calculate electric field of dipole from potential
• Electric Potential Energy
– of charge in external electric field
– later: stored in the electric field itself
• Appendices:
• example calculations of a spherical charge configuration
Text Reference: Chapter 23.5,8
examples: 23.10,12
Equipotentials of a charged sphere
Er
Equipotential
•
•
•
•
The electric field of the charged sphere has spherical symmetry.
The potential depends only on the distance from the center of the
sphere, as is expected from spherical symmetry.
Therefore, the potential is constant along a sphere which is
concentric with the point charge. These surfaces are called
equipotentials.
Notice that the electric field is perpendicular to the equipotential
surface at all points.
Equipotentials
Defined as: The locus of points with the same potential.
•
Example: as we saw, for a spherically symmetric charge distribution, the
equipotentials are spheres centered on the charge.
• GENERAL PROPERTY:
– The electric field is always perpendicular to an equipotential
surface.
• Why??
Along an equipotential surface, there is no change in V.
 
If E  dl
is zero, then this condition is satisfied.
In words, if the dot product of the field vector and the displacement vector
is zero, then these two vectors are perpendicular and there is no potential
difference along this path.
If the electric field is perpendicular to a surface everywhere, the surface is
an equipotential.
Conductors
+
+
+
+
+
+
+
+
+
+
+
+
+
+
• Claim
The surface of a conductor is always an equipotential
surface (in fact, the entire conductor is an equipotential)
• Why??
If surface were not equipotential, there would be an electric
field component parallel to the surface and the charges
would move!!
Charge on Conductors?
• How is charge distributed on the surface of a conductor?
– KEY: Must produce E=0 inside the conductor and E perependicular
(normal) to the surface .
Spherical example (with little off-center charge):
+ + +
+
- -- +
- +
+ -+q - +
+ - +
+ - +
+
+ + +
E=0 inside conducting shell.
charge density induced on
inner surface non-uniform.
charge density induced on
outer surface uniform
E outside has spherical
symmetry centered on spherical
conducting shell.
2
PF06 Question #2:
Two spherical conductors are separated by a large distance.
They each carry the same positive charge Q. Conductor A has a
larger radius than conductor B.
A
B
Compare the potential at the surface of conductor A with
the potential at the surface of conductor B.
a) VA > VB
b) VA = VB
60%
59%
50%
40%
30%
20%
10%
0%
19%
22%
c) VA < VB
PF06:
Question #3-4
B
A
The two conductors are now connected by a wire. How do the
potentials at the conductor surfaces compare now ?
100%
80%
a) VA > VB
b) VA = VB
c) VA < VB
78%
60%
40%
20%
11%
0%
What happens to the charge on conductor A after it is
connected to conductor B ?
60%
50%
a) QA increases
b) QA decreases
c) QA doesn’t change
40%
49%
30%
34%
20%
10%
0%
18%
11%
What you said …
+
+
-∫E• dl
B
+
+
+
A
B
+
+
Right:
+
+
V=kQ/r, so the sphere with the small radius will have the greater potential
VB>VA since took more work to get closer to center when starting from infinity.
The greater surface area of A allows the charge to be distributed more sparesly across the
sphere's surface area. Therefore, the potential at the surface of A would be less. [True because
it would take less work to add a positive test charge from infinity to the larger sphere]
Now that they are connected the charge would disperse again until electrostatic equilibrium
and the potential would be equal
Because potential is proportional to charge, the potential of A must increase so that it is equal
to the potential of B. Since the potential increases, the charge must also increase.
+
Wrong:
dl is larger in sphere A than sphere B. [yes -E*dl>0, but path comes from infinity]
The charge is equal. However, the potential on the first is larger because the radius is larger.
Because V is proportional to Q/r, with Q the same for both, and r, the distance away from the
surface the same for both, Q/r is the same, thus V is the same [in “kQ/r” (potential of point
charge) r is the distance from charge. here potential at surface is related to radius of sphere.]
When the spheres are connected, the potential equalizes. However, the charge doesnt
change. The charge continues to be equal [but charge must be changing to equalize potential]
Adding a wire between the charged spheres would only decrease the overall charge on both
spheres because they would evenly distribute their charge onto the wire. The larger end of this
system, A, would still have greater potential than the smaller side, B [charge need not be even]
when they are connected by a wire, the charges spread out and distribute themselves evenly
[conductors need only be at same potential, not same charge] throughout the two spheres and
the wire. the potential is the same all over the place now. because they are connected, some
charge from A goes into the wire, so it decreases [good point! but remember the wire is a very
small thing, so at the sphere’s potential, it would hold very little charge. the bigger effect is the
charge flowing from B to A, so the net effect is an increase for A]
Lecture 6, Concept 1
A point charge Q is fixed at the
center of an uncharged conducting
spherical shell of inner radius a
and outer radius b.
– What is the value of the potential Va at
the inner surface of the spherical shell?
(1)
(2)
(3)
Hint: remember how potential is defined!
a
Q
b
Eout
Lecture 6, Concept 1
A point charge Q is fixed at the
center of an uncharged conducting
spherical shell of inner radius a
and outer radius b.
– What is the value of the potential Va at
the inner surface of the spherical shell?
(1)
(2)
a
Q
b
(3)
• How to start?? The only thing we know about the potential is
its definition:
• To calculate Va, we need to know the electric field E
• Outside the spherical shell:
• Apply Gauss’ Law to sphere:
• Inside the spherical shell:
E=0
Charge on Conductor Example
• 2 spheres at same potential (wired).
• How are charge, potential, surface field related?
• Which one has biggest charge?
Potentials
equal, so…
• Is surface field proportional to
charge or charge density?
r
Density:
rS

L
Smaller sphere
has the larger
surface charge
density and
local field!
Electric Potential Energy:
Coming Full Circle
• The Coulomb force is a CONSERVATIVE force
(i.e. the work done by it on a particle which
moves around a closed path returning to its
initial position is ZERO.)
• Therefore, a particle moving under the influence of
the Coulomb force is said to have an electric
potential energy defined by:
this “q” is the ‘test
charge” in other
examples...
• The total energy (kinetic + electric potential) is then
conserved for a charged particle moving under the
influence of the Coulomb force.
PF06: Question #6
A
E
C
B
If a negative charge is moved from point A to point C, its electric
potential energy
a) increases
b) decreases
c) doesn’t change
70%
60%
50%
61%
40%
30%
20%
31%
10%
0%
8%
What you said …
My answer…
and
which is negative
since…
Right:
A
E
C
B
Work must be done to get the charge to C from A because it is opposing the electric
field. (A positive charge at A would flow freely towards C)
a neg charge has its greatest potential energy at pt c which is closer to a negative plate [it’s OK
to think of the charge distribution causing the field, as long as you know what it is]
Potential is lower at C from the last preflight. U=q x potential so U is bigger at C
Wrong:
Since E points right, electric potential energy decreases towards the right [The electric potential
decreases as you move to the right. But the potential energy of a negative charge increases.]
The electric field is the same at both spots, so there is no change according to the formula for
potential. [Integrate electric field to get a change in potential, constant E field <> constant V]
Potential Confusion:
Potential energy, potential and potential energy
• Potential energy stored in a static charge
distribution
– work we do to assemble the charges.
More on this later…
• Electrostatic potential at any point in space
– sources give rise to V(r)
• Electric potential energy of a charge in the
presence of a set of source charges
– potential energy of the test charge equals the
potential from the sources times the test charge
Energy Units
MKS:
U = QV
1 Coulomb-Volt
= 1 Joule
elementary particles (e.g. e, p) good unit is “electron-Volt”
qe=1.6x10-19 Coulomb, so
1 eV = 1.6x10-19 J
Accelerators
• Electrostatic: Van de Graaff
protons  1 MeV ( 106 eV)
• Electromagnetic: Fermilab
protons  1TeV ( 1012 eV)
1 mile
E from V?
• We can obtain the electric field E from the
potential V by inverting ourprevious
relation

between E and V: dV   E  dl

r

r  xˆ dx
V
V+dV

dV   E  xˆ dx   E x dx
• Expressed as a vector, E is the negative gradient of V
• Cartesian coordinates:
• Spherical coordinates:
PF06: Question #8
This graph shows the electric
potential at various points
along the x-axis.
At which point(s) is the electric field zero?
60%
A
B
C
D
50%
40%
55%
53%
45%
30%
20%
10%
0%
2%
What you said …
my answer:
the gradient is a derivative w/r to
position. -dV/dx is 0 only at B.
Right:
E is the [negative] derivative of V, so when the slope of V(x) is zero, E is zero
the electric field is zero at point B because the electric potential is not changing
delta V = -Int(E*dl), implies E is the derivative of V
Wrong:
THe potntial is 0 when E is 0, and visa versa
if there is negative potential there is a field of 0. [Potential is not zero to left and right of B]
Since the potential is the integral of the E field, the point where the net area is zero means the field
is zero [potential is graphed here and question asks about field, not the reverse.]
the book, it is given that E=V/d. Since d is never zero, when V is zero, E will also be zero
[but does this formula apply here? this is a special case of a uniform field and the potential difference
across a distance d along the field lines.]
E from V: an Example
• Consider the following electric potential:
• What electric field does this describe?
Cartesian coordinates:
... expressing this as
a vector:
The Bottom Line
If we know the electric field E everywhere,

allows us to calculate the potential function V everywhere (keep
in mind, we often define VA = 0 at some convenient place)
If we know the potential function V everywhere,
allows us to calculate the electric field E everywhere
• Units for Potential! 1 Joule/Coul = 1 VOLT
2
Lecture 6, Concept 2
The electric potential in a region of space is given by
The x-component of the electric field Ex at x = 2 is
(1)Ex=0
(2)Ex>0
(3)Ex<0
Lecture 6, Concept 2
The electric potential in a region of space is given by
The x-component of the electric field Ex at x = 2 is
(1)Ex=0
(2)Ex>0
(3)Ex<0
We know V(x) “everywhere”. To obtain Ex, use
Electric Dipole
The potential is much easier to
calculate than the field since it is
an algebraic sum of 2 scalar
terms.
z
r1
+q
a
r
q
a
r2-r1
-q
• Rewrite this for special case r>>a:
the dipole
moment

Now we can use this potential to calculate the E
field of a dipole!
(remember how messy the direct calculation was?)
r2
Electric Dipole
z
+q
aq
a
-q
• Calculate E in spherical coordinates:
the dipole
moment

r1
r
r2
Dipole Field
y=
z
+q
a q
a
Etot
r
E
q
0
Er
-q
0
p/
p

p
p/
x=
Summary
• Conductors are equipotentials
• Potential Energy
U  qV
• Can get from electric field to potential and potential
to field from:
Reading Assignment: 23.9, 24.1-4
Examples: 24.1,3,7,8
Appendix
Calculate the potential V(r)
at the point shown (r<a)
I
II
uncharged
conductor
III
IV
r a
solid sphere
with total
charge Q
c
b
Calculating Electric Potentials
Calculate the potential V(r)
at the point shown (r < a)
• Where do we know the potential,
and where do we need to know it?
V=0 at r =  ...
III
IV
uncharged
conductor
r a
we need r < a ...
• Determine E(r) for all regions
in between these two points
I
II
c
b
solid sphere
with total
charge Q
• Determine DV for each region by integration
... and so on ...
• Check the sign of each potential difference DV
DV > 0 means we went “uphill”
DV < 0 means we went “downhill”
(from the point of view
of a positive charge)
Calculating Electric Potentials
Calculate the potential V(r)
at the point shown (r < a)
• Look at first term:
III
IV
• Line integral from infinity to “c” has to
be positive, pushing against a force:
Line integral is going “in”
which is just the opposite of
what usually is done
- controlled by limits
• What’s left?
I
II

dl
r a

E
c
1  Q
1 Q



4p eo  r    4p eo c
c
b
Calculating Electric Potentials
I
II
Calculate the potential V(r)
at the point shown (r < a)
• Look at third term:

dl
III
IV

E
• Line integral from “b” to “a”, again has
to be positive, pushing against a force:
Line integral is going “in”
which is just the opposite
of what usually is done
a
1  Q
1 Q(b  a )



4p eo  r   b 4p eo
ab
- controlled by limits
• What’s left?
Previous slide
we have calculated this already
r a
c
b
Calculating Electric Potentials
I
II
Calculate the potential V(r)
at the point shown (r < a)
• Look at last term:

dl
III
IV

E
r a
c
b
• Line integral from “a” to “r”, again has
to be positive, pushing against a force.
• But this time the force doesn’t vary the
same way, since “r ’ ” determines the
amount of source charge
Q
r3
a
3
Q
This is the charge
that is inside “r”
and sources field
• What’s left to do?
• ADD THEM ALL UP!
• Sum the potentials
r
1  Q r  2 
1 Q 
r 2 

1 2
 
3


4p eo  a 2 
4p eo 2a  a 
a
Calculating Electric Potentials
I
II
Calculate the potential V(r)
at the point shown (r < a)
• Add up the terms from I, III and IV:
I
III

dl
III
IV

E
r a
c
b
IV

dl
Potential increase
from moving into
the sphere
The potential difference
from infinity to “a” if
the conducting shell
weren’t there
An adjustment to
account for the fact
that the conductor
is an equipotential,
DV= 0 from c -> b
Calculating Electric Potentials
Summary
The potential as a function of r for all 4
regions is:
I
r > c:
II
b < r < c:
III
a < r < b:
IV
r < a:
I
II
III
IV
r a
c
b