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What are some physics “things” you believe in?*
 Newton’s Laws (classical mechanics)
 conservation of momentum (linear and angular)
 conservation of energy
 laws of thermodynamics
 Maxwell’s equations (E&M theory)
These are the “really big” ideas of classical physics.
*Note: ”believe in” implies more commitment than just “believe” by itself!
Electric Potential and Potential Difference
Energy calculations in our Mechanics segment let us solve
problems that would have been extremely complex had we
tried to solve them using kinematics.
Similarly, energy calculations are going to let us solve complex
E&M problems.
A bit of review:
Consider an object of mass m in a gravitational field. It has
potential energy U(y) = mgy and “feels a gravitational force FG
= GmM/r2, attractive.
y
If released, it gains kinetic
energy and loses potential
energy, but mechanical energy
is conserved: Ef=Ei. The
change in potential energy is
Uf - Ui = -(Wc)if.
Ui = mgyi
yi
x
Uf = 0
What force does Wc? Force due to gravity.
graphic “borrowed” from http://csep10.phys.utk.edu/astr161/lect/history/newtongrav.html
A charged particle in an electric
field has electrical potential
energy.
We’ll write down an equation
later.
It “feels” a force (as given by
Coulomb’s law).
It gains kinetic energy and loses
potential energy if released, and
its mechanical energy is
conserved.
++++++++++++++
+
E
-------------------
We define “electric potential*” or just “potential” per unit
charge at some point a by
OSE: Va = (PE)a /q.
Units for V to be
defined later.
Note that the sign on the charge matters!
This equation is a definition. That’s why it came out of
nowhere. To really make it make sense, we have to define an
electric potential energy.
*Electric potential is not potential energy (we haven’t defined electric potential
energy yet). In Mechanics, we didn’t define a “gravitational potential” like this
because there are not two kinds of gravitational “charge.”
If a charged particle moves (changes position) its electric
potential may change.
We need to be careful about the notation we introduce to
indicate change.
Our textbook author (College Physics) sometimes uses Vab to
mean “the difference when you go from point b to point a.”
And sometimes he uses Vba to mean the same difference.
Very confusing.
To avoid confusion, we’ll use an arrow to show which way
we’re going. A bit of extra work, but worth it. The change,
Vif , means Vf – Vi. Now we are ready for another equation,
which it this point looks like a definition.
You can derive an equation, so you see where it comes from, or you can define something
with an equation. In the latter case, the equation may seem to “come from nowhere.”
It may take work to move a charged particle from point a to
point b. The change in potential is related to the work by
Va – Vb = -(Wba)/q. This gives us another OSE:
OSE:
Wif = q Vif
Vif is potential
difference
What force does Wif ? The electric force, as described by
Coulomb’s law.
The unit of both potential and potential difference is the volt,
abbreviated with a capital V, and is equal to 1J/1C.
A positive plate is at a higher potential than a negative plate,
so a positive charge will move from a place of high potential
to a place of low potential. (What will a negative charge do?)
Potential difference is often called voltage, and is measured in
volts.
Anybody confused yet?
If the answer was “no,” you are too confused to realize you are
confused.
Electric potential:
(also called potential; unit is volt)
Va = (PE)a /q
Electric potential energy:
(unit has to be joules)
not defined yet
Electric potential difference:
(also called potential difference; unit is volt)
(also called voltage)
Vif = Wif /q
The equations for potential and potential difference had better be consistent!
Actually, this isn’t so bad, except several confusingly similar
terms have been introduced at once.
What about consistency (with potential energy in Mechanics)?
Va q = PEa
Va – Vb = -(Wba)/q
Va q – Vb q = -Wba
PEa – PEb = -Wba
The last equation is equivalent to our Mechanics equation
Uf - Ui = -(Wc)if
Looks consistent (with our previous ideas of potential energy) to me.
Comment: two definitions
Va = (PE)a /q
Vif = Wif /q
give us an implicit definition of electrical potential energy:
PEba = -Wba
This is precisely the way we defined potential energy in
Mechanics.
This is what I mean by “the equations for potential difference
had better be consistent (with our equations from
mechanics).”
We can use Coulomb’s law or the definition of the electric field
to calculate the work done in moving a charge from b to a.
All is well… except I need to find a more concise way to
introduce this, and why did we have to go from b to a?
We only define differences in potential, so we must choose a
reference position for where V is zero.
In the process, it would gain KE and
lose PE.
a
+
b
E
-------------------
A positive charge at a would be in a
region of high potential, have a high
potential energy, and would move to
b , a region of low potential, where
it would have a low potential energy.
++++++++++++++
Typically this is “ground,” where a charge would be “happy” to
not move, or infinite separation, where two charges would feel
no force.
Rotate this picture 90º clockwise and you have a picture of an object that would fall
in a gravitational field.
The change in PE of a charge moved from a to b is
= q (Vb - Va )
= q Vab.
This gives us a new OSE
OSE:
PEif= q Vif ,
and the sign on the charge matters, as usual.
a
b
+
E
--------------
= Vb q – Va q
++++++++++
PEab = PEb – PEa
= Va Q – Vb Q
a
b
+
E
--------------
PEba = PEa – PEb
++++++++++
Example: take Va at the positive plate
to be +6V and Vb at the negative plate
to be 0V. Suppose a charge of Q =
+1C is moved from b to a. Then
= Q (Va – Vb )
= (+1C)(6V - 0V)
= +6J.
You had to do work against the electric force, and you increased the potential
energy of the + charge.
In this section, we introduced three new OSE’s:
Va = (PE)a /q
Wif = q Vif
PEif= q Vif
and the sign on the charge “matters.” In other words, don’t
take the absolute value of q, and be sure to include its + or
– sign in the equations.
Example: An electron in
the picture tube of a TV
is accelerated from rest
through a potential
difference Vca = +5000
V.
(a) What is the
change in potential
energy of the
electron?
0V
+5000 V
How do you remember which is anode and which is
cathode? I remember that electrons are “cathode rays,”
because a monitor is a CRT and has a “gun” that
“shoots” electrons. Cathode rays must come from the
cathode. If electrons are exiting the cathode, it must be
negative. Therefore, the anode is positive.
Graphic “borrowed” from http://www.howstuffworks.com. Go there!
(a) What is the change in potential energy of the electron?
Reminder: the steps to reach the solution are
 draw a fully-labeled diagram,*
 OSE,
 replace generic quantities by specifics in OSE,
 solve algebraically,
 numerical answer only at very end.
OSE:
PEif= q Vif
PEca= q Vca = q (Va – Vc)
PEca= (-1.6x10-19 C)(+5000 V – 0 V)
PEca= -8x10-16 J
*Kind of done—see previous slide.
PEca= -8x10-16 J
What is the meaning of the – sign? The electron’s potential
energy has decreased. Total energy is conserved. What
happened to the electron’s kinetic energy?
(b) The potential energy lost by the electron becomes
kinetic energy. What is the speed of the electron as a result
of its acceleration?
OSE:
Ef – Ei = (Wother )
if
“Nooooo… that was last semester. This is this semester. You can‘t…
Yes I can!
OSE:
Ef – Ei = (Wother )
if
There are subtleties here. Wother refers to work done by
non-conservative forces. A conservative force is one for
which you can define a potential.* We have defined a
potential for Coulomb’s Law forces, which tells you they
must be conservative.
Work done by conservative forces gets accounted for in the
potential energies, which in this equation are contained in
the E’s.
*A good way to think of a conservative force is that it doesn’t “rob” you of
useful energy. Energy is “conserved” as you move from initial state to final
state. Of course, energy never disappears, but nonconservative forces are
especially wasteful of useful energy.
PEca= -8x10-16 J
Ef – Ei = (Wother )
0
no non-cons.
forces present
if
0 0
Kf + Uf – Ki – Ui = 0
a
b
½mvf2 = -Uf
bad idea - I combined
two steps here
vf2 = -2 Uf /m
vf = ( -2 Uf /m)½
a the electron is initially at rest
b setting V=0 at the cathode means electron has PE=0 there
vf = ( -2 Uf /m)½
vf = ( -2 PEf /m)½
PE and U mean
the same thing
vf = ( -2 q Vf /m)½
vf = [ -2 (-1.6x10-19 C) (+5000 V) / 9.11x10-31 kg ]½
vf = 4.2x107 m/s
This answer is (slightly) in error because a relativistic
calculation is needed for the velocity of a fast-moving
electron
A note on potential energies:
Last week, the potential energies that went into Ef – Ei =
(Wother ) if were spring and gravitational.
This week, we have discovered a new (to us) potential
energy which results from the electric force (a
conservative force).
Now when we write Ef – Ei = (Wother ) if, we must
include Ugrav, Uspring, and Uelectric.
Why have you been writing PE for electrical potential
energy, instead of Uelec?
Because that’s the way the author of our College Physics
textbook does it.
Wab = q Vab
= FD = +Fd
= qEd

a
+
E
F
d
Vab = Ed
E = Vab/d
This gives us another equation for the electric field.
OSE
D
E=
ΔVif
d
,
away from +
b
--------------
The magnitude of the work done in moving
a charge from a to b in a uniform electric
field E is
++++++++++
Relation Between Electric Potential Electric Field
d
,
away from +
E = |V| / d
E = 50 V / 0.05 m
E = 1000 V/m, to the right
d=0.05 m
E
V = 50 V
--------------
E=
ΔVif
++++++++++
Example: Two parallel plates are charged
to a voltage of 50 V. If the separation
between the plates is 0.050 m, calculate
the electric field between them.
b
a
This is just more stuff to learn. There’s already enough stuff. Why bother?
We started out with charges. That led us to forces due to the
charges – Coulomb’s Law. We introduced the electric field to
help us visualize the forces throughout all of 3D space.
The electric field is a vector quantity. Vectors are a pain to
deal with. Wouldn’t you rather work with scalars?
Sure you would.
We needed to introduce energy, hence potentials and
potential differences. A cool bonus is that potentials are
scalars, and are related to the electric field.
Wouldn’t you rather work with scalar potentials instead of
electric field vectors? Yes, you have to. Sure you would.
If we were doing this “right,” we would define
V(xyz)
Ex = - lim
x 0
x
and similarly for the y and z-components of E.
Calculus would then tell us
dV(xyz)
Ex = dx
and similarly for the y and z-components of E.
We could then elegantly write
E(xyz) = -V(xyz) .
just a fancy way of writing all 3 derivative components at once
I’m not sure that the way I have presented this material,
which is the way our College Physics text presents it, is the
best way to do it.
It is not the way you would present it to a physics major, who
has already had enough calculus so that it is like a second
language.
On the other hand, the presentation here tries to make sense
out of the relationships between electric force, electric field,
and electric potential. The traditional physics approach is
more like a decree from above.
You can see how hyperphysics does it, if you are curious.
Electric potential energy. Work and voltage.
Equipotential Lines
Equipotentials in 2 dimensions are like contour maps.
I got the map on the previous page from
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 3-dimensional
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 some electric field lines I generated using the emfield
program.
Equipotential lines are shown in red.
The Electron Volt, a Unit of Energy
The joule represents too large an energy scale when we
discuss individual electrons or atoms.
An electron volt (eV) is the amount of energy an electron
acquires in being accelerated through a potential difference of
1 volt.
1 eV = 1.6x10-19 joules
The eV is not a unit of energy in the SI (mks) system. If you
are doing calculations involving mks, you should* convert eV
to joules before you calculate, otherwise there is a good
chance you will introduce an error.
*i.e., must, unless you really know what you are doing