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!

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
= GmM/r 2 , attractive.
y
G
If released, it gains kinetic energy and loses potential energy, but mechanical energy is conserved: E
U f
- U i
= -(W c
) f i  f
=E
.
i
. The change in potential energy is y i
U i
= mgy i x
U f
= 0
What force does W c
? 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: V a
= (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 V ab to mean “the difference when you go from point b to point a.”
And sometimes he uses V
Very confusing.
ba to mean the same difference.
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,
 V i  f
, means V f
– V i
. 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
V a
– V b
= -(W b  a
)/q. This gives us another OSE:
OSE: W i  f
= q  V i  f

V i
 f is potential difference
What force does W i  f
Coulomb’s law.
? The electric force, as described by
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.
V a
= (PE) a
/q Electric potential:
(also called potential; unit is volt)
Electric potential energy:
(unit has to be joules)
Electric potential difference:
(also called potential difference; unit is volt)
(also called voltage) not defined yet
 V i  f
= W i  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)?
V a q = PE a
V a
– V b
= -(W b  a
)/q
V a q – V b q = -W b  a
PE a
– PE b
= -W b  a
The last equation is equivalent to our Mechanics equation
U f
- U i
= -(W c
) i  f
Looks consistent (with our previous ideas of potential energy) to me.

Comment: two definitions V a
= (PE) a
/q
 V i  f
= W i  f
/q give us an implicit definition of electrical potential energy:
 PE b  a
= -W b  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.
Typically this is “ground,” where a charge would be “happy” to not move, or infinite separation, where two charges would feel no force.
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.
In the process, it would gain KE and lose PE.
a
+
E b
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
 PE a  b
= PE b
– PE a
= V b q – V a q
= q (V b
- V a
)
= q  V a  b
.
a
+
E b
This gives us a new OSE
OSE:  PE i  f
= q  V i  f
, and the sign on the charge matters, as usual.

Example: take V to be +6V and V a b at the positive plate at the negative plate to be 0V. Suppose a charge of Q =
+1C is moved from b to a. Then
 PE b  a
= PE a
– PE b
= V a
Q – V b
Q
= Q (V a
– V b
)
= (+1C)(6V - 0V)
= +6J.
a
E b
+
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:
V a
= (PE) a
/q
W i  f
= q  V i  f
 PE i  f
= q  V i  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 V c  a
= +5000
V.
0 V
+5000 V
(a) What is the change in potential energy of the electron?
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:  PE i  f
= q  V i  f
 PE c  a
= q  V c  a
= q (V a
– V c
)
 PE c  a
= (-1.6x10
-19 C)(+5000 V – 0 V)
 PE c  a
= -8x10 -16 J
*Kind of done—see previous slide.

 PE c  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: E f
– E i
= (W other
) i  f
“Nooooo… that was last semester. This is this semester. You can‘t…
Yes I can!

OSE: E f
– E i
= (W other
) i  f
There are subtleties here. W other 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.

 PE c  a
= -8x10 -16 J
E f
– E i
= (W other
)
0 i  f
K f
+ U f
– K i a
0 i b
½mv f
2 = -U f no non-cons. forces present bad idea - I combined two steps here v f
2 = -2 U f
/m v f
= ( -2 U f
/m) ½ a the electron is initially at rest b setting V=0 at the cathode means electron has PE=0 there

v f
= ( -2 U f
/m) ½ v f
= ( -2 PE f
/m) ½ PE and U mean the same thing v f
= ( -2 q V f
/m) ½ v f
= [ -2 (-1.6x10
-19 C) (+5000 V) / 9.11x10
-31 kg ] ½ v f
= 4.2x10
7 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 E f
(W other
) i  f were spring and gravitational.
– E i
=
This week, we have discovered a new (to us) potential energy which results from the electric force (a conservative force).
Now when we write E include U grav
, U spring f
– E i
= (W
, and U electric
. other
) i  f
, we must

Why have you been writing PE for electrical potential energy, instead of U elec
?
Because that’s the way the author of our College Physics textbook does it.

Relation Between Electric Potential Electric Field
The
a charge from a to b in a field E is of the work done in moving
electric
W a  b
= q  V a  b a
+
D
E
= F  D = +Fd
F
= qEd d
  V ab
= Ed
E =  V ab
/d
This gives us another equation for the electric field.
OSE E =
ΔV d
, away from + b

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.
E =
ΔV d
, away from +
E = |  V| / d
E = 50 V / 0.05 m
= 1000 V/m, to the right a d=0.05 m
E

V = 50 V b

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)
 x and similarly for the y and z-components of E.
Calculus would then tell us dV(xyz) dx and similarly for the y and z-components of E.
We could then elegantly write 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 .

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 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