1. Electrostatics Potential of Charged Plates

2. Fields and Force + + + + + + + + − − − − − − − −
+ sky g E FE m +q −q Fg FE
− − − − − − − −
– ground
In previous presentations

3. Electric Potential EnergyFg
– ground
+ sky m
− − − − − − − − FE E +q −q
6 m
5 m
4 m
3 m
2 m
1 m
0 m
0.30 m
0.25 m
0.20 m
0.15 m
0.10 m
0.05 m
0.00 m Ug UE
Potential energy is associated with an objects position.
Gravitational potential energy: Position is the objects height
(usually measured from the ground upward)
Electric potential energy: Position is the objects distance from the plates.
(usually measured from the negative plate to the positive plate)

4. Electric Potential EnergyFg
– ground
+ sky m
− − − − − − − − FE E +q −q
6 m
5 m
4 m
3 m
2 m
1 m
0 m
0.30 m
0.25 m
0.20 m
0.15 m
0.10 m
0.05 m
0.00 m Ug UE UE
The energy of objects in uniform gravity and electric fields is very similar.
Charge replaces mass, the electric field replaces the gravity field, and distance replaces height.
Note: The equations for the positive and negative charge are the same.

5. Electric Potential + + + + + + + + − − − − − − − −g Fg
– ground
+ sky m
− − − − − − − − FE E +q −q
6 m
5 m
4 m
3 m
2 m
1 m
0 m
0.30 m
0.25 m
0.20 m
0.15 m
0.10 m
0.05 m
0.00 m Ug UE UE
Electric Potential does not have an equivalent in mechanics.
Electric fields are incredible strong, and even very small changes in distance can cause huge changes in field strength.
Therefore, grouping field E and distance d into one variable is useful in electricity.

6. Electric Potential + + + + + + + + − − − − − − − −g Fg
– ground
+ sky m
− − − − − − − − FE E +q −q
6 m
5 m
4 m
3 m
2 m
1 m
0 m
0.30 m
0.25 m
0.20 m
0.15 m
0.10 m
0.05 m
0.00 m Ug UE UE
Electric Potential is the variable V . Units of volts ( V )
The minus sign in the formula is often ignored. It has to do with the direction of E , which is already known.
Substitute V = Ed into the energy formula and we get the universal equation for electric potential energy

7. Electric Potential + + + + + + + + − − − − − − − −g Fg
– ground
+ sky m
− − − − − − − − FE E +q −q
6 m
5 m
4 m
3 m
2 m
1 m
0 m
0.30 m
0.25 m
0.20 m
0.15 m
0.10 m
0.05 m
0.00 m
, 6 V
, 5 V
, 4 V
, 3 V
, 2 V
, 1 V
, 0 V Ug UE UE
Changes in electric potential cause changes in electric potential energy, just as changes in height cause changes in gravitational potential energy.
Therefore, potential is visualized similar to height in gravity. The potential between plates is measured in a manner similar to height.

8. Potential Difference + + + + + + + + − − − − − − − −g Fg
– ground
+ sky m
− − − − − − − − FE E +q −q
6 m
5 m
4 m
3 m
2 m
1 m
0 m
0.30 m
0.25 m
0.20 m
0.15 m
0.10 m
0.05 m
0.00 m
, 6 V
, 5 V
, 4 V
, 3 V
, 2 V
, 1 V
, 0 V Ug UE UE
A potential difference is a change in potential.
While energy does not require a charge to move, a potential difference does.
A change in potential will also lead to a change in electric potential energy.

9. Work + + + + + + + + − − − − − − − −g Fg
– ground
+ sky m
− − − − − − − − FE E +q −q
6 m
5 m
4 m
3 m
2 m
1 m
0 m
0.30 m
0.25 m
0.20 m
0.15 m
0.10 m
0.05 m
0.00 m
, 6 V
, 5 V
, 4 V
, 3 V
, 2 V
, 1 V
, 0 V
Δd ΔV ΔUE WE Ug UE UE
Δd ΔV ΔUE WE
Δh ΔUg Wg
If energy is changing work is being done.
Work Kinetic Energy Theorem states that work is a change in kinetic energy.
In order for kinetic energy to increase ( +ΔK ) potential energy must decrease ( −ΔU ).

10. Conservation of Energy (objects released from rest)Fg
– ground
+ sky m
− − − − − − − − FE E +q −q
6 m
5 m
4 m
3 m
2 m
1 m
0 m
0.30 m
0.25 m
0.20 m
0.15 m
0.10 m
0.05 m
0.00 m
, 6 V
, 5 V
, 4 V
, 3 V
, 2 V
, 1 V
, 0 V
Δd ΔV ΔUE WE Ug UE UE
Δd ΔV ΔUE WE
Δh ΔUg Wg
If these objects start at rest and a mass moves through a change in height or a charge moves through a potential difference, then

11. Summary of Equations for Uniform Fieldsg Fg
– ground
+ sky m
− − − − − − − − FE E +q −q
6 m
5 m
4 m
3 m
2 m
1 m
0 m
0.30 m
0.25 m
0.20 m
0.15 m
0.10 m
0.05 m
0.00 m
, 6 V
, 5 V
, 4 V
, 3 V
, 2 V
, 1 V
, 0 V
Δd ΔV ΔUE WE Ug UE UE
Δd ΔV ΔUE WE
Δh ΔUg Wg

12. Drawing Equipotential Lines
The object that is the most positive is considered to be at high potential, and the object that is the most negative is considered to be at low potential.
The plates below have a 6 V potential difference (ΔV = 6 V).
When labeling potential start by giving the most negative object zero potential (like making the ground zero meters high).
Then the most positive object will have a potential equal to the potential difference.
Finally fill in the potentials in between.
The most positive object has high potential. The positive plate is always high potential.
The most negative object has low potential. The negative plate is always low potential.
Fields run from high to low potential.
Positive charges move from high to low potential.
Negative charges (opposite charge) move from low the high potential.
− − − − − −
6 V
1 V
2 V
3 V
4 V
5 V
0 V

13. What exactly is Potential ?
Potential can be thought of as electrical pressure (like water pressure). Water pressure from the water company movse water to your home. Electrical pressure from the power company moves electricity to your home.
The term “high voltage” means that there is a “high potential (high likelihood)” that electricity will move from a high to a low potential. Potential by itself cannot harm you. It is a measure of the pressure on the charges and the likelihood that charges will move from an object labeled “high voltage” to you.
(It pretty much means: Don’t get close to me. I have the ability send lightening bolts at any object with less potential than me.)
The size of the lightening bolt (next chapter: current, or flow of charge) determines whether injury or death occurs.

14. Potential
Potential V is an electricity variable that has no mechanics counterpart.
Potential: V is measured in volts V (example: a potential of 2 volts, is written V = 2V)
Potential is commonly called Voltage , but there are many ways to say potential.
Electrostatic potential, electric potential, or just plain potential.
One important version is Potential Difference. This is specifically the change in potential and is the variable ΔV .
A more obscure variant is electromotive force, emf, ε . We will see this version used for batteries and generates that create electric potential.

15. Example 1
The charged plates generate a 20 N/C electric field, and are separated by 15 cm. Determine the speed of the electron exiting the plates.
– – – – – v
Electric field E and distance d are given. You can multiple them together to find potential V , or just use the form of conservation of energy that matches gravity and converts g to E and h to d .

16. Example 2
The charged plates shown have a potential difference of 20 V. Determine the speed of the proton exiting the plates.
++++
– – – – + v
Potential V is given. Use the version of conservation of energy that groups Ed together as V .

17. Accelerated Through a Potential Difference
The previous two example show charges that were “Accelerated through a potential difference.”
There are many ways to convey this well known acceleration.
If the problem says
Electron/proton accelerated through a potential difference
Electron/proton beam
Or any diagram showing a charge moving along field line between two charged plates. (moving along field lines mean moving in a straight line from one plate to the other.
Then it is accelerating through a potential difference.

18. Diagrams of Acceleration thru a Potential Difference
These diagrams vary a lot. It is almost as though they are trying to trick you. Here are some examples showing electrons moving. v +
– –
– – – – – v
Electron Beam v
The equation for all of these is
Problems text says, “accelerated thru a potential difference”

19. What Does the Symbol Mean?
Two of the diagrams contained the symbol
This is the symbol for a battery. It consists of charged plates that release charges. The plates are actually the same size, but they are draw this way to show which plate is positive and which is negative. The large plate is positive, and the small one is negative.
As draw above it is a single cell battery. To make batteries with more potential (voltage) you can link cells together in a series (in line). Car batteries consist of six 2V cells totaling 12V, and could be draw as follows.

20. Example 3 − + − + + + + −− v − − − −
A charged particle accelerates through a potential difference of 40 V in a set of vertical plates. It enters a set of horizontal plates that are 10 cm long with a 20 N/C electric field. + −− − v
− − − − −
a. Determine the sign on all plates. Determine if the accelerating particle is an electron or a proton, and draw its path.
Left Plates: The battery determines the charge on the plates.
The particle must be an electron to be repelled by the left negative plate.
Right Plates: The electric field points down. Upper plate is positive.
The path is straight outside the plates and projectile motion between them.

21. Example 3 − + − + + + + −− v − − − −
A charged particle accelerates through a potential difference of 40 V in a set of vertical plates. It enters a set of horizontal plates that are 10 cm long with a 20 N/C electric field. + −− − v
− − − − −
b. Determine the electric field direction between the left set of plates.
Method 1: Know that the electric field runs from positive to negative, just as gravity runs from the positive sky to the negative ground. Right.
Method 2: Positive (right way) charges follow the field. The positive charge is moving right, so the field is to the right.

22. Example 3 − + − + + + + −− v − − − −
A charged particle accelerates through a potential difference of 40 V in a set of vertical plates. It enters a set of horizontal plates that are 10 cm long with a 20 N/C electric field. + −− − v
− − − − −
c. Determine the speed of the charge exiting the hole in the first set of plates.

23. Example 3 − + − + + + + −− v Δy − − − − Δx
A charged particle accelerates through a potential difference of 40 V in a set of vertical plates. It enters a set of horizontal plates that are 10 cm long with a 20 N/C electric field. + −− − v Δx Δy
− − − − −
d. Determine the vertical deflection in the second set of plates.
This is horizontal projectile motion, just as we did before.
However, it does not involve gravity g = 9.8 .
You must solve for the acceleration of electricity first !

24. Example 3
A charged particle accelerates through a potential difference of 40 V in a set of vertical plates. It enters a set of horizontal plates that are 10 cm long with a 20 N/C electric field.
d. Determine the vertical deflection in the second set of plates.
− − − − Δx Δy

25. Example 3
A charged particle accelerates through a potential difference of 40 V in a set of vertical plates. It enters a set of horizontal plates that are 10 cm long with a 20 N/C electric field.
d. Determine the vertical deflection in the second set of plates.
− − − − Δx Δy
Solve time using the speed from part c .
Solve the vertical deflection using the acceleration and the time.