Download Electric potential energy

Potential of Charged Plates
Fields and Force
+ sky
g
+ + + + + + + +
E
m
Fg
– ground
+q
FE
−q
FE
− − − − − − − −
In previous presentations
g = 9.8 m s2
Fg = mg
E = given or solved for
FE = qE
FE = qE
Electric Potential Energy
+ sky
6m
0.30 m
g
5m
0.25 m
4m
0.20 m
3m
2m
Ug
m
Fg
0.15 m
0.10 m
1m
0.05 m
0m
0.00 m
– ground
+ + + + + + + +
E
UE
+q
FE
FE
−q
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)
Electric Potential Energy
+ sky
6m
0.30 m
g
5m
0.25 m
4m
0.20 m
3m
2m
Ug
m
Fg
0.15 m
0.10 m
1m
0.05 m
0m
0.00 m
– ground
+ + + + + + + +
E
UE
+q
FE
FE
−q
UE
− − − − − − − −
The energy of objects in uniform gravity and electric fields is very similar.
U g = mgh
U E = qEd
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.
Electric Potential
+ sky
6m
0.30 m
g
5m
0.25 m
4m
0.20 m
3m
2m
Ug
m
Fg
0.15 m
0.10 m
1m
0.05 m
0m
0.00 m
– ground
+ + + + + + + +
E
UE
+q
FE
FE
−q
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.
Electric Potential
+ sky
6m
0.30 m
g
5m
0.25 m
4m
0.20 m
3m
2m
Ug
m
Fg
0.15 m
0.10 m
1m
0.05 m
0m
0.00 m
– ground
+ + + + + + + +
E
UE
+q
FE
FE
−q
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
V = -Ed
U E = qEd
U E = qV
Electric Potential
+ sky
6m
0.30 m , 6 V
g
5m
0.25 m , 5 V
4m
0.20 m , 4 V
3m
2m
Ug
m
Fg
0.15 m , 3 V
0.10 m , 2 V
1m
0.05 m , 1 V
0m
0.00 m , 0 V
– ground
+ + + + + + + +
E
UE
+q
FE
FE
−q
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.
Potential Difference
+ sky
6m
0.30 m , 6 V
g
5m
0.25 m , 5 V
4m
0.20 m , 4 V
3m
2m
Ug
m
Fg
0.15 m , 3 V
E
UE
0.05 m , 1 V
0m
0.00 m , 0 V
+q
FE
0.10 m , 2 V
1m
– ground
+ + + + + + + +
FE
−q
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.
DV = V - V0
DU E = qDV
(
DU E = q V - V0
)
Work
+ sky
6m
g
5m
0.25 m , 5 V
4m
0.20 m , 4 V
3m
Δh
ΔUg
Wg
0.30 m , 6 V
2m
Ug
m
Fg
1m
0m
– ground
Δd
ΔV
ΔUE
WE
0.15 m , 3 V
0.10 m , 2 V
+ + + + + + + +
E
UE
+q
FE
FE
−q
UE
Δd
ΔV
ΔUE
WE
0.05 m , 1 V
0.00 m , 0 V
− − − − − − − −
If energy is changing work is being done.
W = DK
Work Kinetic Energy Theorem states that
work is a change in kinetic energy.
W = -DU
In order for kinetic energy to increase ( +ΔK )
potential energy must decrease ( −ΔU ).
WE = -DU E = -qDV
Conservation of Energy (objects released from rest)
+ sky
6m
g
5m
0.25 m , 5 V
4m
0.20 m , 4 V
3m
Δh
ΔUg
Wg
0.30 m , 6 V
2m
Ug
m
Fg
1m
0m
– ground
U g 0 + K0 = U g + K
Δd
ΔV
ΔUE
WE
0.15 m , 3 V
0.10 m , 2 V
+ + + + + + + +
E
UE
+q
FE
FE
−q
UE
0.05 m , 1 V
0.00 m , 0 V
− − − − − − − −
U E 0 + K0 = U E + K
If these objects start at rest and a mass moves through a change in
height or a charge moves through a potential difference, then
1 2
mgDh = mv
2
1 2
qDV = mv
2
Δd
ΔV
ΔUE
WE
Summary of Equations for Uniform Fields
+ sky
6m
g
5m
0.25 m , 5 V
4m
0.20 m , 4 V
3m
Δh
ΔUg
Wg
0.30 m , 6 V
2m
Ug
m
Fg
1m
0m
– ground
Δd
ΔV
ΔUE
WE
0.15 m , 3 V
0.10 m , 2 V
+ + + + + + + +
E
UE
+q
FE
FE
−q
UE
0.05 m , 1 V
0.00 m , 0 V
− − − − − − − −
Fg = mg
FE = qE
U g = mgh
V = -Ed
U E = qV
Wg = -DU g = -mgDh
WE = -DU E = -qDV
1
mgDh = mv 2
2
1
qDV = mv 2
2
Δd
ΔV
ΔUE
WE
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).
1. When labeling potential start by giving the
most negative object zero potential (like
making the ground zero meters high).
2. Then the most positive object will have a
potential equal to the potential difference.
3. 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.
6V
+ + + + + +
5V
4V
3V
2V
1V
0V
− − − − − −
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.
Potential
Potential V is an electricity variable that has no mechanics counterpart.
V = Ed
gh = ?
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.
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.
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 .
UE = K
–
– –
–
–
+
+
+
+
v
1 2
qV = mv
2
1 2
qEd = mv
2
(
)( )( ) (
1.6 ´ 10-19 20 0.15 =
)
1
9.11 ´ 10-31 v 2
2
v = 1.03 ´ 106 m s
Example 2
The charged plates shown have a potential difference of 20 V.
Determine the speed of the proton exiting the plates.
Potential V is given. Use the
version of conservation of
energy that groups Ed
together as V .
+
+ +
+
+
UE = K
1 2
qDV = mv
2
(1.6 ´ 10 ) ( ) (
-19
)
1
20 = 1.67 ´ 10-27 v 2
2
v = 6.19 ´ 104 m s
–
–
–
–
v
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.
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
Problems text says,
“accelerated thru a
potential difference”
The equation for all of
these is
v
1 2
qDV = mv
2
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.
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.
+
+
+
+
+
+
+
+
+
−
−
−
−
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.
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.
+
+
+
+
+
+
+
+
+
−
−
−
−
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.
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.
+
−
− −
−
−
+
+
+
+
+
+
+
+
−
−
−
−
v
c. Determine the speed of the charge exiting the hole in the first set of plates.
1 2
qDV = mv
2
(
1.6 ´ 10
-19
)( ) (
)
1
40 = 9.11 ´ 10-31 v 2
2
v = 3.75 ´ 106 m s
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.
+
+
+
v
Δy
−
−
−
Δx
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 !
−
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.
+
+
+
+
Δy
SF = FE
ma = qE
−
qE
a=
m
−
−
−
Δx
1.6 ´ 10 ) ( 20 )
(
a=
( 9.11 ´ 10 )
-19
-31
a = 3.51´ 1012 m s2
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.
+
+
+
+
Δy
Solve time using the speed
from part c .
Dx = v0x t
(0.10) = (3.75 ´ 10 ) t
−
6
t = 2.67 ´ 10-8 s
−
−
−
Δx
Solve the vertical deflection using the acceleration and the time.
(
)(
1 2 1
Dy = at = 3.51 ´ 1012 2.67 ´ 10-8
2
2
)
2
Dy = 1.25 ´ 10-3 m