A h
A
B
F = mg
GPE = mgΔh
GPE = mgh
A
– mgh
B
GPE = Work (W) required to raise or lower the book.
Where W = (F gravity
)( Δh) h
B

+ + + + + + + + +
ΔEPE = q o
EΔd
ΔEPE = q o
Ed
A
– q o
Ed
B
-W
E(AB)
-W
E(AB)
= q o
Ed
A
– q o
Ed
B
= F e d
A
– F e d
B
A +
F e
= q o
E d
A d
B
B +
F e
= q o
E
-
•Does a proton at rest at point A have more or less potential energy than it would at point B?
More




Much like the book is attracted to the earth due to gravity, two unlike charges are attracted to one another.
Conversely, like charges repel.
It takes positive work to move unlike charges away from one another and negative work to move them closer together.
-q o
F e
 k qq o r
2
E r
EPE

F e r
 kqq o r
+q

+q
-q o
-q o r
A
+q r
B
To change the energy level from
EPE
A to EPE work (W).
B
, it requires positive

W

EPE
B

EPE
A

W
 kqq o r
B
 kqq o r
A
A B

A.
B.
C.
1.
What would happen if the charged particle q was fixed in place and then particle q o was suddenly released from rest?
It would accelerate away from q.
It would accelerate towards q.
It would stay where it is.
-q o
A.
B.
C.
2.
How would the potential energy of this system change?
It would increase.
It would decrease.
It would remain the same.
+q


V

V
B

V
A

EPE
 q o

W q o
Since EPE
 kqq o r

V
 kq r
B
 kq r
A
SI Units: joule/coulomb = 1 volt (V)
Point Charges only


The Electric Potential Difference is equal to the Work required to move a test charge from infinity to a point in an electric field divided by the magnitude of the test charge.
The Electric Potential is the energy per unit of charge
(J/C).

Example 1: Electric Potential

An object with 2.5
 C of charge requires
1.00x10
-3 Joules of energy to move it through an electric field. What is the potential difference through which the charge is moved?

V

W q

0 .
001 J
2 .
5

10

6
C

V

400 J / C

V

400 V

Relationship Between Electric Potential and Distance(point charges)

Consider relationship between V and r.
-W
AB kq kq
V
B
- V
A
= = q o r
B r
A

What happens to V as r
B
•
As r increases, i.e., as r
B goes to  ?
  , V  0.
•
The relationship above reduces to: V = kq/r

The sign of the charge will determine if the electric potential is positive or negative.

When two or more charges are present, the total electric potential is the sum total from all the charges present in the system.



Consider the following system of three point charges.
What is the electric potential that these charges give rise to at some arbitrary point P?
Use superposition to determine V.
V = kQ
1 r
1
+ kQ
2 r
2
+ kQ
3 r
3

Note that the electric potential can be determined from any arbitrary point in space.
Q
3
Q
2 r
2
Q
1 r
1 r
3
P

Electric Potential and Electrical
Potential Energy/Work (point charges)


If we now move a test charge from infinity to point P, we can determine the potential energy of the system or the work required to the test charge to its new location.
EPE

W
 q V o
W
 kq o



Q
1

Q
2

Q r r r
1 2 3
3 

 Q
3
Q
2 r
2
Q
1 r
1
Remember: work = energy.
r
3 q o

Two point charges, + 3.00
µC and -6.10 µC, are separated by
1.00
m. What is the electric potential midway between them?
-6.10 μC +3.00 μC
A B
0.5 m 0.5 m
V total
= V
A
+ V
B
= kq
A
/r
A
+ kq
B
/r
B
V
A
V
B
= (8.99 x 10 9 Nm 2 /C 2 )(-6.10 x 10 -6 C) = -109678 V
0.5m
= (8.99 x 10 9 Nm 2 /C 2 )(3.00 x 10 -6 C) = 53,940 V
0.5m
V total
= -55700 V

Two equal and oppositely charged plates
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ q o
A q o
B
E q o
C
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
Uniform Electric
Field
• Since the electric field is constant, the force acting on a charged particle will be the same everywhere between the plates.
• F e
= q o
E F
A
= F
B
= F
C

W
AB
= F·d
B
F·d
W
AB
= q o
E
 d
A

V =

(EPE) -W
AB q o
= q o
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
A q o
F = q o
E
D
C q o d
A
B q o
F = q o
E d
B
If W
W
CD
AB
= q o
E  d, then what is W
CD
?
= 0 Joules because the force acts perpendicular to the direction of motion.
•Do you remember that W = F·d·cos  ?
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-




From mechanics, W = Fd.
From the previous slide, W = q o
Ed
From the reference table,  V = W/q o
Two equal and oppositely charged plates
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
A q o
F = q o
E
Uniform
Electric
Field d
B
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-

V

W
AB q o

Fd q o

V
 q o
Ed

Ed q o

A spark plug in an automobile engine consists of two metal conductors that are separated by a distance of 0.50 mm. When an electric spark jumps between them, the magnitude of the electric field is 4.8 x 10 7 V/m. What is the magnitude of the potential difference V between the conductors?
V = Ed
V = (4.8 x 10 7 V/m)(5.0 x 10 -4 m)
V = 24,000V d

A proton and an electron are released from rest from a similarly charged plate of a capacitor. The electric potential is 100,000 V and the distance between the two plates is 0.10 mm.
1.
Which charge will have greater kinetic energy at the moment it reaches the opposite plate?
2.
3.
Determine the amount of work done on each particle.
Determine the speed of each particle at the moment it reaches the opposite plate.
4.
5.
Determine the magnitude of the force acting on each particle.
Determine the magnitude of the acceleration of each particle.


+ -
+ -
+ -
+ -
+ p +
-
+ -
+ -
+ -
+ -
+ -
+ e -
-
+ -
+ -
+
+ d -



-
Begin by drawing a picture and listing what is known:
V = 100,000V d = 0.10 mm = 1.0 x 10 -4 m q e
= q p
= 1.6 x 10 -19 C (ignore the sign. We are only interested in magnitude.)


For #1, you could answer #2 first to verify.

The answer is that the kinetic energy of both particles will be the same
• Why?
• because of the formula needed in question #2 applies to both charges, and work = energy.
• Hence: W proton
= W electron q proton
V = q electron
V
W proton
W proton
= W
= W electron electron
= (1.6x10
-19 C)(100,000V)
= 1.6x10
-14 J



Apply the work-energy theorem to determine the final speed of the electron and proton.
W =  KE
Since the initial kinetic energy is equal to 0J:
W = KE f
W = ½ mv f
2

Proton: v f

(
(2)( W ) m proton
)

(2)(1.6 10



27
14
J kg )
)
  6 m s

Electron: v f

(
(2)( W ) m electron
)


 
14

31 kg
J
)
)
  8 m s



Since F = qE, it will be the same for both particles because their charges are the same and the electric field is uniform between two parallel plates.
We also know that W = F  d. Since we know the distance between the plates and the work done to move either charge from one plate to another, we can determine the force as follows:
F

W
 d

14
J

4 m

1.6 10

10
N


Since we have the force acting on each particle, we can now calculate the acceleration of each particle using Newton’s 2 nd Law.
a
 a

F m proton
F m electron



10
N

27 kg
  16 m s
2

10
N

31 kg
 
20 m s
2





Equipotential lines denote where the electric potential is the same in an electric field.
The potential is the same anywhere on an equipotential surface a distance r from a point charge, or d from a plate.
No work is done to move a charge along an equipotential surface. Hence V to B ).
B
= V
A
( The electric potential difference does not depend on the path taken from A
Electric field lines and equipotential lines cross at right angles and point in the direction of decreasing potential.


Parallel Plate Capacitor
Lines of
Equipotential
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Decreasing Electric
Potential / Voltage
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
Electric Field
Lines
Note: Electric field lines and lines of equipotential intersect at right angles.


Point Charge
Lines of
Equipotential
+
Decreasing Electric
Potential / Voltage
Note: Electric field lines and lines of equipotential intersect at right angles.
Electric Field
Lines
Note: A charged surface is also an equipotential surface!

 http://www.cco.caltech.edu/~phys1/java/phys1/
EField/EField.html






Electric potential energy ( U ) is the work required to bring a positive unit charge from infinity to a point in an electric field.
Electric potential ( V ) is the change in energy per unit charge as the charge is brought from one point to another.
The electric field between two charged plates is constant meaning that the force is constant between them as well.
The electric potential between two points is not dependent on the path taken to get there.
Electric field lines and lines of equipotential intersect at right angles.

Note: The force acting on the charge is constant as it moves from one plate to another because the electric field is uniform.
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
A q o
F = q o
E d
A
B q o
F = q o
E d
B
W
AB
W
AB
= EPE
B
= Fd
B
– EPE
A
– Fd
A
W
AB
W
AB
= q o
Ed
B
= q o
E(d
B
– q o
Ed
A
– d
A
) = q o
E  d
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-