1. PHY 102: Lecture 4 4.1 Work/Energy Review
4.2 Electric Potential Energy
4.3 Concept of Electric Potential
4.4 Electric Potential
4.5 Equipotential Surfaces

2. PHY 102: Lecture 4 Electric Potential Energy Electric Potential
4.1
Work / Energy Review

3. Work vs. Energy Energy is a property of matter
Mass, Inertia and Electric Charge are other properties
Matter that has energy can do work by means of a force
Work transfers or changes energy

4. Work – Constant Force Work is a scalar
Work done on an object by a constant force F is:
W = (F cosq)d
F is magnitude of constant force
d is the magnitude of the displacement along a line
q is angle between force and displacement
SI Unit of Work: Newton meter = joule (J)

5. Kinetic Energy KE = Kinetic Energy KE is the energy of a moving object
KE = (1/2)mv2
m is the mass of the object
v is the speed of the object

6. Work – Energy Theorem A net external force does work W on an object
The kinetic energy of the object changes from its initial value of KEi to a final value of KEf
The difference between the two values is equal to the work
W = KEf – KEi = ½ mvf2 – ½ mvi2

7. Work Done by Force of Gravity
An object moves from initial height hi to final height hf
The force of gravity acts on the object
Wgravity = Fd
F = mg = weight
d = hi - hf
Wgravity = mg(hi – hf) = mghi - mghf

8. Gravitational Potential Energy
The gravitational potential energy PE is the energy that an object of mass m has because of its position relative to the surface of the earth
That position is measured by the height h of the object relative to an arbitrary zero level
PE = mgh

9. Conservative Force Version 1 Version 2
A force is conservative when the work it does on a moving object is independent of the path between the object’s initial and final position
Version 2
A force is conservative when it does no net work on an object moving around a closed path starting and finishing at the same point

10. Conservative Forces Nonconservative Forces
Gravity
Spring force
Electrostatic force
Magnetic force
Nonconservative Forces
Friction
Air resistance

11. Principle Conservation of Mechanical Energy
Total mechanical energy (E = KE + PE) of an object remains constant as the object moves
Provided that the net work done by external nonconservative forces is zero, Wnc = 0 J

12. Kinetic Energy vs. Potential Energy
The sum of the kinetic and potential energies at any point is conserved
Kinetic and Potential Energies may be converted or transformed into one another

13. PHY 102: Lecture 4 Electric Potential Energy Electric Potential
4.2
Electric Potential Energy

14. Gravitational Potential Energy
A basketball of mass m falls from point A to point B
The gravitational force is mg in the downward direction
Work done by gravity is
WAB = mghA – mghB
WAB = GPEA – GPEB

15. Electric Potential Energy
Positive test charge moves from point A to B between plates of capacitor
The electric force is q0E in the downward direction
Work done by electric force is
WAB = q0EsA – q0EsB
WAB = EPEA – EPEB
s is distance from negative plate

16. Work by Gravity / Electricity
The gravitational and electrical forces are conservative
The work done in going from point A to point B does not depend on the path taken

17. PHY 102: Lecture 4 Electric Potential Energy Electric Potential
4.3
Concept of Electric Potential

18. Concept of Electric Potential-1
We introduced the concept of the electric field because action at a distance raised concerns and difficulties
The field provides an intermediary through which two charges exert forces on each other
Charge q1 somehow alters the space around it by creating an electric field E1
Charge q2 then responds to the field experiencing force F = q2E1

19. Concept of Electric Potential-2
When we try to understand electric potential energy we face the same kinds of difficulties that action at a distance presented to an understanding of electric force
When we say two charged particles have a potential energy that can be converted to a tangible kinetic energy of motion, where is the energy?
Two positive charges fly apart when you release them gaining kinetic energy
There is no obvious place that the energy had been stored

20. Concept of Electric Potential-3
In defining electric field, we chose to separate the charges that are the source of the field from the charge in the field
The force on charge q is related to the electric field of the source charges by
force on charge q = [charge q] x [alteration of space by source charge or E field]

21. Concept of Electric Potential-4
Try a similar procedure for the potential energy
Electric potential energy is due to the interaction of charge q with other charges
Divide up the potential energy of the system
potential energy of (q + sources) = [charge q] x [potential for interaction of the source charges]

22. Concept of Electric Potential-5
We define electric potential V (or just the potential) as V = Electric Potential Energyq+sources / q
Charge q is used as a probe to determine the electric potential, but the value of V is independent of q
The electric potential is a property of the source charges

23. What Good is the Electric Potential?
The electric potential is an abstract idea
Electric potential depends only on source charges and their geometry
The potential is the “ability” of the source charges to have an interaction if a charge q shows up
This ability, or potential, is present throughout space regardless of whether or not charge q is there to experience it
If we know the electric potential V throughout a region of space, we’ll immediately know the potential energy
The source charges exert influence through the electric potential they establish throughout space
Once we know the potential we can ignore the source charges and work just with the potential
The source charges remain hidden offstage

24. PHY 102: Lecture 4 Electric Potential Energy Electric Potential
4.4
Electric Potential

25. Electric Potential Energy
The electric force and the work it does as the charge moves from A to B depends on the charge q0
It is useful to express this work on a per-unit-charge basis, by dividing both sides by the charge:
WAB/q0 = EPEA /q0- EPEB/q0

26. Electric Potential
The right-hand side of this equation is the difference between two terms
Each of which is an electric potential energy divided by the test charge, EPE/q0
The quantity EPE/q0 is the electric potential energy per unit charge
It is called the electric potential or the potential
It Is referred to with the symbol V
SI Unit: joules/coulomb = volt (V)

27. Work Related to Electric Potential
Work WAB done by the electric force when a charge q0 moves from A to B
VB – VA = EPEB/q0 – EPEA/q0 = -WAB/q0
V = (EPE)/q0 = -WAB/q0

28. Problem 1
Work done by the electric force as the test charge (q0 = +2.0 x 10-6 C) moves from A to B is WAB = +5.0 x 10-5 J
(a) Find value of D(EPE) = EPEB – EPEA
(b) Find potential difference DV = VB – VA
VA has a higher potential then VB

29. Potential Size vs Charge Acceleration - 1
Positive charge accelerates as it moves from A to B because of the electric repulsion from the upper plate and the attraction to the lower plate
Point A has higher electric potential than point B
A positive charge accelerates from a region of higher electric potential toward a region of lower electric potential.
A negative charge accelerates from a region of lower electric potential toward a region of higher electric potential.

30. Potential Size vs Charge Acceleration - 2
Positive charge accelerates as it moves from A to B because of the electric repulsion from the upper plate and the attraction to the lower plate
Positive charge increases in speed and kinetic energy
The increase in kinetic energy comes from a decease in electric potential energy
Point A must have higher potential energy

31. Problem 2 - 1
A particle has a mass of 1.8 x 10-5 kg and a charge of +3.0 x 10-5 C
It is released from rest at point A and accelerated until it reaches point B
The only force acting on the particle is an electrostatic force
The electric potential at A is 25 V greater than that at B (VA – VB = 25 V)
What is speed of the particle at point B?

32. Problem 2 - 2
This is a conservation of energy problem

33. Electric Potential Difference of Point Charge - 1
Positive point charge +q creates an electric potential
There are two locations A and B, at distances rA and rB from the charge
An electrostatic force of repulsion acts on a positive test charge +q0
Coulomb’s law is F = kq0q/r2
When the test charge moves from A to B, work is done by this force
F varies between rA and rB
Work is not the product of force and distance between the points
Work WAB is found using calculus
The result is
WAB = kqq0/rA – kqq0/rB

34. Electric Potential Difference of Point Charge - 2
This result is valid
Whether q is “+” or “-”
Whether q0 is “+” or “-”
Potential difference VB–VA can now be determined by substituting this expression for WAB into the equation
VB – VA = -WAB/q0
VB – VA = kq/rB – kq/rA

35. Electric Potential of Point Charge - 1
Let point B be located farther and farther from the charge q
rB becomes larger and larger
In the limit that rB is infinitely large, the term kq/rB becomes zero
It is customary to set VB equal to zero also
In this limit, the equation becomes VA = kq/rA
It is standard convention to omit the subscripts
Write the potential in the following form
Potential of a point charge V = kq/r

36. Electric Potential of Point Charge - 2
The symbol V does not refer to the potential in an absolute sense
V = kq/r stands for the amount by which the potential at a distance r from a point charge differs from the potential at an infinite distance away
In other words, V refers to potential difference with the arbitrary assumption that the potential at infinity is zero

37. Electric Potential of Point Charge - 3
When q is positive, the value of V = kq/r is also positive
Indicating that the positive charge has everywhere raised the potential above the zero reference value
When q is negative, the potential V is also negative
Indicting that the negative charge has every where decreased the potential below the zero reference value

38. Problem 3 - 1 Use a zero reference potential at infinity
Determine the amount by which a point charge of 4.0 x 10-8 C alters the electric potential at a spot 1.2 m away when the charge is (a) positive and (b) negative

39. Problem 3 - 2
(a)
(b)

40. Electric Potential of More Than 1 Point Charge
Two or more charges are present
The potential due to all charges is obtained by adding together the individual potentials

41. Problem 4 - 1
At locations A and B find the total electric potential due to the two point charges
=+240 V

42. Problem 4 - 1
At locations A and B find the total electric potential due to the two point charges
= 0V
=+0 V

43. Problem 5 - 1 Given three point charges
Initially, they are infinitely far apart
They are then brought together and placed in the corners of an equilateral triangle
Each side of the triangle has a length of 0.50 m
Determine the electric potential energy of the triangular group

44. Problem 5 - 2
The order in which the charges are put on the triangle does not matter
We begin with the charge of +5.0 mC
When this charge is placed at a corner of the triangle, it has no electric potential energy
The total potential V produced by the other two charges is zero since they are infinitely far away
EPE = q0V
V = 0
EPE =0

45. Problem 5 - 3
Once the first charge is in place, the potential at the other corners is
When the +6.0 mC charge is placed at the second corner

46. Problem 5 - 4 The electric potential at the remaining corner is
When the third charge, mC, at the remaining empty corner, the electrical potential energy is

47. Problem 5 - 5
The total potential energy of the three charges is

48. PHY 102: Lecture 4 Electric Potential Energy Electric Potential
4.5
Equipotential Surfaces

49. Equipotential Surfaces
An equipotential surface is a surface on which the electric potential is the same everywhere
The net electric force does no work as a charge moves on an equipotential surface.
VB – VA = -WAB/q0
On the equipotential surface VA = VB Therefore WAB = 0

50. Equipotential Surfaces for Point Charge
Equipotential surfaces that sourround the point charge +q are spherical
As you move from point A to B to C, the electric force does no work because they lie on an equipotential surface
Work is done as you move from point A to D

51. Relation between Electric Field Lines and Equipotential Surfaces
Electric field created by any charge or group of charges is everywhere perpendicular to the associated equipotential surfaces
Electric field points in the direction of decreasing potential
Surface of any conductor is an equipotential surface
The electric field is zero everywhere inside a conductor
The entire conductor can be regarded as an equipotential volume

52. Parallel Plate Capacitor Equipotential Surfaces
Electric field E between the metal plates is perpendicular to them
Electric field E is the same everywhere
The equipotential surfaces are planes that are parallel to the capacitor plates
The capacitor plates are also equipotential surfaces

53. Parallel Plate Capacitor Relation between E and DV
Potential difference between plates is given by
ΔV = VB – VA = -WAB/q0
A is point on positive plate and B is point on negative plate
The work done by the electric force as a positive test charge q0 moves from A to B is WAB = FΔs
F refers to the electric force and s to the displacement along a line perpendicular to the plates
F = q0E
Work becomes WAB = Fs = q0Es
This leads to E = - V / s
The quantity V / s is referred to as the potential gradient
This relationship gives only the component of electric field along the displacement s

54. Problem 6 - 1
The plates of a capacitor are separated by a distance of m
The potential difference between them is DV = VB – VA = -64 V
Between the two labeled equipotential surfaces there is a potential difference of -3.0 V
Find the spacing between the two labeled surfaces

55. Problem 6 - 2 The electric field between the capacitor plates is
E = - DV/DS
E = -64 V/0.032 m
E = 2.0 x 103 V/m
Spacing between labeled equipotential surfaces is
DS = -DV/E
DS = -3.0 V/2.0 x 103 V/m
DS = 1.5 x 10-3 m

56. Capacitor - 1
A parallel plate capacitor consists of two parallel metal plates placed near one another but not touching
In general, a capacitor consists of two conductors of any shape placed near one another without touching
It is common practice to fill the region between the conductors or plates with an electrically insulating material called a dielectric

57. Capacitor - 2 A capacitor stores electric charge
Each capacitor plate carries a charge of the same magnitude
One positive and the other negative
Because of the charges, the electric potential of the positive plate exceeds that of the negative plate by an amount V
When the magnitude q of the charge on each plate is doubled, the magnitude V of the electric potential difference is also doubled
q is proportional to V

58. Relationship between q and V
The magnitude q of the charge on each plate of a capacitor is directly proportional to the magnitude V of the potential difference between the plates
q = CV
C is the capacitance
SI Unit of Capacitance: coulomb/volt = farad (F)

59. Capacitance One farad is an enormous capacitance
Usually smaller amounts, such as a microfarad, are used in electric circuits

60. Dielectric - 1
A dielectric is inserted between the plates of a capacitor
The capacitance increases markedly because of the way in which the dielectric alters the electric field between the plates

61. Dielectric - 2 The region between the charged plates is empty
The field lines point from the positive toward the negative plate

62. Dielectric - 3
The charge on the plates remains constant as the dielectric is inserted
In many materials the molecules possess permanent dipole moments, even though the molecules are electrically neutral
The dipole moment exists because one end of a molecule has slight excess of negative charge while the other end has a slight excess of positive charge
When such molecules are placed between the charged plates of the capacitor, the negative ends are attracted to the positive plate and the positive ends are attracted to the negative plate
As a result, the dipolar molecules tend to orient themselves end-to-end
Left surface of the dielectric is positively charged and the right surface is negatively charged

63. Dielectric - 4
Because of the surface charges on the dielectric, not all the electric field lines generated by the charges on the plates pass through the dielectric
Some of the field lines end on the negative surface charges
And begin again on the positive surface charges
Thus, the electric field inside the dielectric is less strong than the electric field inside the empty capacitor, assuming the charge on the plates remains constant
This reduction in the electric field is described by the dielectric constant , which is the ratio of the field magnitude E0 without the dielectric to the field magnitude E inside the dielectric:
 = E0 / E
Moreover, since the field E0 without the dielectric is greater than the field E inside the dielectric; the dielectric constant is greater than unity

64. Capacitance of Parallel Plate Capacitor
Given a parallel plate capacitor in which the area of each plate is A and the separation between the plates is d
The magnitude of the electric field inside the dielectric is given by as
E = V/d
V is the magnitude of the potential difference between the plates
The electric field inside the dielectric is related to the electric field in the absence of the dielectric
E = E0/ = V/d
Since the electric field within an empty capacitor is
E0 = q / (0A),
it follows that
q / (0A) = V/d
which can be solved for q to give
q = (0A / d)V
A comparison of this express with q = CV reveals that the capacitance C is
Parallel place capacitor filled with a dielectric C = (0A/d)
With C0 representing the capacitance of the empty capacitor ( = 1)
C = C0
In other words, the capacitance with the dielectric present is increased by a factor of  over the capacitance without the dielectric
The relation C = C0 applies to any capacitor, not just to a parallel plate capacitor

65. Energy Storage in Capacitor
Energy = ½ qV
Energy = ½ (CV)V = ½ CV2
Energy = ½ q(q/C) = q2/2C
Energy = ½ CV2 = ½ (0A / d)(Ed)2
Energy Density=Energy/Volume = ½0E2