2. 1. Electric Potential Energy
When an electrostatic force acts between two or more charged particles within a
system of particle, we can assign an electric potential energy π to the system. If the
system changes its configuration from state π to state π, the electrostatic force does
work π on the particles. The resulting change βπ in potential energy of the system
is
βπ = ππ β ππ = βπ.
The work done by the electrostatic force is path independent, since the
electrostatic force is conservative.
We take the reference configuration of a system of charged particles to be that in
which the particles are all infinitely separated from each other. The energy of this
reference configuration is set to be zero.
2
3. 1. Electric Potential Energy
Suppose that several charged particles come together from initially infinite
separations (state π) to form a system of neighboring particles (state π). Let the
initial potential energy ππ be zero and let πβ represents the work done by the
electrostatic force between the particles during the process. The final potential
energy π of the system is
π =βπβ.
3
5. 1. Electric Potential Energy
Example 1: Electrons are continually being knocked out
of air molecules in the atmosphere by cosmic-ray
particles coming in from space. Once released, each
electron experiences an electrostatic force πΉ due to
the electric field πΈ that is produced in the atmosphere
by charged particles already on Earth. Near Earthβs
surface the electric field has the magnitude πΈ = 150 N
/C and is directed downward. What is the change βπ
in the electric potential energy of a released electron
when the electrostatic force causes it to move vertically
upward through a distance π = 520m?
5
6. 1. Electric Potential Energy
The work π done on the electron by the electrostatic
force is
π = πΉ β π = πΉπ cos 0 = πΈππ
= 150 N/C 1.60 Γ 10β19 C 520 m
= 1.2 Γ 10β14 J.
βπ = βπ = β1.2 Γ 10β14 J.
6
7. 2. Electric Potential
The potential energy of a charged particle in an electric field depends on the charge
magnitude. The potential energy per unit charge, however, has a unique value at
any point in an electric field.
For example, a proton that has potential energy of 2.40 Γ 10β17 J in an electric
field has potential per unit charge of
2.40 Γ 10β17 J
1.60 Γ 10β19 C
= 150 J/C.
Suppose we next replace the proton with an alpha particle (charge +2π). The alpha
particle would have electric potential energy of 4.80 Γ 10β17 J. However, the
potential energy per unit charge would be the same 150 J/C.
7
8. 2. Electric Potential
The potential energy per unit charge π/π is a characteristic only of the electric
field. The potential energy per unit charge at a point in an electric field is called the
electric potential π (or simply potential) at that point. Thus,
π
π
π = .
Note that π is a scalar, not a vector.
The electric potential difference βπ between any two points π and π in an electric
field is equal to the difference in potential energy per unit charge between the two
points:
ππ ππ βπ
βπ = ππ βππ =
π
β
π
=
π
.
8
9. 2. Electric Potential
Substituting βπ for βπ we find that
π
βπ = ππ β ππ = β
π
.
The potential difference between two points is the negative of the work done by
the electrostatic force on a unit charge as it moved from one point to the other.
If we set ππ = 0 at infinity as our reference potential energy, π must be zero there
too. The electric potential at any point in an electric field is thus
π π
π πβ
π = = β .
The SI unit for potential π is joule per coulomb or volt (V):
1 volt = 1 joul per coulomb.
9
10. 2. Electric Potential
In terms of this unit, we can write the electric field in a more convenient unit:
1
N
C
= 1
N 1 V 1 J V
C 1 J/C 1 N β m m
= 1 .
We can also define an energy unit that is convenient for energy scale of the atomic
physics; the electron-volt (eV). One electron-volt is the energy equal to the work
required to move a single elementary charge π through a potential difference of
one volt (πβπ). Thus
1 eV = π 1 V = 1.60 Γ 10β19 C 1 J/C = 1.60 Γ 10β19 J.
10
11. 2. Electric Potential
Work Done by an Applied Force
Suppose we move a particle of charge π from point π to point π in an electric field
by applying a force to it. By the work-kinetic energy theorem,
βπΎ = πapp +π,
where πapp is the work done by the applied force and π is the work done by the
electric field.
If βπΎ = 0, we obtainthat
πapp = βπ.
Using that βπ = βπ, we obtainthat
πapp = βπ = πβπ.
11
13. 3. Equipotential Surfaces
An equipotential surface is composed of adjacent points that have the same
electric potential. No net work π is done on a charged particle by an electric field
when the particleβs initial point π and final point π lie on the same equipotential
surface.
13
16. 3. Equipotential Surfaces
Equipotential surfaces are perpendicular to the electric field πΈ.
If πΈ were not perpendicular to an equipotential surface, it would have a component
along the surface. This component would then do work on a charged particle as it
moved along the surface.
16
17. 4. Calculating the Potential from the Field
We can calculate the potential difference
between any two points π and π if we know
the electric field vector πΈ along any path
connecting those points.
Consider the situation shown in the figure.
The differential work ππ done on the
particle by the electric field as it moves
through a differential displacement ππ is
ππ = πΉ β ππ = π0πΈ β ππ .
17
18. 4. Calculating the Potential from the Field
The total work π done between points π
and π is
π
π = π0 πΈ β ππ .
π
Using that π = ββπ = βπ0βπ we obtain
that
π
ππ β ππ = β πΈ β ππ .
π
18
19. 4. Calculating the Potential from the Field
If we set ππ= 0 and rewrite ππas π we get
π
π = β πΈ β ππ .
π
This equation gives us the potential π at
any point π in an electric field relative to
zero potential at point π. If we let point π be
at infinity, this equation gives us the
potential π at any point π relative to zero
potential at infinity.
19
20. 4. Calculating the Potential from the Field
If we set ππ= 0 and rewrite ππas π we get
π
π = β πΈ β ππ .
π
This equation gives us the potential π at
any point π in an electric field relative to
zero potential at point π. If we let point π be
at infinity, this equation gives us the
potential π at any point π relative to zero
potential at infinity.
20
21. 4. Calculating the Potential from the Field
Uniform Field: Letβs find the potential
difference between points π and π in the
case of a uniform electric field. We obtain
π π
ππ β ππ = β πΈ β ππ = β πΈππ cos0
π π
π
= βπΈ ππ =βπΈβπ₯.
π
The is the change in voltage βπ between two
equipotential lines in a uniform field of
magnitude πΈ, separated by distance βπ₯
βπ =βπΈβπ₯.
21
22. 4. Calculating the Potential from the Field
The electric field vector points from higher
potential toward lower potential. This is true
for any electric field.
22
23. 4. Calculating the Potential from the Field
(a) Rightward. (b) Positive: 1,2,3,5. Negative: 4. (c) 3 then 1,2,5 tie, then 4.
πapp = πβπ = βπβπ
23
24. 4. Calculating the Potential from the Field
Example 2:
(a) The figure shows two points π and π in a uniform electric
field πΈ. The points lie on the same electric field line (not
shown) and are separated by a distance π. Find the potential
difference ππβ ππby moving a positive test charge π0 from π to
π along the path shown, which is parallel to the field direction.
πΈ β ππ = πΈππ cos 0 = πΈππ .
π π
ππ β ππ = β πΈ β ππ = βπΈ ππ =βπΈπ.
π π
The potential always decreases along a path that extends in
the direction of the electric field lines.
24
25. 4. Calculating the Potential from the Field
Example 2:
(b) Now find the potential difference ππβ ππby moving
the positive test charge π0 from π to π along the path π
ππ shown in the figure.
ππ β ππ = ππ βππ.
Along path ππ
πΈ β ππ = πΈππ cos 45Β°.
π π
ππ β ππ = β πΈ β ππ = βπΈ cos 45Β° ππ
π
= βπΈ cos 45Β° 2π
π
= βπΈπ.
25
26. 5. Potential Due to a Point Charge
We here want to derive an expression for the electric
0
potential π for the space around a charged particle, relative
to zero potential at infinity.
Todo that, we move a positive test charge π0 from point π to
infinity.
We choose the simplest path connecting the two pointsβa
radial straight line. Evaluating the dot product gives
π ππ
πΈ β ππ = πΈππ cos 0 = πΈ ππ = .
4ππ π2
26
27. 5. Potential Due to a Point Charge
4ππ0
π ππ π
π2 = β
4ππ 0
β ππ
π2
The line integral becomes
π
ππβ ππ= β
π
π
4ππ0 π π
1 β
= β β = β
π
π 1
4ππ0 π
.
Setting ππ= 0 (at β), ππ= π and switching π to π we get
π =
1 π
4ππ0 π
.
Note that π can be positive or negative and that π has the
same sign as π.
27
28. 5. Potential Due to a Point Charge
A positively charged particle produces a positive electric
potential. A negatively charged particle produces a negative
electric potential.
Also, note that the expression for π holds outside or on the
surface of a spherically symmetric charge distribution.
28
29. 6. Potential Due to a Group of Point Charges
The net potential due to π point charges at a given point is
π
=1
π
π =
1
ππ=
4ππ
π
π
=1
ππ
0
π
.
π
Here ππ is the charge (positive or negative) of the πth particle, and ππis the radial
distance of the given point from the πth charge.
An important advantage of potential over electric field is that we sum scalar
quantities instead of vector quantities.
29
31. 6. Potential Due to a Group of Point Charges
Example 3: What is the electric potential at point P,
located at the center of the square of point charges
shown in the figure? The distance π = 1.3 m, and the
charges are
π1 = 12 nC,
π3 = β24 nC,
π2 = 31 nC,
π4 = 17 nC.
31
32. 6. Potential Due to a Group of Point Charges
Example 3: What is the electric potential at point P,
located at the center of the square of point charges
shown in the figure? The distance π = 1.3 m, and the
charges are
π1 = 12 nC,
π3 = β24 nC,
π2 = 31 nC,
π4 = 17 nC.
π
=1
π
4
π = π =
4ππ0
1 π1
+
π2
+
π3
+
π4
π π π π
=
1 π1 + π2 + π3 + π4
4ππ0 π
32
33. 6. Potential Due to a Group of Point Charges
π1 + π2 + π3 +π4 = 12 + 31 β 24 + 17 Γ 10β9 C
= 36 Γ 10β9 C.
2
2
π = π.
109N β m2
C2
3.6 Γ 10β8 C
1.3/ 2m
Substituting gives
π = 8.99 Γ
= 350 V.
33
34. 6. Potential Due to a Group of Point Charges
Example 4: (a) In the figure electrons (of charge βπ) are
equally spaced and fixed around a circle of radius π .
Relative to π = 0 at infinity, what are the electric potential
and electric field at the center πΆ of the circle due to these
electrons?
12
π = ππ= β
π=1
1 12π
4ππ0 π
By the symmetry of the problem, πΈ =0.
34
35. 6. Potential Due to a Group of Point Charges
Example 4: (b) If the electrons are moved along the circle
until they are nonuniformly spaced over a 120Β° arc (see
the figure), what then is the potential at πΆ? How does the
electric field at πΆ change (if at all)?
The potential is not changed because the distances
between πΆ and each charge are not changed.
The electric field is no longer zero, however, because the
arrangement is no longer symmetric. A net field is now
directed toward the charge distribution.
35
36. 7. Potential Due to an Electric Dipole
The potential due to an electric dipole is the net potential
due to the positive and negative charges.
At point π, the net potential is
2
π = ππ= π +
π=1
+ πβ
0
1 π π
= β
4ππ π π
+ β
=
4ππ0
π π β β π +
π + π β
.
36
37. 7. Potential Due to an Electric Dipole
For π β« π, we can make the following approximation (see
the figure):
π β β π + β π cos π ,
and
π β π + β π2.
Substituting in the expression for π gives
π =
4ππ0
π π cosπ
π2
=
4ππ0
1 π cosπ
π2
.
37
38. 7. Potential Due to an Electric Dipole
Point π, π then π.
π =
4ππ0
1 π cosπ
38
π2
41. 8. Calculating the Field from the Potential
Let us see graphically how we can calculate the field from the potential. If we know
the potential π at all points near a charge distribution, we can draw a family of
equipotential surfaces. The electric field lines, sketched perpendicular to those
surfaces, reveal the variation of πΈ.
We want now to write the mathematical equivalence of this graphical procedure.
41
42. 8. Calculating the Field from the Potential
The figure shows cross sections of a family of closely
spaced equipotential surfaces.
The potential difference between each pair of
adjacent surfaces is ππ. The electric field πΈ at any
point π is perpendicular to the equipotential surface
through π.
Suppose that a positive test charge π0 moves
through a displacement ππ from an equipotential
surface to the adjacent surface.
The work ππ done by the electric field on the test
charge is
ππ =βπ0ππ.
42
43. 8. Calculating the Field from the Potential
The work ππ can also be written as
ππ = πΉ β ππ = π0πΈ β ππ = π0πΈππ cos π.
Equating the two expression for ππ yields
βπ0ππ = π0πΈππ cos π ,
or
πΈ cos π = β
ππ
ππ
.
πΈ cos π is the component of πΈ in the direction of
ππ . Thus,
ππ
πΈπ = β
ππ
.
43
44. 8. Calculating the Field from the Potential
The component of πΈ in any direction is the negative
of the rate at which the electric potential changes
with distance in that direction.
If we take ππ to be along the π₯, π¦ and π§ axes, we
obtain the π₯, π¦ and π§ components of πΈ at any point:
ππ
πΈπ₯ = β
ππ₯
;
ππ
πΈπ¦ = β
ππ¦
;
ππ
πΈz = β
ππ§
.
If we know the function π π₯, π¦, π§ , we can find the
components of πΈ at any point.
44
45. 8. Calculating the Field from the Potential
When the electric field πΈ is uniform, we write that
βπ
βπ
πΈ = β .
where π is perpendicular to the equipotential surfaces. The component of the
electric field is zero in any direction parallel to the equipotential surfaces because
there is no change in potential along an equipotential surface.
45
46. 8. Calculating the Field from the Potential
πΈ =β
βπ
βπ
(a) 2, then 3 & 1.
(b) 3.
(c) Accelerates
leftward.
46
47. 8. Calculating the Field from the Potential
Example 5: Starting with the expression for the potential due to an electric dipole,
π =
4ππ0
1 π cosπ
π2
.
derive an expression for the radial component of electric field due to the electric
dipole.
πΈπ = β
ππ
= β
4ππ0
ππ π cos π π 1
= β
4ππ0
π cos π 2
π3
β =
2ππ0
ππ π2
1 π cosπ
π3
.
47
48. 8. Calculating the Field from the Potential
Example 6: The electric potential at any point on the central axis of a uniformly
charged disk is given by
π =
π
2π0
π§2 + π 2 β π§ .
Starting with this expression, derive an expression for the electric field at any point
on the axis of the disk.
π§
ππ§ 2π0 ππ§
ππ π π
πΈ = β = π§ β π§2 +π 2
=
2π0
π π§
π§2 + π 2
1 β .
48
49. 9. Electric Potential Energy of a System of
Charges
The electric potential energy of a system of fixed point charges is equal to the work
that must be done by an external agent to assemble the system, bringing each
charge in from an infinite distance.
We assume that the charges are stationary at both the initial and final states.
50
50. 9. Electric Potential Energy of a System of
Charges
Let us find the electric potential energy of the system
shown in the figure. We can build the system by bringing
π2 from infinity and put it at distance π from π1. The work
we do to bring π2 near π1 is π2π, where π is the potential
that has been set up by π1. Thus, the electric potential of
the pair of point charges shown in the figure is
π = πapp = π2π =
4ππ0 π
1 π1π2
.
π has the same sign as that of π1π2.
The total potential energy of a system of several particles
is the sum of the potential energies for every pair of
particles in the system.
51
51. 9. Electric Potential Energy of a System of
Charges
Example 7: The figure shows three point charges held in
fixed positions by forces that are not shown. What is the
electric potential energy π of this system of charges?
Assume that π = 12 cm andthat
π1 = +π, π2 = β4π, π3 = +2π,
and π = 150nC.
The electric potential energy of the system (energy required
to assemble the system) is
π = π12 + π13 +π23
=
1 π1π2
+
1 π1π3
+
1 π2π3
.
4ππ0 π 4ππ0 π 4ππ0 π
52
52. 9. Electric Potential Energy of a System of
Charges
π =
4ππ0
+ +
1 π β4π 1 π 2π 1 β4π 2π
4ππ0 π π 4ππ0 π
= β
4ππ0π
= 8.99 Γ 1012
10π2 N β m2
C2
10 150 Γ 10β9C
0.12 m
= β17 mJ.
53
53. 9. Electric Potential Energy of a System of
Charges
Example 8: An alpha particle (two protons, two neutrons)
moves into a stationary gold atom (79 protons, 118
neutrons), passing through the electron region that
surrounds the gold nucleus like a shell and headed
directly toward the nucleus. The alpha particle slows until
it momentarily stops when its center is at radial distance π
= 9.23 fm from the nuclear center. Then it moves back
along its incoming path. (Because the gold nucleus is
much more massive than the alpha particle, we can
assume the gold nucleus does not move.) What was the
kinetic energy πΎπof the alpha particle when it was initially
far away (hence external to the gold atom)? Assume that
the only force acting between the alpha particle and the
gold nucleus is the (electrostatic) Coulomb force.
54
54. 9. Electric Potential Energy of a System of
Charges
During the entire process, the mechanical energy of the
alpha particle-gold atom system is conserved.
πΎπ + ππ = πΎπ + ππ.
ππ is zero because the atom is neutral. ππ is the electric
potential energy of the alpha particle-nucleus system. The
electron shell produces zero electric field inside it. πΎπ is
zero.
πΎπ = ππ =
1 2π 79π
4ππ0 9.23 Γ 10β15 m
= 3.94 Γ 10β9 J
= 24.6 MeV.
55
55. 10. Potential of a Charged Isolated Conductor
In Ch. 23, we concluded that πΈ = 0 within an isolated conductor. We then used
Gaussβ law to prove that all excess charge placed on a conductor lies entirely on its
surface. Here we use the first of these facts to prove an extension of the second:
βAn excess charge placed on an isolated conductor will distribute itself on the
surface of that conductor so that all points of the conductorβwhether on the
surface or insideβcome to the same potential. This is true even if the conductor
has an internal cavity and even if that cavity contains a net charge.β
Toprove this fact we use that
π
ππ β ππ = β πΈ β ππ .
π
Since πΈ = 0 for all point within a conductor, ππ= ππfor all possible pairs of points π
and π in the conductor.
56
56. 10. Potential of a Charged Isolated Conductor
The figure shows the potential against radial
distance π from the center from an isolated
spherical conducting shell of radius π = 1.0 m and
charge π = 0.10πC.
For points outside the shell up to its surface, π π
= π π/π. Now assume that we push a test charge
though a hole in the shell. No extra work is required
to do this because πΈ becomes zero once the test
charge gets inside the shell. Thus, the potential at
all points inside the shell has the same value as that
of its surface.
57
57. 10. Potential of a Charged Isolated Conductor
The figure shows the variation of the electric field
with radial distance for the same shell. Everywhere
inside the shell πΈ = 0. Outside the shell πΈ = π π
/π2. The electric field can be obtained from the
potential using πΈ =βππ/ππ.
58
58. 10. Potential of a Charged Isolated Conductor
Spark Discharge from a Charged Conductor
On nonspherical conductors, a surface charge does
not distribute itself uniformly over the surface of
the conductor. At sharp points or sharp edges, the
surface charge densityβand thus the external
electric field, which is proportional to itβmay reach
very high values. The air around such sharp points
or edges may become ionized, producing the
corona discharge.
59
59. 10. Potential of a Charged Isolated Conductor
Isolated Conductor in an External Electric Field
If an isolated conductor is placed in an external
electric field, all points of the conductor still come
to a single potential regardless of whether the
conductor has an excess charge or not. The free
conduction electrons distribute themselves on the
surface in such a way that the electric field they
produce at interior points cancels the external
electric field. Furthermore, the electron distribution
causes the net electric field at all points on the
surface to be perpendicular to the surface.
64