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Essential University Physics
Richard Wolfson
22
Electric Potential
PowerPoint® Lecture prepared by Richard Wolfson
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 22-1
In this lecture you’ll learn
• The concept of electric
potential difference
• Including the meaning of the
familiar term “volt”
• To calculate potential
difference between two points
in an electric field
• To calculate potential
differences of charge
distributions by summing or
integrating over point charges
• The concept of equipotentials
• How charge distributes itself
on conductors
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 22-2
Electric potential difference
• The electric potential difference between two points
describes the energy per unit charge involved in moving
charge between those two points.
B
• Mathematically, VAB  U AB q   A E  dr
where ∆VAB is the potential difference between points A and B,
and ∆UAB is the change in potential energy of a charge q moved
between those points.
• Potential difference is a property
of two points.
• Because the electrostatic field
is conservative, it doesn’t
matter what path is taken
between those points.
• In a uniform field, the potential
difference becomes
VAB   E  r
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 22-3
Clicker question
•
What would happen to the potential difference between
points A and B in the figure if the distance r were
doubled?
A. V would be doubled.
B. V would be halved.
C. V would be quadrupled
D. V would be quartered.
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 22-4
Clicker question
•
What would happen to the potential difference between
points A and B in the figure if the distance r were
doubled?
A. V would be doubled.
B. V would be halved.
C. V would be quadrupled
D. V would be quartered.
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 22-5
The volt and the electronvolt
• The unit of electric potential difference is the volt (V).
• 1 volt is 1 joule per coulomb (J/C).
• Example: A 9-V battery supplies 9 joules of energy to every
coulomb of charge that passes through an external circuit
connected between its two terminals.
• The volt is not a unit of energy, but of energy per charge—that
is, of electric potential difference.
• A related energy unit is the electronvolt (eV), defined as the energy
gained by one elementary charge e “falling” through a potential
difference of 1 volt.
• Therefore 1 eV is 1.6  10–19 J.
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 22-6
Clicker question
•
An alpha particle (charge 2e) moves through a 10-V
potential difference. How much work, expressed in eV,
is done on the alpha particle?
A. 5 eV
B. 10 eV
C. 20 eV
D. 40 eV
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 22-7
Clicker question
•
An alpha particle (charge 2e) moves through a 10-V
potential difference. How much work, expressed in eV,
is done on the alpha particle?
A. 5 eV
B. 10 eV
C. 20 eV
D. 40 eV
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 22-8
Clicker question
•
The figure shows three straight paths AB of the same
length, each in a different electric field. Which one of
the three has the largest potential difference between the
two points?
A. (a)
B. (b)
C. (c)
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 22-9
Clicker question
•
The figure shows three straight paths AB of the same
length, each in a different electric field. Which one of
the three has the largest potential difference between the
two points?
A. (a)
B. (b)
C. (c)
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 22-10
Potential differences in the field of a point
charge
• The point-charge field varies with
position, so potential differences in
the point-charge field must be found
by integrating.
• The result is
VAB
1 1
 kq   
 rA rB 
• Taking the zero of potential at infinity
gives

Vr  V r 
kq
r
for the potential difference between
infinity and any point a distance r from
the point charge.
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 22-11
Clicker question
•
You measure a potential difference of 50 V between two
points a distance 10 cm apart parallel to the field
produced by a point charge. Suppose you move closer
to the point charge. How will the potential difference
over a closer 10-cm interval be different?
A. The potential difference will remain the same.
B. The potential difference will increase.
C. The potential difference will decrease.
D. We cannot find this without knowing how much closer we are.
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 22-12
Clicker question
•
You measure a potential difference of 50 V between two
points a distance 10 cm apart parallel to the field
produced by a point charge. Suppose you move closer
to the point charge. How will the potential difference
over a closer 10-cm interval be different?
A. The potential difference will remain the same.
B. The potential difference will increase.
C. The potential difference will decrease.
D. We cannot find this without knowing how much closer we are.
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 22-13
Potential difference of a charge distribution
• If the electric field of the charge distribution is known,
potential differences can be found by integration as was
done for the point charge on the preceding slide.
• If the distribution consists of point charges, potential
differences can be found by summing point-charge
potentials:

kqi
• For discrete point charges, V P  
ri
i
where V(P) is the potential difference between infinity and a
point P in the electric field of a distribution of point charges q1,
q2, q3,…
k dq
.
• For a continuous charge distribution, V  P   
r
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 22-14
EXAMPLE: A point charge q1 is at
the origin, and a second point
charge q2 is on the x-axis at x
= a. Find the potential
everywhere on the x-axis.
x
The potential is positive everywhere
as charges are both positive
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 22-15
Discrete charges: the dipole potential
• The potential of an electric
dipole follows from
summing the potentials of its
two equal but opposite point
charges:
• For distances r large compared
with the dipole spacing 2a, the
result is
kpcos
V r, 
r2
where p = 2aq is the dipole
moment.
• A 3-D plot of the dipole
potential shows a “hill” for the
positive charge and a “hole”
for the negative charge.
 
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 22-16
Continuous distributions: a ring and a disk
• For a uniformly charged ring of
total charge Q, integration gives
the potential on the ring axis:
k dq k
V  x  
  dq 
r
r
kQ
x2  a2
• Integrating the potentials of
charged rings gives the potential
of a uniformly charged disk:


2kQ
V  x  2
x2  a2  x
a
• This result reduces to the infinitesheet potential close to the disk,
and the point-charge potential far
from the disk.
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 22-17
Potential difference and the electric field
• Potential difference involves an integral over the electric field.
• So the field involves derivatives of the potential.
• Specifically, the component of the electric
field in a given direction is the negative of the
rate of change (the derivative) of potential in
that direction.
• Then, given potential V (a scalar quantity) as a
function of position, the electric field (a vector
quantity) follows from
 V ˆ V ˆ V
E  
i
j
y
z
 x

kˆ 

The derivatives here are partial derivatives,
expressing the variation with respect to one
variable alone.
• This approach may be used to find the field
from the potential.
• Potential is often easier to calculate, since it’s a
scalar rather than a vector.
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 22-18
Equipotentials
• An equipotential is a surface on which the potential is
constant.
• In two-dimensional drawings, we
represent equipotentials by curves similar
to the contours of height on a map.
• The electric field is always perpendicular to the
equipotentials.
• Equipotentials for a dipole:
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 22-19
Clicker question
•
The figure shows cross sections through two
equipotential surfaces. In both diagrams the potential
difference between adjacent equipotentials is the same.
Which of these two could represent the field of a point
charge?
A. (a)
B. (b)
C. neither (a) nor (b)
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 22-20
Clicker question
•
The figure shows cross sections through two
equipotential surfaces. In both diagrams the potential
difference between adjacent equipotentials is the same.
Which of these two could represent the field of a point
charge?
A. (a)
B. (b)
C. neither (a) nor (b)
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 22-21
Charged conductors
• There’s no electric field inside a conductor in electrostatic
equilibrium.
• And even at the surface there’s no field component parallel to the
surface.
• Therefore it takes no work to move charge inside or on the surface
of a conductor in electrostatic equilibrium.
• So a conductor in electrostatic
equilibrium is an equipotential.
• That means equipotential
surfaces near a charged conductor
roughly follow the shape of the
conductor surface.
• That generally makes the
equipotentials closer, and
therefore the electric field
stronger and the charge density
higher, where the conductor
curves more sharply.
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 22-22
Summary
• Electric potential difference describes the work per unit charge
involved in moving charge between two points in an electric field:
B
VAB    E  dr
A
• The SI unit of electric potential is the volt (V), equal to 1 J/C.
• Electric potential always involves two points; to say “the potential at a point”
is to assume a second reference point at which the potential is defined to be
zero.
• Electric potential differences in the field of a point charge follow by
integration: V P  kq r , where the zero of potential is taken at infinity.
• This result may be summed or integrated to find the potentials of charge
distributions.
• The electric field follows from differentiating the potential: E    V iˆ  V
• Equipotentials are surfaces of constant potential.
 x
ˆj  V
y
z

kˆ 

• The electric field and the equipotential surfaces are always perpendicular.
• Equipotentials near a charged conductor approximate the shape of the conductor.
• A conductor in equilibrium is itself an equipotential.
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 22-23
Electric Potential Energy
CHECKPOINT: A proton moves from
point i to point f in a uniform
electric field directed as shown.
Does the electric field do
A. positive or
B. negative work on the proton?
Does the electric potential energy of the
proton
A. increase or
B. decrease?
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Answers:
B.
the field does
negative work
A.
the potential
energy increases
Slide 22-24
Chapter 23 Problem 39
A conducting sphere of radius a is surrounded by a
concentric spherical shell of radius b. Both are
initially uncharged. How much work does it take
to transfer charge from one to the other until
there is charge +Q on the inner sphere and –Q on
the outer shell?
Hint: draw a graph of potential as a function of q,
the charge transferred.
Copyright © 2007 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Slide 22-25