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PHY 184
Spring 2007
Lecture 9
Title: The Electric Potential
1/23/07
184 Lecture 9
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Announcements
 Homework Set 2 done, Set 3 ongoing and Set 4 will
open on Thursday
 Helproom hours of the TAs are listed on the
syllabus in LON-CAPA
• Honors Option students will provide help in the in the
SLC starting this week.
 Remember Clicker’s Law…
Up to 5% (but not more!)
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Review - Potential Energy
 When an electrostatic force acts between charged
particles, assign an electric potential energy, U.
 The difference in U of the system in two different
states, initial i and final f, is
U  U f  Ui
 Reference point: Choose U=0 at infinity.
 If the system is changed from initial state i to the
final state f, the electrostatic force does work, W
U  U f  Ui  W
 Potential energy is a scalar.
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Review - Work
 Work done by an electric field
 
W  F d
so


F  qE
and
 
W  qE  d  qEd cos 
Q is the angle between electric field and displacement
(1) Positive W  U decreases
(2) Negative W  U increases
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Clicker Question
 In the figure, a proton moves
from point i to point f in a
uniform electric field directed
as shown. Does the electric
field do positive, negative or no
work on the proton?
A: positive
B: negative
C: no work is done on the proton
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Clicker Question
 In the figure, a proton moves
from point i to point f in a
uniform electric field directed
as shown. Does the electric
field do positive, negative or no
work on the proton?
B: negative
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Electric Potential
V
 The electric potential, V, is defined as the electric potential
energy, U, per unit charge
U
V
q
 The electric potential is a characteristic of the electric
field, regardless of whether a charged object has been
placed in that field. (because U  q)
 The electric potential energy is an energy of a charged
object in an external electric field (or more precisely, an
energy of the system consisting of the charged object and
the external field).
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Electric Potential Difference
V
 The electric potential difference between an initial
point i and final point f can be expressed in terms
of the electric potential energy of q at each point
Ui U
V  Vf  Vi 


q
q
q
Uf
 Hence we can relate the change in electric
potential to the work done by the electric field on
the charge
W
V  
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e
q
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Electric Potential Difference (2)
 Taking the electric potential energy to be zero at
infinity we have
Explain: i =  , f = x, so that
We,
V 
V = V(x)  0
q
where We, is the work done by the electric field
on the charge as it is brought in from infinity.
 The electric potential can be positive, negative, or
zero, but it has no direction. (i.e., scalar not vector)
 The SI unit for electric potential is
joules/coulomb, i.e., volt.
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The Volt
 The commonly encountered unit joules/coulomb is called the
volt, abbreviated V, after the Italian physicist Alessandro
Volta (1745 - 1827)
1J
1V=
1C
 With this definition of the volt, we can express the units of
the electric field as
[F ] N J/m V
[E ] 
 

[q ] C
C m
 For the remainder of our studies, we will use the unit V/m
for the electric field.
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Example - Energy Gain of a Proton
 A proton is placed between two parallel
conducting plates in a vacuum as shown.
The potential difference between the two
plates is 450 V. The proton is released
from rest close to the positive plate.
 What is the kinetic energy of the proton
when it reaches the negative plate?
The potential difference between the two plates is 450 V.
-
+
= V(+)-V()
The change in potential energy of the proton is U,
and V = U / q (by definition of V), so
U = q V = e[V()V(+)] = 450 eV
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Example - Energy Gain of a Proton (2)
Conservation of energy
K =  U = + 450 eV
initial
final
Because the proton started at rest,
K = 1.6x10-19 C x 450 V = 7.2x10-17 J


Because the acceleration of a charged particle across a potential difference is
often used in nuclear and high energy physics, the energy unit electron-volt (eV)
is common.
An eV is the energy gained by a charge e that accelerates across an electric
potential of 1 volt
1 eV  1.6022 1019 J

The proton in this example would gain kinetic energy of 450 eV = 0.450 keV.
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The Van de Graaff Generator
 A Van de Graaff generator is a device that creates high
electric potential.
 The Van de Graaff generator was invented by Robert J. Van
de Graaff, an American physicist (1901 - 1967).
 Van de Graaff generators can produce electric potentials up
to many 10s of millions of volts.
 Van de Graaff generators can be used to produce particle
accelerators.
 We have been using a Van de Graaff generator in lecture
demonstrations and we will continue to use it.
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The Van de Graaff Generator (2)




The Van de Graaff generator works
by applying a positive charge to a
non-conducting moving belt using a
corona discharge.
The moving belt driven by an
electric motor carries the charge
up into a hollow metal sphere where
the charge is taken from the belt
by a pointed contact connected to
the metal sphere.
The charge that builds up on the
metal sphere distributes itself
uniformly around the outside of the
sphere.
For this particular Van de Graaff
generator, a voltage limiter is used
to keep the Van de Graaff
generator from producing sparks
larger than desired.
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The Tandem Van de Graaff Accelerator
 One use of a Van de
Graaff generator is to
accelerate particles for
condensed matter and
nuclear physics studies.
 Clever design is the
tandem Van de Graaff
accelerator.
 A large positive electric
potential is created by a
huge Van de Graaff
generator.
 Negatively charged C ions
get accelerated towards
the +10 MV terminal (they
gain kinetic energy).
1/23/07
Stripper foil
strips electrons from C
C-1
C+6
Terminal at +10MV
Electrons are stripped from
the C and the now positively
charged C ions are repelled
by the positively charged
terminal and gain more kinetic
energy.
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Example - Energy of Tandem Accelerator
 Suppose we have a tandem Van de Graaff accelerator that
has a terminal voltage of 10 MV (10 million volts). We want
to accelerate 12C nuclei using this accelerator.
 What is the highest energy we can attain for carbon nuclei?
 What is the highest speed we can attain for carbon nuclei?
 There are two stages to the acceleration
• The carbon ion with a -1e charge gains energy
accelerating toward the terminal
• The stripped carbon ion with a +6e charge gains
energy accelerating away from the terminal
15 MV Tandem Van de Graaff at Brookhaven
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Example - Energy of Tandem Accelerator (2)
K   U   q1V  q2  V 
q1  1e
and
q2  6 e
K  7 e  10 MV  70 MeV
1.602  10-19 J
 1.12  1011 J
K  70 MeV 
1 eV
The mass of a 12C nucleus is 1.99 10 -26 kg
1 2
K  mv
2
v
2K

m
2 1.12 10 11 J
7

3.36
10
m/s
-26
1.99 10 kg
v  11% of the speed of light
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Equipotential Surfaces and Lines
 When an electric field is present, the electric potential has a given
value everywhere in space.
V(x) = potential function
 Points close together that have the same electric potential form an
equipotential surface.
i.e, V(x) = constant value
 If a charged particle moves on an equipotential surface, no work is done.
 Equipotential surfaces exist in three
dimensions.
 We will often take advantage
of symmetries in the electric potential
and represent the equipotential surfaces
as equipotential lines in a plane.
Equipotential surface from eight point charges
fixed at the corners of a cube
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General Considerations
 If a charged particle moves perpendicular to electric field
lines, no work is done.
if d  E
 If the work done by the electric field is zero, then the
electric potential must be constant
We
V  
 0  V is constant
q
 Thus equipotential surfaces and lines must always be
perpendicular to the electric field lines.
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Electric field lines and equipotential surfaces
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Constant Electric Field
 Electric field lines: straight lines parallel to E
 Equipotential surfaces (3D): planes perp to E
 Equipotential lines (2D): straight lines perp to E
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Electric Field from a Single Point Charge
 Electric field lines: radial lines emanating from the
point charge.
 Equipotential surfaces (3D): concentric spheres
 Equipotential lines (2D): concentric circles
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Electric Field from Two Oppositely Charged Point Charges
 The electric field lines from two oppositely charge point charges are a
little more complicated.
 The electric field lines originate on the positive charge and terminate on
the negative charge.
 The equipotential lines are always perpendicular to the electric field
lines.
 The red lines represent positive
electric potential.
 The blue lines represent negative
electric potential.
 Close to each charge, the equipotential
lines resemble those from a point
charge.
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ELECTRIC DIPOLE
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Electric Field from Two Identical Point Charges
 The electric field lines from two identical point charges are
also complicated.
 The electric field lines originate on the positive charge and
terminate at infinity.
 Again, the equipotential lines
are always perpendicular to
the electric field lines.
 There are only positive
potentials.
 Close to each charge, the
equipotential lines resemble
those from a point charge.
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TWO POSITIVE CHARGES
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Calculating the Potential from the Field
 To calculate the electric potential from the electric field
we start with the definition of the work dW done on a
particle with charge q by a 
force F over a displacement ds


dW  F  ds
In this case the force is provided by the electric field
 
F = qE
dW  qE  ds
 Integrating the work done by the electric force on the
particle as it moves in the electric field from some initial
point i to some final point f we obtain
W 
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f
i
 
q E  ds
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Calculating the Potential from the Field (2)
 Remembering the relation between the change in electric
potential and the work done …
 …we find
We
V  
q
f 

We
V  Vf  Vi  
   E  ds
i
q
 Taking the convention that the electric potential is zero at
infinity we can express the electric potential in terms of
the electric field as
 


V ( x )   E  ds
i
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( i = , f = x)
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Example - Charge moves in E field
 Given the uniform electric field E,
find the potential difference Vf-Vi by
moving a test charge q0 along the
path icf.
 Idea: Integrate Eds along the path
connecting ic then cf. (Imagine that
we move a test charge q0 from i to c
and then from c to f.)
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Example - Charge moves in E field
c
 
f 

E  ds   E  ds
Vf  Vi   
i
c
c 

 i E  ds  0 (ds perpendicu lar to E)
f 
f

 E  ds   E ds cos(45)  E  distance 
c
c
1
2
distance = sqrt(2) d by Pythagoras
Vf  Vi  Ed
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Clicker Question
 We just derived Vf-Vi for the path i -> c -> f.
What is Vf-Vi when going directly from i to f ?
A:
B:
C:
D:
0
-Ed
+Ed
-1/2 Ed
Quick: V is independent of path.
Explicit: V = -  E . ds =  E ds = - Ed
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