Physics 122B
Electricity and Magnetism
Lecture 11 (Knight: 29.5 to 29.7)
Electric Potential, Equipotential
Surfaces and E = -r V
April 20, 2007
Martin Savage
Lecture 11 Announcements
l Lecture HW Assignment #4 is due at 10 PM, next
Wednesday.
l Uncollected Exam 1 papers may be obtained from Helen
Gribble in room C136 PAB.
l Requests for regrades of Exam 1 should be written on a
separate sheet (see Syllabus) and taken to Helen Gribble in
room C136 PAB. They will be accepted until noon next
Wednesday.
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Physics 122B - Lecture 11
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Midterm 1 Stats
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The Electric Potential
In Chapter 25 we introduced
the concept of an electric field E,
which can be though of as a
normalized force, i.e., E = F/q,
the field E that would produce a
force F on some test charge q.
We can similarly define the
electric potential V as a chargenormalized potential energy, i.e.,
V=Uelec/q, the electric potential V
that would give a test charge q an
electric potential energy Uelec
because it is in the field of some
other source charges.
We define the unit of electric potential as the volt: 1 volt = 1 V =
1 J/C = 1 Nm/C. Other units are: kV=103 V, mV=10-3 V, and mV=10-6 V.
Example: A D-cell battery has a potential of 1.5 V between its terminals.
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What Good is
the Electric Potential?
Like the electric field E, the electric potential V is an abstract idea. It
offers an advantage, however, because it is a scalar quantity while E is a
vector, yet the two can be converted to each other. It is useful because:
l The electric potential V depends only on the charges and their geometries.
The electric potential is the “ability” of the source charges to have an
interaction if a charge q shows up. The potential is present in all space,
whether or not a charge is there to experience it.
l If we know the electric potential V throughout a region of space, we’ll
immediately know the potential energy U = qV of any charge q that enters
that region.
Gaining Potential Energy
Losing Kinetic Energy
Losing Potential Energy
Gaining Kinetic Energy
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Example: Moving Through a
Potential Difference
A proton with a speed of vi = 2.0 x 105 m/s enters a region of space
where source charges have created an electric potential.
What is the proton’s speed after it has moved through a potential
difference of DV = 100 V?
K f  qV f  K i  qVi
K f  Ki  qVi  qV f  Ki  qDV
1
2
What is vf if the proton is
replaced by an electron?
mv f  mvi  qDV
2
1
2
2
2e
2
v f  vi  DV  1.44 105 m/s
m
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2e
v f (electron)  vi 
DV
me
Physics 122B - Lecture 11
2
 5.93 106 m/s
6
The Electric Potential Inside
a Parallel Plate Capacitor
Consider a parallel-plate capacitor with
d  3.0 mm,  =4.42 10-9 C/m2


E   ,  to -    500 N/C, to right 
 0

U elec  U q+sources
 qEs for a charge q,
located at s from the - plate
(with U0=0)
V  U elect / q  Es (V inside a parallel-plate
capaitor with V=0 @ - plate)
DVC  V  V  Ed  1.5 V
DV
d
E  C ; DVC  Ed 
d
0
Note that 1 N/C = 1 V/m.
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Graphical Representations
of Electric Potential
Distance from + plate
DVC
x

V  Es 
(d  x)  DVC 1  
d
 d
This linear relation can be represented as a graph, a set of
equipotential surfaces, a contour plot, or a 3-D elevation graph.
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Field Lines and Contour Lines
Field lines and equipotential
contour lines are the most
widely used representations to
simultaneously show the E field
and the electric potential. The
figure shows the field lines and
equipotential contours for a
parallel plate capacitor.
Remember that for both the
field lines and contours , their
spacing, etc, is a matter of
choice.
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Rules for Equipotentials
1. Equipotentials never intersect
other equipotentials. (Why?)
2. The surface of any static
conductor is an equipotential
surface. The conductor volume
is all at the same potential.
3. Field line cross equipotential
surfaces at right angles. (Why?)
4. Dense equipotentials indicate a
strong electric field. The potential V decreases in the direction
in which the electric field E points, i.e., energetically “downhill”
for a + charge
5. For any system with a net charge, the equipotential surfaces
become spheres at large distances.
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Batteries and Capacitors
How can we arrange for a capacitor
to have a surface charge density of
precisely =4.42 x 10-9 C/m2, as in the
previous example?
In Chapter 28 we introduced
batteries, which are chemical sources
of constant electric potential
difference. By choosing a battery
that supplies a potential difference
E = d/e0, and by arranging the
plate-separation, we can place any
desired charge density  on capacitor
plates.
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Example: A Proton in a Capacitor
A parallel plate capacitor is made of two
2.0 cm diameter disks spaced 2.0 mm
apart. It is charged to a potential
difference of 500 V.
(a) What is the E field in the gap?
(b) How much charge is on each plate?
(c) A proton is shot through a small hole in
the negative plate with a speed of
v=2.0x105 m/s. Does it reach the other
side? If not where is the turning point?
E
E
DVC
(500 V)

 2.5 105 V/m
3
d
(2.0 10 m)

Q

0 0 A
Q   0 EA   0 R 2 E  6.95 10-10 C
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V  DVC (1  x / d )
K f  eV f  Ki  eVi
0  eDVC (1  x f / d )  12 mvi 2  0
m vi 2 d
xf  d 
 1.16 mm
2eDVC
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Choice of the V=0 Point
In the previous
example we assumed
that the negative
plate of the capacitor
was the V=0 point.
0V
However, we could
just as well have
placed the V=0 point
at the right plate, or
half way between the
two plates, since only
the potential
difference DVC
matters in
calculations.
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Question
Which ranking of the potentials at
points a-e is correct? (Ignore edge
effects.)
(a) Va>Vb>Vc>Vd>Ve
(b) Va>Vb=Vc>Vd=Ve
(c) Va=Vb>Vc>Vd=Ve
(d) Va=Vb=Vc=Vd=Ve
(e) Vb>Va>Vc>Ve>Vd
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The Electric Potential
of a Point Charge
U q'+q
1
qq '

4 0 r
U q'+q
1 q
V

q'
4 0 r
Example: q = 1 nC, r = 1 cm;
1 q
V
4 0 r
-9
C)
9
2
2 (1.0  10
 (9.0 10 Nm /C )
(1.0 10-2 m)
 900 V
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Then divide Uq’+q by q’.
15
Visualizing the Potential
of a Point Charge
The potential of a point charge can be
represented as a graph, a set of equipotential
surfaces, a contour map, or a 3-D elevation
graph.
Usually it is represented by a graph or a
contour map, possibly with field lines.
+
Spherical Shells
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Question
Which ranking of the potentials
differences is correct?
(a) DV12>DV23>DV13
(b) DV12<DV23<DV31
(c) DV12<DV23=DV13
(d) DV12=DV23>DV13
(e) DV12=DV23=DV13
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The Electric Potential
of a Charged Sphere
V
1 Q
(sphere of charge Q with r  R)
4 0 r
(same as for point charge Q at center)
1 Q
4 0 R
(potential at surface of sphere)
V0  V ( R) 
Q  4 0 RV0 and V  V0
Q
R
R
r
For a spherical shell of charge or a solid conductor,
V  V0 inside because E drops to zero inside the surface.
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Example: A Proton and
a Charged Sphere
A proton is released from
rest on the surface of
a 1.0 cm diameter sphere
that has been charged
to +1000 V.
(a) What is the charge of
the sphere?
(b) What is the proton’s speed after it travels 1.0 cm from the sphere?
Q  4 0 RV0  (.005 m)(1000 V) /(9.0 109 Nm 2 /C2 )  0.56 nC
K f  eV f  Ki  eVi
1
2
mv f 2 
eR
V0  0  eV0
rf
2eV0  R 
5
vf 
1    3.57 10 m/s
m  rf 
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The Electric Potential
of Many Charges
The principle of superposition allows us to
calculate the potentials created by many point
charges and then add the up. Since the potential V
is a scalar quantity, the superposition of potentials
is simpler than the superposition of fields.
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Physics 122B - Lecture 11
qi
V 
i 4 0 ri
1
20
Example: The Potential
of Two Charges
What is the potential at point p?
p
1 q1
1 q2
1  q1 q2 


  
4 0 r1 4 0 r2 4 0  r1 r2 
 (2.0 10-9 C) (1.0 10-9 C) 
9
 (9.0 10 Vm/C) 


(0.050
m)
(0.040
m)


Vp 
 135 V
Note that:
1/40 = 9.0 x 109 Nm2/C2
= 9.0 x 109 Vm/C,
which, for problems like this,
are more convenient units.
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Example: The Potential
of a Ring of Charge
Find the potential of
a thin uniformly charged
ring of radius R and
charge Q at point P on
the z axis?
rP  R 2  z 2 and dQ 
dV 
Q
Rd
2 R
1
dQ
1
Q

d
2
2
4 0 rP
4 0 2 R  z
VP   dV 
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1
Q
4 0 2 R 2  z 2
2
 d 
0
1
Q
4 0
R2  z 2
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Example: The Potential
of a Disk of Charge
Find the potential of
a uniformly charged disk
of radius R and charge Q
at point P on the z axis?
dV 
1
dQ
4 0
ri 2  z 2
dQ  Q
2 ri dri
2ri dri

Q
 R2
R2
VP   dV 
VP 
Q
2 0 R 2
R
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2
2
ri dri

ri 2  z 2
0
 r z
i
2 0 R 
Q
P
R
2
0

Q
2 0 R
R2  z 2  z
 V0
R
Physics 122B - Lecture 11
R2  z 2  z
R
23
Potential of a Disk of Charge
Von z axis 
Q
2 0 R 2

R2  z 2  z

 Q  R2  z 2  z


2

R
R
0


Von z axis  V0
R2  z 2  z
R
1   R / z  1
2
 V0
R / z
 1  1  R / z 2   1 
2

   1V R  1 Q
 lim V0 
 2 0
z 
R
/
z
z 4 0 z
 




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Example: The Potential of a Dime
A dime (diameter 17.5 mm) is given a charge of Q=+5.0 nC.
+ +
+ + + +
+ + + +
+ +
V0 
(a) What is the potential of the dime at its surface?
(b) What is the potential energy Ue of an electron 1.0 cm
above the dime (on axis)?
Q
2 0 R
Ve  V0
 4(9.0  109 Vm/C)(5.0 109 C) /(.0175 m)  10,300 V
R2  z 2  z
 3,870 V
R
U e  eVe  (1.6 1019 C)(3.87 103 V)  6.2 1016 J
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End of Lecture 11
l Before the next lecture, read Knight, Chapters 30.1
through 30.4.
lLecture HW Assignment #4 is due at 10 PM, next
Wednesday.
lUncollected Exam 1 papers may be obtained from Helen
Gribble in room C136 PAB.
lRequests for regrades of Exam 1 should be written on
a separate sheet (see Syllabus) and taken to Helen
Gribble in room C136 PAB. They will be accepted until
noon next Wednesday.
7/19/2016
Physics 122B - Lecture 11
26