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Chapter 23 Electric Potential
• When a charged particle
moves in an electric field,
the field exerts a force that
can do work on the particle.
• Just as gravitational
potential energy depends
upon the height of the mass
above the earth’s surface,
electric potential energy
depends upon on the
position of the charged
particle in the electric field.
Chapter 23
1
Potential and Kinetic Energy
Work done by a Force on a particle that moves from point a
to point b.
Potential and Kinetic
Energy
K=Kinetic Energy
U= Potential Energy
Chapter 23
2
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Electric Potential Energy (joules)
Electric Potential Energy in a Uniform Electric Field
b 

b
Wa b   F  dl   F cos  dl
a
a
F  q0 E
Wab  Fd  q0 Ed
(Uniform electric field)
where Ua is the potential energy at point a and Ub is the potential energy at point b.
Chapter 23
3
An electric charge moving in an electric field
U decreases
U increases
U decreases
U increases
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Electric Potential Energy
Electric potential energy distance r away from a point charge
rb


rb
rb
ra
ra
Wa b   Fr  dr   Fr cos dr   Fr dr
ra
qq0
qq0  1 1 
   (23-8)
dr 
2
4 0 r
4 0  ra rb 
ra
rb
rb
1
Wab   Fr dr  
ra
Wab 
qq0  1  qq0  1 
 
   U a Ub
4 0  ra  4 0  rb 
joules
where Ua = potential energy at point a and Ub = potential energy at point b
Wa  

As rb
qq0  1 
   U a = Potential energy of q0
4 0  ra 
between point a and
Figure 23-5
infinity.
where U is the potential energy of q0 at any distance r away from a point charge q.
Chapter 23
5
Work out the Integration
qq0
qq0  1 1 
  
dr 
2
4 0 r
4 0  ra rb 
ra
rb
rb
1
Wab   Fr dr  
ra
rb
 4
ra
1
qq0
0
r2
qq0  r 1 
4 0   1 
n
Wab 
dr 
rb

ra
u n 1
 u du  n  1
Recall
qq0
4 0
rb

ra
r
dr
r2

b
qq0  r 1 
4 0   1 
ra
qq0   1  1  qq0  1 1 


  

4 0  rb
ra  4 0  ra rb 
qq0  1  qq0  1 
 
   U a Ub
4 0  ra  4 0  rb 
Chapter 23
6
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Electric Potential Energy is path independent
Figure 23-5
cos dl  dr
rb


rb
rb
Wa b   Fr  dl   Fr cos dl   Fr dr
ra
ra
ra
Chapter 23
7
Potential energy curves—PE versus r
• Graphically, the potential
energy between like
charges increases sharply
to positive (repulsive)
values as the charges
become close.
• Unlike charges have
potential energy
becoming sharply
negative as they become
close (attractive).
• Refer to Example 23.1.
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Electric Potential Energy
Electric potential energy distance r away from many point charges
The total potential
energy is the algebraic
sum of the individual
charge energy
Ua 
q0 q1
qq
qq
q0
 0 2  0 3 
4 0 r1 4 0 r2 4 0 r3 4 0
 q1 q2 q3 
 


r2
r3 
 r1
Chapter 23
9
Chapter 23
10
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Solution
Chapter 23
11
Electrical Potential
• Electric field is force per unit charge
• Flowing from potential energy U on q0
• Electric potential is potential energy per
unit charge U/q0
• Electric potential is also known as
“Potential” units volt
• V = U/q0
Chapter 23
12
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Battery a source of electrical potential
• The potential of a
battery can be
measured between
point a and point b
(the positive and
negative
terminals).
• Moving with the
electrical field
decreases the
electrical potential.
Electric Potential (Voltage) = Potential Energy per unit charge
Wab
a
b
F
Definition Vab 
E
·
·
q0
+q
0
1 volt = 1 joule/coulomb = J/C
Vab 
Va 
Wab U a  U b U a U b



 Va  Vb
q0
q0
q0
q0
Ua
q0
Vb 
volts
b 

Ub
q0
Wab  q0Vab
volts
Note:
1 ev = (1.6 x 10-19) joules
b
Wab   F  dl   q0 E  dl
a
Vab
W
1
 ab 
q0
q0
a
b
b
 q0 E  dl 
 E  dl
a
a
Chapter 23
joules
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Electric Potential (Voltage)
Electric potential distance r away from a point charge
Ua 
Ua 
Va 
1
F a .r a  4 0
qq0
4 0 ra
1
Ua
1 q

q0 4 0 ra
qq0
ra 2
ra
Ub 
qq0
4 0 rb
1
Vb 
Ub
1 q

q0 4 0 rb
Vab  Va  Vb
Chapter 23
15
Electric Potential (Voltage)
Electric potential distance r away from many point charges
The total Electric
Potential is the algebraic
sum of the individual
potentials
Va 
Ua
q3
q1
q2
1




q0
4 0 r1 4 0 r2 4 0 r3 4 0
Chapter 23
 q1 q2 q3 
 
 
 r1 r2 r3 
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Electron-Volts is a Unity of Energy
Wab  U a  Ub  qVa  Vb   qVab
q = e = 1.602 x10-19 C
For Vab = 1 volt
Ua-Ub = 1.602 x10-19 J (energy)
1eV = 1.602 x10-19 J
V = Potential difference
Chapter 23
Fermi Lab in Illinois
Tevatron - record 0.98 TeV
eV =energy in joules
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CERN's Large Hadron Collider (LHC)
reached 3.5 TeV per beam in 2010
1TeV= 1x1012 eV
Chapter 23
18
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LHC running two years at 3.5 TeV
April 2012 LHC operates at 8TeV – a world record
All in the search of the of the subatomic particle that
establishes “mass” (Higgs boson)
2014 14TeV ?
Chapter 23
19
Electric Potential (Voltage)
Example 23-4
Figure 23-13
Va 
q1 = q2 = 12 nC
Ua
q1
q2


 900V
q0
4 0r1 4 0r2
Vb 
Ua
q1
q2


 1930V
q0
4 0r1 4 0r2
Vc 
Va 
Ua
q1
q2


 0V
q0
4 0r1 4 0r2
q1
q2
1
1

 9 x109 x12x109 x(

)  1800 2700  900V
4 0r1 4 0r2
.06 .04
Chapter 23
20
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Chapter 23
21
Example 23.4 Continued
Chapter 23
22
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Movement and Electric Potential difference Ex-23.7 23
A 5.0 ug dust particle with a
charge of 2.0nC moves from
a to b
What is vb ?
K a  U a  Kb  U b
0  U a  Kb  U b Kb  U a  U b
U a  q0Va U b  q0Vb
U a  U b  q0Va  q0Vb  q0 (Va  Vb )
1
2
mvb  q0 (Va  Vb )  q0Vab
2
Kb 
Solve for vb
vb 
2q0 (Va  Vb )
 46m / s
m
Calculate Vb and Va
Vb 
Va 
q1
4 0 r1

q2
4 0 r2
 9 x109 x3x10 9 x(
1
1

)  1350
.01 .02
q1
q2
1
1

 9 x109 x3x10 9 x(

)  1350
4 0 r1 4 0 r2
.01 .02
Va  Vb  1350  (1350)  2700V
See details page 766
Chapter 23
23
A Charged Conducting Sphere – Electric Field and Electric Potential
Inside the sphere E = 0
every where
Inside the sphere V is the same
every where. Outside the field is
same as point charge at center.
No work is done moving a
charge inside the sphere
Vsurface  EsurfaceR
Figure 23-16
Chapter 23
24
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Chapter 23
25
Chapter 23
26
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The Empire State Building is struck up to 500 times
a year…..why?
Chapter 23
27
Ionization and Corona Discharge
For air to be ionized at the surface of the sphere E > 3 x 10 6 V/m
What is V at the surface?
At the surface of a charged conducting sphere
Vsurface  EsurfaceR 
p.768
Vsurface  EsurfaceR
q
4 0 R 2
R
Vm  Em R
q
4 0 R
Vm
 Em  3x106V / m
R
To get more potential energy increase the R. To get
discharge (corona) at low voltage reduce, R.
R
Vm
2m
6MV
Application
0.2m (8 in)
600 KV
10 mm
30KV
Radio antenna top, lighting rod
1 mm
3 KV
Corona wire – (laser printer)
Scientific experiments
Classroom Van de Graaff Generator
Chapter 23
28
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Lightening
Thunder is formed from the shock wave
formed by the rapid heating of the air
along the path of the return stroke, which
reaches some 30,000 degrees K.
The sound of thunder varies depending
on how far you are from the various
parts of the stroke. The sound from a
part of the stroke farther away will get to
you later and be fainter.
29
Chapter 23
Electric Potential Oppositely Charged Parallel Plates
Find the electric potential V due to charged plates
b
b
a
a
Vab   E  dl   E cos dl Ed
From section 22.4

0
  0E
   0Vab / d
E
 coul/m2 charge
Vab = Ed
Uniform Electric Field
Chapter 23
A way of measuring
charge density
 coul/m2 charge
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Difference in Potential between two points outside
an Infinite Line Charge or Charged Conducting Cylinder.
Example 23-10
From section 22.6 The
electric field from a long
wire had only a radial
component
b
Er 
a
1 
2 0 r
Expand the wire to a cylinder
and calculate the potential at
two points a, b. The charge is
on the surface and E=0
inside.
b
b
Vab   E  dl   E cos dl 
a
Vab 
a
rb
 Er dr 
ra

2 0
rb

ra
dr
r
r

ln b
2 0
ra
Chapter 23
31
Chapter 23
32
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Chapter 23
33
Contour Maps & Equipotential Surfaces
Contour lines on a topology map are curves of equal
gravitational potential energy
By analogy Equipotential surfaces are surfaces where the electric
potential is the same
Chapter 23
34
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Equipotential Surfaces
•Field lines are perpendicular to
Equipotential surface (lines)
•E need not be constant over an
Equipotential surface(lines). To the
right, Ea is greater than Eb
b
a
b
Vab 
 E  dl
a
dl 
dV  Edl
dV
E
As E increases dl must
decrease. This means the
Equipotential will closer
Equipotential surfaces and field lines
• Surfaces of equal potential may be drawn any charge or charges
and the field lines they create.
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Equipotential Surfaces
Equipotential Surfaces and Conductors
Figure 23-25
E parallel = 0. Therefore, E is perpendicular to the surface.
This makes the surface of the conductor an equipotential surface.
Chapter 23
37
Potential Gradient
a
b
·
·
E
If cos  = 1 ( = 0o)
b
b
b
a
a
a
b
a
b
a
Vab   E  dl   E cos dl   Edl
Vab   dV    dV
b


dV is the infinitesimal small change in V
b

dV  Edl
a
a
 dV  Edl
E
dV
dl
potential
gradient
Er  
Chapter 23
V
r
potential gradient for
radial field
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The Electrical Potential is useful Technique for finding
the Electric Field
CALCULATE
METHODS
E field caused by a charge
distribution
Add the E fields of point charges
E is a vector requiring a separate a computation for
each component
Calculate potential V and taking the gradient
Often easier because V is a scalar and is an easier
integration
Knowing the potential V is a very useful computational
technique for finding E
Chapter 23
39
Chapter 23
40
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Chapter 23
41
Q23.1
When a positive charge moves in the
direction of the electric field,
A. the field does positive work on it
and the potential energy increases.
E
+q
Motion
E
B. the field does positive work on it
and the potential energy decreases.
C. the field does negative work on it
and the potential energy increases.
D. the field does negative work on it
and the potential energy decreases.
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Q23.2
When a positive charge moves opposite
to the direction of the electric field,
E
Motion
A. the field does positive work on it
and the potential energy increases.
+q
E
B. the field does positive work on it
and the potential energy decreases.
C. the field does negative work on it
and the potential energy increases.
D. the field does negative work on it
and the potential energy decreases.
Q23.3
When a negative charge moves in the
direction of the electric field,
A. the field does positive work on it
and the potential energy increases.
E
–q
Motion
E
B. the field does positive work on it
and the potential energy decreases.
C. the field does negative work on it
and the potential energy increases.
D. the field does negative work on it
and the potential energy decreases.
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Q23.4
E
When a negative charge moves opposite
to the direction of the electric field,
Motion
A. the field does positive work on it
and the potential energy increases.
–q
E
B. the field does positive work on it
and the potential energy decreases.
C. the field does negative work on it
and the potential energy increases.
D. the field does negative work on it
and the potential energy decreases.
Q23.5
The electric potential energy of two
point charges approaches zero as the
two point charges move farther away
from each other.
If the three point charges shown here
lie at the vertices of an equilateral
triangle, the electric potential energy
of the system of three charges is
Charge #2
+q
Charge #1
+q
y
x
–q
Charge #3
A. positive.
B. negative.
C. zero.
D. not enough information given to decide
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Q23.6
The electric potential energy of two
point charges approaches zero as the
two point charges move farther away
from each other.
If the three point charges shown here
lie at the vertices of an equilateral
triangle, the electric potential energy
of the system of three charges is
Charge #2
–q
Charge #1
+q
y
x
–q
Charge #3
A. positive.
B. negative.
C. zero.
D. not enough information given to decide
Q23.7
The electric potential due to a point
charge approaches zero as you move
farther away from the charge.
If the three point charges shown here
lie at the vertices of an equilateral
triangle, the electric potential at the
center of the triangle is
Charge #2
+q
Charge #1
+q
y
x
–q
Charge #3
A. positive.
B. negative.
C. zero.
D. not enough information given to decide
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Q23.8
The electric potential due to a point
charge approaches zero as you move
farther away from the charge.
If the three point charges shown here
lie at the vertices of an equilateral
triangle, the electric potential at the
center of the triangle is
Charge #2
–q
Charge #1
+q
y
x
–q
Charge #3
A. positive.
B. negative.
C. zero.
D. not enough information given to decide
Q23.9
Consider a point P in space where the electric potential is zero.
Which statement is correct?
A. A point charge placed at P would feel no electric force.
B. The electric field at points around P is directed toward P.
C. The electric field at points around P is directed away from P.
D. none of the above
E. not enough information given to decide
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Q23.10
Where an electric field line crosses an equipotential surface, the
angle between the field line and the equipotential is
A. zero.
B. between zero and 90°.
C. 90°.
D. not enough information given to decide
Q23.11
The direction of the electric potential gradient at a certain point
A. is the same as the direction of the electric field
at that point.
B. is opposite to the direction of the electric field
at that point.
C. is perpendicular to the direction of the electric
field at that point.
D. not enough information given to decide
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