Download Electric Potential Energy

Electric Potential
1/22/14
Physics for Scientists & Engineers 2, Chapter 23
1
Announcements
!  First exam is next Tuesday, January 28
•  45 minute exam during lecture time
•  You can bring a 5” by 8” size cheat sheet note card (filled on both
sides) (i.e. ½ letter size piece of paper)
•  Hand-written
•  No computer-printed equation sheets!
January 22, 2014
Physics for Scientists & Engineers 2, Chapter 21
2
Special Symmetries
!  We have applied Gauss’s Law to a point charge and showed
that we get Coulomb’s Law
!  Now let’s look at more complicated distributions of charge
and calculate the resulting electric field
!  We will use a charge density to describe the distribution of
charge
!  This charge density will be different depending on the
geometry
January 22, 2014
Physics for Scientists & Engineers 2, Chapter 22
5
Spherical Symmetry
!  Calculate the electric field from
charge distributed as a spherical
shell
!  Assume that we have a spherical
shell of charge q with radius rS
(gray)
!  We will assume two different
spherical Gaussian surfaces
•  r2 > rS (purple) i.e. outside
•  r1 < rS (red) i.e. inside
January 22, 2014
Physics for Scientists & Engineers 2, Chapter 22
6
Spherical Symmetry
!  Let’s start with the Gaussian surface outside the sphere
of charge, r2 > rS (purple)
!  We know from symmetry
arguments that the electric field
will be radial outside the charged
sphere
!  If we rotate the sphere, the
electric field cannot change
•  Spherical symmetry
!  Thus we can apply Gauss’ Law and get
 
q
1 q
2

∫∫ E ⋅dA = E ( 4π r 2 ) = ε 0 ⇒ E = 4πε 0 r22
January 22, 2014
Physics for Scientists & Engineers 2, Chapter 22
7
Spherical Symmetry
!  Let’s take the Gaussian surface inside the sphere of charge,
r1 < rS (red)
!  The enclosed charge is zero so
 
2
E
⋅d
A
=
E
4
π
r
(
1) = 0 ⇒ E = 0

∫∫
!  We find that the electric field is
zero everywhere inside spherical
shell of charge
!  Thus we obtain two results
•  The electric field outside a spherical
shell of charge is the same as that of
a point charge
•  The electric field inside a spherical shell of charge is zero
January 22, 2014
Physics for Scientists & Engineers 2, Chapter 22
8
Spherical Symmetry: Uniform Distribution
!  Next, let’s calculate the electric field from charge distributed
throughout a spherical volume with
a uniform charge density ρ > 0 and
radius r
!  We will assume two different
spherical Gaussian surfaces
•  r2 > r
•  r1 < r
(outside, red)
(inside, blue)
!  Let’s start with the surface with
r1 < r
!  From the symmetry of the charge distribution, the electric
field is perpendicular to the Gaussian surface everywhere
January 22, 2014
Physics for Scientists & Engineers 2, Chapter 22
9
Spherical Symmetry: Uniform Distribution
!  Gauss’s Law gives us
⎛ 4 3⎞
ρ ⎜ π r1 ⎟
 
⎝3
⎠
q
2

∫∫ E ⋅dA = E ( 4π r1 ) = ε 0 = ε 0
!  Solving for E we find
ρ r1
Einside =
3ε 0
!  The total charge on the sphere is
⎛ 4 3⎞
qt = ρV = ρ ⎜ π r ⎟
⎝3
⎠
!  The enclosed charge is
4 3
π r1
3
volume inside r1
r
q=
qt = 3
qt = 13 qt
4 3
volume
r
πr
3
January 22, 2014
Physics for Scientists & Engineers 2, Chapter 22
10
Spherical Symmetry: Uniform Distribution
!  Gauss’s Law gives us
3
 
q
r
2
t 1
E
⋅d
A
=
E
4
π
r
=
(
)
1

∫∫
ε0 r 3
!  Solving for E we find
qt r1
kqt r1
Einside =
= 3
3
4πε 0r
r
!  Consider r2 > r
 
qt
2

∫∫ E ⋅dA = E ( 4π r2 ) = ε 0
!  Solving for E we find
qt
kqt
Eoutside =
= 2
2
4πε 0r2
r2
January 22, 2014
same as point charge
Physics for Scientists & Engineers 2, Chapter 22
11
Electric Potential
!  An electric field has many similarities to a gravitational field
!  For example, the magnitude of the gravitational force and
the electric force are given by
m1m2
Fg = G 2
r
q1q2
Fe = k 2
r
!  Both gravitational and electrostatic forces depend on the
inverse square of the distance
!  Both gravitational and electrostatic forces are conservative
!  We can define the electric potential energy in analogy with
the gravitational potential energy
1/22/14
Physics for Scientists & Engineers 2, Chapter 23
12
Electric Potential Energy
!  For a conservative force, the work is path-independent
!  When an electrostatic force acts between two or more
charges within a system, we can define an electric potential
energy, U, in terms of the work done by the electric field,
We, when the system changes its configuration from some
initial configuration to some final configuration
•  The change in the electric potential energy is the negative of the
work done by the electric field
ΔU = U f − U i = −We
U i is the initial electric potential energy
U f is the final electric potential energy
1/22/14
Physics for Scientists & Engineers 2, Chapter 23
13
Electric Potential Energy
!  Like gravitational or mechanical potential energy, we must
define a reference point from which to define the electric
potential energy
!  We define the electric potential energy to be zero when all
charges are infinitely far apart
!  We can then write a simpler definition of the electric
potential taking the initial potential energy to be zero,
ΔU = U f − 0 = U = −We,∞
!  The negative sign on the work means
•  If E does positive work then U < 0
•  If E does negative work then U > 0
1/22/14
Physics for Scientists & Engineers 2, Chapter 23
14
Constant Electric Field
!  Let’s look at the electric potential energy when we move a
charge in a constant electric field
!  The work done by a constant force is
 
W = F ⋅d
!  For a constant electric field the force is


F = qE
!  The work done by the electric field on the charge is
 
W = qE ⋅ d = qEd cosθ
1/22/14
Physics for Scientists & Engineers 2, Chapter 23
15
Constant Electric Field - Special Cases
!  If the displacement is in the same
direction as the electric field
W = qEd ⇒ ΔU = −qEd
•  A positive charge loses potential energy
when it moves in the direction of the
electric field
!  If the displacement is in the
direction opposite to the electric
field
W = −qEd so ΔU = qEd
d
d
•  A positive charge gains potential energy
when it moves in the direction opposite
to the electric field
1/22/14
Physics for Scientists & Engineers 2, Chapter 23
E
E
q
q
d
d
q
q
E
E
16
Electric Potential Difference ΔV
!  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
U f U i ΔU
ΔV = Vf −Vi =
− =
q
q
q
!  Hence we can relate the change in electric potential to the
work done by the electric field on the charge
We
ΔV = −
q
!  Taking the electric potential energy to be zero at infinity we
have
We,∞
V =−
q
1/22/14
Physics for Scientists & Engineers 2, Chapter 23
19
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
1/22/14
Physics for Scientists & Engineers 2, Chapter 23
20
Energy Gain of a Proton
!  A proton is placed between two parallel
conducting plates in a vacuum
!  The potential difference between the two
plates is 450 V
!  The proton is released from rest close to
the positive plate
PROBLEM
!  What is the kinetic energy of the proton when it reaches the
negative plate?
SOLUTION
!  The electric potential difference between the plates is 450 V
1/22/14
Physics for Scientists & Engineers 2, Chapter 23
22
Energy Gain of a Proton
!  We can relate the electric potential
difference across the plates to the change
in electric potential energy
ΔU
ΔV =
q
!  All of the electric potential energy lost by
the proton in crossing between the two plates is converted
into kinetic energy
!  Conservation of energy gives us
ΔK + ΔU = 0 ⇒ ΔK = −ΔU = −qΔV = K f − 0 = K f
K f = −qΔV = − (1.602⋅10−19 C )( −450 V ) = 7.21⋅10−17 J
1/22/14
Physics for Scientists & Engineers 2, Chapter 23
23
Batteries
!  A common method of creating an electric potential
difference is a battery.
!  A battery uses chemical processes to maintain a constant
potential difference between its anode (−) and cathode (+).
!  Lithium ion batteries are the focus of modern research.
1/22/14
Chapter 23
24
Battery-Powered Cars
!  The Tesla sports car uses a
53 kW h lithium ion battery
that is charged to 80% of capacity
and discharged to 20% capacity.
!  A typical gasoline powered car carries 50 L of gasoline,
which has an energy content of 34.8 MJ/L.
PROBLEM:
!  How does the available energy in the Tesla sports car
compare with the energy carried by a gasoline-powered car?
SOLUTION:
!  The energy available in the Tesla battery is:
1000 W 3600 s
Eelectric = (60%)(53 kW h)
= 1.14⋅108 J = 114 MJ
1 kW
1h
1/22/14
Chapter 23
25
Battery-Powered Cars
!  The energy available in the gasoline-powered car is:
34.8 MJ
Egasoline = (50 L)
= 1740 MJ
1L
!  So the gasoline-powered car carried 15 times more energy
than the battery-powered car.
!  However, the typical efficiency of a gasoline-powered car
20% while it is 90% for a battery-powered car, so the
comparison of usable energy is:
Eelectric,usable = (90%)(114 MJ ) = 103 MJ
Egasoline,usable = (20%)(1740 MJ ) = 348 MJ
1/22/14
Chapter 23
26
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.
Copyright © The McGraw-Hill Companies, Inc.
Permission required for reproduction or display.
© W. Bauer and G.D. Westfall
1/22/14
Chapter 23
28
The Van de Graaff Generator
!  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.
1/22/14
Chapter 23
Copyright © The McGraw-Hill Companies, Inc.
Permission required for reproduction or display.
© W. Bauer and G.D. Westfall
29
Tandem Van de Graaff Accelerator
!  One use of a Van de Graaff generator is to accelerate particles for
condensed matter and nuclear physics studies.
!  A 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).
!  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.
1/22/14
Chapter 23
30
Tandem Van de Graaff 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.
PROBLEM 1:
!  What is the highest energy we can attain for carbon nuclei?
SOLUTION 1:
!  There are two stages to the acceleration:
•  The carbon ions with a -1e charge gain energy
accelerating toward the terminal.
•  The stripped carbon ions with a +6e charge gain
energy accelerating away from the terminal.
1/22/14
Chapter 23
31
Tandem Van de Graaff Accelerator
!  The kinetic energy gained by each ion is:
ΔK = ΔU = q1ΔV + q2ΔV = K
K = eΔV + 6eΔV = 7eΔV
!  Putting in our numerical values we get:
K = 7 (1.602⋅10−19 C)(10⋅106 V) = 1.12⋅10−11 J
!  Physicists often use electron-volts instead of joules to
express the kinetic energy of accelerated particles:
K = 7e(10⋅106 V) = 7.0⋅107 eV = 70 MeV
PROBLEM 2:
!  What is the highest speed we can attain for carbon nuclei?
1/22/14
Chapter 23
32
Tandem Van de Graaff Accelerator
SOLUTION 2:
!  We can relate speed and kinetic energy:
1
2K
K = mv 2 ⇒ v =
2
m
!  The mass of a 12C nucleus is 1.99·10-26 kg so:
v=
2(1.12⋅10−11 J )
1.99⋅10−26 kg
= 3.36⋅107 m/s
!  This is 11% of the speed of light.
1/22/14
Chapter 23
33