Physics 122B
Electricity and Magnetism
Lecture 10 (Knight: 29.1 to
29.4)
Potential and Potential Energy
April 18, 2007
Martin Savage
Lecture 9 Announcements
 Lecture HW Assignments #3 and #4 has
been posted on the Tycho system.
Assignment #3 is due at 10 PM tonight, and
Assignment #4 is due at 10 PM next
Wednesday.
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Physics 122B - Lecture 10
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Kirchhoff’s Junction Law
 I  I
in
out
The current density is
generally not the same at
all points of the wire
I
i
 0; summed over all the currents to any "junction".
i
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Conductivity and Resistivity
The current density J = nevd is directly proportional to the electron
drift speed vd. Our microscopic conduction model gives vd = eE/m, where
 is the mean time between collisions. Therefore:
 e E  n e 
J  nevd  n e 
E

m
 m 
2
The quantity ne2/m depends only on the properties of the conducting
material, and is independent of how much current density J is flowing.
This suggests a definition:
2
ne 
conductivity:  
so J  J
 E=
m
E
This result is fundamental and tells us three things:
(1) Current is caused by an E-field exerting forces on charge carriers;
(2) Current density J and current I=JA depends linearly on E;
(3) Current density J also depends linearly on . Different materials
have different  values because n and  vary with material type.
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Resistivity and
Conducting Materials
For many applications, it is more convenient to use inverse of conductivity,
which is called the resistivity, denoted by the symbol r:
1
m
resistivity: r   2
 ne 
r

Thus, the current density
is J = E = E/r. Here are the
conducting properties of
common materials:
Units of resistivity are
Wm
Units:
ohms = W = Nm2/CA = Nm2s/C2
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Example:
The Electric Field in a Wire
A 2.0 mm diameter aluminum wire carries a current of 800 mA. What
is the electric field strength inside the wire?
The electric field strength is
I
I

  A  r 2
(0.80 A)

 0.0072 N/C
7
-1 -1
2
(3.5 10 W m ) (.0010 m)
E
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J

Physics 122B - Lecture 10
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Resistors and Resistance
Conducting material that carries current
along its length can form a resistor,
a circuit element characterized by an
electrical resistance R:
R ≡ rL/A
where L is the length of the conductor and A is
its cross sectional area.
R has units of ohms ( W ).
Multiple resistors may be combined in
series, where resistances add, or in parallel,
where inverse resistances add.
I
Rnet
Rnet
For identical resistors can
Series
Connection
[L]:
simply
add
the lengths
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Rnet  R1  R2  R3
For identical resistors can
simply Connection
add the areas
Parallel
[(1/A)]:
Physics 122B - Lecture 10
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1
1
1
 

Rnet R1 R2 R3
7
Superconductivity
The “classical” physics we are studying is an
approximation to quantum mechanics. In the quantum
domain, under certain circumstances (low temperature,
electron pairing) , there may be minimum amount of
energy that an electron can lose in a collision. If the
probable energy loss falls below that minimum, the
system may become a “superconductor”, a material in
which the electrical resistance of the material
vanishes. Superconductivity was discovered in 1911 by
the Dutch physicist Heike Kamerlingh Onnes.
Heike Kamerlingh Onnes
(1853-1926)
Most superconductors exist at only very low temperatures (<20 K), but
in 1986 a new class of “warm” superconductors was discovered that
maintain their superconducting properties up to 125 K. A current initiated
in such a material persists, because there is no electrical resistance to
dissipate the energy.
e.g. a superconducting ring
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Work and Potential Energy
Recall from Physics 121 that Emech= K + U is a
conserved quantity for particles that interact via
conservative forces and that for changes,
DEmech = DK + DU = 0.
The change in potential energy is:
DU = Uf – Ui = -Winteraction forces. If a particle moves a
distance Dr while a constant force F is acting on it, then
the work done is:
W = F·Dr = F Dr cos(q), where q is the angle between the
force F and displacement Dr.
There are three special cases: q=00, q=900, and
q=1800.
If the force is not constant, the work is:
sf
W
 F ds   F ds
s
si
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f
i
Physics 122B - Lecture 10
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The Potential Energy
in Two Uniform Fields
The gravitational field g near the surface of the Earth
is uniform. If a particle moves downward from yi to yf, the
gravitational field will do a positive amount of work:
Wgrav  wDr cos 0  mg y f  yi  mg yi  mg y f
Therefore:
DU grav  U f  U i  Wgrav (i  f )  mg y f  mg yi
U grav  U 0  mg y  Gravitational Potential Energy
Similarly, for displacements s in a uniform electric
field E, with s parallel to E:
Welec  F Dr cos 0  qE s f  si  qEsi  qEs f
DU elec  U f  U i  Welec (i  f )  qEs f  qEsi
U elec  U 0  qEs  Electric Potential Energy
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Charges in an E Field
One difference between a gravity field g and an
electric field E is that a mass m interacting with g is
always positive, while a charge q interacting with E
may be either positive or negative.
However, this is not a problem. A positive charge
gains energy as it moves away from the positive plate
of a parallel plate capacitor, while a negative charge
gains energy as it moves away from the negative
plate of the capacitor. In either case, the charge
gains kinetic energy as its potential energy
decreases.
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Example: Conservation of
Energy inside a Capacitor
A 2.0 cm x 2.0 cm parallel plate
capacitor with a 2.0 mm gap is charged
to ±1.0 nC. First a proton, and then an
electron, are released at the midpoint
of the capacitor.
(a) What is each particle’s change in Uelec
from its release to its collision with a
plate?
(b) What is each particle’s kinetic energy
as it reaches the plate?
d
DU p  U f  U i  (U 0  0)  (U 0  eE )   12 eEd

Q
E


 2.82  105 N/C
2
0 0 A
d
DU e  U f  U i  (U 0  (e) Ed )  (U 0  (e) E )
2
1
17
D
U

D
U


eEd


4.52

10
J
p
e
2
1
  2 eEd
2DU
vf 
 2.33 105 m/s (prot)  9.96 106 m/s (elec)
m
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Question 1
The electric field of a positively charged
rod (end view shown) causes a negative
particle to orbit the rod in a closed circular
path, as shown.
What is the sign of the work done on the
charged particle by the electric field of the
rod?
(A) positive; (B) zero; (C) negative;
(D) not enough information to tell.
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Review of Mass & Spring
The energy of a mass-spring system alternates between potential
energy Usp stored in the spring and kinetic energy K residing in the
moving mass. An energy diagram shows the energy balance vs. position.
xf
Wsp 

xi
xf
Fdx  k  xdx   12 kx f 2  12 kxi 2  12 k ( xi 2  x f 2 )
xi
DUsp  U f  U i  Wsp (i  f )  12 k ( x f 2  xi 2 ); Usp  12 k x 2
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Potential Energy
of Point Charges
The same approach can be applied to
the interaction between two charged
particles. Consider the work by particle
1 on particle 2 as it moves from xi to xf:
xf
Welec 

xf
F1 on 2 dx 
xi

xi
Kq1q2
Kq1q2 Kq1q2
dx



2
x
xf
xi
DU elec  U f  U i  Welec (i  f ) 
U elec
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Kq1q2 Kq1q2

xf
xi
Kq1q2
1 q1q2


x
4 0 x
Physics 122B - Lecture 10
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The Potential Energy of
Like and Unlike Charge Pairs
U elec 
Kq1q2
1 q1q2

r
4 0 r
This approach can be applied to pairs of electrically charged particles,
whether they have the same or opposite charges. However, for like-sign
particles (a) the system energy is positive and decreases with separation,
while for opposite-sign particles (b) the system is typically “bound”, so that
the net energy is negative and increases (closer to zero) with increasing
separation.
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The Electric Force as
a Conservative Force
The electrical force is a “conservative
force”, in that the amount of energy involved in
moving from point i to point f is independent of
the path taken.
This can be demonstrated in the field of a
single point charge by observing that tangential
paths involve no change in energy (because r is
constant). Therefore, an arbitrary path can be
approximated by a succession of radial and
tangential segments, and the tangential
segments eliminated.
What remains is a straight line path from
the initial to the final position of the moving
charge, indicating a net work that will be the
same for all possible paths.
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The Zero of Potential Energy
U elec
Kq1q2
1 q1q2


r
4 0 r
We note that the the equation for electric
potential energy says that Uelec± as
r0 and that Uelec0 as r. This
raises the question of the point at which
the potential energy should be zero.
Only changes in potential energy DU appear
in energy equations and have physical
consequences. Therefore, the point at
which Uelec= 0 is a matter of choice.
Two popular choices:
(1) Uelec0 at r or r “far away”.
(2) Uelec=0 at Earth ground or at some other
“normal” state used for reference.
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Example: Approaching a
Charged Sphere
A proton is fired from far
away at a 1.0 mm diameter glass
sphere that has a charge of
q=+100 nC.
What is the initial speed the
proton must have to just reach
the surface of the glass?
K f  U f  Ki  U i
v
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2 Kq p qs
m p rs

0 K
q p qs
rs
 12 m p v 2  0
2 Keqs
 1.86 107 m/s
m p rs
Physics 122B - Lecture 10
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Example: Escape Velocity
An interaction
between two elementary
particles causes an
electron and a positron
to be shot out back-toback with equal speeds.
What minimum speed
must each particle have
when they are 100 fm
apart in order to escape
each other?
K f  U f  Ki  U i
0  0  me v  m p v  K
1
2
v
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2
1
2
2
qe q p
r
e2
 me v  K
r
2
Ke2
 5.0 107 m/s
me r
Physics 122B - Lecture 10
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Multiple Point Charges
We have established that both energy and electrical forces
obey the principle of superposition, i.e., they can be added linearly
without “cross terms”. Therefore, for multiple point charges:
N
Kqi q j
i<j
ri j
U elec  
Here, “i<j” means that for summing over N particles, the sum
over i runs from 1 to N, and the sum over j runs from i+1 to N for
each value of i. This it a mathematical trick to avoid counting pairs
of point charges twice or having i=j terms, which would give a zero
in the denominator.
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Example: Launching an Electron
Three electrons are spaced 1.0 mm
apart along a vertical line. The
outer two electrons are fixed in
position.
(a) Is the center electron in a point of
stable or unstable equilibrium?
q2 q3
q1q2
(b) If the center electron is displaced
1
2
m
v

0

0

0

K

K
e
2
horizontally by an infinitesimal
r12 i
r23 i


distance, what will be its speed
2
2
when it is very far away?
e
e
U13 is same
K
K
before and after
 r12 i
 r23 i
K f  U f  Ki  U i
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2
v
me

e2
e2 
K
K
  1006 m/s
 r23 i 
  r12 i
Physics 122B - Lecture 10
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The Potential Energy of a Dipole
A dipole with dipole moment p=qd is in a uniform electric field. We are
interested in the work done by the electrical forces as the dipole rotates
by an angle f. Note that when the dipole rotates by an infinitesimal angle
df, the charge is displaced by ds=r df=(½d)df.
dW  F  ds  F ds cos q 

1
2

qd E cos q df 
q  f   / 2 so cos q   sin f

1
2

pE cos q df
dWelec  2dW   pE sin f df
ff
Welec   pE  sin f df  pE (cos f f  cos fi )
fi
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U dipole   pE cos f   p  E
Physics 122B - Lecture 10
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Example: Rotating a Molecule
Water molecules have a permanent electric
dipole moment of p = 6.2 x 10-30 C m. A water
molecule is aligned in an electric field of E = 1.0
x 107 N/C.
How much energy is needed to rotate the
molecule by 900?
U dipole   pE cos f   p  E
DUdipole  U f  Ui   pE cos90  pE cos 0  pE  6.2 1023 J
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End of Lecture 10
 Before the next lecture, read Knight, Chapters
29.5 through 29.7.
 Lecture HW Assignments #3 and #4 has been
posted on the Tycho system. Assignment #3 is
due tonite and #4 is due next week.
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