Transcript Chapter 23

From last time(s)…


Electric charges, forces, and fields
Motion of charged particles in fields.
Today…



Work, energy, and (electric) potential
Electric potential and charge
Electric potential and electric field.
No honors lecture this week
Oct. 4, 2007
1
Forces, work, and energy



Particle of mass m at rest
Apply force to particle - what happens?
Particle accelerates
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Oct. 4, 2007
Stop pushing - what happens?
Particle moves at constant speed
Particle has kinetic energy
2
Work and energy

Work-energy theorem:
 Change in kinetic energy of isolated
particle = work done
 dW  F  ds  Fds cos 

Total work

Oct. 4, 2007
K 
end
end
start
start
 dW   F  ds
3
Electric forces, work, and energy

Consider bringing two positive charges together




They repel each other
Pushing them together requires work
Stop after some distance
How much work was done?
+
Oct. 4, 2007
+
4
Calculating the work


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E.g. Keep Q2 fixed, push Q1 at constant velocity
Net force on Q1 ? Zero
1
Q1Q2
Force from hand on Q1 ? 4  R 2
o
Q1
R
+

xfinal
xinitial

Total work done by hand
Force in
direction of motion
Oct. 4, 2007
+
+
Q2
end
x final
start
x initial
 F  ds  
Q1Q2
ke
dx
2
R  x 
x
Q1Q2 final
 ke
R  x xinitial
5
Conservation of Energy

Work done by hand
Q1Q2
 ke
R x
x final
x initial
Q1Q2
Q1Q2
 ke
 ke
R  x final
R  x initial
Q1Q2
Q1Q2
 ke
 ke
 0 for pos charges
rfinal
rinitial
Where did this energy go?

Energy is stored in the electric field as electric potential energy
Oct. 4, 2007
6
Electric potential energy
of two charges

Define electric potential energy U so that
W external  K  U
Work done
on system
Change in
kinetic energy
Change in
electric potential energy
Q1Q2
 const
Works for a two-charge system if Ur  ke

r

Define: potential energy at infinite separation = 0
Then
Oct. 4, 2007
Q1Q2 
for two charges
Ur  ke
r
Units of Joules
7
Quick Quiz
Two balls of equal mass and equal charge are held
fixed a distance R apart, then suddenly released.
They fly away from each other, each ending up
moving at some constant speed.
If the initial distance between them is reduced by a
factor of four, their final speeds are
A. Two times bigger
B. Four times bigger
C. Two times smaller
D. Four times smaller
E. None of the above
Oct. 4, 2007
8
More About U of 2 Charges

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Like charges  U > 0 and work must be done to bring
the charges together since they repel (W>0)
Unlike charges  U < 0 and work is done to keep the
charges apart since the attract one the other (W<0)
Oct. 4, 2007
9
Electric Potential Energy of single
charge
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Work done to move single charge near charge distribution.
Other charges provide the force, q is charge of interest.
F  Fq1  Fq2  Fq3
q
q1
q2 +
+ +
q3
U

+
 F  ds   F
q1

 Fq 2  Fq 3  ds
 U q1 r1  U q 2 r2   U q 3 r3 
q1q
q2q
q3q
k
k
k
r
r
r
Superposition of individual interactions
Generalize to continuous charge distribution.
Oct. 4, 2007

10
Electric potential
U 
F
Coulomb
 ds 
 qE  ds  q  E  ds
Electric potential U energy proportional to charge q
U /q  V  Electric potential

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Electric potential V usually created by some
charge distribution.
V used to determine electric potential energy U
of some other charge q
V has units of Joules / Coulomb = Volts
Oct. 4, 2007
11
Electric potential of point charge

Consider one charge as ‘creating’ electric
potential, the other charge as ‘experiencing’ it
q
Q
Oct. 4, 2007
Qq
UQq r  ke
r

UQq r
Q
Vq r 
 ke
q
r
12
Electric Potential of point charge


Potential from a point charge
Every point in space has a
numerical value for the electric
potential
kQ
V
r
y
+Q

x
Distance from ‘source’ charge +Q
Oct. 4, 2007
13
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U=qoV
Point B has greater potential
energy than point A
Means that work must be done
to move the test charge qo
from A to B.
This is exactly the work to
overcome the Coulomb
repulsive force.
 F
Electric potential energy=qoV
Potential energy, forces, work
B
Work done = qoVB-qoVA =
Coulomb
B
A
 d
qo > 0
A
Differential form: qodV  FCoulomb  d
Oct. 4, 2007
14
Quick Quiz
Two points in space A and B have electric potential
VA=20 volts and VB=100 volts. How much work
does it take to move a +100µC charge from A to
B?
A. +2 mJ
B. -20 mJ
C. +8 mJ
D. +100 mJ
E. -100 mJ
Oct. 4, 2007
15
V(r) from multiple charges
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Work done to move single charge near charge distribution.
Other charges provide the force, q is charge of interest.
U
q1
q2
q
q1

 Fq 2  Fq 3  ds
 U q1 r1  U q 2 r2   U q 3 r3 
q1q
q2q
q3q
k
k
k
r
r
r
 q1
q2
q3 
 qk  k  k 
 r
r
r 
q3
Superposition of
individual
electric potentials
Oct. 4, 2007
 F  ds   F

 qVq1 r  Vq 2 r  Vq 3 r
V r  Vq1r  Vq 2 r  Vq 3 r
16
Quick Quiz 1

At what point is the electric potential zero for this
electric dipole?
A
x=-a
+Q
x=+a
B
-Q
A. A
B. B
C. Both A and B
D. Neither of them
Oct. 4, 2007
17
Superposition:
the dipole electric potential
x=-a
Superposition of
• potential from +Q
• potential from -Q
x=+a
+Q
-Q
+
=
V in plane
Oct. 4, 2007
18
Electric Potential and Field for a
Continuous Charge Distribution
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If symmetries do not allow an immediate application of the Gauss’ law to
determine E often it is better to start from V!
Consider a small charge element dq
The potential at some point due to this charge element is
To find the total potential, need to
integrate over all the elements
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Oct. 4, 2007
This value for V uses the reference of V = 0 when P is infinitely far
away from the charge distribution
19
Quick Quiz
Two points in space have electric potential VA=200V & VB=150V.
A particle of mass 0.01kg and charge 10-4C starts at point A
with zero speed. A short time later it is at point B.
How fast is it moving?
A. 0.5 m/s
B. 5 m/s
C. 10 m/s
D. 1 m/s
E. 0.1 m/s
Oct. 4, 2007
20
E-field and electric potential

If E-field known,
don’t need to know about charges creating it.
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E-field gives force
From force, find work to move charge q
q
+
+
U 
F
me
 ds 
+
E
+
Felec  qE

 qE  ds  q  E  ds


Electric potential V  U /q   E  ds
Non-constant potential
Oct. 4, 2007

Non-zero E-field
21
Potential of spherical conductor
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Zero electric field in metal -> metal has constant potential
Charge resides on surface,
so this is like the spherical charge shell.
Found E = keQ / R2 in the radial direction.
What is the electric potential of the conductor?
R
V R  V     E  ds
difficult
path



 E  ds
R
Integral along
some path,
from point on
surface to inf.
Oct. 4, 2007
easy
path
Easy because is same
direction as E,
E  ds  E dr  Edr
22
Electric potential of sphere

R
Q
V R  V     E dr   k 2 dr
r

R

Q
Q
 k
k
rR
R
So conducting sphere of
radius R carrying charge Q
is at a potential kQ /R
Conducting spheres connected by conducting wire.

Same potential everywhere.
R1
Oct. 4, 2007
Q1
Q2
R2
But 
not same
everywhere
23
Connected spheres
Since both must be at the same potential,

kQ1 kQ2
Q1 R1



R1
R2
Q2 R2
Charge proportional
to radius
Surface charge densities?
Q
1 R2



2
4R
 2 R1

Electric field?

Oct. 4, 2007

Since E 
2o
Surface charge
density proportional
to 1/R
Local E-field
proportional to 1/R
(1/radius of curvature)
24
Varying E-fields on conductor


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Expect larger electric fields near the small end. Can predict electric field
proportional to local radius of curvature.
Large electric fields at sharp points, just like square
Fields can be so strong that air is ionized and ions accelerated.
Oct. 4, 2007
25
Quick Quiz

Four electrons are added to a long wire. Which of the
following will be the charge distribution?
A)
B)
C)
D)
Oct. 4, 2007
26
Conductors: other geometries
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Rectangular conductor (40
electrons)
Edges are four lines
Charge concentrates at
corners
Equipotential lines closest
together at corners
So potential changes faster
near corners.
So electric field is larger at
corners.
Oct. 4, 2007
27
E-field and potential energy
Oct. 4, 2007
28

What is electric potential energy of isolated charge?
Zero
Oct. 4, 2007
29
The Electric Field




qo dV  FCoulomb  d 
F

dV   Coulomb  d
 qo 
E is the Electric Field
 E  d
It is independent of the test charge,
just like the electric potential
It is a vector, with a magnitude and direction, 
When
 potential arises from other charges,
= Coulomb force per unit charge on a test charge due to
interaction with the other charges.
E
We’ll see later that E-fields in
electromagnetic waves exist w/o charges!
Oct. 4, 2007
30
Electric field and potential
Said before that

dV  E  d
Electric field strength/direction shows how the potential
changesin different directions
For example,


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Potential decreases in direction of local E field at rate  E
Potential increases in direction opposite to local E-field at rate 
potential constant in direction perpendicular to local E-field

Oct. 4, 2007
E  d
0
E


31
Potential from electric field
dV  E  d



Electric field can be used
to find changes in
V
potential
Potential changes largest
in direction of E-field.
Smallest (zero)

perpendicular to

E-field
 Vo
d
d
E

V  Vo  E d

d
V=Vo
 V  Vo  E d

Oct. 4, 2007
32
Quick Quiz 3

Suppose the electric potential is constant
everywhere. What is the electric field?
A) Positive
B) Negative
C) Zero
Oct. 4, 2007
33
V
VB VA
E 

d
d
Electric Potential - Uniform Field
+ V  E ds  V  V 
B
A
B
B
A
A
B
 E  ds
A
   E ds  E  ds  Ex

E ||ds
E cnst

B

A


x
Oct. 4, 2007
Constant E-field corresponds to linearly
increasing electric potential
The particle gains kinetic energy equal to
the potential energy lost by the charge-field
system
34
Electric field from potential



Said before that
dV  E  d
Spell out the vectors:
This works for

dV  Ex dx  Ey dy  Ez dz
dV
dV
dV
Ex  
, Ey  
, Ez  
dx
dy
dz

Usually written

Oct. 4, 2007
dV dV dV 
E  V   ,
, 
dx dy dz 
35
Equipotential lines


Lines of constant potential
In 3D, surfaces of constant potential
Oct. 4, 2007
36
Electric Field and equipotential lines for +
and - point charges

The E lines are directed away
from the source charge

Oct. 4, 2007
A positive test charge would be
repelled away from the positive
source charge
The E lines are directed toward
the source charge
A positive test charge would
be attracted toward the
negative source charge
Blue dashed lines are equipotential
37
Quick Quiz 1
Question: How much work would it take YOU to
assemble 3 negative charges?
1. W = +19.8 mJ
2. W = 0 mJ
3. W = -19.8 mJ
3mC
5m
Likes repel, so YOU will
still do positive work!
Oct. 4, 2007
1mC
5m
5m
2mC
38
Work done to assemble 3 charges
Similarly if they are all positive:

W1 = 0
• W2 = k q1 q2 /r
=(9109)(110-6)(210-6)/5
=3.6 mJ
• W3 = k q1 q3/r + k q2 q3/r
(9109)(110-6)(310-6)/5 + (9109)(210-6)(310-6)/5 =16.2
• W = +19.8 mJ
• WE = -19.8 mJ
• UE = +19.8 mJ
Oct. 4, 2007
5m
q1
1C
3C
5m
mJ
q3
5m
2C
q2
39
Quick Quiz 2
The total work required for YOU to assemble the set of
charges as shown below is:
1. positive
2. zero
3. negative
5m
W1  0
Q(Q)
d
QQ
Q(Q)
W3  k
k
d
d
Q2
T ot al work = k
d
W2  k
Oct. 4, 2007
Q
Q
5m
5m
Q
40
Why U/qo ?

Why is this a good thing?
V=U/qo is independent of the test charge qo

Only depends on the other charges. V arises directly from these
other charges, as described last time.


Last week’s example: electric dipole potential
-Q
+Q
x=-a
Oct. 4, 2007
x=+a
Superposition of
• potential from +Q
• potential from -Q
41
Dipole electric fields

Since most things are neutral, charge separation
leads naturally to dipoles.

Can superpose electric fields from charges just as
with potential

But E-field is a vector,
-add vector components
x=-a
+Q
Oct. 4, 2007
x=+a
-Q
42
Quick Quiz 2
In this electric dipole, what is the direction of the
electric field at point A?
A) Up
A
B) Left
C) Right
D) Zero
Oct. 4, 2007
x=-a
+Q
x=+a
-Q
43
Dipole electric fields
Note properties of E-field lines

+Q
Oct. 4, 2007
-Q
44
Conservative forces


Fg
Conservative Forces: the work done by the force is
independent on the path and depends only on the starting
and ending locations.
It is possible to define the potential energy U
Wconservative  U = Uinitial - Ufinal =
-(Kfinal - Kinitial) = -K
Oct. 4, 2007
=
45
Potential Energy of 2 charges

Consider 2 positive charged particles. The electric force between them is

The work that an external agent should do to bring q2 at a distance rf from
q1 starting from a very far away distance is equal and opposite to the work
done by the electric force.
Charges repel
W>0!
F
W 
Oct. 4, 2007

r12

F  dr  W E
r12
46
Potential Energy of 2 charges
Since the 2 charges repel, the force on q2 due to q1
F12 is opposite to the direction of motion
 The external agent F = -F12 must do positive work!
W > 0 and the work of the electric force WE < 0

r12
W  W E    F  dr    Fdr   
r12


r12

q1q2
ke 2 dr
r
F
dr
r12
 1
q1q2
W  keq1q2   ke
 r 
r12
Oct. 4, 2007
r12
47
Potential Energy of 2 charges



Since WE = -U = Uinitial - Ufinal = = -W  W = U
We set Uinitial = U() = 0 since at infinite distance the force
becomes null
The potential energy of the system is
Oct. 4, 2007
48
More than two charges?
Oct. 4, 2007
49
U with Multiple Charges


If there are more than two
charges, then find U for each pair
of charges and add them
For three charges:
Oct. 4, 2007
50