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From last time(s)…
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Gauss’ law
Conductors in electrostatic equilibrium
Today…
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Oct. 2, 2008
Finish conductors in electrostatic equilibrium
Work, energy, and (electric) potential
Electric potential and charge
Electric potential and electric field.
1
Exam 1 Scores
Class average = 76%
(This is 84/110)
Your score posted
at learn@uw
Curve:
B / BC boundary is 76%
Oct. 2, 2008
2
Conductor in Electrostatic
Equilibrium
In a conductor in electrostatic equilibrium there is no net motion
of charge
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E=0 everywhere inside the conductor
Ein
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Etot =0
Conductor slab in an external field E:
if E  0 free electrons would be
accelerated
These electrons would not be in
equilibrium
When the external field is applied, the
electrons redistribute until they
generate a field in the conductor that
exactly cancels the applied field.
Etot = E+Ein= 0
Oct. 2, 2008
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Conductors: charge on surface
only
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Choose a gaussian surface inside (as close
to the surface as desired)
E=0
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There is no net flux through the gaussian
surface (since E=0)
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Any net charge must reside on the surface
(cannot be inside!)
Oct. 2, 2008
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E-Field Magnitude and Direction
E-field always  surface:
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Parallel component of E would put force on
charges
Charges would accelerate
This is not equilibrium
Apply Gauss’s law at surface
E this surface
 E  EA
E || this surface
 E  0


Oct. 2, 2008
E 0

Tot
E  EA
Qencl  A
 E   /o
this surface

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Summary of conductors
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E  0everywhere inside a conductor
Charge in conductor is only on the surface
E
surface of conductor
---
Oct. 2, 2008
++
+
+
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Electric forces, work, and energy
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Consider positive particle charge q, mass m at rest
in uniform electric field E
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Force on particle from field
Opposite force on particle from hand
Let particle go - it moves a distance d
How much work was done on particle? W  Fd  qEd
1 2
K.E.

mv  W  qEd
How fast is particle moving?
2
v=0

Oct. 2, 2008
+

v>0
+
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Work and kinetic energy
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Work-energy theorem:
 Change in kinetic energy of isolated
particle = work done
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dW  F  ds  Fdscos
Total work
end
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K 
end
 dW   F  ds
start
start
In our case, F  qE

Oct. 2, 2008
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Electric forces, work, and energy
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Same particle, but don’t let go
F  qE
Move particle distance d, keep speed ~0
How much work is done by hand on particle? W  Fd  qEd
What is change in K.E. 
of particle? K.E .  0
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How much force does hand apply?
Conservation of energy? W stored in field as potential energy


+
Oct. 2, 2008
+
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Work, KE, and potential energy
If particle is not isolated,
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Wexternal  K  U
Work done
on system
Change in
kinetic energy
Change in
electric potential energy
Works for constant electric field if U  qE  r

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Only electric potential energy difference

Sometimes a reference
point is chosen
E.g. U r   0 at r  (0,0,0)
 Then U r   qE  r for uniform electric field
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Oct. 2, 2008
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Electric potential V
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Electric potential difference V is the electric
potential energy / unit charge = U/q
For uniform electric field,
U r  qE  r
V r  

 E  r
q
q
This is only valid for a uniform electric field

Oct. 2, 2008
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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. 2, 2008
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Check for uniform E-field
Push particle against E-field, or across E-field
Which requires work?
+
Increasing electric
potential in this direction
Oct. 2, 2008
Constant electric potential
in this direction
+
Decreasing electric
potential in this direction
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Potential from electric field
dV  E  d
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Potential changes
largest in direction of V  V d
o
E-field.
Smallest (zero)
E
perpendicular to


E-field

d
V  Vo  E d

d
V=Vo
 V  Vo  E d

Oct. 2, 2008
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Electric potential: general
U 
F
Coulomb
 ds 
 qE  ds  q  E  ds
Electric potential energy difference U
proportional to charge q that work is done on
U /q  V  Electric potential difference 
 E  ds
Depends only on charges that create E-fields

Electric field usually created 
by some charge
distribution.
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V(r) is electric potential of that charge distribution
V has units of Joules / Coulomb = Volts
Oct. 2, 2008
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Electric potential of point charge

kQ
Electric field from point charge Q is E  2 rˆ
r
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What is the electric potential difference?
V 
end

r final
E  ds 
start
 k
Define V r    0

Oct. 2, 2008

Q
k
dx
 
2
rinitial r
r final
Q
Q
Q
k
k
r rinitial
rinital
rfinal
Then
Q
V r  k
for point charge
r
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
Electric Potential of point charge
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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. 2, 2008
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
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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
 d
B
A
qo > 0
A
Differential form: qodV  FCoulomb  d
Oct. 2, 2008
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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
q2 q
q3q
k
k
k
r
r
r
 q1
q2
q3 
 qk  k  k 
 r
r
r 
q3
Superposition of
individual
electric potentials
Oct. 2, 2008
 F  ds   F

 qVq1 r  Vq 2 r  Vq 3 r
V r  Vq1r  Vq 2 r  Vq 3 r
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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. 2, 2008
20
Superposition:
the dipole electric potential
x=-a
Superposition of
• potential from +Q
• potential from -Q
x=+a
+Q
-Q
+
=
V in plane
Oct. 2, 2008
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