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a) True
b) False
Is your clicker here? 0206DADE 0C962DB7 0D75047C 0EAB5BFE
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iClicker Quiz
HW 4 was due today at 845 a.m.
I have completed at least 50% of the reading and
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A. True
B. False
Quiz 5: Today’s Quiz, these numbers were not registered 0742D095
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245F89F2 33B79317 35D17591 Next time I am publishing names.
Off-axis
(perpendicular)
dipole field
kQr+ k (−Q)r− kQ(r+ − r− )
=
E = E+ + E− = 3 +
r+
r−3
r3
(
) (
)
Q(r+ − r− ) = a ˆi + y ˆj − − a ˆi + y ˆj = Q(2a ˆi ) = −p
For r >> a , r ≈ y
E=−
kp
kp
≈
−
r3
y3
E
p
r+
r−
p
Serway example 23.6
On-axis dipole field
d = 2a
+Q
p
E
-Q
p
E
x
E = E+ + E− =
kQr+ k (−Q)r−
+
3
3
r+
r−
 1
 1
1 
1 
ˆ
ˆ



= kQi  2 − 2  = kQi 
−
2
2 
r− 
 ( x + a) ( x − a) 
 r+
kQˆi
−2
−2
= 2 (1 + a / x ) − (1 − a / x )
x
kQˆi   2a   2a  
≈ 2  1 −  − 1 +  
x 
x  
x 
 − 4a  2k (−2aQˆi ) 2k p
ˆ
= 3
= kQi  3  =
3
x
x
 x 
(
HW 4-1 , see also SG: G8
)
Binomial approximation: (is
used going from 3rd to 4th line)
(1 + s ) p ≈ 1 + ps if
s << 1
Try using the general dipole formula
(only for the purists among you)
3(p ⋅ rˆ )rˆ − p
E=k
r3
Perpendicular case
On-axis case
p = − p ˆi and rˆ = ˆj
3(− pˆi ⋅ ˆj)ˆj − (− pˆi )
E=k
y3
0ˆj + pˆi − kp
=k
= 3
3
y
y
p = − p ˆi and rˆ = ˆi
3(− pˆi ⋅ ˆi )ˆi + pˆi
E=k
x3
− 2kpˆi 2kp
=
= 3
3
x
x
Moments of a distribution (extra details for purists)
(If you tell anyone that I mentioned this, I’ll deny it!)
Monopole moment (scalar)
q = ∫ ρ (r )dV
dQ

r
Dipole moment (vector)
p = ∫ rρ (r ) dV
Quadrupole moment (polar rank-2 tensor)
~
2~
Q = ∫ (3 r ⊗ r − r I ) ρ (r ) dV
Choose origin at the
center of charge
(like center of mass)
The electric field of a charge distribution can be expressed as a series
expansion involving successively higher moments of the distribution.
~
~
qrˆ
3(p ⋅ rˆ )rˆ − p
5(rˆ ⋅ Q ⋅ rˆ ) − Q ⋅ rˆ
E=k 2 +k
+k
+ ...
3
4
r
r
r
There is a region near the origin of a coordinate system that you are told
contains charges. You are not able to see them but have been asked to
learn as much about them as possible. In your experimentation you place a
test charge at three positions a distance d from the origin and determine the
direction of the electric field. Along the positive x axis the field points in
the +x direction. Along the negative x axis it also points in the +x
direction. On the y axis however it points in the negative
x direction. Consider this result and suggest what you might be able to say
about the charges near the origin.
●It would seem from this information that there are at least two charges near
the origin, one positive and one negative.
●It may be said that there is a positive and a negative charge present in the
cluster. The positive charge would be closer to the +x direction than the
negative charge. This would explain the direction of the field staying
consistent in the +x direction on either side of the axis. Along the y axis the
field would be moving in the -x direction because the electric field line from
the positive charge on the right would be moving towards the negative
charge on the left.
●It could be two charges right next to each other. There would be a negative
charge on the left side, and a positive charge on the right. This would cause
a test charge on the right to move right, away from the positive charge, and
the test charge on the left to be attracted towards the positive charge. The
test charge placed on the y axis would be pushed left, because it is slightly
attracted by the negative charge, and slightly repelled by the positive one.
The y-coordinate's would cancel out.
●I believe that the only way for this to be possible is for there to be equal
numbers of positive and negative charges ….
Fields due to individual point charges:
MU 29:11-14, 29
This is true for
electrostatics
1. Field lines start on positive charges
2. Field lines end on negative charges
3. Field lines are closest together in fields of large
magnitude
4. Long distances away from a collection of charges the
pattern of field lines is like that for a point charge.
Electric Field Lines
• The direction of the Electric field is the direction
the force on a positive test charge would be at a
given point if a charge were present.
• The direction of the Electric field lines show the
direction of the field.
• The density of the field lines tells how strong the
field is (how much force a positive test charge
would feel)
• Field lines always start and end on a charge.
Which is a valid representation of field lines when no charges
are present in the viewed area?:
e) More than one are valid
See interactive applet at http://lectureonline.cl.msu.edu/~mmp/applist/applets.htm
E-field perpendicular to conducting surfaces
E-field perpendicular to conducting surfaces
E = σ/ε0 = 4πkσ
Electric field strength at the
surface of a charged conductor
is proportional to local surface
charge density.
Can you identify places where
the field is strongest?
Can a moving charge
cross field lines? Pp.
A. True
B. False
What path will an electron take
in a region of an electric field?
The electricity vs. gravity analogy
Coulomb’s Electric Force Law
Q

r
q
 kQrˆ
E= 2
r

 kqQrˆ
F = qE = 2
r



 qE
F = ma ⇒ a =
m
Newton’s Gravitational Force Law
M

r
m

GMrˆ
g=− 2
r


GmMrˆ
F = mg = −
r2


 
F = ma ⇒ a = g



F = qE = ma

 qE
a=
m
vx = vx 0 + axt = vx 0
 qE 
v y = v y 0 + a yt = 
t { y0 = 0, v y 0 = 0}
 m
 qE  2
y = y0 + v y 0 t + a y t = 
t
 2m 
1
2
2
a = constant
v = v0 + at
x = x0 + v0t + 12 at 2
v − v = 2a ( x − x0 )
2
2
0
F = qE
WE = − ∫ F ⋅ ds = − q ∫ E ⋅ ds
Uniform field :
∫ E ⋅ ds = E ⋅ ∫ ds = E ⋅ s = E x
∆K = WE = qEx
The electricity vs. gravity analogy
Coulomb’s Electric Force Law
Q

r
q
 kQrˆ
E= 2
r

 kqQrˆ
F = qE = 2
r



 qE
F = ma ⇒ a =
m
Newton’s Gravitational Force Law
M

r
m

GMrˆ
g=− 2
r


GmMrˆ
F = mg = −
r2


 
F = ma ⇒ a = g
Future
Flux
Bucket 1 will
catch more
water than
bucket 2
Area vector
1
2
Define an Area vector perpendicular to the surface.
The more that vector is lined up with the rain, the more
rain gets into the bucket.
If this vector is perpendicular to the rain, no rain gets into
the bucket.
Electric Flux
   
Φ = E ⋅ A = E A cos θ
 
Φ = ∫ E ⋅ dA
surface
The net flux through a surface is the number
of lines leaving the surface minus the
number of lines entering the surface.
A cylindrical piece of insulating material is placed in an external electric
field, as shown. The net electric flux passing through the surface of the
cylinder is
A. positive.
B. negative.
C. zero.