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W02D1
Electric Dipoles and Continuous
Charge Distributions
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Announcements
Math Review Tuesday Week Two Tues from 9-11 pm
in 26-152
PS 1 due Week Two Tuesday Tues at 9 pm in boxes
outside 32-082 or 26-152
W02D2 Reading Assignment Course Notes: Chapter
Course Notes: Sections 3.1-3.2, 3.6, 3.7, 3.10
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Outline
Electric Dipoles
Force and Torque on Dipole
Continuous Charge Distributions
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Nature Likes to Make Dipoles
http://youtu.be/EMj10YIjkaY
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Demonstration:
Electric field Lines D16
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Dipole in Uniform Field
r
E  Eφ
i
r
p  2qa(cos φ
i  sin  φ
j)
r
r
r
r
r
Total Net Force: Fnet  F  F  qE  (q)E  0
Torque on Dipole:
r r r r r
 rFpE
  rF sin( )  (2a)(qE)sin( )  pE sin( )
p tends to align with the electric field
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Torque on Dipole
Total Field (dipole + background) shows torque:
• Field lines transmit tension
• Connection between dipole field and constant
field “pulls” dipole into alignment
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Demonstration:
Dipole in a Van de Graaff
Generator D22
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Demonstration:
Bouncing Balloon
Van de Graaf Generator D17
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Concept Question: Dipole in NonUniform Field
E
A dipole sits in a non-uniform
electric field E
Due to the electric field this dipole will
feel:
1.
2.
3.
4.
force but no torque
no force but a torque
both a force and a torque
neither a force nor a torque
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Concept Question Answer: NonUniform Field
E
Answer: 3. both force and torque
Because the field is non-uniform, the forces on
the two equal but opposite point charges do
not cancel.
As always, the dipole wants to rotate to align
with the field – there is a torque on the dipole
as well
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Continuous Charge
Distributions
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Continuous Charge Distributions
Break distribution into parts:
V
Q   qi 
i

dq
V
E field at P due to q
r
r
q
dq
 E  ke 2 rφ  d E  ke 2 rφ
r
r
Superposition:
E P  ?
r
r
r
E   E  d E
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Continuous Sources: Charge Density
R
Volume  V   R2 L
L
w
Area  A  wL
L
Length  L
L
dQ   dV
  Q / V (uniform)
dQ   dA
  Q / A (uniform)
dQ   dL
  Q / L (uniform)
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Group Problem: Charge Densities
A solid cylinder, of length L and radius R, is uniformly
charged with total charge Q.
(a)What is the volume charge density ρ?
(b)What is the linear charge density λ?
(c)What is the relationship between these two densities ρ
and λ?
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Examples of Continuous Sources:
Finite Line of Charge
Length  L
L
dQ   dL
Q

L
E field on
perpendicular
bisector
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Examples of Continuous Sources:
Finite Line of Charge
dQ   dL
Length  L
Q
L

L
E field off axis
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Examples of Continuous Sources:
Finite Line of Charge
dQ   dL
Length  L
Q
L

L
Grass seeds of
total E field
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Concept Question Electric Field of a Rod
A rod of length L lies along the
x-axis with its left end at the
origin. The rod has a uniform
charge density λ. Which of the
following expressions best
describes the electric field at
the point P
x L
r
 dx φ
1. E(P)  ke 
i
3
x 0 ( x  d)
x L
r
 dx φ
4. E(P)  ke 
i
2
x0 ( x  d)
x L
r
 dx φ
2. E(P)  ke 
i
3
x 0 ( x  d)
r
5. E(P)  ke
x L
r
 dx φ
3. E(P)  ke 
i
2
(
x

d)

x 0
r
6. E(P)  ke
L
(d  L)
L
(d  L)
2
2
φ
i
φ
i
r
L
7. E(P)  ke 2 φ
i
d
r
L φ
8. E(P)  ke 2 i
d
9. None of the above.
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Concept Question Electric Field of a Rod:
Answer
A rod of length L lies along the
x-axis with its left end at the
origin. The rod has a uniform
charge density λ. Which of the
following expressions best
describes the electric field at
the point P
x L
r
 dx φ
3. E(P)  ke 
i
2
x 0 ( x   d)
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Group Problem: Line of Charge
Point P lies on perpendicular bisector of uniformly charged line of length
L, a distance s away. The charge on the line is Q. Find an integral
expression for the direction and magnitude of the electric field at P.
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Hint on Line of Charge Group
Problem
ĵ
r̂

î
P
r  s 2  x 2
s


L
2
x
dq   dx 
dx 

L
2
Typically give the integration variable (x’) a
“primed” variable name. ALSO: Difficult integral
(trig. sub.)
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E Field from Line of Charge
Answer:
r
E  ke
Q
φ
j
2
2
1/2
s(s  L / 4)
Limits: s >> L (far away) and s << L (close)
r
Qφ
lim E  ke 2 j
sL
s
Looks like the E field
of a point charge if
we are far away
r
Qφ
 φ Looks like E field of
lim E  2ke
j  2ke j
sL
an infinite charged
Ls
s
line if we are close
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Examples of Continuous Sources:
Ring of Charge
Q

dQ   dL
2 R
E field on the
axis of the ring
of charge
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Examples of Continuous Sources:
Ring of Charge
Q

dQ   dL
2 R
E field off axis
and grass
seeds plot
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Concept Question Electric Field of a Ring
A uniformly charged ring of radius
a has total charge Q. Which of the
following expressions best
describes the electric field at the
point P located at the center of the
ring?
 2
r
 ad φ
1. E(P)  ke 
i
3
a
 0
 2
r
 ad φ
2. E(P)  ke 
i
3
a
 0
r
Qφ
3. E(P)  ke 2 i
a
r
Qφ
4. E(P)  ke 2 i
a
r
r
5. E(P)  0
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Concept Question Electric Field of a Ring:
Answer
A uniformly charged ring of radius
a has total charge Q. Which of the
following expressions best
describes the electric field at the
point P located at the center of the
ring?
r
r
5. E(P)  0
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Demonstration Problem: Ring of Charge
A ring of radius a is uniformly charged with total charge Q.
Find the direction and magnitude of the electric field at the
point P lying a distance x from the center of the ring along
the axis of symmetry of the ring.
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Ring of Charge
1) Think about it
E  0 Symmetry!
2) Define Variables
dq   dl   (a d )
r a x
2
2
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Ring of Charge
3) Write Equation
r
rφ
dE  ke dq 2
r
r
r
r
dE  ke dq 3
r
x
dEx  ke dq 3
r
dq   a d
r  a2  x 2
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Ring of Charge
dq   a d
4) Integrate
r  a2  x 2
x
Ex   dEx   ke dq 3
r
x
 ke 3  dq
r
This particular problem is a very special case because
everything except dq is constant, and
 dq  
2
0
2
 a d   a  d   a2  Q
0
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Ring of Charge
5) Clean Up
x
E x  k eQ 3
r
x
E x  k eQ 2
2 3/2
(a  x )
r
E  k eQ
6) Check Limit a  0
x
φ
i
2
2 3/2
(a  x )
E x  k eQ
k eQ
x
 2
2 3/2
(x )
x
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Group Problem: Uniformly Charged
Disk
(x > 0)
P on axis of disk of charge, x from center
Radius R, charge density .
Find E at P
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Disk: Two Important Limits
r
Answer: E disk

 
x

1
2
2
2 o 
x R



1/2

φ
i


Limits: x >> R (far) and x << R (close)
r
lim Edisk 
xR
1
Qφ
i
2
4 o x
r
 φ
lim Edisk 
i
xR
2 o
Looks like E of a
point charge far
away
Looks like E field of
an infinite charged
plane close up
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Scaling: E for Plane is Constant
1)
2)
3)
4)
Dipole:
E falls off like 1/r3
Point charge: E falls off like 1/r2
Line of charge: E falls off like 1/r
Plane of charge: E constant
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