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W02D1
Electric Dipoles and Continuous
Charge Distributions
1
Announcements
Math Review Tuesday Tues Feb 14 from 9-11 pm in
32-082
PS 1 due Tuesday Tues Feb 14 at 9 pm in boxes
outside 32-082 or 26-152
W02D2 Reading Assignment Course Notes: Chapter
Course Notes: Sections 4.1-4.2, 4.7
Make sure your clicker is registered
2
Outline
Electric Dipoles
Force and Torque on Dipole
Continuous Charge Distributions
3
Nature Likes to Make Dipoles
http://youtu.be/EMj10YIjkaY
4
Demonstration:
Dipole in a Van de Graaff
Generator D22
5
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
6
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
7
Continuous Charge
Distributions
8
Continuous Charge Distributions
Break distribution into parts:
V
Q = å Dqi ®
i
òòò
dq
V
E field at P due to Dq
Dq
dq
DE = ke 2 rˆ ® dE = ke 2 rˆ
r
r
Superposition:
EP  ?
ò
E = å DE ® dE
9
Continuous Sources: Charge Density
R
dQ   dV
Volume = V =
L
w
Area = A = wL
L
Length  L
L
p R2 L
r=
Q
V
dQ   dA
Q
s=
A
dQ   dL
Q
l=
L
10
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?
11
Examples of Continuous Sources:
Finite Line of Charge
Length  L
L
dQ   dL

Q
L
E field on
perpendicular
bisector
12
Examples of Continuous Sources:
Finite Line of Charge
Length  L
L
dQ   dL

Q
L
E field off axis
13
Examples of Continuous Sources:
Finite Line of Charge
Length  L
L
dQ   dL

Q
L
Grass seeds of
total E field
14
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
l dx ˆ
1. E(P) = - ò
i
3
(x - d)
x=0
x=L
l dx ˆ
i
x=L
2. E(P) =
ò
x=0
(x - d)
x=L
3. E(P) = -
ò
x=0
4. E(P) =
5. E(P) = -
3
l dx ˆ
i
(x - d)
l dx ˆ
òx=0 (x - d)2 i
x=L
2
6. E(P) =
lL
(d - x)
lL
(d - x)
2
2
ˆi
7. E(P) = ˆi
lL ˆ
i
d
lL ˆ
8. E(P) = 2 i
d
9. None of the above.
2
15
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
5. E(P) = -
lL
(d - x)
2
ˆi
16
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.
17
Hint on Line of Charge Group
Problem
ˆj
rˆ

ˆi
P
r
s  x
2
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.)
18
E Field from Line of Charge
Answer:
E = ke
Q
s(s + L / 4)
2
2
1/2
ˆj
Limits: s >> L (far away) and s << L (close)
Qˆ
lim E ® ke 2 j
s>> L
s
Q ˆ
lˆ
lim E ® 2ke
j = 2ke j
s<<L
Ls
s
Looks like the E field
of a point charge if
we are far away
Looks like E field of
an infinite charged
line if we are close
19
Examples of Continuous Sources:
Ring of Charge
dQ   dL

Q
2 R
E field on the
axis of the ring
of charge
20
Examples of Continuous Sources:
Ring of Charge
dQ   dL

Q
2 R
E field off axis
and grass
seeds plot
21
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?
1. E(P) = -
q =2p
òq
=0
2. E(P) =
q =2p
òq
=0
l adq ˆ
a3
i
l adq ˆ
a
Qˆ
3. E(P) = - 2 i
a
3
i
Qˆ
4. E(P) = + 2 i
a
5. E(P) = 0
22
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?
5. E(P) = 0
23
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.
24
Ring of Charge
1) Think about it
E = 0 Symmetry!
^
2) Define Variables
(
dq = l dl = l a dj
r 
a  x
2
)
2
25
Ring of Charge
3) Write Equation
rˆ
dE = ke dq 2
r
r
dE = ke dq 3
r
x
dEx = ke dq 3
r
dq = l a dj
r 
a  x
2
2
26
Ring of Charge
dq = l a dj
4) Integrate
x
Ex = ò dEx = ò ke dq 3
r
x
= ke 3 ò dq
r
r 
a  x
2
2
This particular problem is a very special case because
everything except dq is constant, and
dq
ò =
ò
2p
0
2p
l a dj = l a ò dj = l × a2p
Q
0
27
Ring of Charge
5) Clean Up
x
E x = ke Q 3
r
x
E x = ke Q 2
2 3/ 2
(a + x )
6) Check Limit
x
ˆi
E = ke Q 2
2 3/2
(a + x )
a®0
k eQ
x
Ex ® keQ 2 3/ 2 = 2
(x )
x
28
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
29
Disk: Two Important Limits
Answer: E d isk
é
s ê
x
=
12
2
2e o ê
x +R
êë
(
)
1/2
ù
ú ˆi
ú
úû
Limits: x >> R (far) and x << R (close)
lim E disk
x>> R
Qˆ
®
i
2
4pe o x
lim E disk
x<<R
1
s ˆ
®
i
2e o
Looks like E of a
point charge far
away
Looks like E field of
an infinite charged
plane close up
30
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
31
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