Videos
SP212
Ch. 22 - Electric Fields
Millikan Oil Drop
http://www.youtube.com/watch?v=XMfYHag7Liw
Wolfram Demonstration
http://demonstrations.wolfram.com/
DroppingATestChargeIntoAnElectricField/
Maj Jeremy Best USMC
Physics Department, U.S. Naval Academy
January 15, 2016
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SP212
January 15, 2016
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Find the Physics
Figure: techfuels.com
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Details
Figure: wisegeek.com
Figure:
Figure:
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www.projectvida.com
images.yourdictionary.com
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The Electric Field
Fields
Apparently, charged particles exert forces on each other
without touching. How does this action at a distance
come about? We explain this effect by saying that
electric charges create electric fields around themselves.
A field is something that has different values at different
places.
A temperature field
T (x, y , z)
A pressure field
P(x, y , z)
A wind velocity field
~
V(x,
y , z)
Figure: http://weather.unisys.com/upper_air/ua_cont.php?
plot=str&inv=0&t=cur
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SP212
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The Electric Field
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Calculating the Electric Field
The electric field is like the wind velocity field, every
point of the field is a vector. To measure the electric
field, place a positive test charge, q0 at some point.
Measure the force, ~F on that charge. The electric field is
then:
~F =
1 q · q0
r̂r
4π0 r 2
So the field is easy:
~
~E = F
q0
~
~E = F = 1 q r̂r
q0
4π0 r 2
The units of the electric field are N/C
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SP212
We know how to calculate force on a particle q0 due to
another particle q:
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SP212
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The Electric Field
The Electric Dipole
Let’s calculate the field due to two charges of opposite
signs, an electric dipole.
E = E+ + E−
1 q
1 −q
+
=
2
4π0 r+ 4π0 r−2
1
q
q
=
−
4π0 (z − 1/2d)2 (z + 1/2d)2
Electric field lines point away from positive charges
and toward negative charges
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SP212
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SP212
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The Electric Dipole
E=
1
4π0
q
q
−
(z − 1/2d)2 (z + 1/2d)2
q
d
2
2π0 z 3
d 2
1 − 2z
E-Field due to an Electric Dipole
Usually, we are standing much farther away from
the dipole than its separation. That is,
d << 2z, so that ugly denominator just
becomes 1!
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January 15, 2016
qd
2π0 z 3
We often combine the quantity qd into a single
quantity, the electric dipole moment, ~p. Note
that the dipole moment is a vector, and points
from the negative charge to the positive charge.
After some not very simple algebra, we get
E=
E=
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E=
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SP212
1 ~p
2π0 z 3
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Field Due to a Line of Charge
Field Due to a Ring of Charge
Point charges are straightforward. Now we need to
extend our discussion to continuous charge distributions.
To do this, we will need to use charge densities, not just
the total charge on the object
Line of charge λ = q/`
Surface charge σ = q/A
Volume charge ρ = q/V
C/m
C/m2
C/m3
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SP212
Let’s find the electric field ~E due to a ring of
charge q and radius R, at a point on the
central axis of the ring, a distance z from the
center of the ring.
From our definition of charge density, we have
dq = λ ds. Now we apply Coulomb’s Law:
1 dq
4π0 r 2
1
λ ds
=
4π0 (z 2 + R 2 )
dE =
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Field Due to a Ring of Charge
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Field Due to a Ring of Charge
By symmetry, we see that the horizontal
components of the field cancel out ( That’s
important. If you don’t see why, ask. Now).
So our problem just got easier. Let’s only
find the vertical component, dE cos θ.
cos θ = z/r =
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z
(z 2 + R 2 )1/2
Now adding in what we already know about
dE :
dE cos θ =
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Maj Jeremy Best USMC (Physics Department, U.S. Naval Academy)
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zλ
ds
4π0 (z 2 + R 2 )3/2
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Field Due to a Ring of Charge
Field Due to a Ring of Charge
Now, integrate! Note that we’re integrating
over ds, the arc length, and nothing depends
on s!
Z
E = dE cos θ
Z 2πR
zλ
=
ds
4π0 (z 2 + R 2 )3/2 0
zλ(2πR)
=
4π0 (z 2 + R 2 )3/2
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Field Due to a Disk of Charge
Notice that by our definition of charge
density, λ(2πR) = q , the total charge on
the ring. With this, our final answer
becomes:
E Field of Charged Ring
E=
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Field Due to a Disk of Charge
Now let us calculate the field due to a
solid disk of total charge q and radius R,
at a point P, a distance z from the center
of the disk.
First, write a charge element dq in terms
of surface charge density(σ):
Note that our area element is just a ring of
charge, a problem we already solved! We
just need to integrate that solution from
r = 0 → R and we’re done.
dE =
dq = σdA = σ2πr dr
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qz
4π0 (z 2 + R 2 )3/2
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z(σ2πr dr )
4π0 (z 2 + r 2 )3/2
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Field Due to a Disk of Charge
Z
E=
Z
σz R 2
(z + r 2 )−3/2 (2r ) dr
dE =
40 0
u = z 2 + r 2 du = 2r dr
Z
σz
=
u −3/2 du
40
σz −2
√
=
40 u
2σz
2σz
√ −
√
E=
40 z 2 40 z 2 + R 2
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All of that to say:
E−Field due to a charged disk
σ
E=
20
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z
1− √
z2 + R2
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A Point Charge in an Electric Field
In physics, it often important to examine
limiting cases. See what happens when
R → ∞ (an infinite sheet):
!
*
z 0
σ
1 − √ E=
2
2
20
z +R
σ
=
20
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Recall our definition of the electric field: ~E = ~F/q .
This implies why the electric field is so useful:
~F = q ~E
This force is a force like all the others from last
semester! It can go into Newton’s 2nd Law!
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Example
Solution
In this figure, we see a particle of mass
m = 1.3 × 10−10 kg and charge
Q = −1.5 × 10−13 C moving between
two charged plates. The particle is
initially moving along the x-axis velocity
v = 18 m/s.
The length of each plate is L = 1.6 cm. The electric field
between the plates is uniform and directed downward
with magnitude E = 1.4 × 106 N/C. What is the vertical
deflection of the particle at the end of the plates?
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Solution
Because the charge is negative and the
field points downward, the force points
upward . We use Newton’s 2nd Law:
ay = F /m = QE /m
Now we have a projectile motion
problem!
1
y (t) = ay t 2
2
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and x(t) = vx t
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Millikan Oil Drop
We solve for t and substitute everything we know:
t = L/vx
1
y = ay t 2
2
1 L2
= ay 2
2 vx
QEL2
=
2mvx2
= 6.4 × 10−4 m = 0.64 mm
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SP212
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SP212
Sample Problem
Solution
~F = q ~E
In Millikan’s experiment, an oil drop of radius 1.64 µm
and density 0.851 cmg 3 is suspended in chamber C when a
downward Electric Field of 1.92 × 105 NC is applied. Find
the charge on the drop, in terms of e.
We know ~E, and we can find ~F using 211. ~F = m~a
m = 43 πr 3 ρ so ~F = ( 43 πr 3 ρ)(~g)
q=
~F
~E
q = 8.03 × 10−19 C
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A Dipole in an Electric Field
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Microwave
A dipole in an external field does not experience a net
force, but it does experience a torque
We use the tendency of a dipole to align with an
external electric field frequently.
Its called a microwave.
As we discussed earlier, a sinusoidally alternating
electric field is created which causes the water in
food (dipoles) to rotate 180 degrees in alternating
fashion. This frequent movement causes heat,
which warms up other molecules around the water.
~ = ~p × ~E
τ
and energy
U = −~p · ~E
A dipole wants its dipole moment to align with the
external field.
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SP212
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Sample Problem
Solution
An electric dipole consists of charges +2e and −2e
separated by 0.78 nm. It is in an Electric Field of
strength 3.4 × 106 NC . Calculate the magnitude of the
torque on the dipole when the dipole moment is (a)
parallel to, (b) perpendicular to, and (c) antiparallel to
the Electric Field.
~ = ~p × ~E
τ
where p = qd
~ =0
(a) and (c) τ
(b)
~ = 2e dE sin θ
τ
~ = 8.5 × 10−22 N m
τ
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Wiley Plus Homework
Chapter 22: Questions 2, 3, 7, 9. Problems: 2, 7, 19, 22,
24, 34, 38, 44, 48, 59, 62, 83
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