Download The Electric Force

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The Electric Force
Concepts and Principles
The Gravitational Analogy
In introducing the concept of the electric field, I tried to illustrate it by drawing an analogy
r
with the gravitational field, g . This analogy can be extended to electric force and
gravitational force.
From mechanics, the relationship for the gravitational force on an object is:
r
r
F = mg
where
m is the mass of the particle of interest,
and g is the net gravitational field at the location of the particle of interest. (This
field was typically approximated as that of the earth, but should actually be the field
created by all of the massive particles in the universe, other than the particle of
interest.)
r
Mass is the property that allows particles to create gravitational fields and it is also the
property that allows them to interact with other particles’ gravitational fields. This interaction
is termed the gravitational force.
Wouldn’t it be great if charge played the same role with regard to the electric field? Well, it is
great! The electric force on a particle is given by the relation,
r
r
F = qE
where
q is the charge on the particle of interest,
and E is the net electric field at the location of the particle of interest (created by all
of the other charged particles in the universe).
r
All charged particles create electric fields, but this is only half of the story. All charged
particles also interact with other particles’ electric fields. This interaction is termed the
electric force.
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The Electric Force
Analysis Tools
Point Charges
The three charges at right are located as
shown. Each grid square has width a. Find the
net electric force on the negative charge.
-2q
+q
+q
To find the electric force on the negative charge, you must first find the electric field at its
location in space. Thus, this is really just more practice in calculating the electric field!
The electric field at the location of the negative charge will be the vector sum of the electric
field from the left positive charge and the electric field from the right positive charge. Let’s
start by calculating the field from the left charge.
The magnitude of the electric field from the left charge is:
Eleft =
k ( q)
( a ) + (2 a ) 2
kq
= 2
5a
Eleft =
Eleft
kq
r2
2
To determine the direction of this field, notice that
the field vector points in the same direction as a line
connecting the source charge and the point of
interest. This line forms a right triangle with θ given
by:
2a
tan θ =
a
tan θ = 2
2a
+q
θ
a
θ = 63.4 0
2
Thus,
kq
cos 63.4
5a 2
kq
= 2 (0.45)
5a
kq
= 0.089 2
a
kq
sin 63.4
5a 2
kq
= 2 (0.89)
5a
kq
= 0.179 2
a
Eleftx =
Elefty =
Eleftx
Elefty
Eleftx
Elefty
At this point, we could repeat the calculation for the right positive charge, but that would be a
waste of time. Based on the symmetry of the situation, the electric field from the right charge
has exactly the same magnitude as that of the left charge, but with opposite x-direction and
the same y-direction. Thus, when they are summed to yield the total electric field the xcomponents will cancel and the y-components will add resulting in:
E y = Elefty + E righty
kq
kq
+ 0.179 2
2
a
a
kq
E y = 0.358 2
a
E y = 0.179
Thus, the electric field at the location of interest points straight up. The force exerted on the
charge at this location is given by:
r
r
F = qE
Fy = qE y
Fy = (−2q )(0.358
kq
)
a2
kq 2
Fy = −0.715 2
a
The negative charge feels a force directed straight downward.
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Force and Motion
In many applications, oppositely charged parallel plates
are used to “steer” beams of charged particles. In this
example, a proton is injected at 2.0 x 106 m/s into the
space between the plates. The plates are 2.0 cm long.
What charge density is needed on the plates to give the
proton a y-velocity of 2.0 x 105 m/s as it exits the plates?
Since this problem involves the motion of a particle between two distinct events, let’s
complete a motion table.
Event 1: The proton enters Event 2: The proton exits
the device.
the device.
t1 = 0 s
t2 =
r1x = 0 m
r2x = 0.02 m
r1y = 0 m
r2y =
6
v1x = 2 x 10 m/s
v2x =
v1y = 0 m/s
v2y = 2 x 105 m/s
a12x = 0 m/s2
a12y =
Between the plates, the proton will experience an electrical force, and hence acceleration, in
the y-direction. Remember from mechanics that this y-acceleration will not affect the
kinematics of the proton in the x-direction. Thus, in the x-direction the acceleration of the
proton is zero.
Applying the kinematic equations in the x-direction yields:
v 2 = v1 + a12 (t 2 − t1 )
v 2 x = (2 x10 ) + 0(t 2 − 0)
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v 2 x = 2 x10 6 m / s
1
a12 (t 2 − t1 ) 2
2
1
0.02 = 0 + (2 x10 6 )(t 2 − 0) + (0)(t 2 − 0) 2
2
−8
t 2 = 1x10 s
r2 = r1 + v1 (t 2 − t1 ) +
Applying the same kinematic equations in the y-direction yields:
v2 = v1 + a12 (t2 − t1 )
r2 = r1 + v1 (t 2 − t1 ) +
−8
2 x10 = 0 + a y (1x10 − 0)
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1
(2 x10 13 )(1x10 −8 − 0) 2
2
= 0.001m
r2 y = 0 + 0 +
a y = 2 x1013 m / s 2
r2 y
4
1
a12 (t 2 − t1 ) 2
2
Now, using Newton’s Second Law and the relation for electric field from parallel conducting
plates, find the necessary charge density.
F = ma
qE − mg = ma
q(4πkσ ) − mg = ma
ma − mg
σ=
4πkq
σ=
(1.67 x 10 -27 )(2.0 x 1013 ) − (1.67 x 10 -27 )(9.8)
4π (9.0 x 10 9 )(1.6 x 10 -19 )
σ = 1.85µC / m 2
The bottom plate should be made positive and the top plate negative, both with this charge
density. (Note that the force of gravity acting on the proton is completely insignificant
compared to the electrical force. This is generally true and we will typically ignore the force
of gravity acting on individual particles such as protons and electrons.)
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The Electric Force
Activities
6
Determine the direction of the net electric force on each charge.
a.
+q
-q
-q
b.
-q
-q
+q
7
Determine the direction of the net electric force on each charge.
a.
+q
-q
-2q
b.
-q
-2q
+q
8
Determine the direction of the net electric force on each charge.
a.
+q
-q
-q
b.
+q
-q
-q
+q
9
Determine the direction of the net electric force on each charge.
a.
+2q
+q
+q
b.
-q
-q
-q
-q
10
Below are free-body diagrams for three electric charges that lie along a straight line. Determine the relative
positions, with correct spacing, of the three charges.
a.
+q
-q
+q
b.
-q
-q
+q
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Below are free-body diagrams for three electric charges that lie in the same plane. Determine the relative
positions, with correct spacing, of the three charges.
a.
-q
-q
+q
b.
+2q
+q
-2q
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Below are free-body diagrams for three electric charges that lie in the same plane. Determine the relative
positions, with correct spacing, of the three charges.
a.
-2q
-q
+q
b.
+q
+q
-q
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For each of the six combinations of electric charges listed below, rank the combinations on the basis of the
electric force acting on the central charge. Forces pointing to the right are positive.
q1
q1
A 1 nC
B 1 nC
C -1 nC
D 1 nC
E 2 nC
F 2 nC
q2
q3
q2
q3
1 nC
-1 nC
1 nC
2 nC
1 nC
1 nC
1 nC
1 nC
1 nC
-1 nC
2 nC
-2 nC
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest
_____ The ranking can not be determined based on the information provided.
Explain the reason for your ranking:
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For each of the six combinations of electric charges listed below, rank the combinations on the basis of the
electric force acting on the central charge. Forces pointing to the right are positive.
q1
q1
A 1 nC
B 1 nC
C -1 nC
D 2 nC
E 2 nC
F 2 nC
q2
q3
q2
q3
1 nC
1 nC
1 nC
2 nC
1 nC
1 nC
1 nC
2 nC
-4 nC
4 nC
2 nC
-8 nC
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest
_____ The ranking can not be determined based on the information provided.
Explain the reason for your ranking:
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For each of the six combinations of electric charges listed below, the electric force acting on the central
charge is zero. Rank the combinations on the basis of q3.
q1
q2
q3
q1
A
B
C
D
E
F
1 nC
1 nC
-1 nC
2 nC
2 nC
1 nC
q2
1 nC
-1 nC
1 nC
2 nC
1 nC
2 nC
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest
_____ The ranking can not be determined based on the information provided.
Explain the reason for your ranking:
16
For each of the six combinations of electric charges listed below, rank the combinations on the basis of the
electric force acting on the left charge. Forces pointing to the right are positive.
q1
q1
A 1 nC
B 1 nC
C 1 nC
D 1 nC
E -1 nC
F 1 nC
q2
q3
q2
q3
1 nC
-1 nC
1 nC
-1 nC
1 nC
-2 nC
1 nC
-1 nC
-2 nC
4 nC
-4 nC
-4 nC
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest
_____ The ranking can not be determined based on the information provided.
Explain the reason for your ranking:
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For each of the six combinations of electric charges listed below, the electric force acting on the left charge
is zero. Rank the combinations on the basis of q3.
q1
q2
q3
q1
A
B
C
D
E
F
1 nC
1 nC
-1 nC
2 nC
2 nC
1 nC
q2
1 nC
-1 nC
1 nC
2 nC
1 nC
2 nC
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest
_____ The ranking can not be determined based on the information provided.
Explain the reason for your ranking:
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A uniform electric field fills a region of space. In each of the regions, a charged particle (either a proton or
an electron) enters the region with an initial velocity. The subsequent path of the particle is indicated by
sequential velocity vectors. Indicate the direction of the electric field vector in each region of space.
a.
b.
-e
+e
b.
c.
+e
+e
d.
e.
+e
-e
f.
g.
-e
-e
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A sodium ion (Na+) and a chlorine ion (Cl-) are separated by 2.82 x 10-10
m. Find the electric force on each ion.
Na+
Cl-
Qualitative Analysis
On the graphic above, sketch the direction of the electric force on the requested object(s). Explain why the
electric force points in this direction.
Mathematical Analysis
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Find the electric force on each of the atoms in the chain of sodium ions
(Na+) and chlorine ions (Cl-). The interionic separation is 2.82 x 10-10
m.
Cl-
Na+
Cl-
Qualitative Analysis
On the graphic above, sketch the direction of the electric force on the requested object(s). Explain why the
electric force points in this direction.
Mathematical Analysis
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Na+
Find the electric force on each of the atoms in the sodium chloride
“crystal”. The interionic separation is 2.82 x 10-10 m.
Cl-
Na+
Na+
Cl-
Qualitative Analysis
On the graphic above, sketch the direction of the electric force on the requested object(s). Explain why the
electric force points in this direction.
Mathematical Analysis
22
In the H2O molecule illustrated at right, the oxygen atom has an effective
charge of -2e and each hydrogen atom +e. The effective charge separation
between the O and each H is 4.9 x 10-12 m. The molecule forms an angle of
1050. Find the electric force on each of the three atoms.
H
H
O
Qualitative Analysis
On the graphic above, sketch the direction of the electric force on the requested object(s). Explain why the
electric force points in this direction.
Mathematical Analysis
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The two positive charges are separated by a distance 2a. The negative
charge is located at (x, y) = (0, 2a). Find the net electric force on each
charge.
-2q
+q
+q
Qualitative Analysis
On the graphic above, sketch the direction of the electric force on the requested object(s). Explain why the
electric force points in this direction.
Mathematical Analysis
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In the Bohr model of the hydrogen atom, the ground state of hydrogen consists of a stationary proton
encircled by an electron at radius 5.29 x 10-11 m. In the first excited state, the electron's radius is four times
as large. What is the velocity of the electron in the ground state and the first excited state?
Mathematical Analysis
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In alpha decay, an alpha particle (a bound state of 2 protons and 2 neutrons) escapes from a heavy nucleus
and is propelled away due to electrical repulsion. For example, a radon nucleus (atomic number 86) will
spontaneously transform into a polonium nucleus (atomic number 84) and an alpha particle. Immediately
following this transformation, both the polonium nucleus and the alpha particle can be assumed to be at
rest, and separated by approximately 50 x 10-15 m. However, the electrical repulsion between the two
particles will cause the alpha particle to accelerate away from the polonium. Due to its very large mass, the
polonium nucleus can be assumed to remain stationary. What is the initial acceleration of the alpha
particle?
Mathematical Analysis
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In many applications, oppositely charged parallel plates (with small holes
cut for the beam to pass through) are used to accelerate beams of charged
particles. In this example, an electron is injected at 2.0 x 107 m/s into the
space between the plates. If the electron exits the device at 2.5 x 107 m/s,
what is the charge density on each plate? The plates are 5.0 cm apart.
Mathematical Analysis
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In many applications, oppositely charged parallel plates (with small holes
cut for the beam to pass through) are used to accelerate beams of charged
particles. In this example, a proton is injected at 2.0 x 106 m/s into the
space between the plates. The plates have charge density 1.0 µC/m2, with
the left plate positive, and are 5.0 cm apart. What is the velocity of the
proton as it exits the device?
Mathematical Analysis
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In many applications, oppositely charged parallel plates are used to
“suspend” charged particles. In this example, a dust grain with net charge
+1.5 pC is suspended between the plates. The plates have charge density
1.0 µC/m2, with the bottom plate positive. What is the mass of the dust
grain?
Mathematical Analysis
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In many applications, oppositely charged parallel plates are used to
“suspend” charged particles. In this example, an oil droplet of radius 1.8
µm and density 0.85 g/cm3 is suspended between the plates. The plates
have charge density 1.0 µC/m2, with the bottom plate positive. What is the
charge on the drop?
Mathematical Analysis
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In many applications, oppositely charged parallel plates are used to
“steer” beams of charged particles. In this example, a proton is injected at
2.0 x 106 m/s into the space between the plates. The plates have charge
density 1.0 µC/m2, with the bottom plate positive, and are 2.0 cm long.
What is the position and velocity of the proton as it exits the device?
Motion Information
Mathematical Analysis
Event 1:
Event 2:
t1 =
t2 =
r1x =
r2x =
r1y =
r2y =
v1x =
v2x =
v1y =
v2y =
a12x =
a12y =
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In many applications, oppositely charged parallel plates are used to
“steer” beams of charged particles. In this example, a proton is injected at
2.0 x 106 m/s into the space directly between the plates. The plates have
charge density 1.0 µC/m2, with the bottom plate positive, are 0.1 cm apart,
and are very long. What is the position and velocity of the proton as it
strikes the top plate?
Motion Information
Mathematical Analysis
Event 1:
Event 2:
t1 =
t2 =
r1x =
r2x =
r1y =
r2y =
v1x =
v2x =
v1y =
v2y =
a12x =
a12y =
32
In many applications, oppositely charged parallel plates are used to
“steer” beams of charged particles. In this example, an electron is
injected at 2.0 x 107 m/s into the space between the plates. The plates have
charge density 5.0 µC/m2, with the bottom plate positive, and are 2.0 cm
long. 20 cm from the end of the plates is a phosphorescent screen. Where
on the screen does the electron strike?
Motion Information
Event 1:
Event 2:
Event 3:
t1 =
t2 =
t3 =
r1x =
r2x =
r3x =
r1y =
r2y =
r3y =
v1x =
v2x =
v3x =
v1y =
v2y =
v3y =
a12x =
a23x =
a12y =
a23y =
Mathematical Analysis
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