Answer, Key – Homework 8 – David McIntyre
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The due time is Central time.
Chapter 23 problems.
−3.8 µC
7.3
m
q
q
5.1 m
001 (part 1 of 1) 0 points
A solid conducting sphere is given a positive
charge Q.
How is the charge Q distributed in or on
the sphere?
1
3 µC
1. It is uniformly distributed throughout the
sphere.
2. Its density decreases radially outward
from the center.
3. It is concentrated at the center of the
sphere.
Find q .
Correct answer: 0.704413 µC.
Explanation:
Given :
4. Its density increases radially outward
from the center.
5. It is uniformly distributed on the surface
of the sphere only. correct
002 (part 1 of 1) 10 points
Three point charges are located at the vertices
of an equilateral triangle. The charge at the
top vertex of the triangle is given in the figure.
The two charges q at the bottom vertices of
the triangle are equal. A fourth charge 3 µC
is placed below the triangle on its symmetryaxis, and experiences a zero net force from
the other three charges, as shown in the figure
below.
Q1
a
Explanation:
The charge distribution on conductors can
only be on the surface, and since on a spherical
surface every point is like any other surface
point, the charge distribution is uniform.
The electric field is normal to the surface of
a conductor. The conductor is symmetrical
(since it is spherical), so the charge must be
uniform.
Q1 = −3.8 µC ,
Q2 = q ,
Q3 = q ,
Q4 = 3 µC ,
a = 7.3 m , and
d = −5.1 m .
Q2
θ
Q3
d
3 µC
The force on Q4 is due to the Coulomb
forces from Q1 , Q2 , and Q3 . Because Q2 and
Q3 have equal charge, the x-components of
their forces cancel out (by symmetry). Thus
we only need to consider the y-components of
the forces.
Qi Q4
Coulomb’s law tells us Fi = k
is the
r2
ith force on Q4 from the ith charge. The forces
from Q2 and Q3 are equal to each other, and
opposite the direction of the force from Q1 ,
since otherwise they could not cancel. Total
Answer, Key – Homework 8 – David McIntyre
force on Q4 is
F Q4 = k
Q1 Q4
Q2 Q4
+ 2k 2
sin θ
2
r14
r24
4
Our force equation becomes

Q1

0 = k Q4  ³ √
´2
3
a
+
|d|
2
+³
2 Q2
a2
4
+ d2
Rearranging, we get
=
|d|
a2
4
+ d2
003 (part 1 of 1) 0 points
Two identical small charged spheres hang in
equilibrium with the masses shown in the figure. The length of the strings are equal and
the angle (shown in the figure) with the vertical is identical.
6◦
0.02 kg
Find the magnitude of the charge on each
sphere.
Correct answer: 4.43048 × 10−8 C.
Explanation:
Given :

.
i3/2
a2
2
+
d
−Q1 4
h
i2
2 |d| √3
a
+
|d|
2
h
−(−3.8 µC)
2 (5.1 m)
h
i3/2
(7.3 m)2
2
+ (−5.1 m)
4
× h√
i2
3
2 (7.3 m) + (5.1 m)
= 0.704413 µC .
Note: Neither the sign nor the magnitude
of the charge Q4 (given in the problem as
3 µC) enter into this equation.
Actually, “the resultant force on Q4 is zero”
means that the resultant electric field is zero.
Because the electric filed is independent of the
0.02 kg
L = 0.14 m ,
m = 0.02 kg ,
θ = 6◦ .
and
L
Q2 =
´q
test charge Q4 , the answer is also independent
of the sign or magnitude of the charge Q4 .
4m
0.1
with r14 the distance between Q4 and Q1 ,
r24 = r34 the distance between Q4 and either
Q2 or Q3 , and θ indicated in the sketch above.
Remember that this force FQ4 will be set equal
to zero since the problem tells us the forces
are in equilibrium.
Because Q1 , Q2 , and Q3 form an equilateral
triangle, of sides of length a, it can be seen
Ã√
!2
3
a2
2
2
that r14
=
a + |d|
+ d2 .
and r24
=
2
4
Also,
|d|
|d|
sin θ =
=q
.
r24
a2
2
d +
2
q
θ
a
q
m
From the right triangle in the figure above,
we see that
a
sin θ = .
L
m
Therefore,
a = L sin θ
= (0.14 m) sin(6◦ )
= 0.014634 m .
The separation of the spheres is r = 2 a =
0.029268 m . The forces acting on one of the
spheres are shown in the figure below.
Answer, Key – Homework 8 – David McIntyre
T
T cos θ
Fe
θ
θ
T sin θ
a distance much larger than their diameters.
A third identical conducting sphere, C, is uncharged. Sphere C is first touched to A, then
to B, and finally removed. As a result the
electrostatic force between A and B, which
was originally F , becomes
mg
Because the sphere is in equilibrium, the
resultant of the forces in the horizontal and
vertical directions must separately add up to
zero:
X
Fx = T sin θ − Fe = 0
X
Fy = T cos θ − m g = 0 .
From the second equation in the system
mg
, so T can be
above, we see that T =
cos θ
eliminated from the first equation if we make
this substitution. This gives a value
1.
2.
3.
4.
5.
for the electric force.
From Coulomb’s law, the electric force between the charges has magnitude
|Fe | = ke
|q|2
,
r2
where |q| is the magnitude of the charge on
each sphere.
Note: The term |q|2 arises here because the
charge is the same on both spheres.
This equation can be solved for |q| to give
s
|Fe | r2
|q| =
ke
s
(0.0206004 N) (0.029268 m)2
=
(8.99 × 109 N m2 /C2 )
= 4.43048 × 10−8 C .
004 (part 1 of 1) 0 points
Two identical conducting spheres, A and B,
carry equal charge. They are separated by
1
F
2
3
F
4
3
F correct
8
3
F
16
1
F
16
6. 0
7.
Fe = m g tan θ
¡
¢
= (0.02 kg) 9.8 m/s2 tan(6◦ )
= 0.0206004 N ,
3
1
F
8
8. F
1
F
4
Explanation:
Since the two conducting spheres are identical (i.e. same radius), when the spheres touch
the charges redistribute themselves equally
between the two spheres. Let spheres A and
B have an initial charge Q. When an identical uncharged sphere C comes in contact with
sphere A and removed, then by conservation
of charge, each sphere will carry charge
QC1 = QA = Q/2
When sphere C touches sphere B, then each
sphere will carry charge
QC1 + Q B
Q/2 + Q
3
QC2 = Q B =
=
= Q
2
2
4
Hence if the initial force is given by
Q2
Fi = k e 2
d
then the final force is
(3/4 Q)(1/2 Q)
Ff = k e
d2
3
= Fi
8
9.
Answer, Key – Homework 8 – David McIntyre
005 (part 1 of 1) 0 points
Two large, parallel, insulating plates are
charged uniformly with the same positive
areal charge density +σ, which is the charge
per unit area. The permittivity of free space
1
. The magnitude of the resultant
²0 =
4πke
electric field E (where ”outside” stands for
above and below the two plates) is
2σ
between the plates, zero outside.
²0
σ
σ
2.
between the plates,
outside.
2²0
2²0
σ
3.
between the plates, zero outside.
²0
σ
4. Zero between the plates,
outside. cor²0
rect
1.
5. Zero between the plates,
6.
σ
everywhere.
²0
2σ
outside.
²0
7. Zero everywhere.
8. Zero between the plates,
σ
outside.
2²0
σ
between the plates, zero outside.
2²0
2σ
2σ
10.
between the plates,
outside.
²0
²0
9.
I
4
II
III
O
q1
a
q2
x
1. all possibilities: right, left, or zero
2. right correct
3. down
4. up
5. left
6. none of these
Explanation:
The direction of the electric field at a point
P is the direction that a positive charge would
move if placed at P . A positive charge placed
in region II would be attracted to q2 and repelled by q1 . Thus the direction is to the
right.
007 (part 2 of 3) 3 points
Identify the direction of E in region III. (x >
a, along x-axis).
1. left correct
2. none of these
Explanation:
Each plate produces a constant electric field
σ
directed away from the plate for
of E =
2²0
positive charge density, and toward the plate
for negative charge density. Between the two
plates, the two fields cancel each other so that
Enet = 0. Outside the two plates, the fields
σ
add together, so that Enet = .
²0
006 (part 1 of 3) 4 points
Given two charges q1 = 2.01 µC at the origin,
q2 = −5.71 µC, and a = 10 cm in the figure
below. Identify the direction of E in the
region II (0 < x < a, along x-axis).
3. up
4. right
5. down
6. all possibilities: right, left, or zero
Explanation:
In region III, a positive charge would be forced
to the left since |q2 | > |q1 | and q2 is closer to
region III. The effect of q2 dominates and the
direction of the electric field is to the left.
Answer, Key – Homework 8 – David McIntyre
y
008 (part 3 of 3) 3 points
∆θ
Locate the x coordinate such that E = 0.
−− A
−−
−
(Note that the origin O is at q1 .)
r
θ −−
Correct answer: −0.145886 m.
−
−
Explanation:
−
O
−
We have already seen that the electric field is
−
−−
nonzero in regions II and III. Thus the only
−−
candidate is region I (negative x-axis).
−−
B
The point where E = 0 is the point where
the magnitudes are equivalent |E1 | = |E2 |.
∆q is given by
Call the point where this happens x = c.
2 Q ∆θ
Then
1. ∆q =
k |q1 |
k |q2 |
π
=
2
2
c
(c − a)
2. ∆q = Q
⇒
µ
c−a
c
¶2
¯ ¯
¯ q2 ¯
= ¯¯ ¯¯
q1
3. ∆q =
5. ∆q =
−1
= −0.145886 m
II
c
IV
2Q
π
Q
π
Explanation:
The angle of a semicircle is π, thus the
charge on a small segment with angle ∆θ is
∆q =
III
q2
III
x
10. ∆q = 2 π Q
x = −0.145886 m
q1
x
7. ∆q =
We find that E = 0 at
I
I
Q ∆θ
correct
π
Q ∆θ
8. ∆q =
2π
Q
9. ∆q =
2π
a
c = − r¯ ¯
¯ q2 ¯
¯ q1 ¯ − 1
5.71×10−6 C
2.01×10−6 C
II
6. None of these.
Solving for c,
0.1 m
y
4. ∆q = π Q
s¯ ¯
¯ q2 ¯
a
⇒ 1 − = ¯¯ ¯¯
c
q1
c = −q
5
x
a
O 1 of 3) 4 points
009 (part
Consider the setup shown in the figure below, where the arc is a semicircle with radius
r. The total charge Q is negative, and distributed uniformly on the semicircle. The
charge on a small segment with angle ∆θ is
labeled ∆q.
Q ∆θ
.
π
010 (part 2 of 3) 3 points
The magnitude of the x-component of the
electric field at the center, due to ∆q, is given
by
1. ∆Ex = k |∆q| (cos θ) r 2
2. ∆Ex =
k |∆q| cos θ
correct
r2
3. ∆Ex = k |∆q| (sin θ) r 2
Answer, Key – Homework 8 – David McIntyre
4. ∆Ex = k |∆q| r
2
k |∆q|
r2
k |∆q| cos θ
6. ∆Ex =
r
k |∆q| sin θ
7. ∆Ex =
r
5. ∆Ex =
8. ∆Ex = k |∆q| (sin θ) r
9. ∆Ex =
k |∆q| sin θ
r2
10. ∆Ex = k |∆q| (cos θ) r
Explanation:
Negative charge attracts a positive test
charge. At O, ∆E points toward ∆q . According to the sketch, the vector ∆Ex is pointing
along the negative x axis. The magnitude of
the ∆Ex is given by
k |∆q|
cos θ .
∆Ex = ∆E cos θ =
r2
011 (part 3 of 3) 3 points
Given: Q = −41 µC, r = 86 cm, and k =
8.98755 × 109 N m2 /C2 .
Determine the magnitude of the electric
field at O .
Correct answer: 317182 N/C.
Explanation:
By symmetry of the semicircle, the ycomponent of the electric field at the center
is
Ey = 0.
Combining part 1 and part 2,
k |∆q| cos θ
∆Ex =
r2
k |Q|
=
cos θ ∆θ
π r2
Therefore, the magnitude of the electric field
at the center is given by
Z π/2
k |Q|
E = Ex =
cos θdθ
2
−π/2 π r
2 k |Q|
=
.
π r2
6
For the above values, the magnitude is given
by
¢
¡
2 8.98755 × 109 N m2 /C2 |(−41 µC)|
E=
π (86 cm)2
= 317182 N/C .
The direction is along negative x axis.
y
y
∆θ
−− A
II
I
−−
−
x
r
θ −−
III IV
− E
−
x
−
O
−
−
−−
−−
−−
B
012 (part 1 of 1) 10 points
A line of charge starts at x = x0 , where x0
is positive, and extends along the x-axis to
positive infinity. If the linear charge density is given by λ = λ0 x0 /x, where λ0 is a
constant, determine the electric field at the
origin. (Here ı̂ denotes the unit vector in the
positive x direction.)
1.
2.
3.
4.
5.
6.
k λ0
(−ı̂) correct
2 x0
k λ0
(ı̂)
2 x0
k λ0
(ı̂)
x0
k λ20
(ı̂)
2 x0
k λ0
(−ı̂)
x0
k λ0
(ı̂)
2 x20
Explanation:
First we realize that we are dealing with a
continuous distribution of charge (as opposed
to point charges). We must divide the distribution into small elements and integrate.
Answer, Key – Homework 8 – David McIntyre
7
Using Coulomb’s law, the electric field created by each small element with charge dq
is
k dq
dE = 2
x
where
dq = λ dx =
λ0 x0
dx
x
Now we integrate over the entire distribution
(i.e. from x = x0 to x = +∞) and insert our
dq:
Z
k dq
~ =
|E|
x2
Z ∞
dx
=
k λ 0 x0 3
x
x0
k λ0 x0 1 ¯¯∞
=−
¯
2
x 2 x0
k λ0
=
2 x0
Since the distribution is to the right of the
point of interest, the electric field is directed
along the −x axis if λ0 is positive. That is, a
positive charge at the origin would experience
a force in the direction of −ı̂ from this charge
distribution. In fact, the direction of an electric field at a point P in space is defined as
the direction in which the electric force acting
on a positive particle at that point P would
point. So
~ = k λ0 (−ı̂).
E
2 x0
013 (part 1 of 1) 0 points
The diagrams below depict three electric field
patterns. Some of these patterns are physically impossible.
Assume: These electric field patterns are
due to static electric charges outside the regions shown.
(a)
(b)
(c)
Which electrostatic field patterns are physically possible?
1. (b) only correct
2. (b) and (c)
3. (a) and (c)
4. (a) only
5. (a) and (b)
6. (c) only
Explanation:
Electrostatic lines of force do not intersect
one another. Neither do they form a closed
circuit (unless there is a changing magnetic
field present).
014 (part 1 of 1) 0 points
A particle of mass 0.000116 g and charge
36 mC moves in a region of space where the
electric field is uniform and is Ex = 7.2 N/C,
Ey = Ez = 0.
If the initial velocity of the particle is given
by vy = 1.3 × 106 m/s, vx = vz = 0, what is
the speed of the particle at 0.7 s?
Correct answer: 2.03385 × 106 m/s.
Explanation:
Answer, Key – Homework 8 – David McIntyre
Given : m = 0.000116 g = 1.16 × 10−7 kg ,
Ex = 7.2 N/C ,
Ey = E z = 0 ,
vy = 1.3 × 106 m/s ,
vx = v z = 0 ,
t = 0.7 s .
Since the electric field has only an x component, the particle accelerates only in the x
direction
q Ex
ax =
.
m
To determine the x component of the final
velocity, vxf , use the kinematic relation
vxf = vxi + a (tf − ti ) = a tf .
Since ti = 0 and vxi = 0,
q E x tf
m
(0.036 C) (7.2 N/C)(0.7 s)
=
(1.16 × 10−7 kg)
= 1.56414 × 106 m/s .
vxf =
No external force acts on the particle in the y
direction so vyi = vyf = 1.3 × 106 m/s. Hence
the final speed is given by
q
2 + v2
vf = vyf
xf
·
¡
¢2
= 1.3 × 106 m/s
= 2.03385 × 106 m/s .
¸1/2
Note: This is analogous to a particle in a
gravitational field with the coordinates roπ
tated clockwise by (90◦ ).
2
015 (part 1 of 3) 0 points
Given: qe = 1.60218 × 10−19 C .
ı̂
+++++++++
~ = m ~a = q E
~.
F
¡
¢2
+ 1.56414 × 106 m/s
An electron enters the region of a uniform
electric field of 138 N/C, as in the figure.
0.09 m
̂
−−−−−−−−−
3 × 106 m/s
According to Newton’s second law and the
definition of an electric field,
vxf
8
Find the magnitude of the acceleration of
the electron while in the electric field.
Correct answer: 2.42717 × 1013 m/s2 .
Explanation:
Given :
qe = 1.60218 × 10−19 C ,
me = 9.10939 × 10−31 kg ,
E = 138 N/C .
and
F = m a = q E , so
qe E
̂
~a = −
me
(1.60218 × 10−19 C)(138 N/C)
=−
̂
9.10939 × 10−31 kg
= −(2.42717 × 1013 m/s2 ) ̂ ,
and the magnitude of the acceleration of the
electron is 2.42717 × 1013 m/s2 .
016 (part 2 of 3) 0 points
Find the time it takes the electron to travel
through the region of the electric field, assuming it doesn’t hit the side walls.
Correct answer: 3 × 10−8 s.
Explanation:
Given : ` = 0.09 m ,
v0 = 3 × 106 m/s .
The horizontal distance traveled is
` = v0 t
`
t=
v0
0.09 m
=
3 × 106 m/s
= 3 × 10−8 s .
Answer, Key – Homework 8 – David McIntyre
017 (part 3 of 3) 0 points
What is the magnitude of the vertical displacement ∆y of the electron while it is in the
electric field?
Correct answer: 0.0109223 m.
Explanation:
Using the equation for the displacement in
the vertical direction and the results from the
first two parts of the problem, we find that
1 2
at
2
¶
µ
−2.42717 × 1013 m/s2
=
2
−8 2
× (3 × 10 s)
= −0.0109223 m .
∆y =
which has magnitude of 0.0109223 m .
9