Download Document

PHYS 1101 Practice problem set 7, Chapter 25: 12, 37, 46, 49, 60, 74;
Chapter 26: 12, 26, 32, 37, 45, 48, 54, 69, 72
25.12. Model: Model the plastic spheres as point charges.
Visualize:

Solve: (a) The charge q1 = −50.0 nC exerts a force F1 on 2 on q2 = −50.0 nC to the right, and the charge q2 exerts a force

F2 on 1 on q1 to the left. Using Coulomb’s law,
F1 on 2 = F2 on 1 =
K q1 q 2
2
12
r
=
(9.0 × 10
9
)(
)(
N m 2 / C 2 50.0 × 10 −9 C 50.0 × 10 −9 C
( 2.0 × 10
−2
m
)
2
) = 0.0563 N
(b) The ratio is
F1 on 2
w
=
0.0563 N
(2.0 × 10
−3
)(
kg 9.8 m / s 2
)
= 2.87
25.37. Model: Objects A and B are point charges.
Visualize:
Solve: (a) It is given that FA on B = 0.45 N. By Newton’s third law, FB on A = FA on B = 0.45 N.
(b) Coulomb’s law is
Kq A q B
FB on A = FA on B = 0.45 N =
⇒ qB =
(0.45 N )r 2
2K
=
(0.45 N )(10 × 10 −2 m )
(
2 9.0 × 10 9 N m 2 / C 2
r2
=
K (2q B )( q B )
r2
2
)
= 5.0 × 10 −7 C ⇒ q A = 2q B = 1.0 × 10 −6 C
(c) Newton’s second law is FB on A = mAaA. Hence,
aA =
FB on A
mA
=
FA on B
mB
=
0.45 N
0.1 kg
= 4.5 m / s
2
25.46. Model: The charges are point charges.
Visualize: Please refer to Figure P25.46.
Solve: Placing the 1 nC charge at the origin and calling it q1, the −6 nC is q3, the q2 charge is in the first quadrant, and the
q4 charge is in the second quadrant. The net electric force on q1 is the vector sum of the electric forces from the other three
charges q2, q3, and q4. We have
1

 K q1 q2

F2 on 1 = 
, away from q 2 
2
r


(
)(
)(
)
 9.0 × 10 9 N m 2 / C2 1 × 10 −9 C 2 × 10 −9 C


=
,
away
from
q
2
2
−2


5.0 × 10 m


(
(
= 0.72 × 10
−5
N, away from
)
q ) = (0.72 × 10
−5
2
)(
N − cos 45° ˆi − sin 45° jˆ
)

 K q 1 q3

F3 on 1 = 
, toward q 3 
2
r


 9.0 × 10 9 N m 2 / C2 1 × 10 −9 C 6 × 10 −9 C

=
, toward q 3 
2
−2


5.0 × 10 m


(
)(
)(
(
)
)
= 2.16 × 10 −5 N, away from q3 = 2.16 × 10 −5 ˆj N
(
)

 K q1 q4

F4 on 1 = 
, away from q 4  = 0.72 × 10 −5 N cos 45° ˆi − sin 45° ˆj
2
r





−5
−5
−5
= F2 on 1 + F3 on 1 + F4 on 1 = 2.16 × 10 N − 0.72 × 10 N (2sin 45° ) ˆj = 1.14 × 10 ˆj N
)(
(

⇒ Fon 1
[(
) (
)
]
)
25.49. Model: The charged particles are point charges.
Visualize:
Solve: The two 2 nC charges exert an upward force on the 1 nC charge. Since the net force on the 1 nC charge is zero,
the unknown charge must exert a downward force of equal magnitude. This implies that q is a positive charge. The force
of charges 2 on charge 1 is

 K q1 q 2

F2 on 1 = 
, away from q 2 
2
 r12

=
(9.0 × 10 9 N m 2 / C 2 )(1.0 × 10 −9 C)( 2.0 × 10 −9 C)
2
( 0.020 m) + (0.030 m)
2
(cosθ iˆ + sinθ jˆ)
From the figure, θ = tan–1(2/3) = 33.69°. Thus

−5
−5
F2 on 1 = (1.152 × 10 iˆ + 0.768 × 10 ˆj ) N



From symmetry, F3 on 1 is the same except the x-component is reversed. When we add F2 on 1 and F3 on 1 , the x-components cancel
and the y-components add to give


F2 on 1 + F3 on 1 = 1.536 × 10–5 jˆ N

Fq on 1 must have the same magnitude, pointing in the − ˆj direction, so

K q q1
(1.536 × 10 −5 N)(0.020 m) 2
−5
Fq on 1 = 1.536 × 10 N =
⇒
q
=
= 0.68 nC
r2
(9.0 × 10 9 N m 2 / C 2 )(1.0 × 10 −9 C)
A positive charge q = 0.68 nC will cause the net force on the 1 nC charge to be zero.
2
25.60. Model: The charged spheres are point charges.
Visualize:
Each sphere is in static equilibrium when the string makes an angle of 20∞ with the vertical. The three forces acting on
each sphere are the electric force, the weight of the sphere,
force.
 and the tension


Solve: In the static equilibrium, Newton’s first law is Fnet = T + w + Fe = 0 . In component form,
(F )
net
x
⇒ − T sin θ + 0 N +
⇒ T sin θ =
(F )
= Tx + w x + ( Fe )x = 0 N
Kq 2
d
2
=
Kq 2
d2
+ Kq 2
(2 L sin θ )
2
net
= 0N
y
= Ty + wy + ( Fe ) y = 0 N
T cosθ − mg + 0 N = 0 N
T cosθ = + mg
Dividing the two equations and solving for q,
q=
4 sin 2 θ tanθL2 mg
K
=
(
)
2
(
)
4 sin 2 20° tan 20° (1.0 m ) 3.0 × 10 −3 kg (9.8 N / kg)
9
9.0 × 10 N m 2 / C 2
= 746 nC
25.74. Model: The charged balls are point charges.
Visualize:
Because of symmetry and the fact that the three balls have the same charge, the magnitude of the electric force on each
ball is the same. The other forces acting on each ball are the weight of the ball and the tension force.
Solve: The force on ball 3 is the sum of the force from ball 1 and ball 2. We have

 K q1 q3

Kq 2
F1 on 3 = 
, away from q 1  = 2 sin30° iˆ + cos30° ˆj
2
r
 r13

(
)
3

Kq 2
F2 on 3 = 2 − sin30 ° iˆ + cos 30 ˆj
r
(
)

2Kq 2
2 Kq 2
⇒ Fon 3 =
cos30° ˆj ⇒ Fon 3 =
cos 30° = Fon 2 = Fon 1 = Fe
2
r
r2
(
)
2 9.0 × 10 9 N m 2 / C 2 q 2 cos 30°
⇒ Fe =
(0.20 m )2
(
11
= 3.897 × 10
q
2
)N / C
2
The distance l, between one of the balls and the center of the equilateral triangle, is
r
0.10 m
l cos30° = = 0.10 m ⇒ l =
= 0.1155 m
2
cos30°
Thus, the angle made by the string with the plane containing the three balls is
l
0.1155 m
⇒ θ = 81.70°
cosθ =
=
L
0.80 m
From the free-body diagram, we have
− T cosθ + Fe = 0 N
T sin θ − mg = 0 N
mg
mg
=
⇒ tanθ =
Fe
3.897 × 1011 q 2 N / C 2
(
⇒q=
mg
(3.897 × 10
11
2
)
N / C tanθ
)
(3.0 × 10
=
( 3.897 × 10
)
−3
kg ( 9.8 N / kg )
11
N / C 2 tan 81.70°
)
= 1.05 × 10
−7
C = 105 nC
26.12. Model: Assume that the rings are thin and that the charge lies along circle of radius R.
Visualize:
The rings are centered on the z-axis.
Solve: (a) According to Example 26.5, the field of the left (negative) ring at z = 10 cm is
(E )
1 z
=
zQ1
(
2
4πε 0 z + R
2
)
3/ 2
=
(9.0 × 10
9
)
(
N m 2 / C 2 ( 0.10 m ) −20 × 10 −9 C
[(0.10 m )
2
+ (0.05 m )
2 3/ 2
]
) = −1.288 × 10
4
N/C

4
That is, the field is E1 = 1.288 × 10 N / C, left . Ring 2 has the same quantity of charge and is at the same distance, so it

will produce a field of the same strength. Because Q2 is positive, E2 will also point to the left. The net field at the midpoint is
 

4
E = E1 + E 2 = 2.58 × 10 N / C, left
(
)
(
(b) The force is
)


−9
4
−5
F = q E = +1.0 × 10 C 2.576 × 10 N / C, left = 2.58 × 10 N, left
(
)(
) (
)
4
26.26. Model: The infinite charged plane produces a uniform electric field.
Solve: (a) The electric field of a plane of charge with surface charge density η is
E=
 F
 ma 
⇒ η = 2ε 0 E = 2ε 0   = 2ε 0 

q
 q 
η
2ε 0
where m is the mass, q is the charge, and a is the acceleration of the electron. To obtain η we must first find a. From the
2
2
kinematic of motion equation v1 = v0 + 2a Δx ,
a=
⇒η =
7
2 Δx
(
2 8.85 × 10
−12
C
2
(1.0 × 10 m / s )
=
2 (2.0 × 10
/ N m )(9.11 × 10
v12 − v02
2
2
− (0 m / s )
15
= 2.5 × 10 m / s
)
kg )( 2.5 × 10
−2
−31
2
m
15
m /s
2
−19
1.60 × 10 C
(b) Using the kinematic equation of motion v1 = v0 + aΔt ,
Δt =
v1 − v0
=
a
1.0 × 10 7 m / s − 0 m / s
2.50 × 10 15 m / s 2
2
) = 2.52 × 10
= 4.0 × 10
−9
−7
C/m
2
s
26.32. Model: The electric field is that of three point charges q1, q2 and q3.
Visualize: Please refer to Figure P26.32. Assume the charges are in the x-y plane. The −5 nC charge is q1, the bottom 10 nC




charge is q3, and the top 10 nC charge is q2. The net electric field at the dot is E net = E 1 + E 2 + E3 . The procedure will be to find
the magnitudes of the electric fields, to write them in component form, and to add the components.
Solve: The electric field produced by q1 is
E1 =
q1
1
=
2
1
4πε 0 r
( 9.0 × 10
9
)(
) = 112, 500 N / C
N m 2 / C 2 5.0 × 10 −9 C
(0.02 m )2


E1 points toward q1, so in component form E1 = −112, 500 ˆj N / C . The electric field produced by q2 is E2 =


56,250 N / C. E2 points away from q2, so E2 = −56,250 iˆ N / C . Finally, the electric field produced by q3 is
E3 =
q3
1
=
2
3
4πε 0 r
(9.0 × 10
9
)(
N m 2 / C2 10.0 × 10 −9 C
2
( 0.02 m ) + (0.04 m )
2
) = 45,000 N
/C

−1
E3 points away from q3 and makes an angle φ = tan (2 / 4 ) = 26.57° with the x-axis. So,

E3 = − E3 cosφ iˆ + E 3 sinφ ˆj = −40,250 iˆ + 20,130 ˆj N / C
(
Adding these three vectors gives
)




E net = E 1 + E 2 + E3 = −96,500 iˆ − 92,400 ˆj N / C
(
)
This is in component form. The magnitude of the field is
E net =
2
2
Ex + Ey =
and its angle from the x-axis is θ = tan
−1
2
2
( 96,500 N / C ) + ( 92, 400 N / C) = 133, 600 N / C
(E
x

E y = 43.8° . We can also write E net = (133,600 N / C, 43.8 ° below the
)
− x- axis ) .
5
26.37. Model: The electric field of a dipole is that of two opposite charges ±q that make the dipole.
Visualize: Please refer to Figure 26.7.
 The figure shows a dipole aligned on the y-axis, so the x-axis is the bisecting axis.
The field at a point on the x-axis is E dipole = E + + E − .
Solve: From the symmetry of the situation we can see that
 the x-components of the two contributions to the electric field
will cancel, that they have equal y-components, and that E dipole points in the −y-direction. Thus,
(E )
dipole
x
(E )
= 0N /C
dipole
= −2E + sinθ
y

where θ is the angle E + makes with the x-axis. From the geometry of the figure,
s/2
sinθ =
2 1/2
2
x + ( s / 2)
[
]
For a point charge +q, the field is
E+ =
1
q
4πε 0 x 2 + (s 2 )
2
Combining these pieces gives the dipole field at distance x along the bisecting axis:

1
q
s/2
qs
ˆj = −
E dipole = −( 2 )
2
2
2 1/2
2
2
2
4πε 0 x + (s / 2 ) x + ( s / 2 )
4πε 0 x + s / 4
[
(
2
2
If x >> s, then x + s / 4
)
32
(
]
)
3/ 2
ˆj
3
≈ x . Thus

Edipole ≈ −
qsˆj
1
4πε 0 x3

If we note that p = qsˆj and if we replace x with a more general variable r to denote the distance from the dipole, then


1 p
E dipole ≈ −
4πε 0 r 3
This is Equation 26.12.
26.45. Model: The electric field is that of a line charge of length L.
Visualize: Please refer to Figure P26.45. Let the bottom end of the rod be the origin of the coordinate system. Divide the
rod into many small segments of charge Δq and length Δy′. Segment i creates a small electric field at the point P that
makes an angle θ with the horizontal. The field has both x and y components, but Ez = 0 N/C. The distance to segment i
(
2
2 12
)
from point P is x + y ′
.
Solve: The electric field created by segment i at point P is

Ei =

Δq

(cosθ ˆi − sinθ ˆj) = 4πε ( Δq
4π ε ( x + y ′ )
x + y ′ ) 
2
2
2
0
2
0


The net field is the sum of all the Ei , which gives E =

∑E
i
x
x2 + y′2
iˆ −
y′
x 2 + y′ 2

ˆj 


. Δq is not a coordinate, so before converting the sum to an
i
integral we must relate charge Δq to length Δy′. This is done through the linear charge density λ = Q/L, from which we
have the relationship
Q
Δq = λΔ y′ = Δ y ′
L
With this charge, the sum becomes
 Q/ L 
xΔ y ′
E=
∑
4πε 0 i  x 2 + y′ 2

(
)
3 /2
iˆ −
y ′Δ y ′
(x
2
+ y′
2
)
3/ 2

ˆj 


6
Now we let Δy′ → dy′ and replace the sum by an integral from y′ = 0 m to y′ = L . Thus,
 (Q / L)  L
xd y′
∫
E=
2

4πε 0  0 x + y ′ 2
(
=
Q
1
4πε 0 x x + L
2
2
L
)
3 /2
ˆi −
∫
0
iˆ −


(Q / L)  x 
y′
jˆ =

2
2
2

4π ε0


  x x + y′
y′ d y′
(x
2
)
+ y′2
3/ 2
 Q 

1 −
4πε 0  Lx  
1
L

 iˆ −

0

−1

2
 x + y′2

L



0

ˆj 



 ˆj
x +L 
x
2
2
26.48. Model: Assume that the semicircular rod is thin and that the charge lies along the semicircle of radius R.
Visualize:
The origin of the coordinate system is at the center of the circle. Divide the rod into many small segments of charge Δq

and arc length Δs. Segment i creates a small electric field Ei at the origin. The line from the origin to segment i makes an
angle θ with the x-axis.
Solve: Because every segment i at an angle θ above the axis is matched by segment j at angle θ below the axis, the ycomponents of the electric fields will cancel when the field is summed over all segments. This leads to a net field pointing
to the right with
Ex =
∑(E )
i
i
x
=
∑ E cosθ
i
Ey = 0 N / C
i
i
Note that angle θi depends on the location of segment i. Now all segments are at the same distance ri = R from the origin,
so
Δq
Δq
Ei =
=
2
4π ε 0 ri
4πε 0 R 2
The linear charge density on the rod is λ = Q/L, where L is the rod’s length. This allows us to relate charge Δq to the arc
length Δs through
Δq = λ Δs = (Q/L)Δs
Thus, the net field at the origin is
Ex =
∑
i
(Q / L) Δs
4π ε 0 R 2
Q
cosθ i =
4π ε 0 LR 2
∑ cosθ
i
Δs
i
The sum is over all the segments on the rim of a semicircle, so it will be easier to use polar coordinates and integrate over
θ rather than do a two-dimensional integral in x and y. We note that the arc length Δs is related to the small angle Δθ by Δs
= RΔθ, so
Ex =
Q
4πε 0 LR
∑ cosθ
i
Δθ
i
With Δθ → dθ, the sum becomes an integral over all angles forming the rod. θ varies from Δθ = −π/2 to θ = +π/2. So we
finally arrive at
π /2
Q
Q
2Q
π /2
Ex =
cosθ d θ =
sin θ −π /2 =
4πε 0 LR ∫−π /2
4πε 0 LR
4πε 0 LR
7
Since we’re given the rod’s length L and not its radius R, it will be convenient to let R = L/π. So our final expression for

E , now including the vector information, is

1  2π Q  ˆ
E=
i
4πε 0  L2 
(b) Substituting into the above expression,
E=
(9.0 × 10
) (
9
N m 2 / C 2 2π 30 × 10 −9 C
( 0.10 m )
2
) = 1.70 × 10
5
N/C
26.54. Model: Assume that the electric field inside the capacitor is constant, so constant-acceleration kinematic
equations apply.
Visualize: Please refer to Figure P26.54.
Solve: (a) The force on the electron inside the capacitor is




 qE
F = ma = qE ⇒ a =
m

−19
Because E is directed upward (from the positive plate to the negative plate) and q = −1.60 × 10 C , the acceleration of
the electron is downward. We can therefore write the above equation as simply ay = qE/m. To determine E, we must first
find ay. From kinematics,
x1 = x 0 + v 0x (t 1 − t 0 ) +
2
a x (t1 − t 0 ) ⇒ 0.04 m = 0 m + v0 cos 45° (t1 − t 0 ) + 0 m
1
2
⇒ (t 1 − t 0 ) =
(0.04 m )
(5.0 × 10
)
6
m / s cos45°
= 1.1314 × 10 −8 s
Using the kinematic equations for the motion in the y direction,
 t − t0 
 qE   t1 − t 0 
v1 y = v 0 y + a y 1
⇒ 0 m / s = v0 sin45 ° +
 2 
 m  2 
⇒E=−
2 m v 0 sin45°
q (t 1 − t 0 )
=−
(
2 9.1 × 10
(
−31
)(
)
6
kg 5.0 × 10 m / s sin45°
)(
− 1.60 × 10 −19 C 1.1314 × 10 −8 s
= 3550 N /C
)
(
6
)
(b) To determine the separation between the two plates, we note that y0 = 0 m and v 0y = 5.0 × 10 m / s sin45 ° , but at y =
y1, the electron’s highest point, v1y = 0 m/s. From kinematics,
v1y2 = v 20y + 2a y ( y1 − y0 ) ⇒ 0 m 2 / s 2 = v20 sin 2 45° + 2 a y ( y1 − y 0 )
⇒ ( y1 − y 0 ) = −
v20 sin 2 45°
=−
2 ay
v20
4a y
From part (a),
ay =
qE
m
=
(−1.60 × 10
⇒ y1 − y0 = −
−19
)
C (3550 N / C )
9.1 × 10 −31 kg
(
(5.0 × 10
6
m/s
)
14
= − 6.242 × 10 m / s
2
2
4 −6.242 × 10 14 m / s 2
)
= 0.010 m = 1.0 cm
This is the height of the electron’s trajectory, so the minimum spacing is 1.0 cm.
26.69. Model: The long charged wire is an infinite line of charge. The charges on the wire and the plastic rod are
uniform.
Visualize: Please refer to Figure P26.69. The plastic stirrer is located on the x-axis.
Solve: The electric field of the infinite line of charge at a distance x from its axis is
8
Ex =
1
2λ
4πε 0 x
Because the electric field is a function of x, the plastic stirrer experiences an electric field that varies along the length of
the stirrer. We have handled such problems earlier. Take a small segment of the charge on the stirrer and calculate the
electric force due to the line charge on this charge segment Δq. A summation of all such forces on the charge segments Δq
will yield the net force on the stirrer. This procedure is equivalent to integrating the electric force on a small segment Δq
over the length of the stirrer. The force on charge Δq of length Δx at position x due to the infinite line of charge is
ΔF = EΔq = Eλ ′Δx =
1  2 λλ ′Δx 


4πε 0 
x 
In the above expression, λ′ = Q/L is the linear charge density of the stirrer and λ is the linear charge density of the plastic rod.
We have also used the relation Δq Δx = λ ′ . Changing Δx → dx and integrating x from x = 0.02 m to x = 0.08 m, the total
force on the stirrer is
0.08 m
∫
F=
0.02 m
(
 1 
 1 
1 2λ λ ′
0.08 m
 0.08 m 
dx = 
 (2λ λ ′) ln x 0.02 m = 
 (2λ λ ′) ln 
4πε 0 x
0.02 m 
 4πε 0 
 4πε 0 
9
2
2
)(
= 9.0 × 10 N m / C 2 1.0 × 10
−7
−9
 10 × 10 C 
−4
C/m
 ln( 4 ) = 4.16 × 10 N
 0.06 m 
)
26.72. Model: Model the infinitely long sheet of charge with width L as a uniformly charged sheet.
Visualize:
Solve: (a) Consider a point on the x-axis at a distance d from the center of the sheet of charge. (We’ll call this distance d
to begin with, rather than x, to avoid confusion with x as the integration variable.) Once again, let the sheet of charge be
divided into small strips of width Δx. Each strip has a linear charge density λ = ηΔx and acts like a long, charged wire.
Strip i is at distance ri = d − xi from the point of interest, so it contributes the small field

2λ ˆ
2η Δx
ˆi
Ei =
i =
4π ε 0 ri
4πε 0 (d − xi )

In this situation all fields Ei point in the same direction. Their x-components all add to give a net field in the
+x-direction:
Ex =
∑(E )
i
i
x
=
2η
4πε 0
Δx
∑ (d − x )
i
i
9
We’ll replace the sum with an integral from x = −L/2 to x = +L/2, giving
Ex =
2η
4πε 0
L /2
∫
dx
(d − x )
− L/ 2
=
2η
4π ε 0
[− ln( d − x )]
L /2
− L/ 2
=
2η
4π ε 0
2η
[− ln( d − L 2 ) + ln (d + L 2 )] = 4πε
ln
0
 2d + L 
 2d − L 
Now that the integration is complete, we can note that d really is the x-coordinate of the point of interest. Substituting x for
d and changing to vector form, we end up with

2η
2x + L ˆ
E net =
ln
i
4πε 0  2 x − L 
(b) If the point is very distant compared to width of the sheet of charge (x >> L), then the sheet of charge looks like a line
of charge with linear density (charge per unit length) λ = ηL. Write
Ex =
2η
 1 + L /2 x 
2η
[ln(1 + L /2 x ) − ln(1 − L /2 x )]
ln 
=


4πε 0
1 − L /2 x
4πε 0
If x >> L, then L/2x << 1 and we can use the approximation ln(1 + u) ≈ u if u << 1. Thus
Ex ≈
2η  L  L  
2ηL
2λ
− −  =
=

4πε 0  2 x  2 x   4π ε0 x 4πε 0 x
This is the field of a line of charge with λ = ηL, as we expected.
(c) The following table shows the field strength Ex in units of 2η/4πε0 for selected values of z in units of L. A graph of Ex is
shown in the figure above.
2η
E x 4π ε0
zL
0.75
1.0
1.5
2.0
3.0
4.0
1.61
1.10
0.69
0.51
0.34
0.25
10