Download 27.8. Model: The rods are thin. Assume that the charge lies along a

27.8. Model: The rods are thin. Assume that the charge lies along a line.
Visualize:
The electric field of the positively charged glass rod points away from the glass rod, whereas the electric field of
the negatively charged plastic rod points toward the plastic rod. The electric field strength is the magnitude of the
electric field and is always positive.
Solve: Example 27.3 shows that the electric field strength in the plane that bisects a charged rod is
Erod =
Q
1
4πε 0 r r 2 + ( L / 2 )2
The electric field from the glass rod at r = 1 cm from the glass rod is
Eglass = ( 9.0 ×109 N m 2 /C2 )
10 ×10−9 C
( 0.01 m ) ( 0.01 m ) + ( 0.05 m )
2
2
= 1.765 ×105 N/C
The electric fields from the glass rod at r = 2 cm and r = 3 cm are 0.835 × 105 N/C and 0.514 × 105 N/C. The
electric field from the plastic rod at distances 1 cm, 2 cm, and 3 cm from the plastic rod are the same as for the
glass rod. Point P1 is 1.0 cm from the glass rod and is 3.0 cm from the plastic rod, point P2 is 2 cm from both
rods, and point P3 is 3 cm from the glass rod and 1 cm from the plastic rod. Because the direction of the electric
fields at P1 is the same, the net electric field strength 1 cm from the glass rod is the sum of the fields from the
glass rod at 1 cm and the plastic rod at 3 cm. Thus
At 1.0 cm
E = 1.765 ×105 N/C + 0.514 ×105 N/C = 2.3 × 105 N/C
At 2.0 cm
E = 0.835 × 105 N/C + 0.835 ×105 N/C = 1.67 × 105 N/C
At 3.0 cm
E = 0.514 × 105 N/C + 1.765 × 105 N/C = 2.3 × 105 N/C
Assess: The electric field strength in the space between the two rods goes through a minimum. This point is
exactly in the middle of the line connecting the two rods. Also, note that the arrows shown in the figure are not to
scale.
27.12. Model: Assume that the rings are thin and that the charge lies along circle of radius R.
Visualize:
Solve:
cm is
(a) Let the rings be centered on the z-axis. According to Example 27.5, the field of the left ring at z = 10
( E1 ) z =
zQ1
4πε 0 ( z + R
2
)
2 3/ 2
=
( 9.0 ×10
9
N m 2 /C 2 ) ( 0.10 m ) ( 20 ×10−9 C )
2 3/ 2
⎡( 0.10 m ) + ( 0.050 m ) ⎤
⎣
⎦
2
= 1.29 ×104 N/C
G
That is, E1 = (1.29 × 104 N/C, right ) . Ring 2 has the same quantity of charge and is at the same distance, so it will
G
produce a field of the same strength. Because Q2 is positive, E2 will point to the left. The net field at the
G G G
midpoint between the two rings is E = E1 + E2 = 0 N/C.
(b) The field of the left ring at z = 0 cm is ( E1 ) z = 0 N/C. The field of the right ring at z = 20 cm to its left is
( E2 ) z =
( 9.0 ×10
9
N m 2 /C2 ) ( 0.20 m ) ( 20 ×10−9 C )
2 32
= 4.1×103 N/C
⎡( 0.20 m ) + ( 0.050 ) ⎤
⎣
⎦
G G G
⇒ E = E1 + E2 = 0 N/C + (4.1×103 N/C, left)
So the electric field strength is 4.1×103 N/C.
2
27.14. Model: Model each disk as a uniformly charged disk. When the disk is positively charged, the onaxis electric field of the disk points away from the disk.
Visualize:
Solve:
(a) The surface charge density on the disk is
η=
50 ×10−9 C
Q
Q
=
=
= 6.366 ×10−6 C/m 2
A π R 2 π ( 0.050 m )2
From Equation 27.23, the electric field of the left disk at z = 0.10 m is
( E1 ) z =
η ⎡
1
6.366 × 10−6 C/m 2
1
⎤
⎡
⎤
1
−
=
1−
= 38,000 N/C
−12
2
2 ⎢
2
2 ⎥
2 ⎥
2ε 0 ⎢
2
8.85
10
C
/N
m
×
(
)
R
z
1
+
1 + ( 0.050 m /0.10 m ) ⎥⎦
⎣
⎦
⎢⎣
G
Hence, E1 = (38,000 N/C, right) . Similarly, the electric field of the right disk at z = 0.10 m (to its left) is
G G G
G
E2 = (38,000 N/C, left) . The net field at the midpoint between the two disks is E = E1 + E2 = 0 N/C .
(b) The electric field of the left disk at z = 0.050 m is
G
6.366 × 10−6 C/m 2
1
⎡
⎤
1−
= 1.05 × 105 N/C ⇒ E1 = (1.05 × 105 N/C, right )
⎥
2
2 ⎢
−12
2
2 ( 8.85 × 10 C /N m ) ⎢
1 + ( 0.050 m /0.10 m ) ⎥⎦
⎣
G
Similarly, the electric field of the right disk at z = 0.15 m (to its left) is E2 = (1.85 ×104 N/C, left ) . The net field
( E1 ) z =
is thus
The field strength is 8.7 ×104 N/C.
G G G
E = E1 + E2 = ( 8.7 × 104 N/C, right )
27.22. Model: A uniform electric field causes a charge to undergo constant acceleration.
Solve:
Kinematics yields the acceleration of the electron.
v12 = v0 2 + 2aΔx ⇒ a =
v12 − v0 2 (4.0 × 107 m/s) 2 − (2.0 × 107 m/s) 2
=
= 5.0 × 1016 m/s 2
2Δx
2 ( 0.012 m )
The magnitude of the electric field required to obtain this acceleration is
E=
Fnet me a (9.11 × 10−31 kg)(5.0 × 1016 m/s 2 )
=
=
= 2.8 × 105 N/C.
e
e
1.6 × 10−19 C
27.28. Model: The electric field is that of three point charges q1, q2, and q3.
Visualize: Please refer to Figure P27.28. Assume the charges are in the x-y plane. The 5.0 nC charge is q1, the
G
G G G
10 nC charge is q3, and the −5.0 nC charge is q2. The net electric field at the dot is Enet = E1 + E2 + 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: (a) The electric field produced by q1 is
9
2
2
−9
q1 ( 9.0 ×10 N m /C )( 5.0 ×10 C )
=
= 112,500 N/C
2
4πε 0 r12
( 0.020 m )
G
G
E1 points away from q1, so in component form E1 = 112,500iˆ N/C. The electric field produced by q2 is E2 =
G
G
28,120 N/C. E2 points toward q2, so E2 = 28,120ˆj N/C . Finally, the electric field produced by q3 is
E1 =
E3 =
1
9
2
2
−9
q3 ( 9.0 ×10 N m /C )(10 ×10 C )
=
= 45,000 N/C
2
2
4πε 0 r32
( 0.020 m ) + ( 0.040 m )
1
G
E3 points away from q3 and makes an angle φ = tan −1 ( 4 / 2 ) = 63.43° with the x-axis. So,
G
E3 = E3 cos φ iˆ − E3 sin φ ˆj = 20,130iˆ − 40,250ˆj N/C
(
)
Adding these three vectors gives
G
G G G
Enet = E1 + E2 + E3 = 132,600iˆ − 12,130ˆj N/C = 1.33 × 105iˆ − 1.21×104 ˆj N/C
(
)
(
)
This is in component form.
(b) The magnitude of the field is
Enet = Ex2 + E y2 =
(132,600 N/C ) + ( −12,130 N/C )
2
(
and its angle from the x-axis is θ = tan −1 Ex E y
+ x-axis).
)
2
= 133,200 N/C = 1.33 × 105 N/C
G
= 5.2°. We can also write Enet = (1.33 ×105 N/C, 5.2° CW from the
27.46. 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
G
charge Δ q and arc length Δ s. 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
y-components 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 = ∑ ( Ei ) x = ∑ Ei cosθ i
i
E y = 0 N/C
i
Note that angle θ i depends on the location of segment i. Now all segments are at the same distance ri = R from the
origin, so
Ei =
Δq
Δq
=
4πε 0 ri 2 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 Δ q to
the arc length Δ s through
Δ q = λ Δ s = (Q/ L)Δ s
Thus, the net field at the origin is
Ex = ∑
i
( Q / L ) Δ s cosθ
4πε 0 R
2
i
=
Q
∑ cosθi Δ s
4πε 0 LR 2 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 Δ s is related to
the small angle Δθ by Δ s = RΔθ , so
Ex =
Q
∑ cosθ i Δθ
4πε 0 LR i
With Δ θ → dθ , the sum becomes an integral over all angles forming the rod. θ varies from Δ θ = −π/2 to θ =
+π/2. So we finally arrive at
Ex =
π /2
Q
Q
cosθ dθ =
sin θ
∫
−
π
/
2
4πε 0 LR
4πε 0 LR
π /2
−π / 2
=
2Q
4πε 0 LR
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
G
expression for E , now including the vector information, is
G
1 ⎛ 2π Q ⎞ ˆ
E=
⎜
⎟i
4πε 0 ⎝ L2 ⎠
(b) Substituting into the above expression,
E=
( 9.0 ×10
9
N m 2 /C2 ) 2π ( 30 ×10−9 C )
( 0.10 m )
2
= 1.70 ×105 N/C
27.52. Model: The electric field is uniform inside the capacitor, so constant-acceleration kinematic equations
apply to the motion of the proton.
Visualize:
G
Solve: From Equation 27.29, the electric field between the parallel plates E = (η ε 0 ) ˆj. The force on the proton is
G
G
G
G
G qE qη ˆ
qη
=
F = ma = qE ⇒ a =
j ⇒ ay =
m mε 0
mε 0
Using the kinematic equation y1 = y0 + v0 y ( t1 − t0 ) + 12 a y ( t1 − t0 ) ,
2
1 ⎛ qη ⎞
1
2
Δy = y1 − y0 = ( 0 m/s )( t1 − t0 ) + a y ( t1 − t0 ) = ⎜
⎟ ( t1 − t0 )
2 ⎝ mε 0 ⎠
2
To determine t1 − t0, we consider the horizontal motion of the proton. The proton travels a distance of 2.0 cm at a
constant speed of 1.0 × 106 m/s. The velocity is constant because the only force acting on the proton is due to the
field between the plate along the y-direction. Using the same kinematic equation,
Δx = 2.0 × 10−2 m = v0 x ( t1 − t0 ) + 0 m ⇒ ( t1 − t0 ) =
2.0 ×10−2 m
= 2.0 ×10−8 s
1.0 ×106 m/s
−19
−6
−8
2
1 (1.60 × 10 C )(1.0 × 10 C/m )( 2.0 × 10 s )
= 2.2 mm
2
(1.67 ×10−27 kg )(8.85 ×10−12 C2/Nm2 )
2
⇒ Δy =