Download 解答 27.28. Model: The electric field is that of three point charges q1

解答
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 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 .
procedure will be to find the magnitudes of the electric fields, to write them in component form, and to add
components.
Solve: (a) The electric field produced by q1 is
E1 
9
2
2
9
q1  9.0 10 N m /C  5.0 10 C 

 112,500 N/C
2
4 0 r12
 0.020 m 
1
E1 points away from q1, so in component form E1  112,500iˆ N/C. The electric field produced by q2
is E2  28,120 N/C. E2 points toward q2, so E2  28,120ˆj N/C . Finally, the electric field produced by q3
is
E3 
9
2
2
9
q3  9.0 10 N m /C 10 10 C 

 45,000 N/C
2
2
4 0 r32
 0.020 m    0.040 m 
1
E3 points away from q3 and makes an angle   tan 1  4/ 2  63.43 with the x-axis. So,


E3  E3 cos  iˆ  E3 sin  ˆj  20,130iˆ  40,250ˆj N/C
Adding these three vectors gives




Enet  E1  E2  E3  132,600iˆ  12,130ˆj N/C  1.33 105iˆ  1.2110 4 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

2
 133,200 N/C  1.33 105 N/C

and its angle from the x-axis is   tan 1 Ex Ey  5.2. We can also write Enet  (1.33 105 N/C, 5.2 CW
from the  x-axis).
27.54. Model: The parallel plates form a parallel-plate capacitor. The electric field inside a parallel-plate
capacitor is a uniform field, so the electrons will have a constant acceleration.
Visualize:
Solve: (a) The bottom plate should be positive. The electron needs to be repelled by the top plate, so the
top plate must be negative and the bottom plate positive. In other words, the electric field needs to point
away from the bottom plate so the electron’s acceleration a is toward the bottom plate.
(b) Choose an xy-coordinate system with the x-axis parallel to the bottom plate and the origin at the point of
entry. Then the electron’s acceleration, which is parallel to the electric field, is a  ajˆ. Consequently, the
problem looks just like a projectile problem. The kinetic energy K  12 mv02  3.0 1017 J gives an initial
speed v0   2K m 
1/ 2
 8.115 106 m/s. Thus the initial components of the velocity are
vx 0  v0 cos45  5.74 106 m/s
vy 0  v0 sin 45  5.74 106 m/s
What acceleration a will cause the electron to pass through the point (x1, y1)  (1.0 cm, 0 cm)? The
kinematic equations of motion are
1
x1  x0  vx 0t1  axt12  vx 0t1  0.010 m
2
1
1
y1  y0  vy 0t1  a yt1  v y 0t1  at12  0 m
2
2
From the x-equation, t1  x1 vx 0  1.742  109 s. Using this in the y-equation gives
a
2vy 0t1
t12
 6.59 1015 m/s2
But the acceleration of an electron in an electric field is
a
9.111031 kg  6.59 1015 m/s2   37,500 N/C  3.8 104 N/C
Felec qelec E eE
ma


E

m
m
m
e
1.60 1019 C
(c) The minimum separation dmin must equal the “height” ymax of the electron’s trajectory above the bottom plate.
(If d were less than ymax, the electron would collide with the upper plate.) Maximum height occurs at
t  12 t1  8.71 1010 s. At this instant,
1
ymax  vy 0t  at 2  0.0025 m  2.5 mm
2
Thus, dmin  2.5 mm.
28.4. Model: The electric flux “flows” out of a closed surface around a region of space containing a net
positive charge and into a closed surface surrounding a net negative charge.
Visualize: Please refer to Figure EX28.4. Let A be the area in m2 of each of the six faces of the cube.
Solve:
The electric flux is defined as e  E  A  EAcos , where  is the angle between the electric
field and a line perpendicular to the plane of the surface. The electric flux out of the closed cube surface is
out   20 N/C  20 N/C  10 N/C Acos0  50 A N m2/C
Similarly, the electric flux into the closed cube surface is
in  15 N/C  15 N/C  15 N/C Acos180    45 A N m2/C
The net electric flux is
50 A N m2/C   45 A N m2/C  5 A N m2/C.
Since the net electric flux is
positive
(i.e., outward), the closed box contains a positive charge.
28.14. Model: The electric field over the circle in the xy plane is uniform.
Solve:
The area vector of the circle is
2
A   r 2 kˆ    0.015 m  kˆ 
 7.07 10
4
m 2  kˆ
Thus, the flux through the area of the circle is


e  E  A  1500iˆ  1500 ˆj  1500 kˆ N/C   7.07 104 m 2  kˆ
Using iˆ  kˆ  ˆj  kˆ  0 and kˆ  kˆ  1,
 e  1500 N/C   7.07 104 m 2   1.06 N m 2 /C
28.28. Visualize: Please refer to Figure EX28.28.
Solve:
For any closed surface that encloses a total charge Qin, the net electric flux through the closed
surface is e  Qin  0 . In the present case, the conductor is neutral and there is a point charge Q inside the
cavity. Thus Qin  Q and the flux is
e 
Q
0
28.49. Model: The infinitely wide plane of charge with surface charge density  polarizes the infinitely
wide conductor.
Visualize:
Because E  0 in the metal there will be an induced charge polarization. The face of the conductor adjacent to
the plane of charge is negatively charged. This makes the other face of the conductor positively charged. We thus
have three infinite planes of charge. These are P (top conducting face), P (bottom conducting face), and P(plane
of charge).
Solve:
Let 1, 2, and 3 be the surface charge densities of the three surfaces with 2 a negative number.
The electric field due to a plane of charge with surface charge density  is E   2 0 . Because the electric
field inside a conductor is zero (region 2),
EP  EP  EP  0 N/C  
1 ˆ 2 ˆ 3 ˆ
j
j
j  0 N/C  1  2    0 C/m 2
2 0
2 0
2 0
We have made the substitution 3  . Also note that the field inside the conductor is downward from planes
P and P and upward from P. Because 1  2  0 C/m2, because the conductor is neutral, 2  1. The
above equation becomes
1  1    0 C/m 2  1  12   2   12 
We are now in a position to find electric field in regions 1–4.
For region 1,
EP 
 ˆ
j
4 0
EP   
 ˆ
j
EP 
 ˆ
j
4 0
EP 
 ˆ
j
EP  
4 0
 ˆ
j
2 0
The electric field is Enet  EP  EP  EP   20  ˆj.
In region 2, Enet  0 N/C. In region 3,
EP  
 ˆ
j
4 0
EP 
 ˆ
j
2 0
The electric field is Enet   20  ˆj.
In region 4,
EP  
The electric field is Enet    20  ˆj.
 ˆ
j
4 0
EP  
4 0
 ˆ
j
2 0