Download Physics I - Rose

Physics II
Homework VII
CJ
Chapter 26; 4, 14, 25, 36, 45, 48, 50, 55
26.4. Model: The electric field is that of the two charges located on the y-axis.
Visualize: Please refer to Figure Ex26.4. We denote the positive charge by q1 and the negative charge by q2. The
electric field E1 of the positive charge q1 is directed away from q1, but the field E2 is toward the negative charge q2. We will
add E1 and E2 vectorially to find the strength and the direction of the net electric field vector.
Solve: The electric fields from q1 and q2 are



9.0  109 N m2 /C2 1  10 9 C
 1 q1

E1  
, away from q1 along  x axis  
iˆ  3600iˆ N/C
2
2
 4 0 r1

 0.05 m 


 1 q2

E2  
,  below  x -axis 
 4 0 r22



From the geometry of the figure,
10 cm
   63.43
5 cm
tan 
 E2 

 9.0  10
9

N m2 /C2 1  10 9 C
 0.10 m    0.05 m 
2

 Enet  E1  E2  3278iˆ  644 ˆj N/C  Enet 

2
  cos63.43iˆ  sin63.43 ˆj     322iˆ  644ˆj  N/C
 3278 N/C   644 N/C
2
2
 3340 N/C
To find the angle this net vector makes with the horizontal, we calculate
tan 
 Enet y
 Enet x 

644 N/C
   11.1
3278 N/C
Thus, the strength of the net electric field at P is 3341 N/C and E net makes an angle of 11.1 below the x-axis.
26.14. Model: Each disk is a uniformly charged disk. When the disk is charged negatively, the on-axis electric field of the
disk points toward the disk. The electric field points away from the disk for a positively charged disk.
Visualize:
Solve: (a) The surface charge density on the disk is
 
Q
A

Q
R
2

50  109 C
  0.05 m 
2
 6.366  106 C/m2
From Equation 26.22, the electric field of the left disk at z  0.10 m is
 E1 z


 
1
6.366  10 6 C/m 2 
1
1 
 

1
12
2
2 
2
2
2
2 0 
1  R z  2  8.85  10 C /N m  
1   0.05 m 0.10 m 



  38,000 N/C


In other words, E1   38,000 N/C, left  . Similarly, the electric field of the right disk at z  0.10 m (to its left) is
E2   38,000 N/C, left  . The net field at the midpoint between the two rings is E  E1  E2   76,000 N/C, left  .
(b) The force on the charge is



F  qE  1.0  10 9 C  76,000 N/C, left   7.6  105 N, right

Assess: Note that the force on the negative charge is to the right because the electric field is to the left.
26.25. Model: The electric field is uniform, so the electrons will have a constant acceleration.
Visualize:
Solve: A constant-acceleration kinematic equation of motion is
v  v  2a x 
2
1
2
0


5.0  10 7 m/s   0 m/s 
v12  v02
a

 1.042  1017 m/s2
2x
2 1.2  10 2 m

2
2

The net force on the electron in the electric field is F  qE  ma . Thus,
E



9.11  1031 kg 1.042  1017 m/s2
ma

 5.93  105 N/C
q
1.60  10 19 C
Hence, the field strength is 5.93  10 5 N/C.
26.36. Model: The electric field is that of two positive charges.
Visualize:
The figure shows E1 and E2 due to the individual charges. The total field is E  E1  E2 .
Solve:
add. Thus,
(a) From symmetry, the y-components of the two electric fields cancel out. The x-components are equal and
E  2  E1  x iˆ  2E1 cos iˆ
The field strength and angle are
E1 
q
4 0 r
2


q
4 0 x  s 4
2
E
(b) The electric field at position x is
2
cos 


2qx
4 0 x  s2 4
2

3/ 2
iˆ
x

r
x
x  s2 4
2
 9.0  10
E
9
 

N m2 /C2  2  1.0  10 9 C x
 x 2   0.003 m  


2
3/ 2

18x
 x 2   0.003 m 2 


3/ 2
N/C
where x has to be in meters. We can now evaluate E for different values of x:
x (mm)
0
2
4
6
10
x (m)
0.000
0.002
0.004
0.006
0.010
E (N/C)
0
768,000
576,000
358,000
158,000
(c) We can use the values of part (b) as a start in drawing the graph. Also note that E  0 N/C as x  0 m and as
x  . The graph is shown in the figure above.
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 q and length y. Segment i creates a small electric field at the point P
that makes an angle  with the horizontal. The field has both x and y components, but Ez  0 N/C. The distance to segment i

from point P is x 2  y2
Solve:

12
.
The electric field created by segment i at point P is
Ei 

q

cos iˆ  sin ˆj  


4

4  x  y 
 x  y  
q
2
2
2
2
0
0
x
x 2  y 2
iˆ 

ˆj 
x 2  y2 
y
The net field is the sum of all the E i , which gives E   Ei . q is not a coordinate, so before converting the sum to an
i
integral we must relate charge q to length y. This is done through the linear charge density   Q/L, from which we have
the relationship
q  y 
Q
y
L
With this charge, the sum becomes
E

Q/ L 
x y

4 0 i  x 2  y2



iˆ 
3/ 2
x
yy
2
 y 2

3/ 2

ˆj 


Now we let y  dy and replace the sum by an integral from y  0 m to y  L . Thus,
 Q / L   L
E
xdy
4 0  0 x 2  y2


Q

1
4 0 x x  L
2
2
ydy
L

iˆ  
3/ 2
iˆ 
0
x
2
 y2

3/ 2
1  Q 
1 
4 0  Lx  
 

y
ˆj    Q / L   x 

4 0   x 2 x 2  y2

 
L


1
 iˆ  
2

 x  y2
0
L



0

ˆj 



ˆ
j
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 q and
arc length s. Segment i creates a small electric field E i at the origin. The line from the origin to segment i makes an angle 
with the x-axis.
Solve: Because every segment i at an angle  above the axis is matched by segment j at angle  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 cos i
i
Ey  0 N/C
i
Note that angle i depends on the location of segment i. Now all segments are at the same distance ri  R from the origin, so
q
q

2
4 0 ri
4 0 R2
Ei 
The linear charge density on the rod is   Q/L, where L is the rod’s length. This allows us to relate charge q to the arc length
s through
q   s  (Q/L)s
Thus, the net field at the origin is
Ex  
i
 Q / L  s cos
4 0 R2
i

Q
4 0 LR2
 cos s
i
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 
rather than do a two-dimensional integral in x and y. We note that the arc length s is related to the small angle  by s 
R, so
Ex 
Q
 cos i 
4 0 LR i
With   d, the sum becomes an integral over all angles forming the rod.  varies from  /2 to   /2. So we
finally arrive at
Ex 
Q
4 0
 /2
LR  
 /2
cos d 
 /2
Q
2Q
sin  / 2 
4 0 LR
4 0 LR
Since we’re given the rod’s length L and not its radius R, it will be convenient to let R  L/. So our final expression for E ,
now including the vector information, is
E
1  2 Q  ˆ
i
4 0  L2 
(b) Substituting into the above expression,
 9.0  10
E
9
 
N m2 /C2 2 30  10 9 C
 0.10 m 
2
26.50. Model: Assume that the plastic sheets are planes of charge.
Visualize: Please refer to Figure P26.50.
  1.70  10
5
N/C
Solve:
At point 1 the electric field due to the left sheet and the right sheet are

 
Eleft   0 , toward right   0 iˆ
 2 0
 2 0
 3

3
Eright   0 , toward left    0 iˆ
2 0
 2 0

 Enet  Eleft  Eright  
0 ˆ
i
0
At point 2, Eleft   0 2 0  iˆ, Eright    30 2 0  iˆ, and Enet    20  0  iˆ. At point 3,Enet   0  0  iˆ.
26.55. 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
Chapter
6
projectile
v0   2K m 
1/ 2
problem.
The
kinetic
energy
K  21 mv02  3.0  10 17 J
gives
an
initial
speed
 8.115  10 m/s . Thus the initial components of the velocity are
6
vx 0  v0 cos45  5.74  106 m/s
vy0  v0 sin45  5.74  106 m/s
What acceleration a will cause the electron to pass through the point (x1, y1)  (1 cm, 0 cm)? The kinematic equations of
motion are
y1  y0  vy0 t1  21 ayt1  vy0 t1  21 at12  0 m
x1  x0  vx 0 t1  21 ax t12  vx 0 t1  0.01 m
From the x-equation, t1  x1 vx 0  1.742  10 9 s . Using this in the y-equation gives
a
2vy 0 t1
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
Felec qelec E eE
ma


E

 37,500 N/C
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  21 t1  8.71 1010 s . At
this instant,
ymax  vy0 t  21 at 2  0.0025 m  2.5 mm
Thus, dmin  2.5 mm.