Download Kapittel 26

Kapittel 27
Løsning på alle oppgavene finnes på CD. Lånes ut hvis behov.
27.1. Model: The electric field is that of the two charges placed on the y-axis.
Visualize: Please refer to Figure EX27.1. We denote the upper charge by q1 and the lower charge by q2. Because
both the charges are positive, their electric fields at P are directed away from the charges.
Solve: The electric field from q1 is
9
2
2
9
 1 q1
  9.0  10 N m /C  3.0  10 C 
E1  
,

below

x
-axis

cos iˆ  sin  ˆj

2
2
2
 0.050 m    0.050 m 
 4 0 r1



Because tan  5 cm 5 cm  1, the angle   45. Hence,
 1 ˆ 1 ˆ
E1   5400 N/C  
i
j
2 
 2
Similarly, the electric field from q2 is
 1 q2

 1 ˆ 1
E2  
,  above  x -axis    5400 N/C  
i
2
4

r
2
 2
0 2


ˆj 


 1 ˆ
3ˆ
 Enet at P  E1  E2  2  5400 N/C  
 i  7.6 10 i N/C
 2
Thus, the strength of the electric field is 7.6103 N/C and its direction is horizontal.
Assess: Because the charges are located symmetrically on either side of the x-axis and are of equal value, the ycomponents of their fields will cancel when added.
27.3. Model: The electric field is that due to superposition of the fields of the two 3.0 nC charges located on the yaxis.
Visualize: Please refer to Figure EX27.3. We denote the top 3.0 nC charge by q1 and the bottom 3.0 nC charge by q2.
The electric fields ( E1 and E2 ) of both the positive charges are directed away from their respective charges. With
vector addition, they yield the net electric field E net at the point P indicated by the dot.
Solve: The electric fields from q1 and q2 are
9
2
2
9
 1 q1
  9.0 10 N m /C  3.0 10 C  ˆ
E1  
,
along

x
-axis

i  10,800iˆ N/C

2
2
 0.05 m 
 4 0 r1

 1 q2

E2  
,  above  x -axis 
2
4

r
0 2


Because tan  10 cm 5 cm,   tan 1  2  63.43. So,
 9.0 10
9
E2 
N m 2 /C 2  3.0  109 C 
 0.10 m 
2
  0.050 m 
2
 cos63.43 iˆ  sin 63.43 ˆj   966iˆ  1127 ˆj  N/C
The net electric field is thus


Enet at P  E1  E2  11,766iˆ  1127 ˆj N/C
To find the angle this net vector makes with the x-axis, we calculate
tan  
1127 N/C
   5.5
11,766 N/C
Thus, the strength of the electric field at P is
Enet 
11,766 N/C  1127 N/C
2
and E net makes an angle of 5.5 above the x-axis.
2
 11,820 N/C = 1.18103 N/C
Assess: Because of the inverse square dependence on distance, E2  E1. Additionally, because the point P has no
special symmetry relative to the charges, we expected the net field to be at an angle relative to the x-axis.
27.7. Model: We will assume that the wire is thin and that the charge lies on the wire along a line.
Solve:
is
From Equation 27.15, the electric field strength of an infinitely long line of charge having linear charge density 
Eline 
 Eline  r  5.0 cm  
2
1
4 0 5.0 102 m
1 2
4 0 r
Eline  r  10.0 cm  
2
1
4 0 10.0 102 m
Dividing the above two equations gives
 5.0 102 m 
1
3
Eline  r  10.0 cm   
 Eline  r  5.0 cm    2000 N/C  1.00 10 N/C  1000 N/C
2
2
 10.0 10 m 
27.10.
Model: The rod is thin, so assume the charge lies along a line.
Visualize:
Solve: The force on charge q is F  qErod . From Example 27.3, the electric field a distance r from the center of a
charged rod is
Erod 
1
Q
4 0 r r 2   L / 2 2

iˆ 
9.0 109 N m 2 /C 2  40 10 9 C  iˆ
 0.04 m   0.04 m    0.05 m 
2
5
 1.406 105 iˆ N/C
Thus, the force is
F   6.0 109 C 1.406 105 N/C  iˆ  8.4 104 iˆ N
More generally, F  8.4 104 N, away from the rod  .
27.11. 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 27.5, the field of the left (negative) ring at z  10 cm is
 E1  z 
4 0  z  R
2
9.0 10
9
zQ1

2 3/ 2

N m 2 /C2   0.10 m   20 109 C 
2 3/ 2
 0.10 m    0.050 m  


2
 1.288 104 N/C
That is, the field is E1  1.288  104 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
E  E1  E2   2.6 104 N/C, left 
(b) The force is
F  qE   1.0  109 C  2.6  104 N/C, left    2.6 10 5 N, 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  2ax  a 
v12  v0 2 (4.0 107 m/s) 2  (2.0 107 m/s) 2

 5.0  1016 m/s 2
2x
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
27.23. Model: The infinite charged plane produces a uniform electric field.
Solve: (a) The electric field of a plane of charge with surface charge density  is
E

2 0
F
 ma 
   2 0 E  2 0    2 0 

q
 
 q 
where m is the mass, q is the charge, and a is the acceleration of the electron. To obtain  we must first find a. From
the kinematic of motion equation v12  v02  2ax,
a
 
v12  v02 (1.0 107 m/s)2  (0 m/s)2

 2.5 1015 m/s2
2x
2(2.0 102 m)
2(8.85  1012 C2 /N m2 )(9.11 1031 kg)(2.5  1015 m/s2 )
 2.5  107 C/m2
1.60  1019 C
(b) Using the kinematic equation of motion v1  v0  at ,
t 
v1  v0 1.0 107 m/s  0 m/s

 4.0 109 s
a
2.5 1015 m/s2
27.30. Model: The electric field is that of three point charges q1, q2, and q3.
Visualize: Please refer to Figure P27.30. Assume the charges are in the x-y plane. The 10 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 . 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
E1 
9
2
2
9
q1  9.0 10 N m /C 10 10 C 

 100,000 N/C
2
4 0 r12
 0.030 m 
1
E1 points away from q1, so in component form E1  100,000 ˆj N/C . The electric field produced by q2 is
E2  18,000 N/C . E2 points towards q2, so E2  18,000iˆ 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 

 26,471 N/C
2
2
4 0 r32
 0.030 m    0.050 m 
1
E3 points toward q3 and makes an angle   tan 1  3/5  31 with the x-axis. So,


E3  E3 cos  iˆ  E3 sin  ˆj  22,700iˆ  13,634 ˆj N/C
Adding these three vectors gives




Enet  E1  E2  E3  40,700iˆ  86,366 ˆj N/C  4.1104 iˆ  8.6 104 ˆj N/C
This is in component form.
(b) The magnitude of the field is
Enet  Ex2  Ey2 
 40,700 N/C  86,366 N/C
2

2
 95,476 N/C  9.5 104 N/C

and its angle from the x-axis is   tan 1 Ex Ey  25 . We can also write Enet  (9.5  104 N/C, 25 CW from the
 x-axis).
27.38. Model: The fields are those of two infinite lines of charge with linear charge density .
Visualize:
Solve:
The field at distance r from an infinite line of charge is
E
1 2
rˆ
4 0 r
It points straight away from the line. With two perpendicular lines, the field due to the line along the x-axis points in
the y-direction and depends inversely on distance y. Similarly, the field due to the line along the y-axis points in the xdirection and depends inversely on distance x. That is
Ex 
1 2
4 0 x
Ey 
 E  Exiˆ  E y ˆj 
1 2
4 0 y
2  1 ˆ 1 ˆ 
 i  j
4 0  x
y 
The field strength at this point in space is
E  Ex2  E y2 
2
4 0
1/x 2  1/ y 2
27.42. Model: The electric field is that of a line of charge of length L.
Visualize:
The origin of the coordinate system is at the center of the rod. Divide the rod into many small segments of charge q
and length x.
Solve: (a) Segment i creates a small electric field Ei at point P that points to the right. The net field E will point to
the right and have Ey  Ez  0 N/C. The distance to segment i is x, so
Ei   Ei  x 
q
4 0  x  xi 
2
 Ex    Ei  x 
1
q

4 0  x  xi 2
q is not a coordinate, so before converting the sum to an integral we must relate the charge q to length  x. This is
done through the linear charge density   Q/L, from which
q  x 
Q
x
L
With this charge, the sum becomes
Ex 
Q / L
4 0
x
  x  x 
i
2
i
Now we let x  dx and replace the sum by an integral from x   12 L to x   12 L. Thus,
Ex 
Q / L L / 2
4 0
dx
  x  x
2
L / 2

Q / L 
1
Q
1 
Q / L  L
E
iˆ

4 0 x 2  L2 4
4 0  x  x   L / 2
4 0 x 2  L2 4
L/2
The electric field strength at x is
E
1
Q
4 0 x  L2 4
2
(b) For x  L ,
E
1 Q
4 0 x 2
That is, the line charge behaves like a point charge.
(c) Substituting into the above formula
3.0 109 C
E   9.0 109 N m 2 /C2 
 9.8 104 N/C
2
2
2
2
1
3.0

10
m


5.0

10
m

 2

27.45. Model: Assume that the ring of charge is thin and that the charge lies along circle of radius R.
Solve: (a) From Example 27.5, the on-axis field of a ring of charge Q and radius R is
Ez 
1
zQ
4 0  z  R 2 3/ 2
2
For the field to be maximum at a particular value of z, dE dz  0. Taking the derivative,
z  3/ 2  2 z  
dE
Q 
1
1
3z 2


0

 2

3
/
2
5
/
2
3
2
5/ 2
dz 4 0   z  R 2 
 z 2  R2  
 z 2  R2   z 2  R2 

1
(b) The field strength at the point z  R
 Ez max 
Q
4 0
R 2
 R 2 R 
  

2
2
3/ 2

3z 2
R
z
2
2
z R
2
2 is
2
Q
2
3 3 4 0 R
27.49. Motion: The very wide charged electrode is a plane of charge.
Solve:
For a plane of charge with surface charge density , the electric field is
E 
plane z


2 0
   2 0  Eplane  
z
Q
Q

A  R2
 Q  2 R 2 0  Eplane   2  0.0050 m   8.85  1012 C2 /N m 2  1000 N/C   1.39  1012 C  1.39  103 nC
2
z
Assess: Note that the field strength of a plane of charge does not depend on z and is a constant. We did not have to
use the distance z  5.0 cm given in the problem.