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Kapittel 30
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30.12. Model: The electric field is the negative of the slope of the graph of the potential function.
Visualize: Please refer to Figure EX30.12.
Solve: There are three regions of different slope. For 0 cm < x < 1 cm,
V
50 V  0 V

 5000 V/m  Ex  5000 V/m
 x 0.01 m  0 m
For 1 cm < x < 2 cm,
V 50 V   50 V 

 10,000 V/m  Ex  10,000 V/m
dx 0.02 m  0.01 m
For 2 cm < x < 3 cm,
0 V   50 V 
V

 5000 V/m  Ex  5000 V/m
 x 0.03 m  0.02 m
Assess:
V  x and Ex are the negative of each other.
30.13. Visualize:
Solve:
The electric potential difference V between two points in a uniform electric field is
V  x f   V  xi     Ex dx   Ex ( x f  xi )
Choosing xi  1.0 m and xf  1.0 m,
1000 V  (1000 V)  Ex[1.0 m  (1.0 m)]  Ex  1.0 kV/m
Alternatively, xi  1.0 m and xf  1.0 m. For this choice,
1000 V  (1000 V)  Ex [1.0 m  (1.0 m)]  Ex  1.0 kV/m
Assess: The choice of initial and final positions does not change the physical nature of the electric field or the
potential difference.
30.31. Solve: The energy density is uE   5.0 1012 J   0.020 m  0.020 m  0.020 m   6.25 107 J m3 .
uE  12  0 E 2  E 
2uE
0

2  6.25 107 J/m3 
8.85 1012 C2 /N m2
 1.19 kV/m
30.32. Solve: (a) From Equation 30.26, the energy stored in the charged capacitor is
1
1  A 
2
2
U C  C  VC    0   VC 
2
2 d 
12
2
2
1   0.010 m   8.85 10 C /N m 
2
 200 V   1.11107 J
2
0.50 103 m
2

(b) From Equation 30.28, the energy density in the electric field is
1
1  V  1
200 V


3
uE   0 E 2   0  C   8.85 1012 C2 /N m2  
  0.71 J/m
3
2
2  d  2
 0.50 10 m 
2
2
30.33. Model: Assume the capacitor is a parallel-plate capacitor.
Solve: (a) The capacitance is increased by a factor of the dielectric constant  over the equivalent vacuum-filled
capacitor.
C   C0  
0 A
d
  3.7 
8.85 10
12
C2 / N m 2   0.0250 m 
2
0.20  103 m
 1.29  109 F  0.32 nF
We have used  = 3.7 for paper from Table 30.1.
(b) The dielectric strength is the maximum possible electric field in the capacitor before breakdown. In this capacitor,
E
V
V
 16 106 V/m 
 V  3.2 kV
d
 0.20 103 m 
30.34. Model: The electrodes form a parallel-plate capacitor.
Solve: (a) The capacitance of the equivalent vacuum-filled capacitor is
C0 
0 A
d

8.85 10
12
C2 /Nm 2  5.0 103 m 
 0.10 103 m 
2
 2.211012 F
From Table 30.1, the dielectric constant of mylar is   3.1. With the battery attached, the potential difference across
the plates is VC  Vbatt  9.0 V . The charge on the plates is
Q  C V   C0 V   3.1  2.21 1012 F   9.0 V   6.17  1011 C  62 pC
The electric field inside the capacitor is
E
VC 
9.0 V


  90 kV/m
3
d
 0.10 10 m 
(b) With the battery connected, VC  9.0 V. The capacitor is now vacuum-insulated.
E0 
VC
9.0 V

 90 kV/m
d
0.10 103 m
Q  C0 VC   2.211012 F   9.0 V   20 pC
Assess: Since the battery remains connected as the mylar is withdrawn, the potential difference across the plates does
not change.
30.45. Model: The electric field is the negative of the slope of the potential graph.
Visualize: Please refer to Figure P30.45.
Solve: Since the contours are uniformly spaced along the y-axis above and below the origin, the slope method is the
easiest to apply. Point 1 is in the center of a 75 V change (25 V to 100 V) over a distance of 2 cm, so the slope V/ s
is 37.5 V/cm or 3750 V/m. Point 2 has the same potential difference in half the distance. Thus the slope at point 2 is 7500
V/m. The magnitudes of the electric fields at points 1 and 2 are 3750 V/m and 7500 V/m. The directions of the electric
fields are downward at point 1 and upward at point 2, that is, from the higher potential to the lower potential. That is,
E1  3750 V m, down 
E2   7000 V m, up 
30.47. Model: The electric field is the negative of the slope of the graph of the potential function.
Solve: The electric potential in a region of space is V  (150 x2 – 200 y2) V where x and y are in meters. The x- and
y-components are
dV
dV
Ey  
   400 y  V/m
Ex  
   300 x  V/m
dy
dx
At (x,y) = (2.0 m, 2.0 m), Ex  600 V/m and Ey  800 V/m. The magnitude and direction of the electric field are
E  Ex2  E y2 
tan  
Ey
Ex

 600 V/m   800 V/m 
2
2
 1000 V/m
800 V/m 4
    53.1 above the x-axis
600 V/m 3
The electric field points 180°  53° = 127° counterclockwise (ccw) from the +x-axis.