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HW02, Chapter 24:1, 4, 9, 11, 16, 19, 24, 30, 36, and 57. 
 
Due Monday, September 12, 2005. 
 
 
1. An electric field with a magnitude of 3.50 kN/C is applied along the x axis. 
Calculate the electric flux through a rectangular plane 0.350 m wide and 0.700 m 
long assuming that (a) the plane is parallel to the yz plane; (b) the plane is 
parallel to the xy plane; (c) the plane contains the y axis, and its normal makes an 
angle of 40.0° with the x axis. 
 
P24.1 (a) ( ) ( )3 2cos 3.50 10 0.350 0.700 cos0 858 N m CE EAΦ = = × × ° = ⋅θ 
 
(b) 90.0= °θ 0EΦ = 
 
 (c) ( ) ( )3 23.50 10 0.350 0.700 cos40.0 657 N m CEΦ = × × ° = ⋅ 
 
4. Consider a closed triangular box resting within a horizontal electric field 
of magnitude E = 7.80 × 104 N/C as shown in Figure P24.4. Calculate the electric 
flux through (a) the vertical rectangular surface, (b) the slanted surface, and (c) 
the entire surface of the box. 
 
 
 
Figure P24.4 
 
P24.4 (a) ( ) ( )10.0 cm 30.0 cmA =′ 
 ( ) ( )
2 2
,
4
,
2
,
300 cm 0.0300 m
cos
7.80 10 0.0300 cos180
2.34 kN m C
E A
E A
E A
A
EA′
′
′
= =′
Φ = ′
Φ = ×
Φ = − ⋅
θ
° 
 
(b) ( ) ( )4, cos 7.80 10 cos60.0E A EA AΦ = = ×θ ° 
10.0 cm
 30.0 cm
60.0Þ
 
FIG. P24.4 
 
( ) ( ) ( )
( ) ( )
2 2
4 2
,
10.0 cm30.0 cm 30.0 cm 600 cm 0.0600 mcos60.0
7.80 10 0.0600 cos60.0 2.34 kN m CE A
A w  = = = =  °
Φ = × ° = + ⋅
 
 
(c) The bottom and the two triangular sides all lie parallel to E, so for each 
of these. Thus, 
0EΦ =
 
 2 2,total 2.34 kN m C 2.34 kN m C 0 0 0 0EΦ = − ⋅ + ⋅ + + + = . 
 
9. The following charges are located inside a submarine: 5.00 µC, –9.00 µC, 
27.0 µC, and –84.0 µC. (a) Calculate the net electric flux through the hull of the 
submarine. (b) Is the number of electric field lines leaving the submarine greater 
than, equal to, or less than the number entering it? 
 
P24.9 (a) ( ) 6 2in 12 2 2
0
5.00 C 9.00 C 27.0 C 84.0 C 6.89 10 N m C8.85 10 C N mE
q
−
+ − + −Φ = = = − × ⋅∈ × ⋅
µ µ µ µ 2 
 26.89 M N m CEΦ = − ⋅ 
 
(b) Since the net electric flux is negative, more lines enter than leave the surface. 
 
 
11. Four closed surfaces, S1 through S4, together with the charges –2Q, Q, and 
–Q are sketched in Figure P24.11. (The colored lines are the intersections of the 
surfaces with the page.) Find the electric flux through each surface. 
 
 Figure P24.11 
 
P24.11 in
0
E
q
∈Φ = 
 Through S 1
0 0
2
E
Q Q Q= −− +Φ = ∈ ∈ 
 Through S 2
0
0E
Q Q+ − =∈Φ = 
 Through S 3
0 0
2 2
E
Q Q Q Q− + − = −Φ = ∈ ∈ 
 Through S 4 0EΦ = 
 
16. In the air over a particular region at an altitude of 500 m above the ground 
the electric field is 120 N/C directed downward. At 600 m above the ground the 
electric field is 100 N/C downward. What is the average volume charge density 
in the layer of air between these two elevations? Is it positive or negative? 
 
P24.16 Consider as a gaussian surface a box with horizontal area A, lying between 500 and 600 
m elevation. 
 
 
0
q
d⋅ = ∈∫ E A� : ( ) ( ) ( )0
100 m120 N C 100 N C AA A+ + − = ∈
ρ 
 
( ) ( )12 2 2 12 320 N 8.85 10 C N m 1.77 10 C m100 m
−
−× ⋅C= =ρ ×
 
 The charge is po , to produce the net outward flux of electric field. sitive
 
19. An infinitely long line charge having a uniform charge per unit length λ 
lies a distance d from point O as shown in Figure P24.19. Determine the total 
electric flux through the surface of a sphere of radius R centered at O resulting 
from this line charge. Consider both cases, where R < d and R > d. 
 
 
 
 
Figure P24.19 
 
P24.19 If R d≤ , the sphere encloses no charge and in
0
0E
qΦ = =∈ . 
 
 If R , the length of line falling within the sphere is d> 2 22 R d− 
 
 so 
2 2
0
2
E
R d−
∈
λΦ = 
 
24. A solid sphere of radius 40.0 cm has a total positive charge of 26.0 µC 
uniformly distributed throughout its volume. Calculate the magnitude of the 
electric field (a) 0 cm, (b) 10.0 cm, (c) 40.0 cm, and (d) 60.0 cm from the center of 
the sphere. 
 
P24.24 (a) 3 0e
k Q r
E
a
= = 
 
(b) 
( ) ( ) ( )
( )
9 6
3 3
8.99 10 26.0 10 0.100
365 kN C
0.400
ek Q rE
a
−× ×= = = 
 
(c) 
( ) ( )
( )
9 6
2 2
8.99 10 26.0 10
1.46 M N C
0.400
ek QE
r
−× ×= = = 
 
(d) 
( ) ( )
( )
9 6
2 2
8.99 10 26.0 10
649 kN C
0.600
ek QE
r
−× ×= = = 
 The direction for each electric field is radially outw ard . 
 
30. A solid plastic sphere of radius 10.0 cm has charge with uniform density 
throughout its volume. The electric field 5.00 cm from the center is 86.0 kN/C 
radially inward. Find the magnitude of the electric field 15.0 cm from the center. 
 
P24.30 Let ρ represent the charge density. For the field inside the sphere at we have 1 5 cmr =
 
 
3
2 inside 1
1 1
0 0
44 3
q r
E r = =∈ ∈
π ρπ 11
03
r
E = ∈
ρ 
 ( ) ( )12 2 3 5 30 1 2
1
3 8.85 10 C 86 10 N3 4.57 10 C m0.05 m N m
E
r C
−
−× − ×∈= = = − ×ρ . 
 Now for the field outside at 3 15 cmr = 
 
( ) ( ) ( )
( )
3
2 2
3 3
0
3 5 9 2 7
4
3 2 3 2 2
3
3
44 3
0.10 m 4.57 10 C 8.99 10 N m 1.91 10 C4 7.64 10 N C3 m 0.15 m C
76.4 kN C radially inw ard
e
r
E r
k
E
r
− −
= ∈
− × × − ×= = =
=E
π ρπ
π − × 
 
36. An insulating sphere is 8.00 cm in diameter and carries a 5.70-µC charge 
uniformly distributed throughout its interior volume. Calculate the charge 
enclosed by a concentric spherical surface with radius (a) r = 2.00 cm and (b) r = 
6.00 cm. 
 
P24.36 (a) ( )
6
2 3
3 34 43 3
5.70 10 2.13 10 C m
0.0400
Q
a
− −×= = = ×ρ π π 
 
 ( ) ( )33 2 7in 4 42.13 10 0.0200 7.13 10 C 713 nC3 3q r − −   = = × = × =      ρ π π 
 
(b) ( ) ( )33 2in 4 42.13 10 0.0400 5.70 C3 3q r −   = = × =      ρ π π µ 
 
57. A solid, insulating sphere of radius a has a uniform charge density ρ and a 
total charge Q. Concentric with this sphere is an uncharged, conducting hollow 
sphere whose inner and outer radii are b and c, as shown in Figure P24.57. (a) 
Find the magnitude of the electric field in the regions r < a, a < r < b, b < r < c, and r 
> c. (b) Determine the induced charge per unit area on the inner and outer 
surfaces of the hollow sphere. 
 
 Figure P24.57 
 
 
P24.57 (a) ( )2 in
0
4 qd E r⋅ = = ∈∫ E A� π 
 For r a, < 3in 43q r
 =   ρ π 
 
 so 
03
r
E = ∈
ρ . 
 
 For a r and b< < c r< , inq Q= . 
 
 So 2
04
Q
E
r
= ∈π . 
 
 
 
FIG. P24.57 
 
 For b r , c≤ ≤ 0E = , since 0E = inside a conductor. 
 
(b) Let q = induced charge on the inner surface of the hollow sphere. Since 1 0E = 
inside the conductor, the total charge enclosed by a spherical surface of radius 
 must be zero. b r c≤ ≤
 
 Therefore, 1 0q Q+ = and 11 2 24 4
q Q
b b
−= =π πσ . 
 
 Let q = induced charge on the outside surface of the hollow sphere. Since the 
hollow sphere is uncharged, we require 
2
 q q and 1 2 0+ = 12 2 24 4
q Q
c c
= =π πσ .