# Solution manual fundamentals of physics extended, 8th editionch23

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1. The vector area A and the electric field E are shown on the diagram below. The angle
θ between them is 180° – 35° = 145°, so the electric flux through the area is
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2
Φ = E ⋅ A = EA cos θ = (1800 N C ) 3.2 × 10−3 m cos145° = −1.5 ×10−2 N ⋅ m 2 C.

(

)

G
G G
2
2. We use Φ = E ⋅ A , where A = Aj = 140
. m j .

b

g

2
(a) Φ = ( 6.00 N C ) ˆi ⋅ (1.40 m ) ˆj = 0.
2
(b) Φ = ( −2.00 N C ) ˆj ⋅ (1.40 m ) ˆj = −3.92 N ⋅ m 2 C.
2
(c) Φ = ª( −3.00 N C ) ˆi + ( 400 N C ) kˆ º ⋅ (1.40 m ) ˆj = 0 .
¬
¼

(d) The total flux of a uniform field through a closed surface is always zero.

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3. We use Φ = E ⋅ dA and note that the side length of the cube is (3.0 m–1.0 m) = 2.0 m.

z

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(a) On the top face of the cube y = 2.0 m and dA = ( dA ) ˆj . Therefore, we have
G
2
E = 4iˆ − 3 ( 2.0 ) + 2 ˆj = 4iˆ − 18jˆ . Thus the flux is

(

Φ=³

top

)

G G
E ⋅ dA = ³

top

( 4iˆ − 18jˆ ) ⋅ ( dA) ˆj = −18³

dA = ( −18 )( 2.0 ) N ⋅ m 2 C = −72 N ⋅ m 2 C.
2

top

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(b) On the bottom face of the cube y = 0 and dA = dA − j . Therefore, we have

b ge j

E = 4 i − 3 02 + 2 j = 4 i − 6j . Thus, the flux is

c

Φ=³

bottom

h

G G
E ⋅ dA = ³

bottom

( 4iˆ − 6ˆj) ⋅ ( dA) ( −ˆj) = 6³

dA = 6 ( 2.0 ) N ⋅ m 2 C = +24 N ⋅ m 2 C.
2

bottom

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(c) On the left face of the cube dA = ( dA ) −ˆi . So

( )

G
Φ = ³ Eˆ ⋅ dA = ³
left

left

( 4iˆ + E ˆj) ⋅ ( dA) ( −ˆi ) = −4³
y

dA = −4 ( 2.0 ) N ⋅ m 2 C = −16 N ⋅ m 2 C.
2

bottom

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(d) On the back face of the cube dA = ( dA ) −kˆ . But since E has no z component
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E ⋅ dA = 0 . Thus, Φ = 0.

( )

(e) We now have to add the flux through all six faces. One can easily verify that the flux
through the front face is zero, while that through the right face is the opposite of that
through the left one, or +16 N·m2/C. Thus the net flux through the cube is
Φ = (–72 + 24 – 16 + 0 + 0 + 16) N·m2/C = – 48 N·m2/C.

4. The flux through the flat surface encircled by the rim is given by Φ = πa 2 E. Thus, the
flux through the netting is
Φ′ = −Φ = −π a 2 E = −π (0.11 m) 2 (3.0 × 10−3 N/C) = −1.1× 10−4 N ⋅ m 2 /C .

5. We use Gauss’ law: ε 0 Φ = q , where Φ is the total flux through the cube surface and q
is the net charge inside the cube. Thus,
Φ=

q

ε0

=

1.8 ×10−6 C
= 2.0 ×105 N ⋅ m 2 C.
8.85 × 10−12 C2 N ⋅ m 2

6. There is no flux through the sides, so we have two “inward” contributions to the flux,
one from the top (of magnitude (34)(3.0)2) and one from the bottom (of magnitude
(20)(3.0)2). With “inward” flux being negative, the result is Φ = – 486 N⋅m2/C. Gauss’
law then leads to qenc = ε0 Φ = –4.3 × 10–9 C.

7. To exploit the symmetry of the situation, we imagine a closed Gaussian surface in the
shape of a cube, of edge length d, with a proton of charge q = +1.6 ×10−19 C situated at
the inside center of the cube. The cube has six faces, and we expect an equal amount of
flux through each face. The total amount of flux is Φnet = q/ε0, and we conclude that the
flux through the square is one-sixth of that. Thus, Φ = q/6ε0 = 3.01× 10–9 N⋅m2/C.

8. (a) The total surface area bounding the bathroom is
A = 2 ( 2.5 × 3.0 ) + 2 ( 3.0 × 2.0 ) + 2 ( 2.0 × 2.5 ) = 37 m 2 .
The absolute value of the total electric flux, with the assumptions stated in the problem, is
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| Φ |=| ¦ E ⋅ A |=| E | A = (600)(37) = 22 ×103 N ⋅ m 2 / C.

By Gauss’ law, we conclude that the enclosed charge (in absolute value) is
| qenc |= ε 0 | Φ |= 2.0 ×10−7 C. Therefore, with volume V = 15 m3, and recognizing that we
are dealing with negative charges (see problem), the charge density is qenc/V = –1.3 × 10–8
C/m3.
(b) We find (|qenc|/e)/V = (2.0 × 10–7/1.6 × 10–19)/15 = 8.2 × 1010 excess electrons per
cubic meter.

9. Let A be the area of one face of the cube, Eu be the magnitude of the electric field at the
upper face, and El be the magnitude of the field at the lower face. Since the field is
downward, the flux through the upper face is negative and the flux through the lower face
is positive. The flux through the other faces is zero, so the total flux through the cube
surface is Φ = A( EA − Eu ). The net charge inside the cube is given by Gauss’ law:
q = ε 0 Φ = ε 0 A( EA − Eu ) = (8.85 ×10−12 C2 / N ⋅ m 2 )(100 m) 2 (100 N/C − 60.0 N/C)
= 3.54 × 10−6 C = 3.54 µ C.

10. We note that only the smaller shell contributes a (non-zero) field at the designated
point, since the point is inside the radius of the large sphere (and E = 0 inside of a
spherical charge), and the field points towards the − x direction. Thus,
G
E = E (−ˆj) = –

σ2 4πR2 ˆ
q ˆ
j
=

j = – (2.8 × 104 N/C ) ˆj ,
4πεo r2
4πεo (L− x)2

where R = 0.020 m (the radius of the smaller shell), d = 0.10 m and x = 0.020 m.

11. The total flux through any surface that completely surrounds the point charge is q/ε0.

(a) If we stack identical cubes side by side and directly on top of each other, we will find
that eight cubes meet at any corner. Thus, one-eighth of the field lines emanating from
the point charge pass through a cube with a corner at the charge, and the total flux
through the surface of such a cube is q/8ε0. Now the field lines are radial, so at each of
the three cube faces that meet at the charge, the lines are parallel to the face and the flux
through the face is zero.
(b) The fluxes through each of the other three faces are the same, so the flux through each
of them is one-third of the total. That is, the flux through each of these faces is (1/3)(q/8ε0)
= q/24ε0. Thus, the multiple is 1/24 = 0.0417.

12. Eq. 23-6 (Gauss’ law) gives εοΦ = qenclosed .
(a) Thus, the value Φ = 2.0 × 105 (in SI units) for small r leads to qcentral = +1.77 × 10−6 C
or roughly 1.8 µC.
(b) The next value that Φ takes is – 4.0 × 105 (in SI units), which implies
qenc = −3.54 ×10−6 C. But we have already accounted for some of that charge in part (a), so
the result for part (b) is qA = qenc – qcentral = – 5.3 × 10−6 C.
(c) Finally, the large r value for Φ is 6.0 × 105 (in SI units), which implies
qtotal enc = 5.31×10−6 C. Considering what we have already found, then the result is

qtotal enc − q A − qcentral = +8.9 µ C.

13. (a) Let A = (1.40 m)2. Then

(

)( )

Φ = 3.00 y ˆj ⋅ − A ˆj

y =0

(

)( )

+ 3.00 y ˆj ⋅ A ˆj

= ( 3.00 )(1.40 )(1.40 ) = 8.23 N ⋅ m 2 C.
2

y =1.40

(b) The charge is given by

(

)(

)

qenc = ε 0 Φ = 8.85 ×10−12 C2 / N ⋅ m 2 8.23 N ⋅ m 2 C = 7.29 ×10−11 C .

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(c) The electric field can be re-written as E = 3.00 y ˆj + E0 , where E0 = −4.00iˆ + 6.00ˆj is a
constant field which does not contribute to the net flux through the cube. Thus Φ is still
8.23 N⋅m2/C.
(d) The charge is again given by

(

)(

)

qenc = ε 0 Φ = 8.85 ×10−12 C2 / N ⋅ m 2 8.23 N ⋅ m 2 C = 7.29 ×10−11 C .

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14. The total electric flux through the cube is Φ = v³ E ⋅ dA . The net flux through the two

faces parallel to the yz plane is
Φ yz = ³³ [ Ex ( x = x2 ) − Ex ( x = x1 )]dydz = ³
= 6³

y2 =1
y1 = 0

dy ³

z2 = 3

z1 =1

y2 =1
y1 = 0

dy ³

z2 = 3

dz[10 + 2(4) −10 − 2(1)]

z1 =1

dz = 6(1)(2) = 12.

Similarly, the net flux through the two faces parallel to the xz plane is
Φ xz = ³³ [ E y ( y = y2 ) − E y ( y = y1 )]dxdz = ³

x2 = 4

x1 =1

dy ³

z2 = 3

z1 =1

dz[−3 − (−3)] = 0 ,

and the net flux through the two faces parallel to the xy plane is
Φ xy = ³³ [ Ez ( z = z2 ) − Ez ( z = z1 )]dxdy = ³

x2 = 4

x1 =1

dx ³

y2 =1
y1 = 0

dy ( 3b − b ) = 2b(3)(1) = 6b.

Applying Gauss’ law, we obtain
qenc = ε 0 Φ = ε 0 (Φ xy + Φ xz + Φ yz ) = ε 0 (6.00b + 0 + 12.0) = 24.0ε 0

which implies that b = 2.00 N/C ⋅ m .

15. (a) The charge on the surface of the sphere is the product of the surface charge
density σ and the surface area of the sphere (which is 4πr 2 , where r is the radius). Thus,
2

§ 1.2 m ·
−6
−5
2
q = 4πr σ = 4π ¨
¸ ( 8.1×10 C/m ) = 3.7 ×10 C.
2

(b) We choose a Gaussian surface in the form of a sphere, concentric with the conducting
sphere and with a slightly larger radius. The flux is given by Gauss’s law:
Φ=

q

ε0

=

3.66 ×10−5 C
= 4.1× 106 N ⋅ m 2 / C .
8.85 × 10−12 C2 / N ⋅ m 2

16. Using Eq. 23-11, the surface charge density is

σ = Eε 0 = ( 2.3 ×105 N C )( 8.85 ×10−12 C2 / N ⋅ m 2 ) = 2.0 × 10−6 C/m 2 .

17. (a) The area of a sphere may be written 4πR2= πD2. Thus,

σ=

q
2.4 × 10−6 C
=
= 4.5 ×10−7 C/m 2 .
2
2
πD
π (1.3 m )

(b) Eq. 23-11 gives
E=

σ
4.5 × 10−7 C/m 2
=
= 5.1×104 N/C.
ε 0 8.85 ×10−12 C2 / N.m 2

18. Eq. 23-6 (Gauss’ law) gives εοΦ = qenc.
(a) The value Φ = – 9.0 × 105 (in SI units) for small r leads to qcentral = – 7.97 × 10−6 C or
roughly – 8.0 µC.
(b) The next (non-zero) value that Φ takes is +4.0 × 105 (in SI units), which implies
qenc = 3.54 ×10−6 C. But we have already accounted for some of that charge in part (a), so
the result is
qA = qenc – qcentral = 11.5 × 10−6 C ≈ 12 µ C .
(c) Finally, the large r value for Φ is – 2.0 × 105 (in SI units), which implies
qtotal enc = −1.77 ×10−6 C. Considering what we have already found, then the result is
qtotal enc – qA − qcentral = –5.3 µC.

19. (a) Consider a Gaussian surface that is completely within the conductor and surrounds
the cavity. Since the electric field is zero everywhere on the surface, the net charge it
encloses is zero. The net charge is the sum of the charge q in the cavity and the charge qw
on the cavity wall, so q + qw = 0 and qw = –q = –3.0 × 10–6C.

(b) The net charge Q of the conductor is the sum of the charge on the cavity wall and the
charge qs on the outer surface of the conductor, so Q = qw + qs and

(

) (

)

qs = Q − qω = 10 ×10−6 C − −3.0 ×10−6 C = +1.3 × 10−5 C.

20. We imagine a cylindrical Gaussian surface A of radius r and unit length concentric
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q
with the metal tube. Then by symmetry v³ E ⋅ dA = 2πrE = enc .

ε0

A

(a) For r < R, qenc = 0, so E = 0.
(b) For r > R, qenc = λ, so E (r ) = λ / 2π rε 0 . With λ = 2.00 ×10−8 C/m and r = 2.00R =
0.0600 m, we obtain

( 2.0 ×10 C/m )
E=
2π ( 0.0600 m ) ( 8.85 × 10 C
−8

−12

2

/ N⋅m

2

)

= 5.99 ×103 N/C.

(c) The plot of E vs. r is shown below.

Here, the maximum value is

Emax

(

)

2.0 ×10−8 C/m
λ
=
=
= 1.2 × 104 N/C.
−12
2
2
2πrε 0 2π ( 0.030 m ) 8.85 ×10 C / N ⋅ m

(

)

21. The magnitude of the electric field produced by a uniformly charged infinite line is E
= λ/2πε0r, where λ is the linear charge density and r is the distance from the line to the
point where the field is measured. See Eq. 23-12. Thus,

(

)(

)

λ = 2πε 0 Er = 2π 8.85 ×10−12 C2 / N ⋅ m 2 4.5 ×104 N/C ( 2.0 m ) = 5.0 × 10−6 C/m.

22. We combine Newton’s second law (F = ma) with the definition of electric field
( F = qE ) and with Eq. 23-12 (for the field due to a line of charge). In terms of
magnitudes, we have (if r = 0.080 m and λ = 6.0 x 10-6 C/m)
ma = eE =

2πεo r



a=

= 2.1 × 1017 m/s2 .
2πεo r m

23. (a) The side surface area A for the drum of diameter D and length h is given by
A = π Dh . Thus
§
C2 ·
q = σ A = σπDh = πε 0 EDh = π ¨ 8.85 × 10−12
2.3 ×105 N/C ( 0.12 m )( 0.42 m )
2 ¸
N⋅m ¹
= 3.2 ×10−7 C.

(

)

(b) The new charge is
§ A′ ·
§ πD′h′ ·
−7
q′ = q ¨ ¸ = q ¨
¸ = 3.2 ×10 C

(

) « ((12 cm )()(42 cm )) » = 1.4 ×10
ª 8.0 cm 28 cm º

«¬

»¼

−7

C.

24. We reason that point P (the point on the x axis where the net electric field is zero)
cannot be between the lines of charge (since their charges have opposite sign). We
reason further that P is not to the left of “line 1” since its magnitude of charge (per unit
length) exceeds that of “line 2”; thus, we look in the region to the right of “line 2” for P.
Using Eq. 23-12, we have
Enet = E1 + E2 =

λ1
λ2
+
.
2πεo (x + L/2) 2πεo (x − L/2)

Setting this equal to zero and solving for x we find
x=

λ1 − λ 2 L
λ1 + λ 2 2

which, for the values given in the problem, yields x = 8.0 cm.

25. We denote the inner and outer cylinders with subscripts i and o, respectively.

(a) Since ri < r = 4.0 cm < ro,
E (r ) =

λi
5.0 ×10−6 C/m
=
= 2.3 ×106 N/C.
2
2
−12
−2
2πε 0 r 2π (8.85 × 10 C / N ⋅ m ) (4.0 × 10 m)

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(b) The electric field E (r ) points radially outward.
(c) Since r > ro,
λi + λ o
5.0 ×10−6 C/m − 7.0 × 10−6 C/m
E (r ) =
=
= −4.5 ×105 N/C,
−12
−2
2
2
2πε 0 r 2π (8.85 × 10 C / N ⋅ m ) (8.0 ×10 m)
or | E (r ) |= 4.5 ×105 N/C.
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