G

G

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

G G

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 = Aj = 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.

G G

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

G

(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

G

(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 − 6j . 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

G

(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

G

G

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

G G

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

G G

G

| 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 .

G

G

G

(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 .

G G

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 ¹

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

G G

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 =

eλ

2πεo r

a=

eλ

= 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

© A¹

© πDh ¹

(

) « ((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)

G

(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.

G

(d) The minus sign indicates that E (r ) points radially inward.

G

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

G G

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 = Aj = 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.

G G

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

G

(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

G

(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 − 6j . 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

G

(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

G

G

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

G G

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

G G

G

| 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 .

G

G

G

(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 .

G G

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 ¹

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

G G

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 =

eλ

2πεo r

a=

eλ

= 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

© A¹

© πDh ¹

(

) « ((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)

G

(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.

G

(d) The minus sign indicates that E (r ) points radially inward.

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