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CHAPTER 3
3.1. An empty metal paint can is placed on a marble table, the lid is removed, and both parts are discharged (honorably) by touching them to ground. An insulating nylon thread is glued to the center of the lid, and a penny, a nickel, and a dime are glued to the thread so that they are not touching each other. The penny is given a charge of +5 nC, and the nickel and dime are discharged. The assembly is lowered into the can so that the coins hang clear of all walls, and the lid is secured. The outside of the can is again touched momentarily to ground. The device is carefully disassembled with insulating gloves and tools.
a) What charges are found on each of the five metallic pieces? All coins were insulated during the entire procedure, so they will retain their original charges: Penny: +5 nC; nickel: 0; dime: 0. The penny’s charge will have induced an equal and opposite negative charge (-5 nC) on the inside wall of the can and lid. This left a charge layer of +5 nC on the outside surface which was neutralized by the ground connection. Therefore, the can retained a net charge of −5 nC after disassembly.
b) If the penny had been given a charge of +5 nC, the dime a charge of −2 nC, and the nickel a charge of −1 nC, what would the final charge arrangement have been? Again, since the coins are insulated, they retain their original charges. The charge induced on the inside wall of the can and lid is equal to negative the sum of the coin charges, or −2 nC. This is the charge that the can/lid contraption retains after grounding and disassembly.
3.2. A point charge of 20 nC is located at (4,-1,3), and a uniform line charge of -25 nC/m is lies along the intersection of the planes x = −4 and z = 6. a) Calculate D at (3,-1,0): The total flux density at the desired point is
D(3, − 1 , 0) =
20 × 10 −^9
4 π(1 + 9)
[
−ax − 3 az √ 1 + 9
]
point charge
25 × 10 −^9
2 π
[
7 ax − 6 az √ 49 + 36
]
line charge = − 0. 38 ax + 0. 13 az nC/m^2
b) How much electric flux leaves the surface of a sphere of radius 5, centered at the origin? This will be equivalent to how much charge lies within the sphere. First the point charge is at distance from the origin given by Rp =
16 + 1 + 9 = 5.1, and so it is outside. Second, the nearest point on the line charge to the origin is at distance R =
16 + 36 = 7.2, and so the entire line charge is also outside the sphere. Answer: zero. c) Repeat part b if the radius of the sphere is 10. First, from part b, the point charge will now lie inside. Second, the length of line charge that lies inside the sphere will be given by 2√ y 0 , where y 0 satisfies the equation, 16 + y^20 + 36 = 10. Solve to find y 0 = 6.93, or 2y 0 = 13.86. The total charge within the sphere (and the net outward flux) is now
Φ = Qencl = [20 − (25 × 13 .86)] = −326 nC
.
Sol 1.1 DK
3.3. The cylindrical surface ρ = 8 cm contains the surface charge density, ρs = 5e−^20 |z|^ nC/m^2. a) What is the total amount of charge present? We integrate over the surface to find:
Q = 2
0
∫ (^2) π
0
5 e−^20 z^ (.08)dφ dz nC = 20π(.08)
e−^20 z
∞
0
= 0.25 nC
b) How much flux leaves the surface ρ = 8 cm, 1 cm < z < 5cm, 30◦^ < φ < 90 ◦? We just integrate the charge density on that surface to find the flux that leaves it.
Φ = Q′^ =
. 01
30 ◦
5 e−^20 z^ (.08) dφ dz nC =
2 π(5)(.08)
e−^20 z
. 05 . 01 = 9. 45 × 10 −^3 nC = 9.45 pC
3.4. In cylindrical coordinates, let D = (ρaρ + zaz )/
[
4 π(ρ^2 + z^2 )^1.^5
]
. Determine the total flux leaving: a) the infinitely-long cylindrical surface ρ = 7: We use
Φa =
D · dS =
−∞
∫ (^2) π
0
ρ 0 aρ + z az 4 π(ρ^20 + z^2 )^3 /^2
· aρ ρ 0 dφ dz = ρ^20
0
dz (ρ^20 + z^2 )^3 /^2
=
z √ ρ^20 + z^2
∞ 0
where ρ 0 = 7 (immaterial in this case). b) the finite cylinder, ρ = 7, |z| ≤ 10: The total flux through the cylindrical surface and the two end caps are, in this order:
Φb =
∫ (^) z 0
−z 0
∫ (^2) π
0
ρ 0 aρ · aρ 4 π(ρ^20 + z^2 )^3 /^2
ρ 0 dφ dz
∫ (^2) π
0
∫ (^) ρ 0
0
z 0 az · az 4 π(ρ^2 + z^20 )^3 /^2
ρ dρ dφ +
∫ (^2) π
0
∫ (^) ρ 0
0
−z 0 az · −az 4 π(ρ^2 + z 02 )^3 /^2
ρ dρ dφ
where ρ 0 = 7 and z 0 = 10. Simplifying, this becomes
Φb = ρ^20
∫ (^) z 0
0
dz (ρ^20 + z^2 )^3 /^2
∫ (^) ρ 0
0
ρ dρ (ρ^2 + z^20 )^3 /^2
=
z √ ρ^20 + z^2
z 0 0
z 0 √ ρ^2 + z^20
ρ 0 0
z 0 √ ρ^20 + z 02
z 0 √ ρ^20 + z 02
where again, the actual values of ρ 0 and z 0 (7 and 10) did not matter.
3.5. Let D = 4xyax + 2(x^2 + z^2 )ay + 4yzaz C/m^2 and evaluate surface integrals to find the total charge enclosed in the rectangular parallelepiped 0 < x < 2, 0 < y < 3, 0 < z < 5 m: Of the 6 surfaces to consider, only 2 will contribute to the net outward flux. Why? First consider the planes at y = 0 and 3. The y component of D will penetrate those surfaces, but will be inward at y = 0 and outward at y = 3, while having the same magnitude in both cases. These fluxes
b) By using Gauss’s law, calculate the value of Dr on the surface r = 1 mm: The gaussian surface is a spherical shell of radius 1 mm. The enclosed charge is the result of part a. We thus write 4πr^2 Dr = Q, or
Dr =
Q
4 πr^2
4. 0 × 10 −^9
4 π(.001)^2
= 3. 2 × 10 −^4 nC/m^2
3.8. Use Gauss’s law in integral form to show that an inverse distance field in spherical coordinates, D = Aar /r, where A is a constant, requires every spherical shell of 1 m thickness to contain 4 πA coulombs of charge. Does this indicate a continuous charge distribution? If so, find the charge density variation with r. The net outward flux of this field through a spherical surface of radius r is
D · dS =
∫ (^2) π
0
∫ (^) π
0
A
r
ar · ar r^2 sin θ dθ dφ = 4πAr = Qencl
We see from this that with every increase in r by one m, the enclosed charge increases by 4πA (done). It is evident that the charge density is continuous, and we can find the density indirectly by constructing the integral for the enclosed charge, in which we already found the latter from Gauss’s law:
Qencl = 4πAr =
∫ (^2) π
0
∫ (^) π
0
∫ (^) r
0
ρ(r′) (r′)^2 sin θ dr′^ dθ dφ = 4π
∫ (^) r
0
ρ(r′) (r′)^2 dr′
To obtain the correct enclosed charge, the integrand must be ρ(r) = A/r^2.
3.9. A uniform volume charge density of 80 μC/m^3 is present throughout the region 8 mm < r < 10 mm. Let ρv = 0 for 0 < r < 8 mm. a) Find the total charge inside the spherical surface r = 10 mm: This will be
Q =
∫ (^2) π
0
∫ (^) π
0
. 008
(80 × 10 −^6 )r^2 sin θ dr dθ dφ = 4π × (80 × 10 −^6 )
r^3 3
∣∣.^010
. 008 = 1. 64 × 10 −^10 C = 164 pC
b) Find Dr at r = 10 mm: Using a spherical gaussian surface at r = 10, Gauss’ law is written as 4πr^2 Dr = Q = 164 × 10 −^12 , or
Dr (10 mm) =
164 × 10 −^12
4 π(.01)^2
= 1. 30 × 10 −^7 C/m^2 = 130 nC/m^2
c) If there is no charge for r > 10 mm, find Dr at r = 20 mm: This will be the same computation as in part b, except the gaussian surface now lies at 20 mm. Thus
Dr (20 mm) =
164 × 10 −^12
4 π(.02)^2
= 3. 25 × 10 −^8 C/m^2 = 32.5 nC/m^2
3.10. Volume charge density varies in spherical coordinates as ρv = (ρ 0 sin πr)/r^2 , where ρ 0 is a constant. Find the surfaces on which D = 0.
3.11. In cylindrical coordinates, let ρv = 0 for ρ < 1 mm, ρv = 2 sin(2000πρ) nC/m^3 for 1 mm < ρ < 1 .5 mm, and ρv = 0 for ρ > 1 .5 mm. Find D everywhere: Since the charge varies only with radius, and is in the form of a cylinder, symmetry tells us that the flux density will be radially-directed and will be constant over a cylindrical surface of a fixed radius. Gauss’ law applied to such a surface of unit length in z gives: a) for ρ < 1 mm, Dρ = 0, since no charge is enclosed by a cylindrical surface whose radius lies within this range.
b) for 1 mm < ρ < 1 .5 mm, we have
2 πρDρ = 2π
∫ (^) ρ
. 001
2 × 10 −^9 sin(2000πρ′)ρ′^ dρ′
= 4π × 10 −^9
[
(2000π)^2
sin(2000πρ) −
ρ 2000 π
cos(2000πρ)
]ρ
. 001
or finally,
Dρ =
10 −^15
2 π^2 ρ
[
sin(2000πρ) + 2π
[
1 − 103 ρ cos(2000πρ)
] ]
C/m^2 (1 mm < ρ < 1 .5 mm)
a) Calculate the total charge in the region 0 < ρ < ρ 1 , 0 < z < L, where 1 < ρ 1 < 2 mm: We find Q =
∫ L
0
∫ (^2) π
0
∫ (^) ρ 1
. 001
4 ρ ρ dρ dφ dz =
8 πL 3
[ρ^31 − 10 −^9 ] μC
where ρ 1 is in meters.
b) Use Gauss’ law to determine Dρ at ρ = ρ 1 : Gauss’ law states that 2πρ 1 LDρ = Q, where Q is the result of part a. Thus
Dρ(ρ 1 ) =
4(ρ^31 − 10 −^9 ) 3 ρ 1
μC/m^2
where ρ 1 is in meters.
c) Evaluate Dρ at ρ = 0.8 mm, 1 .6 mm, and 2.4 mm: At ρ = 0.8 mm, no charge is enclosed by a cylindrical gaussian surface of that radius, so Dρ(0.8mm) = 0. At ρ = 1.6 mm, we evaluate the part b result at ρ 1 = 1.6 to obtain:
Dρ(1.6mm) =
4[(.0016)^3 − (.0010)^3 ]
= 3. 6 × 10 −^6 μC/m^2
At ρ = 2.4, we evaluate the charge integral of part a from .001 to .002, and Gauss’ law is written as 2 πρLDρ =
8 πL 3
[(.002)^2 − (.001)^2 ] μC
from which Dρ(2.4mm) = 3. 9 × 10 −^6 μC/m^2.
3.16. In spherical coordinates, a volume charge density ρv = 10e−^2 r^ C/m^3 is present. (a) Determine D. (b) Check your result of part a by evaluating ∇ · D.
3.17. A cube is defined by 1 < x, y, z < 1 .2. If D = 2x^2 yax + 3x^2 y^2 ay C/m^2 :
a) apply Gauss’ law to find the total flux leaving the closed surface of the cube. We call the surfaces at x = 1.2 and x = 1 the front and back surfaces respectively, those at y = 1. 2 and y = 1 the right and left surfaces, and those at z = 1.2 and z = 1 the top and bottom surfaces. To evaluate the total charge, we integrate D · n over all six surfaces and sum the results. We note that there is no z component of D, so there will be no outward flux contributions from the top and bottom surfaces. The fluxes through the remaining four are
Φ = Q =
D · n da =
1
1
2(1.2)^2 y dy dz ︸ ︷︷ ︸ front
1
1
−2(1)^2 y dy dz ︸ ︷︷ ︸ back
1
1
− 3 x^2 (1)^2 dx dz ︸ ︷︷ ︸ left
1
1
3 x^2 (1.2)^2 dx dz ︸ ︷︷ ︸ right
= 0.1028 C
b) evaluate ∇ · D at the center of the cube: This is
∇ · D =
[
4 xy + 6x^2 y
]
c) Estimate the total charge enclosed within the cube by using Eq. (8): This is
Q
= ∇ · D
center ×^ ∆v^ = 12.^83 ×^ (0.2)
(^3) = 0. 1026 Close!
3.18. State whether the divergence of the following vector fields is positive, negative, or zero: (a) the thermal energy flow in J/(m^2 − s) at any point in a freezing ice cube; (b) the current density in A/m^2 in a bus bar carrying direct current; (c) the mass flow rate in kg/(m^2 − s) below the surface of water in a basin, in which the water is circulating clockwise as viewed from above.
3.19. A spherical surface of radius 3 mm is centered at P (4, 1 , 5) in free space. Let D = xax C/m^2. Use the results of Sec. 3.4 to estimate the net electric flux leaving the spherical surface: We use Φ
= ∇ · D∆v, where in this case ∇ · D = (∂/∂x)x = 1 C/m^3. Thus
π(.003)^3 (1) = 1. 13 × 10 −^7 C = 113 nC
3.20. Suppose that an electric flux density in cylindrical coordinates is of the form D = Dρ aρ. Describe the dependence of the charge density ρv on coordinates ρ, φ, and z if (a) Dρ = f (φ, z); (b) Dρ = (1/ρ)f (φ, z); (c) Dρ = f (ρ).
3.21. Calculate the divergence of D at the point specified if a) D = (1/z^2 )
[
10 xyz ax + 5x^2 z ay + (2z^3 − 5 x^2 y) az
]
at P (− 2 , 3 , 5): We find
∇ · D =
[
10 y z
10 x^2 y z^3
]
(− 2 , 3 ,5)
b) D = 5z^2 aρ + 10ρz az at P (3, − 45 ◦, 5): In cylindrical coordinates, we have
∇ · D =
ρ
∂ρ
(ρDρ) +
ρ
∂Dφ ∂φ
∂Dz ∂z
[
5 z^2 ρ
]
(3,− 45 ◦,5)
c) D = 2r sin θ sin φ ar + r cos θ sin φ aθ + r cos φ aφ at P (3, 45 ◦, − 45 ◦): In spherical coordi- nates, we have
∇ · D =
r^2
∂r
(r^2 Dr ) +
r sin θ
∂θ
(sin θDθ ) +
r sin θ
∂Dφ ∂φ
=
[
6 sin θ sin φ +
cos 2θ sin φ sin θ
sin φ sin θ
]
(3, 45 ◦,− 45 ◦)
3.22. (a) A flux density field is given as F 1 = 5az. Evaluate the outward flux of F 1 through the hemispherical surface, r = a, 0 < θ < π/2, 0 < φ < 2 π. (b) What simple observation would have saved a lot of work in part a? (c) Now suppose the field is given by F 2 = 5zaz. Using the appropriate surface integrals, evaluate the net outward flux of F 2 through the closed surface consisting of the hemisphere of part a and its circular base in the xy plane. (d) Repeat part c by using the divergence theorem and an appropriate volume integral.
leads to the continuity equation, ∇ · (ρmU) = −∂ρm/∂t. (a) Explain in words the physical interpretation of this equation. (b) Show that
s ρmU^ ·^ dS^ =^ −dM/dt, where^ M^ is the total mass of the gas within the constant closed surface, S, and explain the physical significance of the equation.
3.27. Let D = 5. 00 r^2 ar mC/m^2 for r ≤ 0 .08 m and D = 0. 205 ar /r^2 μC/m^2 for r ≥ 0 .08 m (note error in problem statement). a) Find ρv for r = 0.06 m: This radius lies within the first region, and so
ρv = ∇ · D =
r^2
d dr
(r^2 Dr ) =
r^2
d dr
(5. 00 r^4 ) = 20r mC/m^3
which when evaluated at r = 0.06 yields ρv (r = .06) = 1.20 mC/m^3.
b) Find ρv for r = 0.1 m: This is in the region where the second field expression is valid. The 1/r^2 dependence of this field yields a zero divergence (shown in Problem 3.23), and so the volume charge density is zero at 0.1 m.
c) What surface charge density could be located at r = 0.08 m to cause D = 0 for r > 0. 08 m? The total surface charge should be equal and opposite to the total volume charge. The latter is
Q =
∫ (^2) π
0
∫ (^) π
0
0
20 r(mC/m^3 ) r^2 sin θ dr dθ dφ = 2. 57 × 10 −^3 mC = 2. 57 μC
So now ρs = −
[
4 π(.08)^2
]
= − 32 μC/m^2
3.28. Repeat Problem 3.8, but use ∇ · D = ρv and take an appropriate volume integral.
3.29. In the region of free space that includes the volume 2 < x, y, z < 3,
D =
z^2
(yz ax + xz ay − 2 xy az ) C/m^2
a) Evaluate the volume integral side of the divergence theorem for the volume defined above: In cartesian, we find ∇ · D = 8xy/z^3. The volume integral side is now
∫
vol
∇ · D dv =
2
2
2
8 xy z^3
dxdydz = (9 − 4)(9 − 4)
= 3.47 C
b. Evaluate the surface integral side for the corresponding closed surface: We call the surfaces at x = 3 and x = 2 the front and back surfaces respectively, those at y = 3 and y = 2 the right and left surfaces, and those at z = 3 and z = 2 the top and bottom surfaces. To evaluate the surface integral side, we integrate D · n over all six surfaces and sum the results. Note that since the x component of D does not vary with x, the outward fluxes from the front and back surfaces will cancel each other. The same is true for the left
and right surfaces, since Dy does not vary with y. This leaves only the top and bottom surfaces, where the fluxes are: ∮ D · dS =
2
2
− 4 xy 32
dxdy ︸ ︷︷ ︸ top
2
2
− 4 xy 22
dxdy ︸ ︷︷ ︸ bottom
= 3.47 C
3.30. Let D = 20ρ^2 aρ C/m^2. (a) What is the volume charge density at the point P (0. 5 , 60 ◦, 2)? (b) Use two different methods to find the amount of charge lying within the closed surface bounded by ρ = 3, 0 ≤ z ≤ 2.
3.31. Given the flux density
D =
r
cos(2θ) aθ C/m^2 ,
use two different methods to find the total charge within the region 1 < r < 2 m, 1 < θ < 2 rad, 1 < φ < 2 rad: We use the divergence theorem and first evaluate the surface integral side. We are evaluating the net outward flux through a curvilinear “cube”, whose boundaries are defined by the specified ranges. The flux contributions will be only through the surfaces of constant θ, however, since D has only a θ component. On a constant-theta surface, the differential area is da = r sin θdrdφ, where θ is fixed at the surface location. Our flux integral becomes ∮ D · dS = −
1
1
r
cos(2) r sin(1) drdφ ︸ ︷︷ ︸ θ=
1
1
r
cos(4) r sin(2) drdφ ︸ ︷︷ ︸ θ= = −16 [cos(2) sin(1) − cos(4) sin(2)] = − 3 .91 C
We next evaluate the volume integral side of the divergence theorem, where in this case,
∇ · D =
r sin θ
d dθ
(sin θ Dθ ) =
r sin θ
d dθ
[
r
cos 2θ sin θ
]
r^2
[
cos 2θ cos θ sin θ
− 2 sin 2θ
]
We now evaluate: ∫
vol
∇ · D dv =
1
1
1
r^2
[
cos 2θ cos θ sin θ
− 2 sin 2θ
]
r^2 sin θ drdθdφ
The integral simplifies to ∫ (^2)
1
1
1
16[cos 2θ cos θ − 2 sin 2θ sin θ] drdθdφ = 8
1
[3 cos 3θ − cos θ] dθ = − 3 .91 C
