EE324
HW#6
Spring12
Problem 4.3 Find the total charge contained in a cone defined by R ≤ 2 m and
0 ≤ θ ≤ π /4, given that ρv = 10R2 cos2 θ (mC/m3 ).
Solution: For the cone of Fig. P4.3, application of Eq. (4.5) gives
Q=
Z 2π Z π /4 Z 2
φ =0 θ =0
R=0
10R2 cos2 θ R2 sin θ dR d θ d φ
¶ ¯¯2 ¯¯π /4 ¯¯2π
−2 5
¯ ¯ ¯
=
R φ cos3 θ ¯ ¯ ¯
¯ ¯ ¯
3
R=0 θ =0 φ =0

à √ !3 
2 
128π 
1−
= 86.65 (mC).
=
3
2
µ
z
2m
π/4
0
y
x
Figure P4.3: Cone of Problem 4.3.
Problem 4.8 An electron beam shaped like a circular cylinder of radius r0 carries a
charge density given by
¶
µ
−ρ0
ρv =
(C/m3 )
1 + r2
where ρ0 is a positive constant and the beam’s axis is coincident with the z-axis.
(a) Determine the total charge contained in length L of the beam.
(b) If the electrons are moving in the +z-direction with uniform speed u, determine
the magnitude and direction of the current crossing the z-plane.
Solution:
(a)
Q=
Z r0 Z L
r=0 z=0
ρv d V =
¶
Z r0 Z L µ
−ρ0
2π r dr dz
1 + r2
r
dr = −πρ0 L ln(1 + r02 ).
= −2πρ0 L
2
1
+
r
0
r=0 z=0
Z r0
(b)
J = ρv u = −ẑ
I=
Z
J · ds
uρ0
1 + r2
Z r0 Z 2π µ
(A/m2 ),
¶
uρ0
=
−ẑ
· ẑr dr d φ
1 + r2
r=0 φ =0
Z r0
r
= −2π uρ0
dr = −π uρ0 ln(1 + r02 ) (A).
2
0 1+r
Current direction is along −ẑ.
Problem 4.14 A line of charge with uniform density ρℓ = 8 (µ C/m) exists in air
along the z-axis between z = 0 and z = 5 cm. Find E at (0,10 cm,0).
Solution: Use of Eq. (4.21c) for the line of charge shown in Fig. P4.14 gives
′
1
′ ρl dl
R̂
,
2
4πε0 l ′
R′
R′ = ŷ 0.1 − ẑz
Z
E=
Z 0.05
(ŷ0.1 − ẑz)
dz
[(0.1)2 + z2 ]3/2
z=0
"
# ¯0.05
¯
ŷ10z + ẑ
8 × 10−6
¯
p
=
¯
4πε0
(0.1)2 + z2 ¯
=
1
4πε0
(8 × 10−6 )
z=0
= 71.86 × 103 [ŷ 4.47 − ẑ 1.06] = ŷ 321.4 × 103 − ẑ 76.2 × 103
z
5 cm
dz
R' = ^y0.1 - ^zz
0
y
10 cm
x
Figure P4.14: Line charge.
(V/m).
Problem 4.20 Three infinite lines of charge, ρl1 = 3 (nC/m), ρl2 = −3 (nC/m), and
ρl3 = 3 (nC/m), are all parallel to the z-axis. If they pass through the respective points
(0, −b), (0, 0), and (0, b) in the x–y plane, find the electric field at (a, 0, 0). Evaluate
your result for a = 2 cm and b = 1 cm.
Solution:
y
(0,b)
ρl3
E1
R3
ρl2
E2
P
x
(a,0)
E3
(0,-b)
ρl1
Figure P4.20: Three parallel line charges.
ρl1 = 3 (nC/m),
ρl2 = −3 (nC/m),
ρl3 = ρl1 ,
E = E1 + E2 + E3 .
Components of line charges 1 and 3 along y cancel and components along x add.
Hence, using Eq. (4.33),
E = x̂
2ρl1
ρl2
cos θ + x̂
.
2πε0 R1
2πε0 a
√
a
and R1 = a2 + b2 ,
with cos θ = √
2
2
a +b
¸
·
1
x̂ 3
2a
−
× 10−9
E=
2πε0 a2 + b2 a
For a = 2 cm and b = 1 cm,
E = x̂ 1.62 (kV/m).
(V/m).
Problem 4.22
Given the electric flux density
D = x̂2(x + y) + ŷ(3x − 2y) (C/m2 )
determine
(a) ρv by applying Eq. (4.26).
(b) The total charge Q enclosed in a cube 2 m on a side, located in the first octant
with three of its sides coincident with the x-, y-, and z-axes and one of its
corners at the origin.
(c) The total charge Q in the cube, obtained by applying Eq. (4.29).
Solution:
(a) By applying Eq. (4.26)
ρv = ∇ · D =
∂
∂
(2x + 2y) + (3x − 2y) = 0.
∂x
∂y
(b) Integrate the charge density over the volume as in Eq. (4.27):
Q=
Z
V
∇ · D dV =
Z 2 Z 2 Z 2
0 dx dy dz = 0.
x=0 y=0 z=0
(c) Apply Gauss’ law to calculate the total charge from Eq. (4.29)
Q=
Z
D · ds = Ffront + Fback + Fright + Fleft + Ftop + Fbottom ,
¯
Z 2 Z 2
¯
¯
(x̂2(x + y) + ŷ(3x − 2y))¯ · (x̂ dz dy)
Ffront =
¯
y=0 z=0
x=2
¯

¯
¶ ¯¯2
µ
Z 2 Z 2
¯
¯2
1
¯
¯
¯
2
2(x + y)¯
=
dz dy = 2z 2y + y ¯  ¯
= 24,
¯
¯
¯
2
y=0 z=0
x=2
z=0
y=0
¯
Z 2 Z 2
¯
¯
Fback =
(x̂2(x + y) + ŷ(3x − 2y))¯ · (−x̂ dz dy)
¯
y=0 z=0
x=0
 ¯ ¯
¯
Z 2 Z 2
¯
¯2
¯2
¯
¯
¯
2
=−
2(x + y)¯
dz dy = − zy ¯  ¯
= −8,
¯
¯
¯
y=0 z=0
x=0
z=0
y=0
¯
Z 2 Z 2
¯
¯
(x̂2(x + y) + ŷ(3x − 2y))¯ · (ŷ dz dx)
Fright =
¯
x=0 z=0
y=2
¯

¯
¶ ¯¯2
µ
Z 2 Z 2
¯2
¯
3 2
¯ ¯
¯

x − 4x ¯
= −4,
(3x − 2y)¯
=
dz dx = z
¯
¯
¯
¯
2
x=0 z=0
n
y=2
z=0
x=0
¯
¯
¯
(x̂2(x + y) + ŷ(3x − 2y))¯ · (−ŷ dz dx)
Fleft =
¯
x=0 z=0
y=0

¯
¯
µ
¶ ¯¯2
Z 2 Z 2
¯
¯2
3
¯
¯
¯
2
dz dx = − z
(3x − 2y)¯
=−
x ¯ ¯
= −12,
¯
¯
¯
2
x=0 z=0
y=0
x=0
z=0
¯
Z 2 Z 2
¯
¯
(x̂2(x + y) + ŷ(3x − 2y))¯ · (ẑ dy dx)
Ftop =
¯
x=0 z=0
z=2
¯
Z 2 Z 2 ¯
¯
0¯ dy dx = 0,
=
x=0 z=0 ¯
z=2
¯
Z 2 Z 2
¯
¯
(x̂2(x + y) + ŷ(3x − 2y))¯ · (ẑ dy dx)
Fbottom =
¯
x=0 z=0
z=0
¯
Z 2 Z 2 ¯
¯
0¯ dy dx = 0.
=
x=0 z=0 ¯
Z 2 Z 2
z=0
Thus Q =
Z
D · ds = 24 − 8 − 4 − 12 + 0 + 0 = 0.
n
Problem 4.24 Charge Q1 is uniformly distributed over a thin spherical shell of
radius a, and charge Q2 is uniformly distributed over a second spherical shell of
radius b, with b > a. Apply Gauss’s law to find E in the regions R < a, a < R < b,
and R > b.
Solution: Using symmetry considerations, we know D = R̂DR . From Table 3.1,
ds = R̂R2 sin θ d θ d φ for an element of a spherical surface. Using Gauss’s law in
integral form (Eq. (4.29)),
Z
D · ds = Qtot ,
n
S
where Qtot is the total charge enclosed in S. For a spherical surface of radius R,
Z 2π Z π
φ =0 θ =0
(R̂DR ) · (R̂R2 sin θ d θ d φ ) = Qtot ,
DR R2 (2π )[− cos θ ]π0 = Qtot ,
Qtot
DR =
.
4π R2
From Eq. (4.15), we know a linear, isotropic material has the constitutive relationship
D = ε E. Thus, we find E from D.
(a) In the region R < a,
Qtot = 0,
E = R̂ER =
R̂Qtot
4π R2 ε
= 0 (V/m).
(b) In the region a < R < b,
Qtot = Q1 ,
E = R̂ER =
R̂Q1
4π R2 ε
(V/m).
(c) In the region R > b,
Qtot = Q1 + Q2 ,
E = R̂ER =
R̂(Q1 + Q2 )
4π R2 ε
(V/m).
Problem 4.25 The electric flux density inside a dielectric sphere of radius a
centered at the origin is given by
D = R̂ρ0 R
(C/m2 )
where ρ0 is a constant. Find the total charge inside the sphere.
Solution:
Q=
Z
D · ds =
n
S
Z π Z 2π
θ =0
= 2πρ0 a
¯
¯
R̂ρ0 R · R̂R sin θ d θ d φ ¯¯
φ =0
3
Z π
0
2
R=a
sin θ d θ = −2πρ0 a cos θ |π0 = 4πρ0 a3
3
(C).
Problem 4.27 An infinitely long cylindrical shell extending between r = 1 m and
r = 3 m contains a uniform charge density ρv0 . Apply Gauss’s law to find D in all
regions.
Solution: For r < 1 m, D = 0.
For 1 ≤ r ≤ 3 m,
Z
r̂ Dr · ds = Q,
n
S
Dr · 2π rL = ρv0 · π L(r2 − 12 ),
D = r̂ Dr = r̂
ρv0 π L(r2 − 1)
ρv0 (r2 − 1)
= r̂
,
2π rL
2r
For r ≥ 3 m,
Dr · 2π rL = ρv0 π L(32 − 12 ) = 8ρv0 π L,
4ρv0
D = r̂ Dr = r̂
,
r ≥ 3 m.
r
1 ≤ r ≤ 3 m.
z
1m
L
3m
r
Figure P4.27: Cylindrical shell.