ECE331 HW6 soln

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Problem 4.1 A cube 2 m on a side is located in the first octant in a Cartesian
coordinate system, with one of its corners at the origin. Find the total charge
contained in the cube if the charge density is given by ρv = xy2 e−2z (mC/m3 ).
Solution: For the cube shown in Fig. P4.1, application of Eq. (4.5) gives
Q=
Z
V
=
µ
ρv d V =
Z 2 Z 2 Z 2
x=0 y=0 z=0
¶ ¯¯2 ¯¯2 ¯¯2
¯ ¯
2 3 −2z ¯
−1
x y e
12
¯
¯
¯
¯
¯
¯
xy2 e−2z dx dy dz
x=0 y=0 z=0
8
= (1 − e−4 ) = 2.62 mC.
3
z
2m
2m
0
2m
x
Figure P4.1: Cube of Problem 4.1.
y
Problem 4.5 Find the total charge on a circular disk defined by r ≤ a and z = 0 if:
(a) ρs = ρs0 cos φ (C/m2 )
(b) ρs = ρs0 sin2 φ (C/m2 )
(c) ρs = ρs0 e−r (C/m2 )
(d) ρs = ρs0 e−r sin2 φ (C/m2 )
where ρs0 is a constant.
Solution:
(a)
Q=
Z
ρs ds =
Z a Z 2π
r=0 φ =0
ρs0 cos φ r dr d φ = ρs0
(b)
Q=
(c)
¯a Z µ
¶
r2 ¯¯ 2π 1 − cos 2φ
dφ
ρs0 sin φ r dr d φ = ρs0 ¯
2 0 0
2
φ =0
µ
¶¯
sin 2φ ¯¯2π π a2
ρs0 a2
=
φ−
¯ = 2 ρs0 .
4
2
0
Z a Z 2π
r=0
¯2π
¯a
¯
r2 ¯¯
sin φ ¯¯ = 0.
¯
2 0
0
Q=
2
Z a Z 2π
r=0 φ =0
ρs0 e r dr d φ = 2πρs0
−r
Z a
0
re−r dr
£
¤a
= 2πρs0 −re−r − e−r 0
= 2πρs0 [1 − e−a (1 + a)].
(d)
Q=
Z a Z 2π
r=0 φ =0
Z a
= ρs0
r=0
ρs0 e−r sin2 φ r dr d φ
re−r dr
Z 2π
φ =0
sin2 φ d φ
= ρs0 [1 − e−a (1 + a)] · π = πρs0 [1 − e−a (1 + a)].
Problem 4.6 If J = ŷ4xz (A/m2 ), find the current I flowing through a square with
corners at (0, 0, 0), (2, 0, 0), (2, 0, 2), and (0, 0, 2).
Solution: Using Eq. (4.12), the net current flowing through the square shown in Fig.
P4.6 is
¯2 ¯2
¯
Z 2 Z 2
Z
¯
¡ 2 2 ¢ ¯¯ ¯¯
¯
(ŷ4xz)¯ · (ŷ dx dz) = x z ¯ ¯ = 16 A.
I = J · ds =
¯ ¯
¯
x=0 z=0
S
x=0 z=0
y=0
z
2m
J
0
2m
x
Figure P4.6: Square surface.
y
Problem 4.10 A line of charge of uniform density ρℓ occupies a semicircle of
radius b as shown in Fig. P4.10. Use the material presented in Example 4-4 to
determine the electric field at the origin.
z
ρl
y
x
b
Figure P4.10: Problem 4.10.
Solution: Since we have only half of a circle, we need to integrate the expression for
dE1 given in Example 4-4 over φ from 0 to π . Before we do that, however, we need
to set h = 0 (the problem asks for E at the origin). Hence,
¯
¯
ρl b (−r̂ b + ẑ h)
¯
dE1 =
φ
d
¯
4πε0 (b2 + h2 )3/2
h=0
−r̂ ρl
=
dφ
4πε0 b
Z π
−r̂ ρl
E1 =
dE1 = −
.
4ε0 b
φ =0
Problem 4.11 A square with sides of 2 m has a charge of 40 µ C at each of its four
corners. Determine the electric field at a point 5 m above the center of the square.
Solution: The distance |R| between any of the charges and point P is
p
√
|R| = 12 + 12 + 52 = 27.
¸
·
R2
R3
R4
R1
Q
+
+
+
E=
4πε0 |R|3 |R|3 |R|3 |R|3
¸
·
Q −x̂ − ŷ + ẑ5 x̂ − ŷ + ẑ5 −x̂ + ŷ + ẑ5 x̂ + ŷ + ẑ5
+
+
+
=
4πε0
(27)3/2
(27)3/2
(27)3/2
(27)3/2
5Q
5 × 40 µ C
1.42
= ẑ
= ẑ
=
× 10−6 (V/m) = ẑ 51.2 (kV/m).
3/2
3/2
πε0
(27) πε0
(27) πε0
z
P(0,0,5)
R2
R3
Q3(-1,-1,0)
R4
R1
Q2(-1,1,0)
y
Q4(1,-1,0)
Q1(1,1,0)
x
Figure P4.11: Square with charges at the corners.
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 4.25 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 4.28 If the charge density increases linearly with distance from the origin
such that ρv = 0 at the origin and ρv = 4 C/m3 at R = 2 m, find the corresponding
variation of D.
Solution:
ρv (R) = a + bR,
ρv (0) = a = 0,
ρv (2) = 2b = 40.
Hence, b = 20.
ρv (R) = 20R (C/m3 ).
Applying Gauss’s law to a spherical surface of radius R,
Z
D · ds =
n
S
DR · 4π R2 =
Z
ρv d V ,
V
Z R
0
DR = 5R2
20R · 4π R2 dR = 80π
(C/m2 ),
D = R̂ DR = R̂ 5R2
(C/m2 ).
R4
,
4
Problem 4.30 A square in the x–y plane in free space has a point charge of +Q at
corner (a/2, a/2), the same at corner (a/2, −a/2), and a point charge of −Q at each
of the other two corners.
(a) Find the electric potential at any point P along the x-axis.
(b) Evaluate V at x = a/2.
Solution: R1 = R2 and R3 = R4 .
y
a/2
-Q
Q
R3
-a/2
R1
P(x,0)
a/2
x
R4
R2
-Q
-a/2
Q
Figure P4.30: Potential due to four point charges.
Q
−Q
−Q
Q
Q
+
+
+
=
V=
4πε0 R1 4πε0 R2 4πε0 R3 4πε0 R4 2πε0
with
r³
a ´2 ³ a ´2
+
,
2
2
r³
a ´2 ³ a ´2
R3 =
+
.
x+
2
2
R1 =
At x = a/2,
R1 =
a
,
2
x−
µ
1
1
−
R1 R3
¶
√
a 5
,
R3 =
2 µ
¶
0.55Q
Q
2
2
=
V=
−√
.
2πε0 a
πε0 a
5a
Problem 4.33 Show that the electric potential difference V12 between two points in
air at radial distances r1 and r2 from an infinite line of charge with density ρℓ along
the z-axis is V12 = (ρℓ /2πε0 ) ln(r2 /r1 ).
Solution: From Eq. (4.33), the electric field due to an infinite line of charge is
E = r̂Er = r̂
ρl
.
2πε0 r
Hence, the potential difference is
V12 = −
Z r1
r2
E · dl = −
Z r1
r̂ρl
r2
µ ¶
ρl
r2
.
· r̂ dr =
ln
2πε0 r
2πε0
r1
Problem 4.35 For the electric dipole shown in Fig. 4-13, d = 1 cm and |E| = 4
(mV/m) at R = 1 m and θ = 0◦ . Find E at R = 2 m and θ = 90◦ .
Solution: For R = 1 m and θ = 0◦ , |E| = 4 mV/m, we can solve for q using Eq. (4.56):
E=
qd
(R̂2 cos θ + θ̂θ sin θ ).
4πε0 R3
Hence,
¶
qd
2 = 4 mV/m at θ = 0◦ ,
|E| =
4πε0
10−3 × 8πε0 10−3 × 8πε0
q=
=
= 0.8πε0
d
10−2
µ
(C).
Again using Eq. (4.56) to find E at R = 2 m and θ = 90◦ , we have
E=
1
0.8πε0 × 10−2
(R̂(0) + θ̂θ) = θ̂θ
3
4πε0 × 2
4
(mV/m).
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