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Problem 4.48 With reference to Fig. 419, findE1 if E2 = x̂3− ŷ2+ ẑ2 (V/m), ε1 = 2ε0, ε2 = 18ε0, and the boundary has a surface charge density ρs = 3.54×10−11 (C/m2). What angle doesE2 make with thezaxis? Solution: We know thatE1t = E2t for any 2 media. Hence,E1t = E2t = x̂3− ŷ2. Also, (D1−D2) · n̂ = ρs (from Table 4.3). Hence,ε1(E1 · n̂)− ε2(E2 · n̂) = ρs, which gives
E1z = ρs+ ε2E2z
ε1 =
3.54×10−11 2ε0
+ 18(2)
2 =
3.54×10−11 2×8.85×10−12 +18= 20 (V/m).
Hence,E1 = x̂3− ŷ2+ ẑ20 (V/m). Finding the angleE2 makes with thezaxis:
E2 · ẑ = E2cosθ , 2 = √
9+4+4cosθ , θ = cos−1 (
2√ 17
)
= 61◦.
Problem 4.56 Figure P4.56(a) depicts a capacitor consisting of two parallel, conducting plates separated by a distanced. The space between the plates contains two adjacent dielectrics, one with permittivityε1 and surface areaA1 and another with ε2 andA2. The objective of this problem is to show that the capacitanceC of the configuration shown in Fig. P4.56(a) is equivalent to two capacitances in parallel, as illustrated in Fig. P4.56(b), with
C = C1 +C2 (19)
where
C1 = ε1A1
d (20)
C2 = ε2A2
d (21)
To this end, proceed as follows:
(a) Find the electric fieldsE1 andE2 in the two dielectric layers.
(b) Calculate the energy stored in each section and use the result to calculateC1 andC2.
(c) Use the total energy stored in the capacitor to obtain an expression forC. Show that (19) is indeed a valid result.
(a)
(b)
ε1
A1 A2
ε2d
+
− V
C1 C2 V
+
−
Figure P4.56: (a) Capacitor with parallel dielectric section, and (b) equivalent circuit.
Solution:
V +

(c)
E1 E2
ε2
d
ε1
Figure P4.56: (c) Electric field inside of capacitor.
(a) Within each dielectric section,E will point from the plate with positive voltage to the plate with negative voltage, as shown in Fig. P456(c). FromV = Ed,
E1 = E2 = V d
.
(b)
We1 = 1 2
ε1E21 · V = 1 2
ε1 V2
d2 ·A1d =
1 2
ε1V2 A1 d
.
But, from Eq. (4.121),
We1 = 1 2
C1V 2.
HenceC1 = ε1 A1 d
. Similarly,C2 = ε2 A2 d
.
(c) Total energy is
We = We1 +We2 = 1 2
V2
d (ε1A1 + ε2A2) =
1 2
CV2.
Hence,
C = ε1A1
d +
ε2A2 d
= C1 +C2.
Problem 5.12 Two infinitely long, parallel wires are carrying 6A currents in opposite directions. Determine the magnetic flux density at pointP in Fig. P5.12.
I2 = 6 AI1 = 6 A
0.5 m
2 m
P
Figure P5.12: Arrangement for Problem 5.12.
Solution:
B = φ̂φφ µ0I1
2π(0.5) +φ̂φφ
µ0I2 2π(1.5)
= φ̂φφ µ0 π
(6+2) = φ̂φφ 8µ0 π
(T).
Problem 5.14 Two parallel, circular loops carrying a current of 40 A each are arranged as shown in Fig. P5.14. The first loop is situated in thex–y plane with its center at the origin, and the second loop’s center is atz = 2 m. If the two loops have the same radiusa = 3 m, determine the magnetic field at:
(a) z = 0
(b) z = 1 m
(c) z = 2 m
z = 2 m
0
z
y
x
a
a
I
I
Figure P5.14: Parallel circular loops of Problem 5.14.
Solution: The magnetic field due to a circular loop is given by (5.34) for a loop in the x–y plane carrying a currentI in the+φ̂φφdirection. Considering that the bottom loop in Fig. is in thex–y plane, but the current direction is along−φ̂φφ,
H1 = −ẑ Ia2
2(a2 + z2)3/2 ,
wherez is the observation point along thezaxis. For the second loop, which is at a height of 2 m, we can use the same expression butz should be replaced with(z−2). Hence,
H2 = −ẑ Ia2
2[a2 +(z−2)2]3/2 .
The total field is
H = H1 +H2 = −ẑ Ia2
2
[
1
(a2 + z2)3/2 +
1
[a2 +(z−2)2]3/2 ]
A/m.
(a) At z = 0, and witha = 3 m andI = 40 A,
H = −ẑ 40×9 2
[
1 33
+ 1
(9+4)3/2
]
= −ẑ10.5 A/m.
(b) At z = 1 m (midway between the loops):
H = −ẑ 40×9 2
[
1
(9+1)3/2 +
1
(9+1)3/2
]
= −ẑ11.38 A/m.
(c) At z = 2 m,H should be the same as atz = 0. Thus,
H = −ẑ10.5 A/m.
Problem 4.49 An infinitely long conducting cylinder of radiusa has a surface charge densityρs. The cylinder is surrounded by a dielectric medium withεr = 4 and contains no free charges. The tangential component of the electric field in the regionr ≥ a is given byEt = −φ̂φφcosφ/r2. Since a static conductor cannot have any tangential field, this must be cancelled by an externally applied electric field. Find the surface charge density on the conductor.
Solution: Let the conducting cylinder be medium 1 and the surrounding dielectric medium be medium 2. In medium 2,
E2 = r̂Er − φ̂φφ 1 r2
cosφ ,
with Er, the normal component ofE2, unknown. The surface charge density is related to Er. To findEr, we invoke Gauss’s law in medium 2:
∇ ·D2 = 0,
or 1 r
∂ ∂ r
(rEr)+ 1 r
∂ ∂φ
(
− 1 r2
cosφ )
= 0,
which leads to ∂ ∂ r
(rEr) = ∂
∂φ
(
1 r2
cosφ )
= − 1 r2
sinφ .
Integrating both sides with respect tor,
∫ ∂ ∂ r
(rEr) dr = −sinφ ∫
1 r2
dr
rEr = 1 r
sinφ ,
or
Er = 1 r2
sinφ .
Hence,
E2 = r̂ 1 r2
sinφ .
According to Eq. (4.93), n̂2 · (D1−D2) = ρs,
wheren̂2 is the normal to the boundary and points away from medium 1. Hence, n̂2 = r̂ . Also,D1 = 0 because the cylinder is a conductor. Consequently,
ρs = −r̂ ·D2r=a = −r̂ · ε2E2r=a
= −r̂ · εrε0 [
r̂ 1 r2
sinφ ]∣
∣
∣
∣
r=a
= −4ε0 a2
sinφ (C/m2).
Problem 4.50 If E = R̂150 (V/m) at the surface of a 5cm conducting sphere centered at the origin, what is the total chargeQ on the sphere’s surface?
Solution: From Table 43,̂n · (D1−D2) = ρs. E2 inside the sphere is zero, since we assume it is a perfect conductor. Hence, for a sphere with surface areaS= 4πa2,
D1R = ρs, E1R = ρs ε0
= Q
Sε0 ,
Q = ERSε0 = (150)4π(0.05)2ε0 = 3πε0
2 (C).
Problem 4.58 The capacitor shown in Fig. P4.58 consists of two parallel dielectric layers. Use energy considerations to show that the equivalent capacitance of the overall capacitor,C, is equal to the series combination of the capacitances of the individual layers,C1 andC2, namely
C = C1C2
C1 +C2 (22)
where
C1 = ε1 A d1
, C2 = ε2 A d2
(a) Let V1 andV2 be the electric potentials across the upper and lower dielectrics, respectively. What are the corresponding electric fieldsE1 and E2? By applying the appropriate boundary condition at the interface between the two dielectrics, obtain explicit expressions forE1 andE2 in terms ofε1, ε2, V, and the indicated dimensions of the capacitor.
(b) Calculate the energy stored in each of the dielectric layers and then use the sum to obtain an expression forC.
(c) Show thatC is given by Eq. (22).
(a)
(b)
V +
−
C1
C2
+
−
d1
d2 V
A
ε1
ε2
Figure P4.58: (a) Capacitor with parallel dielectric layers, and (b) equivalent circuit (Problem 4.58).
Solution:
+
 V E1
E2
ε1
ε1
V1 +

V1 +

d1
d2
Figure P4.58: (c) Electric fields inside of capacitor.
(a) If V1 is the voltage across the top layer andV2 across the bottom layer, then
V = V1 +V2,
and
E1 = V1 d1
, E2 = V2 d2
.
According to boundary conditions, the normal component ofD is continuous across the boundary (in the absence of surface charge). This means that at the interface between the two dielectric layers,
D1n = D2n
or ε1E1 = ε2E2.
Hence,
V = E1d1 +E2d2 = E1d1 + ε1E1
ε2 d2,
which can be solved forE1:
E1 = V
d1 + ε1 ε2
d2 .
Similarly,
E2 = V
d2 + ε2 ε1
d1 .
(b)
We1 = 1 2
ε1E21 · V 1 = 1 2
ε1
V
d1 + ε1 ε2
d2
2
·Ad1 = 1 2
V2 [
ε1ε22Ad1 (ε2d1 + ε1d2)2
]
,
We2 = 1 2
ε2E22 · V 2 = 1 2
ε2
V
d2 + ε2 ε1
d1
2
·Ad2 = 1 2
V2 [
ε21ε2Ad2 (ε1d2 + ε2d1)2
]
,
We = We1 +We2 = 1 2
V2 [
ε1ε22Ad1 + ε21ε2Ad2 (ε1d2 + ε2d1)2
]
.
But