# Eletromagnetismo - Hayt - 7ª Ed - Soluções - chapter06 7th solution, Manuais, Projetos, Pesquisas de Eletrônica

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Chapter6 7th

CHAPTER 6.

6.1. Atomic hydrogen contains 5.5× 1025 atoms/m3 at a certain temperature and pressure. When an electric field of 4 kV/m is applied, each dipole formed by the electron and positive nucleus has an effective length of 7.1 × 1019 m. a) Find P: With all identical dipoles, we have

P = Nqd = (5.5× 1025)(1.602× 1019)(7.1× 1019) = 6.26× 1012 C/m2 = 6.26 pC/m2

b) Find r: We use P = 0χeE, and so

χe = P

0E =

6.26 × 1012 (8.85 × 1012)(4 × 103) = 1.76 × 10

4

Then r = 1 + χe = 1.000176.

6.2. Find the dielectric constant of a material in which the electric flux density is four times the polarization.

First we use D = 0E + P = 0E + (1/4)D. Therefore D = (4/3)0E, so we identify r = 4/3.

6.3. A coaxial conductor has radii a = 0.8 mm and b = 3 mm and a polystyrene dielectric for which r = 2.56. If P = (2)aρ nC/m2 in the dielectric, find: a) D and E as functions of ρ: Use

E = P

0(r − 1) =

(2) × 109aρ (8.85 × 1012)(1.56) =

144.9 ρ

aρ V/m

Then

D = 0E + P = 2 × 109aρ

ρ

[ 1

1.56 + 1

] =

3.28 × 109aρ ρ

C/m2 = 3.28aρ

ρ nC/m2

b) Find Vab and χe: Use

Vab = ∫ 0.8

3

144.9 ρ

= 144.9 ln (

3 0.8

) = 192 V

χe = r − 1 = 1.56, as found in part a. c) If there are 4 × 1019 molecules per cubic meter in the dielectric, find p(ρ): Use

p = P N

= (2 × 109)

4 × 1019 aρ = 5.0 × 1029

ρ aρ C · m

1

6.4. Consider a composite material made up of two species, having number densities N1 and N2 molecules/m3 respectively. The two materials are uniformly mixed, yielding a total number density of N = N1 + N2. The presence of an electric field E, induces molecular dipole moments p1 and p2 within the individual species, whether mixed or not. Show that the dielectric constant of the composite material is given by r = fr1 + (1− f)r2, where f is the number fraction of species 1 dipoles in the composite, and where r1 and r2 are the dielectric constants that the unmixed species would have if each had number density N .

We may write the total polarization vector as

Ptot = N1p1 + N2p2 = N ( N1 N

p1 + N2 N

p2

) = N [fp1 + (1 − f)p2] = fP1 + (1 − f)P2

In terms of the susceptibilities, this becomes Ptot = 0 [fχe1 + (1 − f)χe2]E, where χe1 and χe2 are evaluated at the composite number density, N . Now

D = r0E = 0E + Ptot = 0 [1 + fχe1 + (1 − f)χe2]︸ ︷︷ ︸ r

E

Identifying r as shown, we may rewrite it by adding and subracting f :

r = [1 + f − f + fχe1 + (1 − f)χe2] = [f(1 + χe1) + (1 − f)(1 + χe2)] = [fr1 + (1 − f)r2] Q.E.D.

6.5. The surface x = 0 separates two perfect dielectrics. For x > 0, let r = r1 = 3, while r2 = 5 where x < 0. If E1 = 80ax − 60ay − 30az V/m, find: a) EN1: This will be E1 · ax = 80 V/m. b) ET1. This has components of E1 not normal to the surface, or ET1 = 60ay − 30az V/m.

c) ET1 = √

(60)2 + (30)2 = 67.1 V/m.

d) E1 = √

(80)2 + (60)2 + (30)2 = 104.4 V/m.

e) The angle θ1 between E1 and a normal to the surface: Use

cos θ1 = E1 · ax E1

= 80

104.4 ⇒ θ1 = 40.0

f) DN2 = DN1 = r10EN1 = 3(8.85 × 1012)(80) = 2.12 nC/m2.

g) DT2 = r20ET1 = 5(8.85 × 1012)(67.1) = 2.97 nC/m2.

h) D2 = r10EN1ax + r20ET1 = 2.12ax − 2.66ay − 1.33az nC/m2.

i) P2 = D2 − 0E2 = D2 [1 (1/r2)] = (4/5)D2 = 1.70ax − 2.13ay − 1.06az nC/m2.

j) the angle θ2 between E2 and a normal to the surface: Use

cos θ2 = E2 · ax E2

= D2 · ax D2

= 2.12√

(2.12)2 = (2.66)2 + (1.33)2 = .581

Thus θ2 = cos1(.581) = 54.5.

2

6.6. The potential field in a slab of dielectric material for which r = 1.6 is given by V = 5000x. a) Find D, E, and P in the material.

First, E = −∇V = 5000ax V/m. Then D = r0E = 1.60(5000)ax = 70.8ax nC/m2. Then, χe = r − 1 = 0.6, and so P = 0χeE = 0.60(5000)ax = 26.6ax nC/m2.

b) Evaluate ρv, ρb, and ρt in the material. Using the results in part a, we find ρv = ∇·D = 0, ρb = −∇ · P = 0, and ρt = ∇ · 0E = 0.

6.7. Two perfect dielectrics have relative permittivities r1 = 2 and r2 = 8. The planar interface between them is the surface x−y+2z = 5. The origin lies in region 1. If E1 = 100ax+200ay− 50az V/m, find E2: We need to find the components of E1 that are normal and tangent to the boundary, and then apply the appropriate boundary conditions. The normal component will be EN1 = E1 · n. Taking f = x− y + 2z, the unit vector that is normal to the surface is

n = ∇f |∇f | =

16

[ax − ay + 2az]

This normal will point in the direction of increasing f , which will be away from the origin, or into region 2 (you can visualize a portion of the surface as a triangle whose vertices are on the three coordinate axes at x = 5, y = 5, and z = 2.5). So EN1 = (1/

6)[100 200 100] =

81.7 V/m. Since the magnitude is negative, the normal component points into region 1 from the surface. Then

EN1 = 81.65 (

16

) [ax − ay + 2az] = 33.33ax + 33.33ay − 66.67az V/m

Now, the tangential component will be ET1 = E1 EN1 = 133.3ax + 166.7ay + 16.67az. Our boundary conditions state that ET2 = ET1 and EN2 = (r1/r2)EN1 = (1/4)EN1. Thus

E2 = ET2 + EN2 = ET1 + 1 4 EN1 = 133.3ax + 166.7ay + 16.67az − 8.3ax + 8.3ay − 16.67az

= 125ax + 175ay V/m

6.8. Region 1 (x ≥ 0) is a dielectric with r1 = 2, while region 2 (x < 0) has r2 = 5. Let E1 = 20ax − 10ay + 50az V/m. a) Find D2: One approach is to first find E2. This will have the same y and z (tangential)

components as E1, but the normal component, Ex, will differ by the ratio r1/r2; this arises from Dx1 = Dx2 (normal component of D is continuous across a non-charged interface). Therefore E2 = 20(r1/r2)ax − 10ay + 50az = 8ax − 10ay + 50az. The flux density is then

D2 = r20E2 = 400 ax − 500 ay + 2500 az = 0.35ax − 0.44ay + 2.21az nC/m2

b) Find the energy density in both regions: These will be

we1 = 1 2 r10E1 · E1 =

1 2 (2)0

[ (20)2 + (10)2 + (50)2

] = 30000 = 26.6 nJ/m3

we2 = 1 2 r20E2 · E2 =

1 2 (5)0

[ (8)2 + (10)2 + (50)2

] = 66600 = 59.0 nJ/m3

3

6.9. Let the cylindrical surfaces ρ = 4 cm and ρ = 9 cm enclose two wedges of perfect dielectrics, r1 = 2 for 0 < φ < π/2, and r2 = 5 for π/2 < φ < 2π. If E1 = (2000)aρ V/m, find: a) E2: The interfaces between the two media will lie on planes of constant φ, to which E1

is parallel. Thus the field is the same on either side of the boundaries, and so E2 = E1.

b) the total electrostatic energy stored in a 1m length of each region: In general we have wE = (1/2)r0E2. So in region 1:

WE1 = ∫ 1

0

π/2 0

∫ 9 4

1 2 (2)0

(2000)2

ρ2 ρ dρ dφ dz =

π

2 0(2000)2 ln

( 9 4

) = 45.1µJ

In region 2, we have

WE2 = ∫ 1

0

∫ 2π π/2

∫ 9 4

1 2 (5)0

(2000)2

ρ2 ρ dρ dφ dz =

15π 4

0(2000)2 ln (

9 4

) = 338µJ

6.10. Let S = 100 mm2, d = 3 mm, and r = 12 for a parallel-plate capacitor. a) Calculate the capacitance:

C = r0A

d =

120(100 × 106) 3 × 103 = 0.40 = 3.54 pf

b) After connecting a 6 V battery across the capacitor, calculate E, D, Q, and the total stored electrostatic energy: First,

E = V0/d = 6/(3 × 103) = 2000 V/m, then D = r0E = 2.4 × 1040 = 0.21 µC/m2

The charge in this case is

Q = D · n|s = DA = 0.21 × (100 × 106) = 0.21 × 104 µC = 21 pC

Finally, We = (1/2)QV0 = 0.5(21)(6) = 63 pJ.

c) With the source still connected, the dielectric is carefully withdrawn from between the plates. With the dielectric gone, re-calculate E, D, Q, and the energy stored in the capacitor.

E = V0/d = 6/(3 × 103) = 2000 V/m, as before. D = 0E = 20000 = 17.7 nC/m2

The charge is now Q = DA = 17.7 × (100 × 106) nC = 1.8 pC. Finally, We = (1/2)QV0 = 0.5(1.8)(6) = 5.4 pJ.

d) If the charge and energy found in (c) are less than that found in (b) (which you should have discovered), what became of the missing charge and energy? In the absence of friction in removing the dielectric, the charge and energy have returned to the battery that gave it.

4

6.11. Capacitors tend to be more expensive as their capacitance and maximum voltage, Vmax, increase. The voltage Vmax is limited by the field strength at which the dielectric breaks down, EBD. Which of these dielectrics will give the largest CVmax product for equal plate areas: (a) air: r = 1, EBD = 3 MV/m; (b) barium titanate: r = 1200, EBD = 3 MV/m; (c) silicon dioxide: r = 3.78, EBD = 16 MV/m; (d) polyethylene: r = 2.26, EBD = 4.7 MV/m? Note that Vmax = EBDd, where d is the plate separation. Also, C = r0A/d, and so VmaxC = r0AEBD, where A is the plate area. The maximum CVmax product is found through the maximum rEBD product. Trying this with the given materials yields the winner, which is barium titanate.

6.12. An air-filled parallel-plate capacitor with plate separation d and plate area A is connected to a battery which applies a voltage V0 between plates. With the battery left connected, the plates are moved apart to a distance of 10d. Determine by what factor each of the following quantities changes: a) V0: Remains the same, since the battery is left connected.

b) C: As C = 0A/d, increasing d by a factor of ten decreases C by a factor of 0.1.

c) E: We require E × d = V0, where V0 has not changed. Therefore, E has decreased by a factor of 0.1.

d) D: As D = 0E, and since E has decreased by 0.1, D decreases by 0.1.

e) Q: Since Q = CV0, and as C is down by 0.1, Q also decreases by 0.1.

f) ρS : As Q is reduced by 0.1, ρS reduces by 0.1. This is also consistent with D having been reduced by 0.1.

g) We: Use We = 1/2CV 20 , to observe its reduction by 0.1, since C is reduced by that factor.

6.13. A parallel plate capacitor is filled with a nonuniform dielectric characterized by r = 2 + 2 × 106x2, where x is the distance from one plate. If S = 0.02 m2, and d = 1 mm, find C: Start by assuming charge density ρs on the top plate. D will, as usual, be x-directed, originating at the top plate and terminating on the bottom plate. The key here is that D will be constant over the distance between plates. This can be understood by considering the x-varying dielectric as constructed of many thin layers, each having constant permittivity. The permittivity changes from layer to layer to approximate the given function of x. The approximation becomes exact as the layer thicknesses approach zero. We know that D, which is normal to the layers, will be continuous across each boundary, and so D is constant over the plate separation distance, and will be given in magnitude by ρs. The electric field magnitude is now

E = D

0r =

ρs 0(2 + 2 × 106x2)

The voltage beween plates is then

V0 = ∫ 103

0

ρs dx

0(2 + 2 × 106x2) =

ρs 0

14 × 106

tan1 ( x √

4 × 106 2

)∣∣∣103 0

= ρs 0

1 2 × 103

(π 4

)

Now Q = ρs(.02), and so

C = Q

V0 =

ρs(.02)0(2 × 103)(4) ρsπ

= 4.51 × 1010 F = 451 pF

5

6.14. Repeat Problem 6.12 assuming the battery is disconnected before the plate separation is increased: The ordering of parameters is changed over that in Problem 6.12, as the progression of thought on the matter is different.

a) Q: Remains the same, since with the battery disconnected, the charge has nowhere to go.

b) ρS : As Q is unchanged, ρS is also unchanged, since the plate area is the same.

c) D: As D = ρS , it will remain the same also.

d) E: Since E = D/0, and as D is not changed, E will also remain the same.

e) V0: We require E × d = V0, where E has not changed. Therefore, V0 has increased by a factor of 10.

f) C: As C = 0A/d, increasing d by a factor of ten decreases C by a factor of 0.1. The same result occurs because C = Q/V0, where V0 is increased by 10, whereas Q has not changed.

g) We: Use We = 1/2CV 20 = 1/2QV0, to observe its increase by a factor of 10.

6.15. Let r1 = 2.5 for 0 < y < 1 mm, r2 = 4 for 1 < y < 3 mm, and r3 for 3 < y < 5 mm. Conducting surfaces are present at y = 0 and y = 5 mm. Calculate the capacitance per square meter of surface area if: a) r3 is that of air; b) r3 = r1; c) r3 = r2; d) region 3 is silver: The combination will be three capacitors in series, for which

1 C

= 1 C1

+ 1 C2

+ 1 C3

= d1

r10(1) +

d2 r20(1)

+ d3

r30(1) =

103

0

[ 1

2.5 +

2 4

+ 2 r3

]

So that

C = (5 × 103)0r3

10 + 4.5r3 Evaluating this for the four cases, we find a) C = 3.05 nF for r3 = 1, b) C = 5.21 nF for r3 = 2.5, c) C = 6.32 nF for r3 = 4, and d) C = 9.83 nF if silver (taken as a perfect conductor) forms region 3; this has the effect of removing the term involving r3 from the original formula (first equation line), or equivalently, allowing r3 to approach infinity.

6.16. A parallel-plate capacitor is made using two circular plates of radius a, with the bottom plate on the xy plane, centered at the origin. The top plate is located at z = d, with its center on the z axis. Potential V0 is on the top plate; the bottom plate is grounded. Dielectric having radially-dependent permittivity fills the region between plates. The permittivity is given by (ρ) = 0(1 + ρ/a). Find: a) E: Since  does not vary in the z direction, and since we must always obtain V0 when

integrating E between plates, it must follow that E = −V0/daz V/m. b) D: D = E = [0(1 + ρ/a)V0/d] az C/m2. c) Q: Here we find the integral of the surface charge density over the top plate:

Q = ∫ S

D · dS = ∫ 2π

0

a 0

−0(1 + ρ/a)V0 d

az · (az) ρ dρ dφ = 2π0V0

d

a 0

(ρ + ρ2/a)

= 5π0a2

3d V0

d) C: We use C = Q/V0 and our previous result to find C = 50(πa2)/(3d) F.

6

6.17. Two coaxial conducting cylinders of radius 2 cm and 4 cm have a length of 1m. The region between the cylinders contains a layer of dielectric from ρ = c to ρ = d with r = 4. Find the capacitance if a) c = 2 cm, d = 3 cm: This is two capacitors in series, and so

1 C

= 1 C1

+ 1 C2

= 1

2π0

[ 1 4

ln (

3 2

) + ln

( 4 3

)] ⇒ C = 143 pF

b) d = 4 cm, and the volume of the dielectric is the same as in part a: Having equal volumes requires that 32 22 = 42 − c2, from which c = 3.32 cm. Now

1 C

= 1 C1

+ 1 C2

= 1

2π0

[ ln

( 3.32 2

) +

1 4

ln (

4 3.32

)] ⇒ C = 101 pF

6.18. (a) If we could specify a material to be used as the dielectric in a coaxial capacitor for which the permittivity varied continuously with radius, what variation with ρ should be used in order to maintain a uniform value of the electric field intensity?

Gauss’s law tells us that regardless of the radially-varying permittivity, D = (aρs/ρ)aρ, where a is the inner radius and ρs is the presumed surface charge density on the inner cylinder. Now

E = D 

= aρs ρ

aρ

which indicates that  must have a 1dependence if E is to be constant with radius.

b) Under the conditions of part a, how do the inner and outer radii appear in the expression for the capacitance per unit distance? Let  = g/ρ where g is a constant. Then E = aρs/g aρ and the voltage between cylinders will be

V0 = a b

aρs g

aρ · aρ dρ = aρs g

(b− a)

where b is the outer radius. The capacitance per unit length is then C = 2πaρs/V0 = 2πg/(b− a), or a simple inverse-distance relation.

6.19. Two conducting spherical shells have radii a = 3 cm and b = 6 cm. The interior is a perfect dielectric for which r = 8.

a) Find C: For a spherical capacitor, we know that:

C = 4πr0 1 a − 1b

= 4π(8)0(

1 3 16

) (100)

= 1.92π0 = 53.3 pF

b) A portion of the dielectric is now removed so that r = 1.0, 0 < φ < π/2, and r = 8, π/2 < φ < 2π. Again, find C: We recognize here that removing that portion leaves us with two capacitors in parallel (whose C’s will add). We use the fact that with the dielectric completely removed, the capacitance would be C(r = 1) = 53.3/8 = 6.67 pF. With one-fourth the dielectric removed, the total capacitance will be

C = 1 4 (6.67) +

3 4 (53.4) = 41.7 pF

7

6.20. Show that the capacitance per unit length of a cylinder of radius a is zero: Let ρs be the surface charge density on the surface at ρ = a. Then the charge per unit length is Q = 2πaρs. The electric field (assuming free space) is E = (aρs)/(0ρ)aρ. The potential difference is evaluated between radius a and infinite radius, and is

V0 = a ∞

aρs 0ρ

aρ · aρ dρ → ∞

The capacitance, equal to Q/V0, is therefore zero.

6.21. With reference to Fig. 6.9, let b = 6 m, h = 15 m, and the conductor potential be 250 V. Take  = 0. Find values for K1, ρL, a, and C: We have

K1 =

[ h +

√ h2 + b2

b

]2 =

[ 15 +

√ (15)2 + (6)2

6

]2 = 23.0

We then have

ρL = 4π0V0 lnK1

= 4π0(250)

ln(23) = 8.87 nC/m

Next, a = √ h2 − b2 =

√ (15)2 (6)2 = 13.8 m. Finally,

C = 2π

cosh1(h/b) =

2π0 cosh1(15/6)

= 35.5 pF

6.22. Two #16 copper conductors (1.29-mm diameter) are parallel with a separation d between axes. Determine d so that the capacitance between wires in air is 30 pF/m.

We use C

L = 60 pF/m =

2π0 cosh1(h/b)

The above expression evaluates the capacitance of one of the wires suspended over a plane at mid-span, h = d/2. Therefore the capacitance of that structure is doubled over that required (from 30 to 60 pF/m). Using this,

h

b = cosh

( 2π0 C/L

) = cosh

( 2π × 8.854

60

) = 1.46

Therefore, d = 2h = 2b(1.46) = 2(1.29/2)(1.46) = 1.88 mm.

6.23. A 2 cm diameter conductor is suspended in air with its axis 5 cm from a conducting plane. Let the potential of the cylinder be 100 V and that of the plane be 0 V. Find the surface charge density on the:

a) cylinder at a point nearest the plane: The cylinder will image across the plane, producing an equivalent two-cylinder problem, with the second one at location 5 cm below the plane. We will take the plane as the zy plane, with the cylinder positions at x = ±5. Now b = 1 cm, h = 5 cm, and V0 = 100 V. Thus a =

√ h2 − b2 = 4.90 cm. Then

K1 = [(h + a)/b]2 = 98.0, and ρL = (4π0V0)/ lnK1 = 2.43 nC/m. Now

D = 0E = − ρL 2π

[ (x + a)ax + yay (x + a)2 + y2

(x− a)ax + yay (x− a)2 + y2

]

8

6.23a. (continued) and

ρs,max = D · (ax) ∣∣∣ x=h−b,y=0

= ρL 2π

[ h− b + a

(h− b + a)2 − h− b− a

(h− b− a)2 ]

= 473 nC/m2

b) plane at a point nearest the cylinder: At x = y = 0,

D(0, 0) = −ρL 2π

[ aax a2

− −aax a2

] = −ρL

2π 2 a ax

from which ρs = D(0, 0) · ax =

ρL πa

= 15.8 nC/m2

6.24. For the conductor configuration of Problem 6.23, determine the capacitance per unit length. This is a quick one if we have already solved 6.23. The capacitance per unit length will be C = ρL/V0 = 2.43 [nC/m]/100 = 24.3 pF/m.

6.25 Construct a curvilinear square map for a coaxial capacitor of 3-cm inner radius and 8-cm outer radius. These dimensions are suitable for the drawing. a) Use your sketch to calculate the capacitance per meter length, assuming R = 1: The

sketch is shown below. Note that only a 9sector was drawn, since this would then be duplicated 40 times around the circumference to complete the drawing. The capacitance is thus

C .= 0

NQ NV

= 0 40 6

= 59 pF/m

b) Calculate an exact value for the capacitance per unit length: This will be

C = 2π0

ln(8/3) = 57 pF/m

9

6.26 Construct a curvilinear-square map of the potential field about two parallel circular cylinders, each of 2.5 cm radius, separated by a center-to-center distance of 13cm. These dimensions are suitable for the actual sketch if symmetry is considered. As a check, compute the capacitance per meter both from your sketch and from the exact formula. Assume R = 1.

Symmetry allows us to plot the field lines and equipotentials over just the first quadrant, as is done in the sketch below (shown to one-half scale). The capacitance is found from the formula C = (NQ/NV )0, where NQ is twice the number of squares around the perimeter of the half-circle and NV is twice the number of squares between the half-circle and the left vertical plane. The result is

C = NQ NV

0 = 32 16

0 = 20 = 17.7 pF/m

We check this result with that using the exact formula:

C = π0

cosh1(d/2a) =

π0

cosh1(13/5) = 1.950 = 17.3 pF/m

10

6.27. Construct a curvilinear square map of the potential field between two parallel circular cylin- ders, one of 4-cm radius inside one of 8-cm radius. The two axes are displaced by 2.5 cm. These dimensions are suitable for the drawing. As a check on the accuracy, compute the capacitance per meter from the sketch and from the exact expression:

C = 2π

cosh1 [(a2 + b2 −D2)/(2ab)]

where a and b are the conductor radii and D is the axis separation.

The drawing is shown below. Use of the exact expression above yields a capacitance value of C = 11.50 F/m. Use of the drawing produces:

C .=

22 × 2 4

0 = 110 F/m

11

6.28. A solid conducting cylinder of 4-cm radius is centered within a rectangular conducting cylinder with a 12-cm by 20-cm cross-section.

a) Make a full-size sketch of one quadrant of this configuration and construct a curvilinear- square map for its interior: The result below could still be improved a little, but is nevertheless sufficient for a reasonable capacitance estimate. Note that the five-sided region in the upper right corner has been partially subdivided (dashed line) in anticipation of how it would look when the next-level subdivision is done (doubling the number of field lines and equipotentials).

b) Assume  = 0 and estimate C per meter length: In this case NQ is the number of squares around the full perimeter of the circular conductor, or four times the number of squares shown in the drawing. NV is the number of squares between the circle and the rectangle, or 5. The capacitance is estimated to be

C = NQ NV

0 = 4 × 13

5 0 = 10.40

.= 90 pF/m

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6.29. The inner conductor of the transmission line shown in Fig. 6.14 has a square cross-section 2a × 2a, while the outer square is 5a × 5a. The axes are displaced as shown. (a) Construct a good-sized drawing of the transmission line, say with a = 2.5 cm, and then prepare a curvilinear-square plot of the electrostatic field between the conductors. (b) Use the map to calculate the capacitance per meter length if  = 1.60. (c) How would your result to part b change if a = 0.6 cm?

a) The plot is shown below. Some improvement is possible, depending on how much time one wishes to spend.

b) From the plot, the capacitance is found to be

C .=

16 × 2 4

(1.6)0 = 12.80 .= 110 pF/m

c) If a is changed, the result of part b would not change, since all dimensions retain the same relative scale.

6.30. For the coaxial capacitor of Problem 6.18, suppose that the dielectric is leaky, allowing current to flow between the inner and outer conductors, while the electric field is still uniform with radius.

a) What functional form must the dielectric conductivity assume? We must have constant current through any cross-section, which means that J = I/(2πρ)aρ A/m2, where I is the radial current per unit length. Then, from J = σE, where E is constant, we require a 1dependence on σ, or let σ = σ0, where σ0 is a constant.

b) What is the basic functional form of the resistance per unit distance, R? From Problem 6.18, we had E = aρs/g aρ V/m, where ρs is the surface charge density on the inner con- ductor, and g is the constant parameter in the permittivity,  = g/ρ. Now, I = 2πρσE = 2πaρsσ0/g, and V0 = aρs(b− a)/g (from 6.18). Then R = V0/I = (b− a)/(2πσ0).

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6.30c) What parameters remain in the product, RC, where the form of C, the capacitance per unit distance, has been determined in Problem 6.18? With C = 2πg/(b− a) (from 6.18), we have RC = g/σ0.

6.31. A two-wire transmission line consists of two parallel perfectly-conducting cylinders, each hav- ing a radius of 0.2 mm, separated by center-to-center distance of 2 mm. The medium sur- rounding the wires has r = 3 and σ = 1.5 mS/m. A 100-V battery is connected between the wires. Calculate: a) the magnitude of the charge per meter length on each wire: Use

C = π

cosh1(h/b) =

π × 3 × 8.85 × 1012 cosh1 (1/0.2)

= 3.64 × 109 C/m

Then the charge per unit length will be

Q = CV0 = (3.64 × 1011)(100) = 3.64 × 109 C/m = 3.64 nC/m

b) the battery current: Use

RC = 

σ ⇒ R = 3 × 8.85 × 10

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(1.5 × 103)(3.64 × 1011) = 486 Ω

Then I =

V0 R

= 100 486

= 0.206 A = 206 mA

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