Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

(a) Suppose you imbedded some free charge in a piece of glass. About how long would it take for the charge to flow to the surface?

(b) Silver is an excellent conductor, but it’s expensive. Suppose you were designing a microwave experiment to operate at a frequency of1010Hz. How thick would you make the silver coatings?

(c) Find the wavelength and propagation speed in copper for radio waves at role="math" localid="1655716459863" 1MHz. Compare the corresponding values in air (or vacuum).

Short Answer

Expert verified

Answer

(a) The time in which the charge flows to the surface is 20s.

(b) The thickness of the silver coating is role="math" localid="1655716514781" 6.34×10-4mm.

(c) The wavelength and propagation speed in copper for radio waves is 0.4mmand 400msrespectively, and the wavelength for radio waves in air is 300m, and the propagation speed in copper is losses gradually with respect to the penetration depth.

Step by step solution

01

Expression for the characteristic time and the relativity permittivity of the medium:

Write the expression for the characteristic time.

ζ=εσ …… (1)

Here, ζis the characteristic time, εis the permittivity of the medium, and σis the conductivity.

Write the expression for the relativity permittivity.

εr=εε0ε=εrε0 …… (2)

Here, ε0is the Permittivity of the free space.

02

Determine the time in which the charge flow to the surface:

(a)

Substitute the value of equation (2) in equation (1).

ζ=εrε0σ …… (3)

Write the expression for the relation between relative permittivity and refractive index of the medium.

εr=n2

Here, n is the refractive index of glass n=1.5.

Substitute n=1.5, ε0=8.85×10-12C2/N.mand σ=10-12Ω-1·m-1in equation (3).

ζ=n2ε0σζ=1.52×8.85×10-12C2/N·m10-12Ω-1·m-1ζ=19.91s20sζ=20s

Therefore, the time in which the charge flows to the surface is 20s.

03

Determine the thickness of the silver coating:

(b)

Write the expression for the skin depth.

d=1k …… (4)

Here,kis the imaginary constant.

Write the expression for the value ofk.

d=12πfμ1ρ2d=2ρ2πfμd=ρπfμ

For σ>>>ωε0, the above equation becomes,

k=ωμσ2 …… (5)

Substitute the value of equation (5) in equation (4).

d=12πfμ1ρ2d=2ρ2πfμd=ρπfμ

Here, ρis the resistivity of silver 1.59×10-8Ω·m-1and f is the frequency.

Substitute ρ=1.59×10-8Ω·m-1, f=1010Hz, and μ=4π×10-7H/min the above expression.

d=1.59×10-8Ω·m-1π×1010Hz×4π×10-7H/md=6.34×10-7m×103mm1md=6.34×104mm

Therefore, the thickness of the silver coating is 6.34×104mm.

04

Determine the wavelength of radio waves:

(c)

Write the expression for the wavelength of a radio wave.

λ=2πk

Here,kis the wavenumber.

Substitute the value of equation (5) in the above expression.

λ=2πωμσ2λ=2π22πf1ρμ

Substitute f=1MHz, ρ=1.59×10-8Ω·m-1and μ=4π×10-7H/min the above expression.

λ=2π22π×1MHz×106Hz1MHz×11.59×10-8Ω·m-14π×10-7H/mλ=4×10-4m×103mm1mλ=0.4mm

05

Determine the propagation speed for radio waves:

Write the expression for propagation speed.

v=λf

Substitute λ=4×10-4mand f=1MHzin the above expression.

v=4×10-4×1MHz×106Hz1MHzv=400m/s

Calculate the wavelength in a vacuum.

λ=cfλ=3×108ms1MHz×106Hz1MHzλ=300m

The speed of radio waves is much lesser than the speed of light, which means the propagation speed in copper is losses gradually with respect to the penetration depth.

Therefore, the wavelength and propagation speed in copper for radio waves is 0.4mmand 400msrespectively, and the corresponding values in the air is 300m, and the propagation speed in copper is losses gradually with respect to the penetration depth.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) Show that the skin depth in a poor conductor σ<<ωεis (εσ)2μ(independent of frequency). Find the skin depth (in meters) for (pure) water. (Use the static values of ε,μand σ; your answers will be valid, then, only at relatively low frequencies.)

(b) Show that the skin depth in a good conductor (σ<<ωε)is λ2π(where λ is the wavelength in the conductor). Find the skin depth (in nanometers) for a typical metal (σ>>Ωm107-1)in the visible range (ω1015/s), assuming ε=ε0and μμ0. Why are metals opaque?

(c) Show that in a good conductor the magnetic field lags the electric field by 45°, and find the ratio of their amplitudes. For a numerical example, use the “typical metal” in part (b).

Suppose string 2 is embedded in a viscous medium (such as molasses), which imposes a drag force that is proportional to its (transverse) speed:

Fdrag=-Yftz.

(a) Derive the modified wave equation describing the motion of the string.

(b) Solve this equation, assuming the string vibrates at the incident frequency. That is, look for solutions of the form f~(z,t)=eiωtF~(z).

(c) Show that the waves are attenuated (that is, their amplitude decreases with increasing z). Find the characteristic penetration distance, at which the amplitude is of its original value, in terms of Υ,T,μand ω.

(d) If a wave of amplitude A , phase δ,= 0 and frequencyω is incident from the left (string 1), find the reflected wave’s amplitude and phase.

Question: Show that the standing wave fz,t=Asinkzcoskvtsatisfies the wave equation, and express it as the sum of a wave traveling to the left and a wave traveling to the right (Eq. 9.6).

Calculate the reflection coefficient for light at an air-to-silver interface (μ1=μ2=μ0,ε=ε0,σ=6×107(Ωm)-1)at optical frequencies(ω=4×1015/s).

(a) Show directly that Eqs. 9.197 satisfy Maxwell’s equations (Eq. 9.177) and the boundary conditions (Eq. 9.175).

(b) Find the charge density, λ(z,t), and the current,I(z,t) , on the inner conductor.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free