Help   |   Contact Us

 

Lecture 16: Electromagnetic Waves and Conductors

OCW Scholar

« Previous | Next »

Lecture Topics

A vial sits in a clear vase turned upside down.
  • Boundary Conditions for EM Waves at Conductors
  • Transmission Lines
  • Polarizers
  • Photons
  • Radiation Pressure

Learning Objectives

By the end of this lecture, you should:

  • understand the concept of skin depth and that it is small in good conductors.
  • know the boundary conditions at the surface of ideal conductors.
  • solve for generation of a standing EM wave by reflection from a conducting plate.
  • describe transmission lines (wire pair and coaxial).
  • explain reflection conditions at the end of transmission lines.
  • explain polarization of EM radiation passing through a conducting grid.
  • describe the photon picture of light.
  • relate radiation pressure to the Poynting vector.
  • give examples in which radiation pressure plays a role.

Lecture Activities

Check Yourself

  • The radio-frequency conductivity of sea water is about 5 (Ω m)-1, which is about 10 million times less than that of copper. There are two interesting quick calculations one can do with this number. The first is to assume that it holds also at optical frequencies and calculate the skin depth. It should be clear something is wrong. The second is to calculate the skin depth at a reasonable radio frequency of 100 kHz and see what problems would arise for communicating with submarines.

› View/hide answer

The skin depth is in this case \[\delta = \sqrt {\frac{2}{{\omega {\mu _0}\sigma }}} = \sqrt {\frac{2}{{2\pi f(4\pi \times {{10}^{ - 7}})(5)}}} = \frac{{225}}{{\sqrt f }}m.\] Light is mentioned in the lecture as having f = 5 × 1014 Hz, so the formula implies a skin depth of \[\delta = \frac{{225}}{{\sqrt {5 \times {{10}^{14}}} }} = {10^{ - 5}}m.\] Anyone who has seen the bottom through many meters of clear sea water will know this cannot be true and that some other physical process must take over at optical frequencies. On the other hand, at the given radio frequency, we get \[\delta = \frac{{225}}{{\sqrt {100 \times {{10}^3}} }} = 0.7m.\] A submarine can only communicate at this frequency if it is on the surface, which defeats most of the advantages of being a submarine. In practice, submarine communication is done with huge antennas operating at extremely low frequencies.

 

  • What does all the discussion about skin depth actually mean? How does the concept of skin depth affect the treatment of EM wavs at the interface between two materials?

› View/hide answer

Since, apart from some cases where we would need to be careful, as discussed in the previous question, the skin depth is very small for good conductors, we can treat their surfaces as sharp boundaries, easier to handle mathematically than the real situation where the fields decrease in a complicated fashion (but quickly). We will make very little error in doing that even for time-varying fields, as long as the wavelength is long compared to the skin depth.

 

  • What assumptions do we make in applying Gauss' Law for electric fields to get the boundary condition for a changing electric field?

› View/hide answer

In addition to the assumption about a small skin depth, we assume that the conductor can supply any required charge to meet the boundary condition, essentially instantaneously.

 

  • A potential application of radiation pressure is “solar sailing.” In this application, a reflective Mylar film creates a force from incident solar light. What is the maximum acceleration a Mylar film of areal density 10 g/m2 could generate near the Earth?

› View/hide answer

The radiation pressure is \[P = \frac{{ < S > }}{c} = \frac{{1400}}{{3 \times {{10}^8}}} = 5 \times {10^{ - 6}}N/m^2.\] Twice this pressure would be the force on one square meter (which has mass 10 g = 0.01 kg), due to the reflection. The acceleration would be \[a = \frac{F}{m} = 2\frac{{5 \times {{10}^{ - 6}}}}{{0.01}} = {10^{ - 3}}ms^{-2}.\] Although this is a very small acceleration, if consistently applied as needed, and to a body already in the balance of major forces that allows it to be in orbit, it could allow "fuel-free" travel. This calculation, however, neglects the mass of support structures and other parts of the solar spacecraft. Including these would make the acceleration less.

 

  • Discuss qualitatively the outgoing pulses discussed near minute 48 of the lecture (100 ns duration voltage pulses traveling in a dielectric-filled coaxial cable) in terms of surface charge on the central conductor and outer shell.

› View/hide answer

Even though the cable is wound up, any local section of it is like a cylinder since the cable diameter would be much less than the size of the coil. A 100 ns long pulse, positive on the center wire, results in a radially outward electric field. The pulse travels at .65c so its length is \[l = v\Delta t = .65(3 \times {10^8})(100 \times {10^{ - 9}}) = 19.5m,\] much less than the cable’s length. The radially outward electric field corresponds to a positive surface charge on the inner conductor, a negative surface charge on the outer conductor, and polarization of the dielectric between them. In between pulses, there is no surface charge, nor dielectric polarization. This pattern of surface charge and polarization moves along the cable.

 

« Previous | Next »