equation for intensity of light epsilon naught

Equation for intensity of light epsilon naught for free#
Equation for intensity of light epsilon naught free#

| Home | General | Memb Svcs | Publications | Press | Technical | Internet | Index |Įntire website copyright © The SETI League, Inc. Like the impedance discrepancy, the difference is, for all practical purposes, nonexistent. Using more precise values for mu and epsilon, we get a more precise value of c = 2.998 x 10 8 m/s. Using the approximations above yields the generally accepted value of c = 3 x 10 8 m/s.

Equation for intensity of light epsilon naught free#

Since a VSWR of 1 : 1 represents a perfect impedance match, I think you will agree that the discrepancy is negligible.Īs an aside, I should point out that the values of permittivity and permeability of free space also allow us to calculate the speed of light. But, how close did we come, in practical terms? If you match an antenna to an impedance of 120 pi ohms, instead of the more precise value you cited in your question, an impedance mismatch will arise, resulting in a voltage standing wave ratio of 1.00069 : 1. Which is why our estimate of 120 pi ohms is similarly accurate to only a few hundredths of a percent. It turns out the values cited are not precise, but are accurate to within a few hundredths of a percent. Of course, this value is only as accurate as our approximations of free space permittivity and permeability. The relevant dimensions for Z o would be the square root of ohms Where epsilon represents permittivity (in Farads per meter), mu is permeability (in Henries per meter), Z means impedance, and the subscript o indicates free space.

So, if we know the permittivity and permeability of free space, we can determine the impedance to which an electromagnetic wave propagating in free space is subject. Similarly, an electrical current results in the creation of a magnetic field, the intensity of which is inversely proportional to the square root of the permeability of the medium in which it propagates. But, how do we find where the actual impedance of free space falls within this rather wide range?Ĭonsider that an electrical potential gives rise to an electrostatic field, the intensity of which varies inversely with the square root of the permittivity of the medium in which it propagates. This is because power is the product of potential and current, the potential across the short would be zero (yielding a zero product), and the current through an open is zero (also resulting in a zero product). (You probably know of this relationship as as Ohm's Law.)įor propagation of electromagnetic energy, the impedance has to be somewhere between zero ohms and infinity. Electromotive force is a manifestation of electrical potential energy (we even call voltage 'electrical potential'), and current is the rate of flow of charged particles, which is a manifestation of kinetic energy. One way to define impedance is as the ratio of potential to kinetic energy in a dynamic system. Since we can consider free space to be, for all practical purposes, infinite in extent, any antenna will appear to be terminated in this calculable impedance. The characteristic (or surge) impedance of any transmission medium is the impedance we would encounter if looking into an infinitely long segment of that medium. Why the disprepancy?Īlthough one might rightly argue that the discrepancy is neglibible, you're wise to question the exact value, because this is the impedance to which one end (the radiating end) of our antennas must be matched. In the engineering literature, I find: 376.730313462 ohms (not exactly 120 pi). The SETI League's literature sets the impedance of free space at 120 pi, or about 377 ohms.

Equation for intensity of light epsilon naught for free#

I am a little confused about the value for free space characteristic imepdance.