Lec12 - units, then in a frequently more convenient system...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: units, then in a frequently more convenient system of units. 1 4 = k = 9 10 9 N m 2 C 2 Therefore P = 2 3 9 10 9 Nm 2 C 2 (10 10- 9 C) 2 5 4 m 4 s 4 1 2 m 2 (3 10 8 m s ) 3 = 14 , 000 10- 33 W = 1 . 4 10- 29 W Alternatively, e 2 4 = 1 . 44 eV nm and the given charge is q/e = 6 . 25 10 10 , e being the electron charge. Substituting in P we get P = 8 . 68 10- 11 eV s- 1 which equals 1 . 4 10- 29 W. Notice that is these units e 2 / (4 ) is at order unity. It is therefore particularly convenient to use in physical systems in which the energy scale is at the order of the eV and the length scale is at the order of the nm. This length scale is about atomic size, and the eV is the energy scale for atomic transitions. For this reason, this alternative system of units is very convenient to use in atomic physics. Energy density in an electromagnetic wave In the case of mechanical waves in a string, we found that the radiated power (which is the energy density per unit time) scales with the square of the waves amplitude. We therefore expect that in an electromagnetic wave the energy density also scales with the square of E and B . Indeed, in a general electromagnetic field (i.e., not necessarily that of a wave, with the electric and magnetic waves are possibly independent fields), the energy density is given by E = 1 2 E 2 + 1 2 1 B 2 . The energy density in the electric part is most easily found by looking at the energy in a capacitor, and the student is referred to such a derivation. This expression is correct for any electromagnetic field, and in particular for that of an electromagnetic wave. The difference is that now E and B are no longer independent, but rather must satisfy E = cB , as we found above. Also, c 2 = 1 / ( ). Put together, E = 1 2 ( E 2 + c 2 B 2 ). For an electromagnetic wave E = cB , so that the energy densities stored in the E and B fields are equal, despite that fact that E appears to be much larger than B . By how much larger? By a factor of c , and c is very large. Or is it?...
View Full Document

This note was uploaded on 07/19/2011 for the course PH 113 taught by Professor Liorburko during the Spring '09 term at University of Alabama - Huntsville.

Page1 / 4

Lec12 - units, then in a frequently more convenient system...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online