20190624_065453338 test.pdf

# This follows from a gausss law argument consider the

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region. This follows from a Gauss’s-law argument. Consider the pillbox shown in Fig. 24, A B C D Figure 24 which is located at the kink in the field at point C . The long sides of the box are oriented in the tangential direction, and also in the direction perpendicular to the page. The short sides are chosen to be infinitesimally small, so E θ contributes essentially nothing to the flux. The flux is therefore due only to the radial component, so we conclude that E transition r = E outside r . (And likewise, a similar pillbox at point B tells us that E transition r = E inside r , but we won’t need this. E transition r varies slightly over the transition region, but the change is negligible if Δ t is small.) We know that E outside r = q 4 π² 0 r 2 = q 4 π² 0 ( cT ) 2 . (80) Eq. (79) then gives E θ = vT sin θ c Δ t E r = vT sin θ c Δ t · q 4 π² 0 ( cT ) 2 = q sin θ 4 π² 0 c 2 ( cT ) · v Δ t = qa sin θ 4 π² 0 rc 2 (using a = v/ Δ t and r = cT ) (81) So the field inside the transition region is given by ( E r , E θ ) = q 4 π² 0 1 r 2 , a sin θ rc 2 (82) Both of the components in the parentheses have units of 1 / m 2 , as they should. We will explain below why this field leads to an electromagnetic wave, but first some remarks. Remarks: 1. The location of the E θ field (that is, the transition region) propagates outward with speed c . 2. E θ is proportional to a . Given this fact, you can show that the only way to obtain the right units for E θ using a , r , c , and θ , is to have a function of the form, af ( θ ) /rc 2 , where f ( θ ) is an arbitrary function of θ . And f ( θ ) happens to be sin θ (times q/ 4 π² 0 ). 3. If θ = 0 or θ = π , then E θ = 0. In other words, there is no radiation in the forward or backward directions. E θ is maximum at θ = ± π/ 2, that is, in the transverse direction.

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8.7. RADIATION FROM A POINT CHARGE 27 4. The acceleration vector associated with Fig. 22 points to the left, since the charge was de- celerating. And we found that E θ has a rightward component. So in general, the E θ vector is E θ ( r , t ) = - q 4 π² 0 · a ( t 0 ) rc 2 , (83) where t 0 t - r/c is the time at which the kink at point C in Fig. 22 was emmitted, and where a is the component of a that is perpendicular to the radial direction. In other words, it is the component with magnitude a sin θ that you “see” across your vision if you are located at position r . This is consistent with the previous remark, because a = 0 if θ = 0 or θ = π . Note the minus sign in Eq. (83). 5. Last, but certainly not least, we have the extremely important fact: For sufficiently large r , E r is negligible compared with E θ . This follows from the fact that in Eq. (82), E θ has only one r in the denominator, whereas E r has two. So for large r , we can ignore the “standard” radial part of the field. We essentially have only the new “strange” tangential field. By “large r ,” we mean a/rc 2 1 /r 2 = r c 2 /a . Or equivalently ra c . In other words, ignoring
• Summer '08
• KimGreenhalgh
• Acceleration, Magnetic Field, Electromagnetic field

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