Extra Problem A7 for 7D [just one this week]. Please upload your solution to the course website by 11:59 Tuesday 6/6/2023 A) An accelerating charged particle will produce an electromagnetic wave (EM wave). It should be sensible that EM waves carry energy outward from the accelerating charged particle. In addition, the energy is transported very quickly, at the speed of light, with velocity c. 50 the energy that ows radially outward though an imaginary spherical shell of radius r is characterized by energy per unit time (or power). The power emitted by an accelerating charge can be calculated (and you will see how to do it in 1123), but it is surprisingly complicated. But there is another way to get an approximate formula for EM power emitted by an accelerating charge. In this rst part, I ask you develop a formula using dimensional analysis (that is, putting relevant quantities together in such a way to get the right units). Power has units of energy/time or equivalently, since work = energy, Force*length/time, or in SI units, N*m/s. What can possibly be used to calculate the power? Well, the acceleration, a, for one. Charge q must be another, and the EM wave travels at constant velocity c, which must be included since transporting energy through a spherical surface must depend on the velocity of the wave. There is more power if the same energy is transported at high velocity than at slow velocity. This is pretty much it in terms of constants and essential physics of the problem (charge must accelerate...) The quantity q is contained in Coulombs force law, F=q2/(4atgor2) where F has units of newtons,N . So you got Coulombs Force Law, c, and a. The idea is convert these quantities into SI units (c has m/s, 3 has m/sz). Coulombs Law has q in Coulombs,C, and the constants (411280) and the SI units 1/m2 from the 1/r2. So notice that q2/(4Tcaol has units of N*m2. So start your formula with q1/(4msa) with units N*m2. This way, you have the required N unit for power, and m2. Next you need to gure out how to make the rest of the formula, using c and a, to give the nal required units of Nm/s in the power expression. Let me give you a start. Try to nd an exponent of c and a different exponent of a that gives the required units. In other words, you want Nm2 (from qZ/(4rtsg) part of the formula ) multiplied by clak to give units of Nm/s. The N part is already done. So we need to nd mlciak=m/s=mls'1. Rewrite c and a in terms of units, you get m2(m/s)l(m/sz)k= m2(m15'1)l(mls'2)Emmi\")sH'm. This last expression has to have units m15'1. Solve for] and k. Your formula for power P is then P: cjakq2/(4itau) . BTW, this expression is very close to the \"right" answer. Look up \"Larmor Formula". Another BTW using dimensional analysis such as outlined in this problem is a very useful starting point before trying a more complicated calculation. It is a common technique. b) The expression for Power in (a) is very useful for estimating the EM power emitted or collected by a dipole antenna, energy loss by particles in an circular accelerator, and many other phenomena. However, lets apply it to an atomic scale problem such as the hydrogen atom. Lets model the hydrogen atom as an electron circling about a proton (which is stationary). The electron has charge 1.6x10'19 C. Assume it is traveling in a circular path with a radius approximately given the size of the atom or r=10'1 m. For simplicity, you can estimate the velocity of the electron from the fact that its kinetic energy is K=13.6 eV (this is binding energy ofthe electron to the proton in the hydrogen atom). Look up the mass of the electron (in SI units!) and convert eV to Joules and assume that K=(1/2)mv2. Find the velocity, v, of the electron in SI units. What is the fraction f= v/c? c) nd the acceleration of the electron, a (use formulas from 7C for circular motion to get acceleration). d) Using your formula in part (a), compute the EM power radiated per second, P, in SI units. e) How much time is required for the electron to radiate all of its kinetic energy? Does this make sense? What does this tell you about using classical physics (in particular, Newton's Laws and Maxwell's Equations) to understand the physics ofthe hydrogen atom? {aside comment: Comment: In part e, you should have found a very short time to lose all the energy. Hydrogen atoms are stable and can live for billions of years. So what is wrong with the calculation? You might think it is the fact that we did not use Special Relativity. But that isn't it. The velocity of the electron is not very close to the speed of light, so special relativity is not so important. The problem is elsewhere, and this will be the topic of 61A