Jackson 4.7 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell

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Jackson 4.7 Homewok obem Soution D. Chistophe S. Baid Univesity of Massachusetts Lowe ROBLEM: A ocaized distibution of chage has a chage density ρ()= 6 e sin θ (a) Make a mutipoe expansion of the potentia due to this chage density and detemine a the nonvanishing mutipoe moments. Wite down the potentia at age distances as a finite expansion in Legende poynomias. (b) Detemine the potentia expicity at any point in space and show the soution nea the oigin, coect to incusive, Φ() [ 4 (cosθ)] (c) If thee exists at the oigin a nuceus with a quadupoe moment Q = -8 m, detemine the magnitude of the inteaction enegy, assuming that the unit of chage in ρ() above is the eectonic chage and the unit of ength is the hydogen Boh adius a = ϵ ħ /m e =.59 m. Expess you answe as a fequency by dividing by ank's constant h. The chage density in this pobem is that fo the m=± states of the p eve in hydogen, whie the quadupoe inteaction is of the same ode as found in moecues. SOLUTION: (a) The genea soution in tems of a mutipoe expansion of a ocaized distibution in spheica hamonics is:,, = 4 = m= 4 q Y m, m whee q m = Y * m ', ' ' x' d x ' Fo cases whee thee is azimutha symmety in the chage distibution, this educes to:,, = 4 = 4 q cos whee q = 4 cos ' ' ', ' sin ' d ' d '

Let us now pug in the chage density fo this case: q = 6 π + ( x')( x' )dx ' ' +4 e ' d ' Use the identity easiy found in an intega tabe: x n e x dx=n! if n is a positive intege, which is the case hee. q = 6 π + (+4)! (x ')( x ' )dx ' We can use the identity: x ' = (x ')+ (x ') which is easiy deived fom the definition of the two specific Legende poynomias. q = 64 π π + 4 π (+4)! [ (x ') (x ')dx '+ '] ( x') ( x')dx Due to othogonaity, evey q is zeo except when = o = : q = q = 5 π We can now find the fina soution to the potentia: Φ(,θ,ϕ)= [ q + π 5 q cos θ ] Φ(,θ,ϕ)= [ + 9 cos θ ] o in tems of Legende poynomias: Φ(,θ,ϕ)= [ 6 ] Note that this soution is exact as ong as the obsevation point is beyond the edge of the chage density. Thee is no fa-away imitation to this soution because a highe ode tems ae zeo. Unfotunatey, in this pobem, thee is no ceay defined edge to the chage density. This soution becomes exact once e - is so sma so as to be negigibe.

(b) The mutipoe expansion is ony vaid if the obsevation point is extena to a oca chage distibution. To find the potentia within the chage distibution, we must use Couomb's aw: = 4 x' x x' d x' π π 6 ' e ' sin θ' ' sin θ ' d ' d θ ' d ϕ' x x ' Expand the denominato in spheica hamonics π π 6 ' 4 e ' sin θ ' 4 π = m= + Due to azimutha symmety, ony the m = tems wi suvive. = (cosθ) (x ')( x ' )d x ' < Y * + m(θ', ϕ')y m (θ,ϕ)sin θ' d ' d θ' d ϕ' < ' 4 e ' d ' + Again use the identity: x ' = (x ')+ (x ') (cosθ)[ = (x ') (x ')d x '+ Due to othogonaity, a tems dop out except = and = : [ 4 ' 4 e ' d ' 4 5 (cosθ) ' 4 e ' < d ' ] We must beak each intega into two cases, when ' < and ' [ 4 ' 4 e ' d '+ 4 4 5 (cosθ) ' e ' d ' ( x') ( x')d x'] ' 4 e ' ' 6 e ' d ' 4 5 (cos θ) ' e ' d ' ] [ 4 [e ' ( ' 4 4 ' ' 4 4 ' 4)] + [e ' ( ' ' 6 ' 6)] < + d ' 4 5 (cosθ) ( [e ' ( ' 6 6 ' 5 ' 4 ' 6 ' 7 ' 7)] + [ ' ( '+)] ) ]

Φ= [ 4[ ( +6 +8+ 4 )+ 4 ] (cosθ) [ e ( 5 6 7 7 7 )+ ] This soution is exact and appies eveywhee. We can take the imit as e - becomes zeo to check the soution obtained using mutipoe moments. We find: Φ(,θ,ϕ)= [ 6 (cos(θ)) ] This matches the fa-away soution found using the mutipoe expansion. We can aso take the imit cose to the oigin. We must fist expand the exponentia into a Tayo seies: e = 6 4 4 5... Afte distibuting out a tems, we can thow out tems, 4 and highe because they contibute vey itte nea the oigin. Φ [ 4 [ 68 4 +6 +8+4 9 +4 + 4 ] (cosθ) [ 5 + + 6 +68+6 7 7 6 7 7 6 6 7 ]] Most of these tems cance out and the fina soution cose to the oigin educes to: Φ [ 4 (cosθ)] (c) We ae to assume that the chage density of the pevious pats, and its associated potentia, is the aveage chage distibution of a singe eecton bound to a hydogen nuceus. The nuceus of is so sma that is can be consideed as contained at the oigin, so that we ony need to use the expession fo the eecton's potentia nea the oigin (the ast equation of pat b). If we convet fom units of eementay chage and Boh adius engths to SI units, the potentia of the eecton nea the oigin becomes: Φ a [ 4 ( /a ) (cosθ) ] The inteaction enegy is the tota potentia enegy between the inteacting eecton's potentia and nuceus' chage: ]

W = ρ nuc Φ eec d x W = ρ nuc ϵ a [ 4 ( /a ) (cosθ) ] d x W = 4 π ϵ a [ 4 q nuc ρ nuc (z )d 4 a x] W = ϵ a [ 4 q nuc e 4 a Q zz, nuc] The pobem does not expicity ask fo the monopoe moment inteaction of the nuceus. We dop the fist tem and find: W h = αc Q zz, nuc 48πa whee the zeo-enegy fine stuctue constant is α= e ϵ ħ c 7. W MHz h