12.1.1&12.1.3
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8/3/2019 12.1.1&12.1.3 http://slidepdf.com/reader/full/12111213 1/6 EXERCISE 8.1.1 Replacing z by z+1 in the Euler integral, we can write Γz+1=0∞e-ttz dt=limM→∞0Me-ttz dt Let us integrate by parts, calling tz=u, e-tdt=dv; then we get du=ztz-1 dt , and v=-e-t , Γz+1=limM→∞tz-e-t0M-0M-e-tztz-1 dt =limM→∞-Mze-M+z0Me-ttz-1 dt But since e-∞=0, the first term of the above expression vanishes. Hence, what remains is Γz+1=z0∞e-ttz-1 dt , from which we obtain the basic relation Γz+1=zΓz . Now since Γ1=0∞e-tdt=1, we conclude for integer n, Γz+1=zΓz=z(z-1)Γz-1 =z(z-1)⋯1Γ1=n! EXERCISE 8.1.4 1
Transcript of 12.1.1&12.1.3
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EXERCISE 8.1.1
Replacing z by z+1 in the Euler integral, we can write
Γz+1=0∞e-ttz dt=limM→∞0Me-ttz dt
Let us integrate by parts, calling tz=u, e-tdt=dv; then we get
du=ztz-1 dt , and v=-e-t ,
Γz+1=limM→∞tz-e-t0M-0M-e-tztz-1 dt
=limM→∞-Mze-M+z0Me-ttz-1 dt
But since e-∞=0, the first term of the above expression vanishes. Hence,what remains isΓz+1=z0∞e-ttz-1 dt ,
from which we obtain the basic relation
Γz+1=zΓz .
Now sinceΓ1=0∞e-tdt=1,
we conclude for integer n,
Γz+1=zΓz=z(z-1)Γz-1
=z(z-1)⋯1Γ1=n!
EXERCISE 8.1.4
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Starting with the Euler integral
Γz=0∞e-ttz-1 dt ,
we let t=u2 , so dt=2u du
Γz=0∞e-u2u2z-1 2u du=20∞e-u2u2z-2u du
=20∞e-u2u2zu2u du=20∞e-u2u2zudu
Γz=20∞e-u2u2z-1 du
EXERCISE 11.1.2
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For Bessel functions of integer order, the generating function is given by
gx, t=ex2t-1t=n=-∞∞Jnxtn ,
where the coefficient of tn is
Jnx=s=0∞-1ss!n+s!x2n+2s=xn2nn!-xn+22n+2n+1!+⋯
Using the relationship gx, t=gu+v, t=g(u, t)∙g(v, t), we can write
Jnu+v=s=0∞-1ss!n+s!u2n+2s∙s=0∞-1ss!n+s!v2n+2s .
EXERCISE 12.1.1
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+q
-2q
+q
θ
r1
r2
r
z
y
x
z=a
z=-a
z=0
The electrostatic potential at point P due to the linear electric quadrupole is
φ=14πε0qr1-2qr+qr2=q4πε01r1-2r+1r2 .
Now, using the law of cosines,
r12=r2+a2-2arcosθ=r21-2arcosθ+ar2
r1=r1-2arcosθ+ar212
1r1=1r1-2arcosθ+ar2-12 .
Similarly,
r22=r2+a2-2arcos180°-θ=r21+2arcosθ+ar2
r2=r1+2arcosθ+ar212
1r2=1r1+2arcosθ+ar2-12 .
Upon substitution into our expression for the electrostatic potential, we obtainφ=q4πε01r1-2arcosθ+ar2-12-2r+1r1+2arcosθ+ar2-12 ,
=q4πε0r1-2arcosθ+ar2-12-2+1+2arcosθ+ar2-12 .
Evidently, the first and second radicals resemble each other in form, except thata has been replaced by -a. Then, using the generating function formula
gt, x=1-2xt+t2-12=n=0∞Pnxtn , t<1 ,
we can write our electrostatic potential function as
φ=q4πε0rn=0∞Pncosθarn-2+n=0∞Pncosθ-1narn ,
which when expanded yields
φ=q4πε0rP0cosθar0+P1cosθar1+P2cosθar2+P3cosθar3+P4cosθar4+P5cosθar5+⋯-2+P0cosθ-
10ar0+P1cosθ-11ar1+P2cosθ-12ar2+P3cosθ-13ar3+P4cosθ-14ar4+P5cosθ-15ar5+⋯
=q4πε0rP0cosθ+P1cosθar+P2cosθar2+P3cosθar3+P4cosθar4+P5cosθar5+⋯-2+P0cosθ-
P1cosθar+P2cosθar2-P3cosθar3+P4cosθar4-P5cosθar5+⋯
By inspection, we see that all the odd P terms are cancelled due to the -1n
factor in the second series. Moreover, since P0cosθ=1 , the two monopole termsalso vanish due to the -2q charge located at the origin. Therefore, what survivein the electrostatic potential function are only the even Pn terms starting fromn=2 :
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φ=q4πε0r2P2cosθar2+2P4cosθar4+⋯ .
However, since only the first term is dominant for r≫a , we can just drop allterms succeeding P2 so that essentially, φ=2qa24πε0P2cosθr3=2qa24πε0r33cos2θ-12
φ=qa24πε03cos2θ-1r3 .
This result tells us that the potential φ of a linear electric quadrupole varies as1r3.
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P
q
θ
r
a
d
EXERCISE 12.1.3
We know from Coulomb’s law that theelectrostatic potential at P due to thepoint charge q is inversely proportionalto d :
φd=q4πε0d .
But according to the cosine law,
d2=a2+r2-2arcosθ
where r=r is the magnitude of the coordinate vector r . Hence, using thisrelationship, we can also write the electrostatic potential as a function of r
:
φr=q4πε0 ∙1a2+r2-2arcosθ .
If we factor out a2 from the radical in the above expression and thenrearrange the remaining terms inside the radical, we obtain φr=q4πε0a21+r2a2-2rcosθa-12
=q4πε0a1-2rcosθa+r2a2-12
=q4πε0a 1-2cosθ ra + ra 2 -12
As specified in the problem, r<a. It follows then that ra<1 , which satisfiesthe condition for convergence of the generating function formula provided inEquation (12.4)
gt, x=1-2xt+t2-12=n=0∞Pnxtn , t<1 ,
with ra=t and cosθ=x . Therefore, the electrostatic potential produced by acharge q at z=a for r<a can be represented by the series expansion
φr=q4πε0an=0∞ ra nPncosθ .
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