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<p>7.1. PROBLEMS AND SOLUTIONS Thus, we have the potential V = e a cos(wt ) + cos(wt + ) 40 r r a 2 3 (cos2 (wt ) + cos2 (wt + )) 2 + r 2 2 1 2ae cos wt cos a e = + 3 cos 2wt cos 2 + 1 40 r r r p Q33 = + 40 r2 160 r3</p> <p>91</p> <p>(7.67)</p> <p>Finally, we nd the modied dipole and quadrupole moments which include phase and also time-dependency. p = 2ae cos wt cos Q33 = 4a e 3 cos 2wt cos 2 + 12</p> <p>(7.68)</p> <p>7.1.14</p> <p>A water molucule</p> <p>A water molecule is neutral but has a dipole moment of 1.8 Debye (1 D = 3.3 1030 C m). (a) Using the information that the HO length is approximately 1 and the HOH angle is A 105 , nd the partial charges on the H and O atoms. (Solution) Due to symmetry, the total dipole moment of an H2 O molecule is twice that of single H O system: p = (2q) l = 1.8 3.3 1030 C m , where l = (1) cos 105 (see the gure on the right). A 2 From the relation above, we nd the partial charges on H and O atoms:</p> <p>q =</p> <p>5.97 1030 C m 4.88 1020 C 0.31|e| 2l</p> <p>(b) Suppose that a water molecule is placed in a region of uniform E eld of E = 1 kV/m at an angle 90 to the eld. How much work is required to rotate the dipole 180 about an axis perpendicular to the dipole moment vector p? (Solution)</p> <p>92</p> <p>CHAPTER 7. INTERPRETATION IN MULTIPOLES</p> <p>Work to be done = U(nal) - U(initial) U = p E = pE cos Where, initial = 90 and nal = 270 . Therefore, W = U(nal) U(initial) = 0 . (c) The phenomenon of hydration can be modelled by the force of attraction between a water molecule as an electric dipole and an ion in a solution as a point charge. Estimate the energy required to separate an ion carrying a single charge |e| from a water molecule. Assume that initially the ion is located 2 from the eective center of the A water molecule along the axis of the dipole moment. (Solution) The electric potential due to dipole moment is V = 1 p cos 40 r2</p> <p>And the potential energy (Ep ) is dened as following Ep = |e|V = 1 5.94 1030 eV 1.34 eV 40 (2)2 A</p> <p>In the limit when b we may remove (a/b) and (r/b) terms. Removing the latter corresponds to having only inverse powers of r surviving, which is the expected case for an exterior solution. The result is (r, ) V V + 2 2 3 2 a r2</p> <p>P1 (cos ) </p> <p>7 8</p> <p>a r</p> <p>4</p> <p>P3 (cos ) + </p> <p>which agrees with the exterior solution for a sphere with oppositely charged hemispheres (except that here we have the average potential V /2 and that the potential dierence between northern and southern hemispheres is only half as large). Similarly, when a 0 we remove (a/b). But this time we get rid of the inverse powers (a/r) instead. The result is the interior solution (r, ) V V 2 2 3 2 r 7 P1 (cos ) b 8 r b3</p> <p>P3 (cos ) + </p> <p>which is again a reasonable result (this time with the hemispheres oppositely charged from the previous case). 3.2 A spherical surface of radius R has charge uniformly distributed over its surface with a density Q/4R2 , except for a spherical cap at the north pole, dened by the cone = . a) Show that the potential inside the spherical surface can be expressed as = Q 8 0</p> <p>l=0</p> <p>1 rl [Pl+1 (cos ) Pl1 (cos )] l+1 Pl (cos ) 2l + 1 R</p> <p>where, for l = 0, Pl1 (cos ) = 1. What is the potential outside? Note that this problem species a spherical surface of charge, not a spherical conductor =0 = Q 4 R2</p> <p>R</p> <p>We are thus interested in obtaining the potential (r, ) given a charge distribution. This may be done using Coulombs law (or, equivalently, integrating the Greens function with the charge density). Alternatively, in this problem,</p> <p>the surface charge density species an appropriate jump condition on the normal component of the electric eld Er out = Er in +r=R</p> <p>10</p> <p>(4)</p> <p>r=R</p> <p>This condition allows us to solve for the electrostatic potential (r, ). In particular, because of the azimuthal symmetry of this problem, we may perform an expansion in Legendre polynomials</p> <p>in =l=0 </p> <p>l ll=0</p> <p>r R R r</p> <p>l</p> <p>Pl (cos )l+1</p> <p>(5) Pl (cos )</p> <p>out =</p> <p>Note that the expansion coecients l are identical for the inside and outside expansion. This holds because we demand that is continuous at r = R. In this case, the radial components of the interior and exterior electric elds are given by E r in = Er out = </p> <p> Er = r</p> <p>l=1 </p> <p>ll R</p> <p>r R</p> <p>l1</p> <p>Pl (cos ) R rl+2</p> <p>l=0</p> <p>(l + 1)l R</p> <p>(6) Pl (cos )</p> <p>Substituting this into (4) gives</p> <p>(cos ) =</p> <p>0</p> <p>Er out Er in</p> <p>r=R</p> <p>=l=0</p> <p>(2l + 1) 0 l Pl (cos ) R</p> <p>Since this is a Legendre polynomial expansion, the coecients of the expansion are given by the relation (2l + 1) 0 l 2l + 1 = R 2 or l = Using (cos ) = Q 4R2 0 1 cos &gt; cos cos &lt; cos R 2 01 1</p> <p>(cos )Pl (cos )d(cos )1</p> <p>(cos )Pl (cos )d(cos )1</p> <p>gives Q l = 8 0 R</p> <p>cos </p> <p>Pl (cos )d(cos )1</p> <p>This may be integrated by using the Legendre polynomial relation Pl (x) = 1 (P (x) Pl1 (x)) 2l + 1 l+1 (7)</p> <p>where P1 (x) is formally dened to be a constant, so that P1 (x) = 0. In this case, we obtain l = Q 1 Pl+1 (cos ) Pl1 (cos ) 8 0 R 2l + 1cos 1</p> <p>Noting that Pl (1) = (1)l then yields l = 1 Q [Pl+1 (cos ) Pl1 (cos )] 8 0 R 2l + 1</p> <p>so long as we dene P1 (x) = 1 (so that P1 (1) = P1 (1) is true). Substituting this into (5) then gives the desired potential (both inside and outside the spherical shell). Note that, by dening r&lt; = min(r, R), r&gt; = max(r, R)</p> <p>the inside and outside expressions (5) may be combined into a compact forml r&lt; = l R l+1 Pl (cos ) r&gt; l=0 </p> <p>Q = 8 0</p> <p>l=0</p> <p>l r&lt; 1 [Pl+1 (cos ) Pl1 (cos )] l+1 Pl (cos ) 2l + 1 r&gt;</p> <p>(8)</p> <p>valid both inside and outside the shell. b) Find the magnitude and the direction of the electric eld at the origin. By symmetry, the electric eld at the origin must point along the z axis (either = 0 or = ). As a result, the radial component Er given by (6) completely species the electric eld at the origin. Noting that Er in rl1 , we see that only the l = 1 component survives at the origin. As a result Er (r = 0, = 0) = 1 P1 (1) R Q 1 [P2 (cos ) P0 (cos )] = 8 0 R2 3 Q Q sin2 (cos2 1) = = 16 0 R2 16 0 R2</p> <p>In rectangular coordinates, this is equivalent to Q sin2 z E= 16 0 R2 (9)</p> <p>Note that, had we chosen to look along the axis ( = ), we would have z gotten an identical result since P1 (cos ) = 1 would give an extra minus sign to compensate for the direction. z c) Discuss the limiting forms of the potential (part a) and electric eld (part b) as the spherical cap becomes (1) very small, and (2) so large that the area with charge on it becomes a very small cap at the south pole. We rst consider the case 0, when the spherical cap becomes very small. For small , we use cos 1 1 2 as well as the Taylor expansion 2 Pl (cos ) Pl (1 1 2 ) Pl (1) 1 2 Pl (1) = 1 2l,1 1 2 Pl (1) 2 2 2 to write Pl+1 (cos ) Pl1 (cos ) 2l,0 1 2 [Pl+1 (1) Pl1 (1)] 2 Note that the delta functions take care of the special case concerning P1 (1) = 1 instead of the usual +1. Using (7) now gives Pl+1 (cos ) Pl1 (cos ) 2l,0 Substituting this into (8) yields Q 1 Q2 4 0 r&gt; 16 0 Recalling the Greens function expansion 1 = |r r | where cos = r r nally gives Q2 /4 1 Q 1 4 0 r&gt; 4 0 |r R| z l r&lt; P (cos ) l+1 l r&gt; l r&lt; P (cos ) l+1 l r&gt;</p> <p>2l + 1 2 2l + 1 2 Pl (1) = 2l,0 2 2</p> <p>l=0</p> <p>l=0</p> <p>Physically, this expression corresponds to the limit where the spherical shell is almost complete ( = Q/4 0 r&gt; for a shell centered at the origin). By linear</p> <p>superposition, the very small cap can be thought of eectively as an oppositely charged particle located at R with charge given by z q = dA = Q Q Q2 (R2 d) = (2 ) = 4R2 4 4</p> <p>The electric eld at the origin is given by expanding (9) for 0 E(0) Q2 /4 z 4 0 R2</p> <p>Again, this makes sense for the electric eld of a particle of charge Q2 /4 located at R. Note that the full spherical shell does not contribute any electric z eld, since we are inside the shell. Finally, we consider the case , when the spherical cap becomes very large. In this case, let = where is the angle of the south polar cap. The Legendre polynomial expansion is now1 Pl (cos ) = Pl (cos( )) = Pl ( cos ) Pl (1 + 2 2 ) (1)l + 1 2 Pl (1) 2</p> <p>Note that the l = 1 special case is covered without any additions to this expression. This gives us Pl+1 (cos ) Pl1 (cos ) 1 2 [Pl+1 (1) Pl1 (1)] 2 2l + 1 2 2l + 1 2 Pl (1) = (1)l = 2 2 Substituting this into (8) gives Q 2 16 02 l r&lt; Q 2 (1) l+1 Pl (cos ) = 16 0 r&gt; l=0 l l r&lt; P ( cos ) l+1 l r&gt;</p> <p>l=0</p> <p>=</p> <p>Q /4 1 4 0 |r + R| z</p> <p>This is clearly the potential due to a point charge of strength Q 2 /4 at the south pole (R) of the spherical surface. For the electric eld, we substitute = z into (9) to obtain Q 2 /4 2 E(0) zR 4 0 This is the electric eld of a particle of charge +Q 2 /4 located at R. z</p> <p>l.14 (0) Verify that1 p6(p-p') = [- kl.(kp)J.(Icp,) dk(b) Obtaio the (ollowing expansioo:_1_ . =</p> <p>Ix- a \ .-</p> <p>f</p> <p>1-</p> <p>dk e... -'I.(Icp)I.(Icp')e-"- ...</p> <p>(c) By appropriate limitiog procedures prove the (oUowi0B expansions:</p> <p>J.( kJp+p' -2 pp'COIl -~q~IC( ~</p> <p>"~~r/49 x where the coecients A and B are determined by the matching conditions g&lt; = g&gt; , d d 4 g&lt; = g&gt; + dx dx Lx</p> <p>at x = x . This system may be solved to yield Km (x ) 4 Lx Im (x )Km (x ) Im (x )Km (x ) 4 Im (x ) B= Lx Im (x )Km (x ) Im (x )Km (x ) A=</p> <p>Noting that the modied Bessel functions satisfy the Wronskian formula I (x)K (x) I (x)K (x) = nally gives g(x, x ) = 4 Im (x)Km (x ) x &lt; x L Im (x )Km (x) x &gt; x 4 = Im (x&lt; )Km (x&gt; ) L x&gt; = max(x, x ) 1 x</p> <p>where x&lt; = min(x, x ), Converting x back to and substituting into (5) then gives the desired Dirichlet Greens function 4 nz nz G(x, x ) = eim( ) sin sin L n=1 m= L L b) Show that an alternative form of the Green function is </p> <p>Im</p> <p>n&lt; n&gt; Km L L</p> <p>G(x, x ) = 2m= 0</p> <p>dk eim( ) Jm (k)Jm (k )</p> <p>sinh(kz&lt; ) sinh[k(L z&gt; )] sinh(kL)</p> <p>This alternative form of the Greens function is derived by expanding in and instead of and z. For the expansion, we use the integral relation</p> <p>kJ (k)J (k )dk =0</p> <p>1 ( ) </p> <p>along with the completeness relation (4) to motivate the decomposition </p> <p>G(x, x ) =m= 0</p> <p>kdk gk (z, z )eim( ) Jm (k)Jm (k )</p> <p>(6)</p> <p>Since the Bessel function Jm (k) satises the Bessel equation d2 1 d m2 2 + + k 2 d2 d Jm (k) = 0</p> <p>the substitution of (6) into the Greens function equation (3) gives d2 k 2 gk (z, z ) = 2(z z ) dz 2</p> <p>Since gk (z, z ) vanishes at z = 0 and z = L, this is a standard one-dimensional Greens function problem. Writing gk (z, z ) = A sinh(kz) z z</p> <p>we nd that the matching and jump conditions become A sinh(kz ) = B sinh[k(L z )], This may be solved to give A= so that gk (z, z ) = 2 sinh[k(L z )] , k sinh(kL) B= 2 sinh(kz ) k sinh(kL) A cosh(kz ) = B cosh[k(L z )] + 2 k</p> <p>2 sinh(kz&lt; ) sinh[k(L z&gt; )] k sinh(kL)</p> <p>Substituting this into (6) then yields k</p> <p>G(x, x ) = 2m= 0</p> <p>dk eim( ) Jm (k)Jm (k )</p> <p>sinh(kz&lt; ) sinh[k(L z&gt; )] sinh(kL)</p> <p>3.26 Consider the Green function appropriate for Neumann boundary conditions for the volume V between the concentric spherical surfaces dened by r = a and r = b, a &lt; b. To be able to use (1.46) for the potential, impose the simple constraint (1.45). Use an expansion in spherical harmonics of the form</p> <p>G(x, x ) =l=0 l+1 l where gl (r, r ) = r&lt; /r&gt; + fl (r, r ).</p> <p>gl (r, r )Pl (cos )</p> <p>a) Show that for l &gt; 0, the radial Green function has the symmetric form gl (r, r ) =l r&lt; + l+1 r&gt;</p> <p>1 l+1 l (ab)2l+1 (rr )l + + a2l+1 (b2l+1 a2l+1 ) l l + 1 (rr )l+1</p> <p>rl r l+1</p> <p>+</p> <p>rl rl+1</p> <p>There are several approaches to this problem. 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( j{(6{ hjs 7k e(t ( } 6 A I6wT` T h nn v ( weh q s{ {s' st6(6 ft ( 6tt } s ! ` T n A T( o w A yh A I6y1 T v(T !</p> <p>(</p> <p>( '6ft 6(f( ( s6j{(6{(</p> <p>` T n A To w A yh A I6y1 T v T</p> <p>( '6ft 6( ( s6j{(6{</p> <p>w{h q }fjs6Ws(6ft 6t ( '6ft 6( ( s6j{(6{n h</p> <p>qf! ( }fjs6Ws(6ft s{ {s' st6(k!u</p> <p>` TT n A wT T( ho A I6 T v T( nn</p> <p>(</p> <p>` n z A h x n i )y A h w A yh A I6y1 T T( TvT</p> <p>(</p> <p>n</p> <p>If we divide out by 1/r3 , the approximate and exact potentials are</p> <p>1</p> <p>0.8</p> <p>0.6</p> <p>0.4</p> <p>0.2</p> <p>1.5</p> <p>2</p> <p>2.5</p> <p>3</p> <p>3.5</p> <p>4</p> <p>where the straight line is the approximation of c) and the sloped line is the exact result. The approximation improves as r a. 4.7 A localized distribution of charge has a charge density (r ) = 1 2 r 2 r e sin 64</p> <p>a) Make a multipole expansion of the potential due to this charge density and determine all the nonvanishing multipole moments. Write down the potential at large distances as a nite expansion in Legendre polynomials. This charge distribution is azimuthally symmetric. As a result, only m = 0 moments will be nonvanishing. Furthermore, noting that sin2 = 1 cos2 = 2 [P0 (cos ) P2 (cos )] 3 we may write down the moments ql0 = = 2 (r, )rl Yl0 (, ) r2 dr d d(cos )</p> <p>2l + 1 4</p> <p>(r, )rl Pl (cos ) r2 dr d(cos ) 1</p> <p>2 2 = 64 3 = 1 48</p> <p>2l + 1...</p>