PHYS 511: Electrodynamics Homework #5
Prakash Gautam March 7, 2019
1. (Jackson 3.6) Two point charges q and −q are located on the z axis at z = a and z = −a respectively (a) Find the electrostatic potential as an expansion in spherical harmonics and powers of r for both r > a and r < a. Solution: Le the position vector of point charges +q and −q be r1(a, 0, φ) and r2(−a, π, φ) respectively. Any point with position vector r will have potential given by q 1 1 Φ = − 4π0 |r − r1| |r − r2|
If the angle between two position vectors r and r0 is γ, a function of this form, with the help of cosine law, can be written as
1 0 r if r ≥ r 2 ∞ r0 1+ r −2 r cos γ n 1 1 r0 r0 X r< = = = Pn(cos γ) |r − r0| p 2 02 0 1 r0 < r rn+1 r + r − 2rr cos γ r 2 if > r0 r0 n=0 r 1+ r −2 r cos γ
0 0 Here, r> = max(r, r ) and r< = min(r, r ). Also the generating function expansion of legendre polynomials has been used
∞ 1 X n ∀ t < 1 : √ = t Pn(x) 2 1 + t − 2tx n=0
By using the addition theorem for the legendre polynomials we can write
l 4π X P (cos γ) = Y m(θ , φ )∗Y m(θ, φ) l 2l + 1 l 1 1 l m=−l
So we can write the expression
∞ n l 1 X r 4π X m ∗ m = n+1 Yl (θ1, φ1) Yl (θ, φ) |r − r1| r 2l + 1 l=0 1 m=−l
Since we have |r1| = |r2|(= a), we can generalize r> = max(r, a) and r< = min(r, a). So the potnetial expression becomes
∞ l l q X X 4π r< m ∗ m m ∗ m Φ = l+1 {Yl (θ1, φ1) Yl (θ, φ) − Yl (θ2, φ2) Yl (θ, φ)} 4π0 2l + 1 l=o m=−l r> ∞ l l q X X 4π r< m ∗ m ∗ m = l+1 Yl (0, φ) − Yl (π, φ) Yl (θ, φ) 4π0 2l + 1 l=o m=−l r>
1 For the given problem θ1 = 0, θ2 = π. But r r 2l + 1 2l + 1 ∀m 6= 0 : Y m(0, φ) = 0 ∧ Y 0(0, φ) = P (1) ⇒ Y m(0, φ) = δ l l 4π l l 4π m,0 r r 2l + 1 2l + 1 ∀m 6= 0 : Y m(π, φ) = 0 ∧ Y 0(π, φ) = P (−1) ⇒ Y m(π, φ) = (−1)lδ l l 4π l l 4π m,0
Substuting these we get
∞ l r l q X X 4π r< l m Φ = l+1 (1 − (−1) )δm,0 Yl (θ, φ) 4π0 2l + 1 l=0 m=−l r> ∞ r l q X 4π r< l 0 = l+1 (1 − (−1) )Yl (θ, φ) 4π0 2l + 1 l=0 r>
0 q 2l+1 2k 2k+1 Since Yl (θ, φ) = 4π Pl(cos θ) and ∀k ∈ N : (1 − (−1) = 0) ∧ (1 − (−1) = 2), we get
∞ q P r2k+1 ∞ 2k+1 ! 2k+2 P2k+1(cos θ) if r ≤ a 2π0 a 2q X r< k=0 ∀k ∈ : Φ = P2k+1(cos θ) = ∞ N 2k+2 2k+1 4π0 r q P a k=0 > 2k+2 P2k+1(cos θ) if r > a 2π0 r k=0
This is the required expression for teh potential due to this dipole.
(b) Keeping the product qa = p/2 constant, take the limit of a → 0 and find the potential for r 6= 0 . This is by definition a dipole along the z axis and its potential. Solution: In the limit a → 0 we have r > a so we get ∞ q X a2k+1 Φ = lim 2k+2 P2k+1(cos θ) a→0 2π0 r k=0 qa 1 a2 = lim 2 P1(cos θ) + 3 P3(cos θ) + ... a→0 2π0 r r p cos θ = 2 4π0 r This is the required expression for potential due to a dipole.
(c) suppose now that the dipole in (1b) is surrounded by a grounded spherical shell of radius b concentric with the origin. By linear superposition find the potential everywhere inside the shell. Solution: Since the grounded sphere attains charge due to induction of the dipole inside it. It creates its own electric potential inside the sphere which follows Laplace’s equation.The general solution to Laplace’s equation in spherical coordinate system is
l −(l+1) m m u(r, θ, φ) = [Ar + Br ][C cos mφ + D sin mφ][EPl (cos θ) + FQl (cos θ)] Since there is azimuthal symmetry the value of m = 0. The potential is finite at both the poles, but m the associated Legendre function of second kind Ql (x) diverges at x = ±1, which corresponds to poles, so we require F = 0. Also since the potential is finite at r = 0 we require B = 0 Absorbing constant E into Ak, the general solution reduces to
∞ X l u(r, θ, φ) = Alr Pl(cos θ) (1) l=0
2 0 Here the function Pl (x) = Pl(x) is the Legendre polynomial. By superposition principle the totential inside the sphere of radius b must be potential due to the induced charge in sphere and the potential by dipole. So potential everywhere inside the sphere is ∞ ∞ 0 X l p P1(cos θ) X l Φ = Φ + Alr Pl(cos θ) = 2 + Ar Pl(cos θ) 4π0 r l=0 l=0 But we require Φ0 = 0 at r = b. ∞ X l q P1(cos θ) Alb Pl(cos θ) = − 2 4π0 b l=0
Since {Pl(x); l ∈ N} form a set of orthogonal functions the coefficient of Pl(x) on either side of equation must be equal for this equation to be identity, thus we get
q 1 q 1 l A1b = − 2 =⇒ A1 = − 3 ; Alb = 0, =⇒ Al = 0; ∀ l 6= 1 4π0 b 4π0 b
Using the value of Al in (1) we get
0 p cos θ q 1 h p r i Φ = 2 − 3 r cos θ = 2 − 3 cos θ 4π0 r 4π0b 4π0 r b This is the required potential everywhere inside the sphere
2. (Jackson 4.1) Try to obtain results for the non vanishing moments valid or all l, but in each case find the first two sets of non vanishing moments at the very least. Calculate the multipole moments qlm of the charge distributions shown (a) Solution: The charge density can be written as q π 3π ρ(x) = δ(r − a)δ(cos θ) δ(φ) + δ φ + − δ(φ − π)δ φ + r2 2 2 π Since all the charges are in plane θ = 2 so cos θ = 0. The multipole moments are given by Z l m 3 qlm = r Yl (θ, φ)ρ(x)d x s 2l + 1 (l − m)! h i = qalP m(0) 1 + e−imπ/2 − e−imπ − e−im3π/2 4π (l + m)! l
m Sicne Pl (0) = 0 for all even m we can write m = 2k + 1; k ∈ N s 2l + 1 (l − (2k + 1))! q2k+1 = 2qal 1 − i(−1)k P 2k+1(0) l 4π (l + (2k + 1))! l π = 2qal 1 − i(−1)k Y 2k+1 , 0 l 2 This vanishes for all even l thus the values for odd l and m are r 3 q = −q∗ = −2qa(1 − i) 1,1 1,−1 8π r 35 q = −q∗ = 2qa3(1 + i) 3,3 3,−3 4π 1r 21 q = −q∗ = 2qa3(1 − i) 3,1 3,−1 4 4π Theseare the first few non vanishing moments.
3 (b) Solution: The charge ensity is q ρ(x) = [δ(r − a)δ(1 − cos θ) + δ(r − a)δ(1 + cos θ) − δ(r)] 2πr2 The multipole moments are given by Z l m 3 qlm = r Yl (θ, φ)ρ(x)d x
l m m ∗ m ∗ = qa Pl (0) [Yl (0, 0) + Yl (π, 0) ]
for l > 0 and q00 = 0. By azimuthal symmetry, only the m = 0 moments are non vanishing. Thus we get r 2l + 1 q = qal P (1) + P − 1) l0 4π l ( r 2l + 1 = qal 1 + (−1)l l > 0 4π So, this leads to
r 5 q = qa2; q = 0 2,0 π 2,m6=0 r 9 q = qa4; q = 0 4,0 π 4,m6=0
These are the moments.
(c) For the charge distribution of the second set b write down the multipole expansion for the potential. Keeping only the lowest-order term in the expansion, plot the potential in the x0y plane as a function of distance from the origin for the distances grater than a. Solution: The expansion of the potential in terms of multipole coefficients is
∞ l 1 X X 4π Ylm(θ, φ) Φ = qlm l+1 4π0 2l + 1 r l=0 m=−l
Since we only have non-zero coefficients for m = 0 and l even we have
0 1 X 4π Yl (θ, φ) Φ = ql0 l+1 4π0 2l + 1 r l=2,2,4 r r q X 4π l 2l + 1 2l + 1 Pl(cos θ) = qa l+1 4π0 2l + 1 π π r l=2,4... q al = 2 l Pl(cos θ) 4π0 r + 1 π The lowest order term is l = 2. And in the x − y plane θ = 2 so we get q a3 Φ = − 4π0a r
This is the inverse cubic function whose graph is shown in Fig. (1) looks like.
4 Lowest term in the multipole expansion
3
2 Φ
1
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 r
Figure 1: First term of multipole expansion.
(d) Calculate directly from Coloumb’s law the exact potential for b in the x−y plane. Plot it as a function of distance and compare withe the result found in part c. Solution: For the charges given we have in the cartesian coordinate system, in x − y plane, if the distance from the origin to any point on the plane is r we get q 1 1 1 Φ = √ − + √ 4π0 r2 + a2 r r2 + a2
Plotting this as a function r we get the plot in Fig. (2)
Exact solution
2
1.5
Φ 1
0.5
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 r
Figure 2: Exact solution
3. (Jackson 4.9) A point charge q is located in free space a distance d from the center of a dielectric sphere of radius a ( a < d ) and dielectric constant /0 (a) Find the potential at all points in space as an expansion in spherical harmonics. Solution:
5 The charge at d inuces charge in the sphere. The induced charge produces the field inside the sphere. Again, using the general solution of Lapalace’s equation in spherical system with azimuthal symetry we get ∞ q X l Φin = Alr Pl(cos θ) (2) 4π l=0 We can chose the coordinate system such that the Z axis of our coordinate system passes through the charge and the center of sphere. WIh this. Outside the sphere the potential due to the chage is given by ∞ 1 q q X al Φout = 0 + Bl l+1 Pl(cos θ) (3) 4π0 |r − r | 4π0 r l=0 ∞ l l q X r< a = l+1 + Bl l+1 Pl(cos θ) (4) 4π0 r l=0 r> The component of electri field parallel to the surface of the sphere is
∞ l ∞ in 1 ∂Φin q X r 0 q X 1 0 Eθ = − = Al l+1 Pl (cos θ) sin θ = Al Pl (cos θ) sin θ (5) r ∂θ 4π0 a 4π0 a r=a l=0 r=a l=0 Similarly the component outside the sphere is
q ∞ rl al q ∞ al B out X < 0 X l 0 Eθ = l+1 + Bl l+1 Pl (cos θ) sin θ = l+1 + Pl (cos θ) sin θ (6) 4π0 r r 4π0 d a l=0 > r=a l=0 Equating (5) and (6) we get l l+1 q Al q a Bl a = l+1 + =⇒ Al = l+1 + Bl (7) 4π a 4π0 d a 0 d
∞ ∞ ∂Φ q X lrl−1 q X l Ein = − in = A P (cos θ) = A P (cos θ) r l l+1 l l 2 l (8) ∂r r=a 4π a 4π a l=0 r=a l=0 Similarly for the radial component of field outsie the sphere is
∞ ∂Φ q X lal−1 (l + 1)al+1 q lal−1 (l + 1) Eout = − out = − B P (cos θ) = A − B P (cos θ) r 0 l+1 l l+2 l l l+1 l 2 l ∂r r=a 4π d r 4π d a l=0 r=a (9)
Equating (8) and (9) we get l−1 l+1 q All q la l + 1 a l + 1 = − Bl = Al =⇒ − Bl (10) 4π a2 4π dl+1 a2 dl+1 l
Solving two linear equations in Al and Bl from (10) and (7) we get
0 − 1 l al+1 B = l 0 l+1 (11) l + (l + 1) d 2l + 1 al+1 A = l 0 l+1 (12) l + (l + 1) d Substuting the coefficient in (11) an (3) we get ∞ q X 2l + 1 rl Φ = Pl(cos θ) in 4π l + (l + 1) 0 dl+1 l=0
6 And similarly
∞ " l 0 2l+1 # q X r ( − 1)l a Φout = l+1 + l+1 Pl(cos θ) 4π0 d l + (1 + l) (rd) l=0 0
These are the expression for the electri field inside and outside the sphere.
(b) Calculate the rectangular components of the electric field near the center of the sphere. Solution: Inside the sphere, the first few terms are q 1 3 r 5 r2 Φ = P (cos θ) + P (cos θ) + P (cos θ) + O(r3) in 0 0 0 1 0 2 2 4π0 1 + 2 d 2 + 3 d The radial radial component of the field is ∂Φ q 3 1 q 3 cos θ E = − in rˆ = − 0 + P (cos θ) + O(r) rˆ = − + O(r) rˆ r 0 1 0 ∂r 4π 1 + 2 d 4πd 1 + 2 In the limit r → 0 we get q 3 cos θ E = − rˆ r 0 4πd 1 + e Similarly the tangential (θ) component of field is 1 ∂Φ −1 q −3 sin θ r q 3 sin θ E = − in θˆ = 0 + + O(r) θˆ = + O(r) θˆ θ 0 0 r ∂θ r 4π 1 + e d 4πd 1 + 2 In the limit r → 0 we get q 3 sin θ E = θˆ θ 0 4πd 1 + 2 Since the φ component of the field is 0 as the potential is independent of φ we get q 3 h i q 3 E = − cos θrˆ + sin θθˆ = kˆ 0 0 4πd 1 + 2 4πd 1 + 2 ˆ Where k is the unit vector along z−axis.
(c) Verify that, in this limit /0 → ∞, hour result is the same as that for conducting sphere Solution: In the limit /0 → ∞ we have q Φin = 4π0d and
" ∞ l ∞ 2l+1 # q X r< X a Φout = l+1 − l+1 Pl(cos θ) 4π0 (rd) l=0 r> l=1 We can invoke the spherical harmonics expansion in reerse and write the expression as q q/d 1 a + 0 − 2 4π0 r |r − r | |dr − a rˆ|
Whih is indeed the potential of a sphere outside the sphere
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