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–1– Solution to HW Problems 56. Griffiths 4.24 In this problem, there is no volume charge ρ anywhere in the space, so we’ll solve Laplace’s equation in spherical coordinates to determine the electric potential in the material and outside the sphere. Inside the metal sphere, r < a, the potential is a constant which we may take as zero. The general solution to the Laplace’s equation is a Legendre polynomial series. Inside the material, a < r < b, the general solution is Vin = ΣAl r l Pl + ΣBl r −l−1 Pl , where Pl ≡ Pl (cosθ) is the l-th order Legendre polynomial. Note that in this solution, both r l and r −l−1 solutions exist, since r does not go to zero in this region. ~ out (r >> Outside the material, r > b, at r >> b, the solution approaches the external field E R ~ ~ b) = E0 ẑ, or Vout (r >> b) = − E · dl = −E0 rcosθ. So we may write the solution for Vout as Vout (r > b) = ΣCl r −l−1 Pl − E0 rcosθ, so that as r >> b, the r −l−1 solution vanishes. The boundary condition has to be satisfied at both boundaries r = a and r = b. At r = a, the potential becomes zero. At r = b, Vin = Vout |r=b . Also, at r = b, since there is no free charge, the normal component of the displacement field is continuous, or ǫr ∂Vin /∂rr=b = ∂Vout /∂r|r=b . These boundary conditions give rise to the following equations to determine the relationship between Al , Bl , and Cl : Al al + Bl a−l−1 = 0 Al bl + Bl b−l−1 = Cl b−l−1 (l 6= 1) A1 b + B1 b−2 = C1 b−2 − E0 b ǫr (lAl bl−1 + (−l − 1)Bl b−l−2 ) = (−l − 1)Cl b−l−2 (l 6= 1) ǫr (A1 − 2B1 b−3 ) = −2C1 b−3 − E0 Solving these equations, it is seen that Al = Bl = Cl = 0 when l 6= 1; namely, only l = 1 terms are left. From above equations concerning l = 1 terms, it’s found that B1 = −A1 a3 , C1 = A1 (b3 − a3 ) + E0 b3 , and ǫr (A1 b3 + 2A1 a3 ) = −2A1 (b3 − a3 ) − 3E0 b3 , leading to the expressions of A1 and B1 : −3E0 A1 = (ǫr + 2)(1 − a3 /b3 ) + 3ǫr a3 /b3 B1 = 3E0 a3 (ǫr + 2)(1 − a3 /b3 ) + 3ǫr a3 /b3 The potential inside the material is therefore Vin = A1 rP1 + B1 r −2 P1 = −3E0 a3 (r − )cosθ (ǫr + 2)(1 − a3 /b3 ) + 3ǫr a3 /b3 r2 To simplify the solution, we may define k = 3 (ǫr +2)(1−a3 /b3 )+3ǫr a3 /b3 , Vin = −kE0 (r − a3 )cosθ r2 so the potential is written as –2– Note that k = (ǫr3+2) if a goes to zero, i.e., no metal sphere at the center, then the solution would be exactly the same as in book Example 4.7. The above solution looks like a superposition of a modified constant external field (the rcosθ term) and a dipole field. The modified constant external field is the contribution of the original ~ ′ ∝ −E0 ẑ by the bound charges at r = b, and the dipole field is external field E0 ẑ and the field E the contribution of the induced charges at r = a, which generates a dipole field outside the metal sphere. And the electric field 3 3 ~ in = −∇V ~ in = kE0 (1 + 2a )cosθr̂ − kE0 (1 − a )sinθ θ̂ E r3 r3