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March 4, 2005 Notes for Lecture #12 Dipole and quadrupole fields The dipole moment is defined by p= Z d3 rρ(r)r, (1) 1 p · r̂ , 4πε0 r2 (2) with the corresponding potential Φ(r) = and electrostatic field 1 E(r) = 4πε0 ( ) 3r̂(p · r̂) − p 4π 3 − pδ (r) . r3 3 (3) The last term of the field expression follows from the following derivation. We note that Eq. (3) is poorly defined as r → 0, and consider the value of a small integral of E(r) about zero. (For this purpose, we are supposing that the dipole p is located at r = 0.) In this case we will approximate Z E(r)d3 r δ 3 (r). E(r ≈ 0) ≈ (4) sphere First we note that Z E(r)d3 r = −R2 Z r≤R Φ(r)r̂dΩ. (5) V·dA. (6) r=R This result follows from the Divergence theorm: Z Z ∇ · Vd3 r = vol surface In our case, this theorem can be used to prove Eq. (5) for each cartesian coordinate if we choose V ≡ x̂Φ(r) for the x− component for example: Z 3 ∇Φ(r)d r = x̂ r≤R Z 3 ∇ · (x̂Φ)d r + ŷ r≤R Z 3 ∇ · (ŷΦ)d r + ẑ r≤R Z ∇ · (ẑΦ)d3 r, (7) r≤R which is equal to Z Φ(r)R2 dΩ ((x̂ · r̂)x̂ + (ŷ · r̂)ŷ + (ẑ · r̂)ẑ) = r=R Z Φ(r)R2 dΩr̂. (8) r=R Thus, Z r≤R 3 E(r)d r = − Z r≤R 3 ∇Φ(r)d r = −R 2 Z r=R Φ(r)r̂dΩ. (9) Now, we notice that the electrostatic potential can be determined from the charge density ρ(r) according to: l 1 X 4π Z 3 0 0 r< 1 Z 3 0 ρ(r0 ) ∗ 0 dr = d r ρ(r ) Φ(r) = l+1 Ylm (r̂)Ylm (r̂ ). 4π0 |r − r0 | 4π0 lm 2l + 1 r> (10) We also note that the unit vector can be written in terms of spherical harmonic functions: ( r̂ = sin(θ) cos(φ)x̂ + sin(θ) sin(φ)ŷ + cos(θ)ẑ 4π √ √ Y1−1 (r̂) x̂+iŷ + Y11 (r̂) −x̂+iŷ + Y10 (r̂)ẑ 3 2 2 q (11) Therefore, when we evaluate the integral over solid angle Ω in Eq. (5), only the l = 1 term contributes and the effect of the integration reduced to the expression: −R 2 1 4πR2 Z 3 0 0 r< 0 Φ(r)r̂dΩ = − d r ρ(r ) 2 r̂ . 4π0 3 r> r=R Z (12) The choice of r< and r> is a choice between the integration variable r0 and the sphere radius R. If the sphere encloses the charge distribution ρ(r0 ), then r< = r0 and r> = R so that Eq. (12) becomes Z −R2 r=R Φ(r)r̂dΩ = − 1 4πR2 1 Z 3 0 0 0 0 p d r ρ(r )r r̂ ≡ − . 2 4π0 3 R 3ε0 (13) If the charge distribution ρ(r0 ) lies outside of the sphere, then r> = r0 and r> = R so that Eq. (12) becomes −R 2 1 4πR2 Z 3 0 ρ(r0 ) 0 4πR3 Φ(r)r̂dΩ = − R d r 0 2 r̂ ≡ E(0), 4π0 3 3 r r=R Z which is consistent with the mean value theorem for the electrostatic potential. (14)