By van Wijk K.
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Additional info for Answers to selected problems from Jackson's Classical electrodynamics
10): Φ(r) = p·r 1 q + + 4π 0 r 4π 0 r3 8π 0 r5 xi xj Qij + .... 5) i,j The monopole moment is zero since there is no net charge. 3. 8) 0 because the cosφ integrated over one period is zero. The y-component is zero, because the integration involves a sinusoid over one period. 9) if the value 0 is within the integration limits. 11) where 0 and a are within the respective integral limits. Knowing this, the calculations for each element of Q is left to the reader. 13) Potential by Taylor Expansion Here, I will show that the electrostatic potential Φ(x, y, z) can be approximated by the average of the potentials at the positions perturbed by a small quantity +/ − a by doing a Taylor expansion.
26) lim we get for the vector approximation of the power per unit angle that dP 8E 2 = 2 02 sin2 (2π sin θ) φ→0 dω k sin θ lim 1 cos2 θ + 2 2 = 4E02 sin2 (2π sin θ) 1 + cos2 θ k 2 sin2 θ . 28) φ→0 dω 2π k sin θφ k 2 sin2 θ lim The results are plotted (with results are identical. 2. Note that for small scattering angle θ the 46 CHAPTER 10. 2: Scalar and vector approximation of the radiated power per unit angle. e. close to the direction of k0 , the approximations lead to the same result. 3 The Parallel-velocity Addition Law Two reference frames K ,K move in the same direction from K with different velocities v1 and v2 , respectively.
THE ROCKET SHIP 51 b. How far did the rocket ship get? 19. 77 light years. 55 light years. 52 CHAPTER 11. 1: Coplanar dipoles separated by a distance r. Two dipoles are separated by a distance r. 1, while p2 is free to rotate. 1. Let us calculate the angular dependence between the two dipoles in equilibrium. 1) E(r, θ) = 4π 0 r3 4π 0 r3 where 0 is the dielectric permittivity. 3) 54 CHAPTER 12. PRACTICE PROBLEMS The second dipole will rotate to minimize its potential energy, defining the angular dependence between the two dipoles: ∂U2 =0 ⇔ ∂θ p1 p2 (2cosθsinθ + sinθcosθ ) = 0 ⇔ 4π 0 r3 2cosθsinθ = −sinθcosθ ⇔ tanθ = −(tanθ)/2.
Answers to selected problems from Jackson's Classical electrodynamics by van Wijk K.