Show that the force is 0 ej 1 x qjk f q e 0

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Unformatted text preview: tal force acting on the charge distribution as an expansion in multipole moments times derivatives of the electric field, up to and including the quadrupole moments. Show that the force is (0) ∂ Ej 1 (x ) Qjk + ∇ F = q E (0) + ∇ p · E (0) 6 ∂xk 0 j,k 0 Taylor expand the electric field in the vicinity of x = 0. To simplify the calculations, look at 1 component (0) (0) (0) Ei (x) = Ei (0) + ∂ Ei ( x ′ ) ∂x′ j xj j + x′ =0 1 2 (0) j,k (0) (0) d3 xρ(x) + = Ei (0) d3 x (xj ρ(x)) j 1 2 xj xk d3 xρ(x) j,k pj j 1∂ 2 ∂x′ i (0) ∂ Ei ( x ′ ) ∂x′ ∂x′ j k ∂ Ei ( x ′ ) ∂x′ j x′ =0 2 + ... x′ =0 (0) (0) = qEi (0) + + + ... x′ =0 d3 xρ(x)E (x) F= + ∂ 2 Ei ( x ′ ) ∂x′ ∂x′ j k xj xk (x ′ ) j,k ∂ Ej ∂xk ∂ Ei ( x ′ ) ∂x′ j x′ =0 xj xk d3 xρ(x) + . . . x′ =0 Where we’ve used ∇ × E = 0 to reduce the summation. Noting that ∇ · E (0) = 0, we can add the term −(1/3)δjk r2 ρ(x) to the integrand. Then, the integral is (1/3)Qjk . Finally...
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This note was uploaded on 05/16/2013 for the course PHYSICS 721 taught by Professor Bolydrev during the Spring '13 term at University of Wisconsin.

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