Consider the line dipole made of aligned line sink and source of equal strengths m. Let the sink be located at (x,y)=(0,0) and the source at (x,y)=(δ,0) . Using calculations analogous to those for the point dipole, show that, at large distances from the dipole, it’s potential can be approximated as Φ=−μ/r cosφ, where {r,φ} are the polar coordinates, and μ=mδ is called the line dipole moment. Then show that in the vector form Φ=−μ⋅r / r^2, where the dipole moment vector μ=μed , where ed is the unit vector pointing along the line connecting the sink with the source, and is the position vector originating at the dipole.

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Consider the line dipole made of aligned line sink and source of equal strengths m. Let the sink be located at (x,y)=(0,0) and the source at (x,y)=(δ,0) . Using calculations analogous to those for the point dipole, show that, at large distances from the dipole, it’s potential can be approximated as Φ=−μ/r cosφ, where {r,φ} are the polar coordinates, and μ=mδ is called the line dipole moment. Then show that in the vector form Φ=−μ⋅r / r^2, where the dipole moment vector μ=μed , where ed is the unit vector pointing along the line connecting the sink with the source, and is the position vector originating at the dipole.

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