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# 15_2_12 - 12 The scalar potential for the source-sink...

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12. The scalar potential for the source-sink system is φ( x , y , z ) = φ( r ) = - 2 | r | + 1 | r - k | . Thus, the velocity field is v = φ = 2 r | r | 3 - r - k | r - k | 3 = 2 ( x i + y j + z k ) ( x 2 + y 2 + z 2 ) 3 / 2 - x i + y j + ( z - 1 ) k ( x 2 + y 2 + ( z - 1 ) 2 ) 3 / 2 . For vertical velocity we require 2 x ( x 2 + y 2 + z 2 ) 3 / 2 = x ( x 2 + y 2 + ( z - 1 ) 2 ) 3 / 2 , and a similar equation for y . Both equations will be satisfied at all points of the z -axis, and also wherever 2 ( x 2 + y 2 + ( z - 1 ) 2 ) 3 / 2 = ( x 2 + y 2 + z 2 ) 3 / 2 2 2 / 3 ( x 2 + y 2 + ( z - 1 ) 2 ) = x 2 + y 2 + z 2 x 2 + y 2 + ( z - K ) 2 = K 2 - K , where K = 2 2 / 3 /( 2 2 / 3 - 1 ) . This latter equation represents a sphere, S , since K 2 - K > 0. The velocity is vertical at all points on S , as well as at all points on the z -axis. Since the source at the origin is twice as strong as the sink at ( 0 , 0 , 1 ) , only half the fluid it emits will be sucked into the sink. By symmetry, this half will the half emitted into
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