12.The scalar potential for the source-sink system isφ(x,y,z)=φ(r)= -2|r|+1|r-k|.Thus, the velocity field isv=φ=2r|r|3-r-k|r-k|3=2(xi+yj+zk)(x2+y2+z2)3/2-xi+yj+(z-1)k(x2+y2+(z-1)2)3/2.For vertical velocity we require2x(x2+y2+z2)3/2=x(x2+y2+(z-1)2)3/2,and a similar equation fory. Both equations will be satisfied at all points of thez-axis, andalso wherever2(x2+y2+(z-1)2)3/2=(x2+y2+z2)3/222/3(x2+y2+(z-1)2)=x2+y2+z2x2+y2+(z-K)2=K2-K,whereK=22/3/(22/3-1). This latter equation represents a sphere,S, sinceK2-K>0.The velocity is vertical at all points onS, as well as at all points on thez-axis.Since the source at the origin is twice as strong as the sink at(0,0,1), only half thefluid it emits will be sucked into the sink. By symmetry, this half will the half emitted into
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