 Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2012/k/k3/d/solution

c = a^{2} / b

is singular;

b

lies outside. Squaring the velocity gives

(w^{'})^{2} = \frac{A ^{2}}{( z - b ) ^{4}} - \frac{2 A B}{( z - b ) ^{2} ( z - c ) ^{2}} + \frac{B ^{2}}{( z - c ) ^{4}} .

The last term has no residue, and the middle term has residue

res ((w^{'})^{2}; c) = \frac{d}{d z} (- \frac{2 A B}{( z - b ) ^{2}})_{z = c} = \frac{4 A B}{( c - b ) ^{3}} = - \frac{4 m ^{2} a ^{2} b}{( b ^{2} - a ^{2} ) ^{3}},

because

A B = m^{2} a^{2} / b^{2}

and

c - b = - (b^{2} - a^{2}) / b

. The residue theorem and part (a) therefore yield

F_{x} - i F_{y} = - π ρ res ((w^{'})^{2}; c) = \frac{4 π ρ m ^{2} a ^{2} b}{( b ^{2} - a ^{2} ) ^{3}} .

Thus

F_{y} = 0

and

F_{x} > 0

. The force points towards the planar potential doublet and is independent of its orientation, since the factors

e^{i α}

and

e^{- i α}

cancel in

A B

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