 Solution

ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2013-resit/7/iii/solution

Here

q_{1} = 3.8 q_{3}

and

q_{2} = - 1.9 q_{3}

. The part of the electrostatic potential energy that changes is

U (f) = \frac{k _{e} q _{3}^{2}}{L} (\frac{3.8}{1 + f} - \frac{1.9}{1 - f}) .

At the two endpoints,

U (- 0.9) = 37 k_{e} q_{3}^{2} / L

and

U (0.9) = - 17 k_{e} q_{3}^{2} / L

. The fixed-source interaction energy is unchanged, and conservation of energy gives

\frac{1}{2} m v^{2} = U (- 0.9) - U (0.9) = \frac{54 k _{e} q _{3}^{2}}{L}, v = \frac{108 k _{e} q _{3}^{2}}{m L} .

The force points toward increasing

f

throughout this interval: the left charge repels

q_{3}

and the right charge attracts it, so release from rest reaches the stated endpoint without a turning point.

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