 Solution

ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2019/9/a/ii/solution

For the negative test electric charge, use the same change in the fixed point charge's electric potential:

W_{-} = (- q) [ϕ (d) - ϕ (2 d)] = - \frac{k _{e} q ^{2}}{2 d} = - \frac{q ^{2}}{8 π ϵ _{0} d} .

The external agent extracts energy, since the final Coulomb potential energy is lower. Again the path must avoid the fixed point charge, and the work done by the electric field is

- W_{-}

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