Past exam of the physics 2 paper of the University of Oxford / 2019 / 9 / a

The fixed point charge produces electric potential $\varphi (r)=k_{e}q/r$, where $k_e=1/(4\pi {ϵ}_0)$. The external quasistatic work is the increase in Coulomb potential energy:The endpoints have radii $2d$ and $d$, irrespective of their opposite coordinate signs. One must use a path avoiding the singular point charge at the origin. The electric field is conservative on such paths, so the endpoint calculation suffices; a literal straight path through the origin is inadmissible. The work done by the electric field has the opposite sign.

For the negative test electric charge, use the same change in the fixed point charge's electric potential: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 $-W_{-}$.