= Solution
For the displayed gauge field,
$$
i_KF=d\Phi,
\qquad
\Phi=\frac{Q}{r^{d-3}},
$$
which already vanishes at infinity. A static future-directed particle has $U=K/\sqrt f$, so
$$
E(r)=m\sqrt{f(r)}+\frac{qQ}{r^{d-3}}.
$$
If $qQ<0$, the first term tends to zero at $r_+$ while the second remains negative. Lowering the particle sufficiently close to the horizon therefore produces $E<0$. Dropping it into the hole reduces the black-hole mass by $|E|$, while the external agent receives the corresponding positive work: this is charged-particle <black-hole energy extraction>.
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