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Differentiate $E=-m{K}_a{U}^a+q\Phi$ along the worldline. The symmetric contraction $U^a{U}^b{\nabla }_a{K}_b$ vanishes by the Killing equation. Using the Lorentz force and $K^a{F}_{ab}={\nabla }_b\Phi$, the remaining force term cancels $q{U}^a{\nabla }_a\Phi$. Hence

Killing equation is ${\nabla }_{(a}{K}_{b)}=0$. For the stress-energy current from a Killing vector $J^a=T^a{}_b{K}^b$,The first term vanishes on shell, while the second contracts the symmetric tensor $T^{ab}$ with the antisymmetric derivative selected by Killing's equation. Thus $\boxed{{\nabla }_a{J}^a=0}$.

For $M>0$ and $M\ge |Q|$, put $r_{\pm }=M\pm \sqrt{M^2-Q^2}$. The stationary Killing vector field normalized at infinity generates the outer Killing horizon. In an ingoing null chart its covector is $dr$ on that horizon; the Killing equation then gives ${\nabla }_{k}k=\frac{1}{2}{f}^{\prime }(r_{+})k$, where $f=(r-r_{+})(r-r_{-})/r^2$. Thus the positive nonextremal surface gravity is $(r_{+}-r_{-})/(2r_{+}^2)$. It tends to zero when $M=|Q|$, where the horizon becomes a degenerate Killing horizon. The past horizon branch has the opposite signed nonaffinity for the same time-translation field.

For a Killing vector field and the curvature convention $[{\nabla }_{\mu },{\nabla }_{\nu }]V^{\alpha }=R^{\alpha }{}_{\beta \mu \nu }{V}^{\beta }$,The identity follows by combining the Ricci identity, Killing equation, and first Bianchi identity.