= Solution
An <electromagnetic field tensor> obeys the homogeneous <Maxwell equations> $dF=0$. Together with its invariance under the <Killing vector field> $V$, <Cartan's magic formula> gives $d(\iota_VF)=\mathcal L_VF-\iota_VdF=0$. The <local potential of a closed differential one-form> construction, the one-form <Poincaré lemma>, therefore supplies a <scalar field> $\Phi$ on a sufficiently small <contractible> neighborhood with
$$
\boxed{\iota_VF=d\Phi.}
$$
For an explicit local proof, write $\iota_VF=\alpha_i(x)dx^i$ on a coordinate ball centered at zero and take $\Phi(x)=\int_0^1 x^i\alpha_i(sx)\,ds$. Since $d\alpha=0$, differentiating under the integral gives $\partial_j\Phi=\int_0^1[\alpha_j(sx)+s x^i\partial_i\alpha_j(sx)]ds=\alpha_j(x)$. There need not be a global potential when this closed one-form has nonzero periods. The homogeneous Maxwell equation is essential; an arbitrary invariant two-form would not suffice.
Use <proper time> for the massive particle, let $u$ be its tangent, and assume $m\ne0$. The <Killing equation> makes $u^au^b\nabla_aV_b=0$. The <Lorentz force> equation then implies
$$
\frac d{d\tau}(V_au^a)=\frac qm V_aF^a{}_bu^b=\frac qm V^aF_{ab}u^b=\frac qm\frac{d\Phi}{d\tau}.
$$
Consequently the local conserved quantity is
$$
\boxed{mV_au^a-q\Phi=\text{constant}.}
$$
The sign follows from contracting $F$ in its first argument, exactly as in the source. Adding a constant to $\Phi$ merely shifts the conserved quantity. This is a <charged-particle conserved quantity from a Killing symmetry>.
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