= Solution
Use <natural units> and the <Minkowski metric> $g_{\mu\nu}=\operatorname{diag}(1,-1,-1,-1)$, with $A^\mu=(\Phi,\mathbf A)$ and $A_\mu=(\Phi,-\mathbf A)$. Then $F_{0i}=E_i$, $F^{i0}=E_i$, and $F^{ij}=-\epsilon_{ijk}B_k$. The <electromagnetic field tensor> is an <antisymmetric tensor>, which gives gives
$$
\delta\mathcal L=-\tfrac12F^{\mu\nu}\delta F_{\mu\nu}=-F^{\mu\nu}\partial_\mu\delta A_\nu.
$$
<integration by parts> in the <action>, with the <variation> vanishing on its boundary, yields $\delta S=\int d^4x\,(\partial_\mu F^{\mu\nu})\delta A_\nu$. Independent <variations> of the four components therefore give \b[$\partial_\mu F^{\mu\nu}=0$]. The time component is $\nabla\cdot\mathbf E=0$; the spatial components are $\nabla\times\mathbf B-\partial_t\mathbf E=0$. These are the two source-free <Maxwell equations> obtained dynamically. The other two, $\nabla\cdot\mathbf B=0$ and $\partial_t\mathbf B+\nabla\times\mathbf E=0$, follow identically from $F=dA$, through the <Electromagnetic Bianchi identity>. They are not additional <Euler-Lagrange field equations>.<Quantum Field Theory by David Tong>[chapter 1, §§1.1 and 1.1.3: potential conventions and the Maxwell variation]{ref}
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