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In part II, lecture 18, of Feynman's Lectures on Physics, on Table 18-1 Feynman writes Ampère's law as $$ c^2 \nabla \times \vec{B} = \frac{\vec{j}}{\epsilon_0} + \frac{\partial \vec{E}}{\partial t}. $$

What unit system is this in? It's not obviously Gaussian, which states, $$ c \nabla \times \vec{B} = 4 \pi \vec{j} + \frac{\partial \vec{E}}{\partial t}. $$

Nor does it appear to be formulated in SI, which states,

$$ \frac{1}{\mu_0} \nabla \times \vec{B} = \vec{j} + \epsilon_0 \frac{\partial \vec{E}}{\partial t}. $$

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    $\begingroup$ This appears to be using SI convention, taking advantage of the relation $\epsilon_0\mu_0 c^2 = 1$ $\endgroup$
    – By Symmetry
    Commented Oct 21, 2019 at 17:49
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    $\begingroup$ @BySymmetry Shouldn't that be an answer instead of a comment? $\endgroup$
    – Emilio Pisanty
    Commented Oct 21, 2019 at 18:03

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Multiply $$ c^2 \nabla \times \vec{B} = \frac{\vec{j}}{\epsilon_0} + \frac{\partial \vec{E}}{\partial t} $$ through by $\epsilon_0$, and substitute in the identity $$ c^2 = \frac1{\epsilon_0\mu_0} , $$ and it will be quickly revealed to be in the SI form you quote.

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