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Maxwell's Equations written with usual vector calculus are

$$

abla \cdot E=\rho/\epsilon_0 \qquad

abla \cdot B=0$$ $$

abla\times E=-\dfrac{\partial B}{\partial t} \qquad

abla\times B=\mu_0j+\dfrac{1}{c^2}\dfrac{\partial E}{\partial t}$$

now, if we are to translate into differential forms we notice something: from the first two equations, it seems that $E$ and $B$ should be $2$-forms. The reason is simple: we are taking divergence, and divergence of a vector field is equivalent to the exterior derivative of a $2$-form, so this is the first point.

The second two equations, though, suggests $E$ and $B$ should be $1$-forms, because we are taking curl. Thinking of integrals, the first two we integrate over surfaces, so the integrands should be $2$-forms and the second two we integrate over paths and so the integrands should be $1$-forms.

In that case, how do we represent $E$ and $B$ with differential forms, if in each equation they should be a different kind of form?