Static electric field which admits no potential

Conservative condition for (static) electric field $$\mathcal{E}$$ is usually defined as $$\mathcal{E}$$ being closed (curl-free). Now this clearly holds when for the given manifold $$X$$ we have $$H_\text{dR}^{1}(X)\simeq 0$$, but consider the case when it is not so.

For general $$k$$-manifolds:

Consider smooth $$k$$-manifold $$X$$ s.t. $$H_\text{dR}^1(X)\not\simeq 0$$. Denote by $$\Phi$$ the set of all closed 1-forms which are not exact. Let $$\mathcal{E}=\mathcal{E_0}\varphi$$ for some $$\mathcal{E_0}\in\mathbf{R}$$ and $$\varphi\in \Phi$$. This gives charge density for this setting (let units be s.t. $$\epsilon_0 = 1$$): $$\rho = \star d\star \mathcal{E}$$ And, by definition of $$\varphi$$, we have $$d\mathcal{E} = 0$$. Then (by non-exactness) there exists a closed 1-path $$\gamma\colon [0,1]\rightarrow X$$ s.t. $$\int_\gamma \mathcal{E} \neq 0$$ Which evidently violates the condition of static electric field being conservative.

An example on $$S^1$$:

Suppose one is given $$S^1$$, and the electric field $$\mathcal{E}\in\Gamma(TS^1)$$ is a smooth section, defined by: $$\mathcal{E}((x,y)) ~\stackrel{\text{proj.2.}}{=}~ \mathcal{E}_0 \frac{x~dy-y~dx}{x^2+y^2}$$ Evidently the value of $$\mathcal{E}$$ is a closed 1-form, furthermore $$\mathcal{E}$$ is divergence-free.

Assuming that there are no external fields, and all fields are time-independent, we have $$\mathcal{E}$$ as a solution to Maxwell's equations: $$d\mathcal{E} = 0\qquad \star d\star\mathcal{E} = 0 = \rho$$ Thus $$\rho$$ is well defined, bounded and constant. However $$\mathcal{E}((x,y)) = \mathcal{E}_0\mathrm{vol}_{S^1}$$ is not exact, moreover we have: $$\int_{S^1}\mathcal{E} = 2\pi \mathcal{E_0} \neq 0\qquad \text{by choosing}~\mathcal{E}_0\neq 0$$ hence $$\mathcal{E}$$ admits no potential and is thus non-conservative.

Are Maxwell's equations restricted on (smooth) manifolds which have trivial de Rham cohomology, or is there some other condition I'm missing?