 Solution

Let $D:\mathcal{J}\to \mathbf{Fld}$ be connected and nonempty. The limit of the underlying diagram of commutative rings is the subringIf $x\in L$ is nonzero in one component, it is nonzero in every component: field homomorphisms are injective, and connectedness propagates this fact along zigzags. Hence the componentwise inverses $(x_j^{-1})$ are defined and compatible. Thus $L$ is a field. Since the inclusion $\mathbf{Fld}\hookrightarrow \mathbf{CRng}$ is full, the same cone is limiting in the category of fields.

If $\mathcal{J}$ is disconnected, choose two components and use the constant field ${\mathbb{F}}_2$ on one and ${\mathbb{F}}_3$ on the other. There is no cone in fields because its apex would map to fields of two different characteristics. Hence $\mathbf{Fld}$ does not have all limits of any disconnected shape.