 Solution

A time-independent conservative force defines a conservative vector field on a domain when the work between two points is independent of path, equivalently when every closed-path work integral vanishes. Define a potential energy $U$ by $\boxed{\mathbf{F}=-\nabla U}$ and $U(B)-U(A)=-{\int }_A^B\mathbf{F}\cdot d\mathbf{r}$. The additive constant is arbitrary; only differences affect work. A vanishing curl is necessary, and is sufficient on a simply connected domain under the usual smoothness assumptions. Local zero curl by itself need not guarantee a single-valued global potential on a domain with holes.

For a test mass outside a point or spherical mass, $U=-GMm/r$ with zero at infinity reproduces the inward gravitational force. Near a surface with approximately uniform downward gravity, $U=mgz$ when $z$ increases upward. A linear spring has $U=k{x}^2/2$. In electrostatics the electric potential is energy per unit charge, so $U=qV$ and $\mathbf{E}=-\nabla V$; potential and potential energy have different units and roles.

For motion under a time-independent conservative force, $d(K+U)/dt=0$. nonconservative forces change the mechanical energy by their work, and an explicitly time-dependent potential can exchange energy with the external agency changing it. A time-varying magnetic field can give a circulating electric field, so not every electric field is electrostatic or conservative. Conversely, zero magnetic work on a charge does not turn a velocity-dependent magnetic force into an ordinary position-potential force.