Discussion
For a conservative field with potential \( \phi \), the integral \( \displaystyle\int_{A}^{B} \mathbf{F} \cdot d\mathbf{r} \) equals
Answer: D
With \( \mathbf{F} = \nabla\phi \) the integrand becomes an exact differential, \( \nabla\phi \cdot d\mathbf{r} = d\phi \), so
\[ \int_{A}^{B} \mathbf{F} \cdot d\mathbf{r} \;=\; \phi(B) - \phi(A). \]
It vanishes only when the two endpoints share a potential value, which for a closed path is automatic but for an open one is not
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