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Three of these four hold for a conservative field. The exception is

  • A.its divergence vanishes everywhere
  • B.its circulation round every closed path vanishes
  • C.it has a scalar potential
  • D.its line integral depends only on the endpoints

Answer: A

Conservativeness constrains the curl, not the divergence: the field \( x\mathbf{i} + y\mathbf{j} \) has potential \( (x^{2}+y^{2})/2 \) and divergence two. The other three are equivalent restatements of conservativeness on a simply connected region

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