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The relation between the electric field and the potential is

  • A.\( \mathbf{E} = -\nabla V \)
  • B.\( \mathbf{E} = \nabla V \)
  • C.\( V = -\nabla \cdot \mathbf{E} \)
  • D.\( \mathbf{E} = \nabla^2 V \)

Answer: A

The minus sign makes the field point from high potential to low, which is the direction a positive charge is pushed. The magnitude of the field is the greatest rate of change of potential with distance, which is why volts per metre is its unit.

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