Discussion
The field corresponding to a potential \( V = 3x^2 \) is
Answer: A
Correct answer: \( -6x\,\mathbf{a}_x \)
\[ \mathbf{E} = -\nabla V = -\frac{dV}{dx}\mathbf{a}_x = -6x\,\mathbf{a}_x. \] The minus sign is not decoration: the field points from high potential to low, that is downhill, while the gradient points uphill.
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