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The relation between the electric field \[\vec{E}\] and the potential gradient is expressed as:

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Question

The relation between the electric field \[\vec{E}\] and the potential gradient is expressed as:

Options

  • \[\vec{E} = -\frac{dV}{dr}\]

  • \[\vec{E} = +\frac{dV}{dr}\]

  • \[\vec{E} = -\frac{d^2V}{dr^2}\]

  • \[\vec{E} = V \cdot r\]

MCQ
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Solution

The electric field is the negative of the potential gradient: \[\vec{E} = -\frac{dV}{dr}\]. The negative sign indicates that the field points from high potential to low potential. A greater E means the potential changes faster (denser equipotential surfaces), while a weaker E means the potential changes slowly (surfaces farther apart).

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