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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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