NEET Physics Electric Potential MCQ Question
Consider a diagram showing two closely spaced equipotential surfaces A and B with potentials V and V + dV, respectively, and a point P on surface B. If a unit positive charge is moved perpendicularly from surface B to surface A against the electric field E, what is the correct expression for the magnitude of the electric field E?
|E| = -dV/dl
|E| = dV/dl
|E| = dl/dV
|E| = -dl/dV
Correct Answer
Detailed Explanation
The NCERT context explains that the work done in moving a unit positive charge from surface B to A is equal to the potential difference VA - VB. This leads to the expression |E| = -dV/dl.
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