Physics-Electric Potential

NEET Physics Electric Potential MCQ Question

Type: MCQ-diagram based-Hard-Class 12

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?

Question diagram
A

|E| = -dV/dl

B

|E| = dV/dl

C

|E| = dl/dV

D

|E| = -dl/dV

Correct Answer

Option A

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.

Found an issue with this question?