Figure shows three points A, B and C in a region of uniform electric field \(\overrightarrow{E}\). The line AB is perpendicular and BC is parallel to the field lines. Then which of the following holds good? (VA, VB and VC represent the electric potential at points A, B and C respectively)
In a uniform electric field, the electric potential decreases in the direction of the field. Since point B is at a lower potential than point A (due to the perpendicular direction of the line AB to the field lines), the potential at point B will be less than at point A. As point C lies along the field lines, the potential at C will be the lowest among the three. Therefore, the potential decreases as we move from A to B to C.
The correct option is (B) : VA=VB > VC
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In a uniform electric field, the electric potential difference between two points is given by: \[ \Delta V = - \int \vec{E} \cdot d\vec{r} \] Since the electric field is uniform, the potential decreases along the direction of the electric field. The key observation here is that: 1. Line AB is perpendicular to the field lines, which means there is no potential difference between A and B because the displacement between the two points is perpendicular to the field lines. Therefore, \(V_A = V_B\).
2. Line BC is parallel to the field lines, and in the direction of the field, the potential decreases.
Thus, \(V_C < V_B\) because point C is farther along the electric field lines from point B.
Thus, the correct relation is: \[ V_A = V_B > V_C \]
