A uniform electric field pointing in positive X direction exists in a region. Let O be the origin, A be the point on X axis at \( x = +2\,\mathrm{cm} \) and B be the point on Y axis at \( y = +1\,\mathrm{cm} \). Then potential at the point O, A and B satisfy :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( V_O \gt V_A \)
In a uniform field the potential decreases along the field direction, and points on a line perpendicular to the field lie on the same equipotential surface.
\( \displaystyle V_A - V_O = -E\,(x_A - x_O) = -E\times0.02\,\mathrm{m} \lt 0 \)
B lies on the y-axis, which is perpendicular to the field, so it is on the same equipotential plane \( x = 0 \) as O.
\( V_O \gt V_A \)
\( V_O = V_B \)
Option (1) \( V_O \gt V_A \) — Correct. A is 2 cm further along the field than O, so its potential is lower.
Option (2) \( V_O \lt V_A \) — Incorrect. The potential falls, not rises, in the direction of the field.
Option (3) \( V_O \gt V_B \) — Incorrect. O and B are both on the plane \( x = 0 \), so \( V_O = V_B \).
Option (4) \( V_O \lt V_B \) — Incorrect. Same reason: O and B are at equal potential.
NCERT (2026-27):
Relation Between Field and Potential – “… We thus arrive at two important conclusions concerning the relation between electric field and potential: (i) Electric field is in the direction in which the potential decreases steepest.”