Electric Potential & Equipotentials question from JEE Main 2025 (2 Apr, Shift 2)

Answer: (d)

1. Field between the plates is uniform and perpendicular to them: $E = \frac{V}{0.10\,\text{m}}$.
2. Only the component of AB along the field matters: $CB = \sqrt{5^2 - 3^2} = 4\,\text{cm}$ (AC is parallel to the plates).
3. $V_{AB} = E\times 0.04 = \frac{V}{0.1}\times 0.04 = \frac{2}{5}V$
4. So the answer is (d) $\frac{2}{5}V$.