JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Electrostatics

The field between the parallel plates is uniform and perpendicular to them, so the potential depends only on the distance measured across the gap. Points A and C lie on the same line parallel to the plates and so sit at the same potential; the drop from A to B therefore equals the drop across the horizontal separation CB. In the right triangle the vertical side $AC=3$ cm and the hypotenuse $AB=5$ cm give $CB=\sqrt{5^2-3^2}=4$ cm. With the full gap $d=10$ cm holding a potential difference $V$, the field is $E=\frac{V}{10}$ per cm, so $V_{AB}=E\times 4=\frac{4}{10}V=\frac{2}{5}V$.