Q 12-02-206JEE MainJEE Main 2025 (2 Apr, Shift 2)Easy
Two large plane parallel conducting plates are kept $10\ \text{cm}$ apart as shown in the figure. The potential difference between them is $V$. The potential difference between the points A and B (shown in the figure) is:
Answer: (B) $\dfrac25V$
Between the plates the field is uniform, $E = \dfrac{V}{0.10\ \text{m}}$, and perpendicular to the plates (horizontal).
In the right triangle ACB, $AC = 3\ \text{cm}$ (parallel to the plates) and $AB = 5\ \text{cm}$, so $CB = 4\ \text{cm}$ (along the field). Only the displacement along the field changes the potential:
$$V_{AB} = E\times CB = \frac{V}{10}\times4 = \frac{2V}{5}$$
Solution by Sreeraj P, M.Sc Physics