JEE Main 2025 Apr 2 Shift 2, Physics Q9: Electric Potential

JEE Main2025Apr 2, Shift 2Physics

Two large plane parallel conducting plates are kept $10$ cm apart as shown in figure. The potential difference between them is $V$ . The potential difference between the points $A$ and $B$ (shown in the figure) is :

A
$\frac{1}{4} V$
B
$\frac{2}{5} V$
C
$\frac{3}{4} V$
D
$1 V$

Solution

The electric field $E$ between large parallel plates is uniform and directed perpendicular to the plates. Its magnitude is $E = \frac{V}{d} = \frac{V}{10 cm}$ .

Only the component of displacement along $E$ contributes to the potential difference. From the geometry $A$ and $B$ are separated by $4$ cm in the direction perpendicular to the plates (the horizontal leg $C B$ ); the $3$ cm vertical leg is parallel to the plates and contributes nothing.

$∴ V_{A B} = E \cdot 4 cm = \frac{V}{10} \times 4 = \frac{2 V}{5}$ .

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JEE Main 2025 Apr 2 Shift 2, Physics Q9: Electric Potential

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