JEE Main 2025 Apr 4 Shift 2, Physics Q7: Coulomb's Law And Electric Field

By symmetry, the field at the centre produced by a quarter arc (e.g. $AB$) lies along the bisector of that arc and has magnitude $E$.

Resolve the field from arc $AB$ along the two perpendicular diameters $OA$ and $OB$: each component has magnitude $\frac{E}{\sqrt{2}}$.

Arc $ABC$ is two adjacent quarter arcs: $AB$ and $BC$. Each contributes a field $E$ along its own bisector. The bisector of $AB$ is at $45^{\circ }$ between $OA$ and $OB$, and the bisector of $BC$ is at $45^{\circ }$ between $OB$ and $OC$. Their components along $OA$ and $OC$ are equal and opposite (along the vertical $AC$ direction) and cancel; their components along $OB$ both have magnitude $\frac{E}{\sqrt{2}}$ and add.

Net field $=2\cdot \frac{E}{\sqrt{2}}=\sqrt{2}E$, directed along $OB$ (away from the charged arc).

$\therefore$ Magnitude of field due to arc $ABC=\sqrt{2}E$.