JEE Physics 2025 Question with Solution

Given: The electric field at $O$ due to arc $AB$ is $E$. The ring is uniformly charged.

Find: The magnitude of electric field at $O$ due to arc $ABC$.

Arc $ABC$ is made of two equal quarter-arcs, $AB$ and $BC$. Since the charge distribution is uniform, arc $BC$ also produces an electric field of magnitude $E$ at $O$.

The field due to arc $AB$ acts along the bisector of that arc, and the field due to arc $BC$ acts along the bisector of arc $BC$. These two directions are perpendicular to each other.

So, the net field is the vector sum of two perpendicular vectors of equal magnitude $E$:

Therefore, the magnitude of the electric field at $O$ due to arc $ABC$ is $\sqrt{2}E$. The correct option is B.

Given: Arc $AB$ gives field $E$ at $O$.

Find: Field due to arc $ABC$.

By symmetry, arc $BC$ contributes the same magnitude $E$. The two resultants are at right angles because the two quarter-arcs are adjacent and subtend equal $90^\circ$ sectors.

Hence, treat them as perpendicular vectors:

Therefore, the correct option is B.