Q 12-01-163JEE MainJEE Main 2025 (4 Apr, Shift 2)Medium
A metallic ring is uniformly charged as shown in the figure. AC and BD are two mutually perpendicular diameters. The electric field due to arc AB at O is $E$ in magnitude. What would be the magnitude of the electric field at O due to arc ABC?
Answer: (B) $\sqrt2E$
By symmetry, the field of quarter arc AB at O points away from the arc's midpoint, along the bisector of angle AOB. Quarter arc BC gives a field of the same magnitude $E$ along the bisector of angle BOC.
These two bisectors are $90^\circ$ apart, so the field of the half ring ABC is
$$E_{ABC} = \sqrt{E^2 + E^2} = \sqrt2E$$
Solution by Sreeraj P, M.Sc Physics