A uniformly charged circular ring has mutually perpendicular diameters AC and BD. The magnitude of the electric field at the centre O due to arc AB is E. The magnitude of the electric field at O due to arc ABC is:

Question

G1 Physics · Accepted Answer

For a uniformly charged arc of radius r subtending angle φ at the centre, the field magnitude along its symmetry axis is proportional to 2sin(φ/2). Arc AB is a quarter circle, so φ = 90°: E ∝ 2sin45° = √2. Arc ABC is a semicircle, so φ = 180°: E' ∝ 2sin90° = 2. Therefore, E'/E = 2/√2 = √2. Hence, E' = √2 E. Correct option: C.