Question

A metallic ring is uniformly charged as shown in the figure. AC and BD are two mutually perpendicular diameters. Electric field due to arc AB to O is ‘E’ magnitude. What would be the magnitude of electric field at ‘O’ due to arc ABC?

• A. \( 2E \)
• B. \( \sqrt{2}E \)
• C. \( E/2 \)
• D. Zero

Hint:

The electric field at the center of a uniformly charged arc can be calculated based on the symmetry and the fraction of the total circle covered by the arc.

Answer 1

The Correct Option is B

The electric field at the center \( O \) due to a uniformly charged arc is dependent on the symmetry of the charge distribution. 
1. Electric Field Due to a Uniformly Charged Arc: The electric field at the center of a uniformly charged arc is directed radially, and its magnitude is proportional to the charge distribution along the arc. 
2. Contribution of Arc ABC: Since the arc ABC is one-fourth of the full circle, the electric field at the center due to arc ABC will be proportional to the total electric field due to a half-circle, which is \( E \). 
Thus, the magnitude of the electric field at the center due to arc ABC will be \( \sqrt{2}E \). 
Thus, the correct answer is (2).