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

To find the electric field at point 'O' caused by arc 'ABC', we must consider the electric field from each arc segment.

1. Electric Field from Arc 'AB':
   • The electric field at 'O' from arc 'AB' is designated as \( E \).
   • This electric field vector is directed along the angle bisector of the arc 'AB' at the center 'O'.
2. Electric Field from Arc 'BC':
   • Arc 'BC' is symmetrical to arc 'AB' with respect to line 'OC'.
   • Consequently, the electric field at 'O' from arc 'BC' also has a magnitude of \( E \) and is directed along the angle bisector of arc 'BC' at 'O'.
3. Total Electric Field at 'O':
   • The electric fields from arcs 'AB' and 'BC' are orthogonal since arcs 'AB' and 'BC' are separated by a 90-degree angle.
   • The resultant electric field is the vector sum of these perpendicular electric fields.
   • The magnitude of the resultant electric field, \( E_{\text{net}} \), is calculated as:

Therefore, the magnitude of the electric field at 'O' due to arc 'ABC' is \( \sqrt{2}E \).