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 determine the electric field at point 'O' due to the arc 'ABC', we need to analyze the contribution of each segment of the arc to the net electric field.

1. Electric Field Due to Arc 'AB':
   • Given that the electric field at 'O' due to arc 'AB' is \( E \).
   • This electric field vector points along the bisector of the angle subtended by the arc 'AB' at the center 'O'.
2. Electric Field Due to Arc 'BC':
   • The arc 'BC' is symmetric to 'AB' about the line 'OC'.
   • Hence, the electric field due to arc 'BC' at 'O' also has magnitude \( E \) and points along the bisector of the angle subtended by arc 'BC' at 'O'.
3. Net Electric Field at 'O':
   • The electric fields due to arcs 'AB' and 'BC' are perpendicular to each other because 'AB' and 'BC' are separated by a right angle.
   • Therefore, the resultant electric field is the vector sum of these perpendicular electric fields.
   • The magnitude of the resultant electric field \( E_{\text{net}} \) is given by:

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

Answer 2

To determine the electric field at point \( O \) due to the arc \( ABC \), we start by analyzing the contributions from arcs \( AB \) and \( BC \).

1. The problem states that the electric field at \( O \) due to arc \( AB \) is of magnitude \( E \).
2. Since the ring is uniformly charged, the contribution from arc \( BC \) (which is symmetrical to \( AB \) but on the opposite side) will also produce an electric field of magnitude \( E \) at \( O \), but in the perpendicular direction to \( E \).
3. Therefore, the total electric field at \( O \) due to the entire arc \( ABC \) is the vector sum of the electric fields due to arcs \( AB \) and \( BC \).

Since these fields are perpendicular to each other and of the same magnitude, we use the Pythagorean theorem to find the resultant electric field:

\(E_{\text{resultant}} = \sqrt{E^2 + E^2} = \sqrt{2E^2} = \sqrt{2}E\)

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

The correct answer is: \(\sqrt{2}E\)