JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2025
- Hard
- MCQ
A metallic ring is uniformly charged as shown in figure. AC and BD are two mutually perpendicular diameters. Electric field due to \(\operatorname{arc} \mathrm{AB}\) to ' O ' is ' E ' is magnitude. What would be the magnitude of electric field at ' O ' due to arc ABC ?
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(\sqrt{2} \mathrm{E}\)
Step-by-step Solution
Detailed explanation
Let the metallic ring be divided into four identical quadrants: arc AB, arc BC, arc CD, and arc DA.
Since the charge distribution is uniform, the magnitude of the electric field at the center 'O' due to each individual quadrant is equal. We are given that the electric field due to arc AB has a magnitude of \( E \).
The electric field vector for any uniformly charged arc points along its angle bisector, away from the positive charge. - Let the electric field due to arc AB be \( \vec{E}_{AB} \) (magnitude = \( E \)). - Let the electric field due to arc BC be \( \vec{E}_{BC} \) (magnitude = \( E \)).
Because AC and BD are mutually perpendicular diameters, the quadrants AB and BC sit 90° apart. Consequently, their respective electric field vectors, \( \vec{E}_{AB} \) and \( \vec{E}_{BC} \), are also perpendicular to each other.
The total electric field at 'O' due to the combined arc ABC is the vector sum of the fields from arc AB and arc BC. Since the two vectors are perpendicular, we calculate the resultant magnitude using the Pythagorean theorem:
\( E_{ABC} = \sqrt{|\vec{E}_{AB}|^2 + |\vec{E}_{BC}|^2} \)
\( E_{ABC} = \sqrt{E^2 + E^2} \)
\( E_{ABC} = \sqrt{2E^2} = \sqrt{2}E \)
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