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JEE Main 2025 (4 Apr, Shift 2) · Electric Field & Field Lines

Electric Field & Field Lines question from JEE Main 2025 (4 Apr, Shift 2)

Q9JEE Main 2025 · 4 APR · SHIFT 2Medium

Electric Field & Field Lines

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

Figure

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Answer: (d)

  1. Quarter arc AB gives field EE at O along the bisector of AB (pointing away from the arc).
  2. Arc ABC = arc AB + arc BC; arc BC gives an equal field EE along its own bisector.
  3. The two bisectors are 90∘90^\circ apart, so Enet=E2+E2=2EE_{net} = \sqrt{E^2 + E^2} = \sqrt{2}E.
  4. So the answer is (d) 2E\sqrt{2}E.
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