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

Solution

Physics sir · 4 steps

Split arc ABC into two quarter arcs; their fields at O are equal and at right angles, so add them as vectors.

1. Break the arc. Arc ABC is made of arc AB and arc BC. Both are quarter circles carrying the same charge, so each gives a field of magnitude E at O.
2. Find directions. The field of a uniformly charged arc at the centre lies along the arc's line of symmetry. For positive charge it points away from the arc. Arc AB pushes O towards the lower left, and arc BC pushes O towards the upper left.
3. Angle between them. The symmetry lines of AB and BC bisect two neighbouring quadrants, so they are 45° + 45° = 90° apart. The two fields are perpendicular.

   E_(ABC) = √(E² + E²) = √2 E

4. So. The field at O due to arc ABC is √2E, which is option (d). It points along BO⃗, away from B.

The idea to keep. Fields from pieces of a charged ring add as vectors, not as numbers. A quarter arc gives √2kλ/R and a half ring gives 2kλ/R, so their ratio is √2.

Answer from the NTA final answer key. Solution by PrepWiser.