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A metallic ring is uniformly charged as shown in the figure [a positively charged ring centred at O with A at the top, B at the right, C at the bottom and D at the left]. AC and BD are two mutually perpendicular diameters. The electric field at O due to the arc AB has magnitude E. What would be the magnitude of the electric field at O due to the arc ABC?

Asked in JEE Main 4th April 2nd Shift 2025 · Ring, arc and disc

Figure: Ring, arc and disc
Answer: (2) √2 E

Step-by-step solution

Arc ABC is a semicircle, made of the two quarter arcs AB and BC, and each quarter contributes a field of magnitude E at O.

For a positively charged arc the field at the centre points away from the arc, along its symmetry axis: the contribution of AB (upper right quadrant) points towards the lower left, that of BC (lower right quadrant) towards the upper left.

Those two symmetry axes are 90° apart, so the two contributions of magnitude E are perpendicular.

E_ABC=√E²+E²=√2 E, directed along OD, away from the semicircle.

Why the other options are wrong

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