A uniformly charged semicircular arc ABCD produces electric field magnitude E₀ at its centre O. The arcs AB, BC and CD subtend equal angles at O. The end arcs AB and CD are removed. The electric field magnitude at O due to the remaining arc BC is:
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Question

A uniformly charged semicircular arc ABCD produces electric field magnitude E₀ at its centre O. The arcs AB, BC and CD subtend equal angles at O. The end arcs AB and CD are removed. The electric field magnitude at O due to the remaining arc BC is:

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G1 Physics · Accepted Answer

The full semicircle subtends 180°. Since AB, BC and CD are equal arcs, each subtends 60°. For a uniformly charged arc symmetric about its bisector, E ∝ 2 sin(θ/2), where θ is the total angle of the arc. Full semicircle: E₀ ∝ 2sin90° = 2. Middle arc BC: E B C E_BC E B ​ C ∝ 2sin30° = 1. Therefore, E B C E_BC E B ​ C = E₀/2.