A thin semicircular ring of radius ‘r’ has a positive charge distributed uniformly over it. The net field E at the centre ‘O’ is

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q2π2∈0r2j¯

q4π2∈0r2j¯

−q4π2∈0r2j¯

−q2π2∈0r2j¯

answer is D.

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Detailed Solution

E=14π∈0qsin⁡π/2r2π/2E=qsin⁡π/22π2∈0r2(−j→)

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