Uniformly charged ring is shown is figure, E due to ring is maximum at
A
Centre
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B
∞
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C
R√2 distance from centre
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D
R distance from centre
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Solution
The correct option is CR√2 distance from centre By graph (theoretical) OR E=kQx(R2+x2)32 But dEdx=0 dEdx=ddx(kQx(R2+x2)32)=0 x.ddx(R2+x2)32−(R2+x2)32ddx(x)(R2+x2)3=0 Solve ⇒x=R√2