The maximum electric field intensity on the axis of a uniformly charged ring of charge q and radius R will be :

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14πε0q33R2

14πε02q3R2

14πε02q33R2

14πε03q22R2

answer is C.

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Detailed Solution

E=kqx(R2+x2)3/2, for maximum E, dEdx=0⇒kq(1)(R2+x2)3/2+-32kqx(R2+x2)5/2(2x)=0⇒R2+x2=3x2⇒x2=R22⇒Emax=kq(R/2)R2+R223/2=2kq33R2

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