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Q.A uniformly charged disc of radius RR has charge density σ\sigma. The electric field at a point on its axis at distance xx from centre is given by (JEE Main 2021)

a
σ2ϵ0(1−xx2+R2)\dfrac{\sigma}{2\epsilon_0} \left(1 - \dfrac{x}{\sqrt{x^2 + R^2}}\right)
b
σ2ϵ0(xx2+R2)\dfrac{\sigma}{2\epsilon_0} \left(\dfrac{x}{\sqrt{x^2 + R^2}}\right)
c
σϵ0(1−xx2+R2)\dfrac{\sigma}{\epsilon_0} \left(1 - \dfrac{x}{\sqrt{x^2 + R^2}}\right)
d
σR22ϵ0(x2+R2)\dfrac{\sigma R^2}{2\epsilon_0 (x^2 + R^2)}

Correct Answer: Option A

The correct solution involves applying the fundamental concept to derive the final value step by step...

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