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Correct option is D)

The electric field due to the charged disc for an axial point, $E_{d}=2ϵ_{0}σ (1−x_{2}+R_{2} x )=2ϵ_{0}σ (1−cosθ_{0})$

Force on q, $F=qE_{d}=2ϵ_{0}qσ (1−cosθ_{0}).....(1)$

Consider a ring of radius $y$ and thickness $dy$.

The flux through this ring,$dϕ=(Ecosθ)2πydy$ (as only horizontal component contribute flux and flux due to vertical component is zero because $E_{y}j^ .i^=0$ )

Total flux,$ϕ=∫_{0}(4πϵ_{0}r_{2}q cosθ)2πydy$

$⇒ϕ=2ϵ_{0}q ∫_{0}x_{2}+y_{2}1 x_{2}+y_{2} x ydy$

$ϕ=2ϵ_{0}qx ∫_{0}(x_{2}+y_{2})_{3/2}y dy=2ϵ_{0}q [1−x_{2}+R_{2} x ]$

$⇒ϕ=2ϵ_{0}q (1−cosθ_{0})....(2)$

From (1), (2) becomes,$ϕ=σF $

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