The electric field due to the charged disc for an axial point, Ed=σ2ϵ0(1−x√x2+R2)=σ2ϵ0(1−cosθ0)
Force on q, F=qEd=qσ2ϵ0(1−cosθ0).....(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 Ey^j.^i=0 )
Total flux,ϕ=∫R0(q4πϵ0r2cosθ)2πydy
⇒ϕ=q2ϵ0∫R01x2+y2x√x2+y2ydy
ϕ=qx2ϵ0∫R0y(x2+y2)3/2dy=q2ϵ0[1−x√x2+R2]
⇒ϕ=q2ϵ0(1−cosθ0)....(2)
From (1), (2) becomes,ϕ=Fσ