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Question

A point charge placed on the axis of a uniformly charged disc experiences a force F due to the disc. If the surface charge density on the disc is σ, the electric flux through the disc due to the point charge will be :
  1. 2πFσ
  2. 2Fσ
  3. Fσ
  4. F2πσ

A
F2πσ
B
2Fσ
C
Fσ
D
2πFσ
Solution
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The electric field due to the charged disc for an axial point, Ed=σ2ϵ0(1xx2+R2)=σ2ϵ0(1cosθ0)
Force on q, F=qEd=qσ2ϵ0(1cosθ0).....(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=q2ϵ0R01x2+y2xx2+y2ydy
ϕ=qx2ϵ0R0y(x2+y2)3/2dy=q2ϵ0[1xx2+R2]
ϕ=q2ϵ0(1cosθ0)....(2)
From (1), (2) becomes,ϕ=Fσ

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