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 :
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Correct option is D)
The electric field due to the charged disc for an axial point, Ed=2ϵ0σ(1−x2+R2x)=2ϵ0σ(1−cosθ0)
Force on q, F=qEd=2ϵ0qσ(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 Eyj^.i^=0 )