A point charge placed on the axis of a uniformly charged disc experiences a force f due to the disc. If the charge density on the disc `sigam` is, the electric flux through the disc, due to the point charge will be

Options

A. `(2 pi f)/(sigma)`
B. `(f)/(2 pi sigma)`
C. `(f^(2))/(sigma)`
D. `(f)/(sigma)`

Answer

D

Explanation

Force on `q = Eq = (sigma)/(2epsilon_(0)) (1 - cos theta_(0)) q = f`

Consider a ring of radius y and thickness dy.
Flux through this ring `d phi = (E cos theta) 2pi ydy`
Total flux `= underset(0)overset(theta_(0))int (1)/(4 pi epsilon_(0)) (q)/(r^(2)) 2pi ydy cos theta`
`(q)/(2epsilon_(0)) (1 - cos theta_(0))`
So, `phi = (f)/(sigma)`