A point charge Q is located on the axis of a disc of radius R at a distance b from the plane of the disc (figure). Show that if one-fourth of the electric flux from the charge passes through the disc, then `R=sqrt3b`.

Options

A. `sqrt(5)b`
B. `sqrt(2)b`
C. `sqrt(3)b`
D. `2sqrt(3)b`

Answer

c

Explanation

Total flux associated with the charge `Q` for solid angle `4p is (Q)/(epsilon_(0)`
Solid angle substended by disc at `Q`
`Omega= 2pi(1-cos alpha)= 2pi[1-(b)/(sqrt(b^(2)+R^(2)))]`
Flux through disc `= (Q Omega)/(epsilon_(0) 4pi)`
It is given to be `=(Q)/(4 epsilon_(0))`
So, `(Q)/(4epsilon_(0))=(Q Omega)/(epsilon_(0)4pi)`
Solve to get `R= sqrt(3)b`