A charged shell of radius R carries a total charge Q. Given `phi` as the flux of electric field through a closed cylindrical surface of height h, radius r & with its centre same as that of the shell. Here centre of cylinder is a point on the axis of the cylinder which is equidistant from its top & bottom surfaces. which of the followintg are correct.

Options

A. if hgt2R & rgtR then `phi=Q/(epsilon_(0))`
B. If `h lt (8R)/5` & `r=(3R)/5` then `phi=0`
C. If h gt 2R & `r=(4R)/5` then `phi=Q/(5epsilon _(0))`
D. If hgt2R & `r=(3R)/5` then `phi=Q/(5epsilon_(0))`

Answer

A, B, D

Explanation

(a)`" h" gt "2R" " r gt R`

`phi=Q/(epsilon_(0))` clearly from Gauss's law
(b) suppose `h=(8R)/5" r=(3R)/5`

`phi=0`
So for `h lt (8R)/5 " " phi=0`
(C) for `h=2R" " r=(4R)/5`

Shaded charge `=2pi(1-cos53^(@)xx(Q))/(4pi)`
`:. =(Q)/(5)`
`:. q_("enclosed")=(2Q)/(5)`
`:. phi = (2Q)/(5_(epsilon_(0)))`
`:.` for ` h gt 2R r = (4R)/(5)" ":. phi = (2Q)/(5_(epsilon_(0)))`
(d) like option (C) for `h=2R " "r=(3R)/5`
`q_("enclosed")=2xx2pi(1-cos37^(@)) Q/(4pi)=Q/5`
`:. phi=Q/(5epsilon_(0))`