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 ` h gt 2R ` and `r = 4R //5` then `phi = Q // 5 in_(0)`
B. If `h gt 2R and r = 3R// 5` then `phi = Q // 5 in_(0)`
C. If `h gt 2R and r gt R` then `phi = // in_(0)`
D. If `h lt 8 R// 5` and `r = 3R//5` then `phi = 0`

Answer

B, C, D

Explanation

For `hgt2R` and the sphere will completely be inside the cylinder.

`thereforeQ_("enclosed")=Q`
`rArrphi=Q/(in_(0))` (C) is correct
For `r=(3r)/5` and `h/2=(4R)/5`
`r^(2)+(h/2)^(2)=(9R^(2)+86R^(2))/(25)=R^(2)`
`rArr` For `r=(3R)/5` and `hltR/5`, the sphere will completely be outside the cylinder.
`thereforeQ_("enclosed")=0rArrphi=0` (D) is correct
For `hgt2R` and `r=(3R)/5, sintheta=r/R=3/5`
Area of curved surface of sphere enclosed by cylinder
`S=2[2pi(1-costheta)R^(2)]`
`S=4piR^(2)(1-4/5) thereforeQ_("enclosed")=Q/(4piR^(2))S=Q/5`
`rArrphi=Q/(5in_(0))therefore` (B) is correct
Similarly, for `hgt2R,r=(4R)/5 , sintheta=4/5,S=2[2pi(1-costheta)R^(2)]=4piR^(2)(1-3/5)`
`thereforeQ_("enclosed")=Q/(4piR^(2))S=(2Q)/5rArrphi=(2Q)/(5in_(0))therefore` (A) is wrong