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

$I f h > 2 R a n d r > R t h e n ϕ = Q / Є o$

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$I f h > 2 R a n d r = 4 R / 5 t h e n ϕ = Q / 5 Є o$

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$I f h < 8 R / 5 a n d r = 3 R / 5 t h e n ϕ = 0$

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$I f h > 2 R a n d r = 3 R / 5 t h e n ϕ = Q / 5 Є o$

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Solution

The correct option is D
$I f h > 2 R a n d r = 3 R / 5 t h e n ϕ = Q / 5 Є o$

Step 1 : Given data
Radius of the shell $= R$
Total charge $= Q$
Flux of electric field $= ϕ$
Height of cylindrical surface $= h$
Radius of cylindrical surface $= r$

Step 2 : To find the values of $h, r$ and $ϕ$
$(a) h > 2 R r > R$
According to Gauss Law
$ϕ = \frac{Q}{ε_{0}}$
Suppose
$h = \frac{8 R}{5} r = \frac{3 R}{5}$
$h < \frac{8 R}{5} ϕ$
Then
$h = 2 R r = \frac{4 R}{5}$
Shaded charge
$= 2 π (1 - \cos 53^{0}) \times \frac{Q}{4 π} = \frac{Q}{5} q_{e n c l o s e d} = \frac{2 Q}{5} ϕ = \frac{2 Q}{5 ε_{0}}$
Step3: For $h > 2 R$
$r = \frac{4 R}{5} ϕ = \frac{2 Q}{ε_{0}} q_{e n c l o s e d} = 2 \times 2 π (1 - \cos 37^{0}) \frac{Q}{4 π} = \frac{Q}{5} ϕ = \frac{Q}{5 ε_{0}}$
Therefore the correct answer is Option A, B, D

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