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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 and with its center same as that of the shell. Here, center of the cylinder is a point on the axis of the cylinder which is equidistant from its top and bottom surfaces. Which of the following option(s) is /are correct?
[ $ε_{0}$ is the permittivity of free space]

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$I f h > 2 R a n d r = 3 R / 5 t h e n ϕ = Q / 5 ε_{0}$

$I f h > 2 R a n d r > R t h e n ϕ = Q / ε_{0}$

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

$I f h > 2 R a n d r > 4 R / 5 t h e n ϕ = Q / 5 ε_{0}$

answer is A, B, C.

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Detailed Solution

For option (A): $(h > 2 R r = \frac{3 R}{5})$

$\sin α = \frac{r}{R} = \frac{3}{5} \cos α = \frac{4}{5} Q_{e n c l o s e d} = \frac{Q}{5} ∴ ϕ = \frac{Q}{5 \in_{0}}$

For option (B): $h < \frac{8 R}{5}, r = \frac{3 R}{5}$

$Q_{e n c l o s e d} = 0 a n d ϕ = 0$

Question Image

For option (C): $(h > 2 R a n d r > R)$

$Q_{e n c l o s e d} = Q ∴ ϕ = \frac{Q}{\in_{0}}$

Question Image
For option (D): $h > 2 R r = \frac{4 R}{5}$

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$Q_{e n c l o s e d} = Q [1 - \cos α] \sin α = \frac{r}{R} = \frac{3}{5} a n d \cos α = \frac{3}{5} Q_{e n c l o s e d} = \frac{2 Q}{5}$

$∴ ϕ = \frac{2 Q}{5 \in_{0}}$

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