A charged shell of radius carries a total charge . Given as the flux of electric field through a closed cylindrical surface of height , radius and with its center same as that of the shell. Here, center of the cylinder is a point on the axis of the cyl

Question

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 [2019]

[ $ε_{0}$ is the permittivity of free space]

Smart Achievers · Accepted Answer

(1, 2, 3)

(1) $Q$

$\sin θ = \frac{3 R / 5}{R} = \frac{3}{5}$ $= 37 °$

$q_{in} = Q (1 - \cos 37 °) = Q [1 - \frac{4}{5}] = \frac{Q}{5}$

$From Gauss's theorem$ $Q = \frac{q_{in}}{ε_{0}}$

$∴$ $ϕ = \frac{Q}{5 ε_{0}}$

(2) $For h > 2 R and r > R$

$ϕ = \frac{q_{in}}{ε_{0}} = \frac{Q}{ε_{0}}$

(3) $For h < \frac{8}{5} R and r = \frac{3}{5} R$

$ϕ = \frac{q_{in}}{ε_{0}} = 0$

(4) $For h > 2 R and r > \frac{4}{5} R$

$\sin θ = \frac{4 R / 5}{R} = \frac{4}{5} = 0.8$ $\Rightarrow$ $θ = 53 °$

$q_{in} = Q (1 - \cos θ) = Q (1 - \frac{3}{5}) = \frac{2 Q}{5}$

$∴$ $ϕ = \frac{q_{in}}{ε_{0}} = \frac{2 Q}{5 ε_{0}}$

Solution

1 If $h > 2 R$ and $r = \frac{3 R}{5}$ , then $ϕ = \frac{Q}{5 ε_{0}}$

2 If $h > 2 R$ and $r > R$ , then $ϕ = \frac{Q}{ε_{0}}$

3 If $h < \frac{8 R}{5}$ and $r = \frac{3 R}{5}$ , then $ϕ = 0$

4 If $h > 2 R$ and $r > \frac{4 R}{5}$ , then $ϕ = \frac{Q}{5 ε_{0}}$

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Solution

1 If h>2R and r=3R5, then ϕ=Q5ε0

2 If h>2R and r>R, then ϕ=Qε0

3 If h<8R5 and r=3R5, then ϕ=0

4 If h>2R and r>4R5, then ϕ=Q5ε0

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1 If $h > 2 R$ and $r = \frac{3 R}{5}$ , then $ϕ = \frac{Q}{5 ε_{0}}$

2 If $h > 2 R$ and $r > R$ , then $ϕ = \frac{Q}{ε_{0}}$

3 If $h < \frac{8 R}{5}$ and $r = \frac{3 R}{5}$ , then $ϕ = 0$

4 If $h > 2 R$ and $r > \frac{4 R}{5}$ , then $ϕ = \frac{Q}{5 ε_{0}}$