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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 $r$ 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 ? $\epsilon_0$ is the permittivity of free space]
$(1)$ If $h >2 R$ and $r > R$ then $\Phi=\frac{ Q }{\epsilon_0}$
$(2)$ If $h <\frac{8 R }{5}$ and $r =\frac{3 R }{5}$ then $\Phi=0$
$(3)$ If $h >2 R$ and $r =\frac{4 K }{5}$ then $\Phi=\frac{ Q }{5 \epsilon_0}$
$(4)$ If $h >2 R$ and $r =\frac{3 K }{5}$ then $\Phi=\frac{ Q }{5 \epsilon_0}$
$1,2,3$
$1,2,4$
$1,2$
$1,3$
Solution
For option $(1)$, cylinder encloses the shell, thus option is correct For option $(2)$,
(image
cylinder perfectly enclosed by shell, thus $\phi=0$, so option is correct. For option $(3)$
(image)$\phi=\frac{2 \times Q}{2 \epsilon_0}\left(1-\cos 53^{\circ}\right)=\frac{2 Q}{5 \epsilon_0}$
For option $(4)$ :
Flux enclosed by cylinder $=\phi=\frac{2 Q }{2 \epsilon_0}\left(1-\cos 37^{\circ}\right)=\frac{ Q }{5 \epsilon_0}$
