A charged shell of radius R carries a total charge Q . Given Φ as the flux of electric field through a…

1. If $\mathrm{h}>2\mathrm{R}$ and $\mathrm{r}>\mathrm{R}$ then $\mathrm{\Phi }=\frac{\mathrm{Q}}{{\in }_0}$
2. If $\mathrm{h}>2\mathrm{R}$ and $\mathrm{r}=\frac{3\mathrm{R}}{5}$ then $\mathrm{\Phi }=\frac{\mathrm{Q}}{5{\in }_0}$
3. If $\mathrm{h}<\frac{8\mathrm{R}}{5}$ and $\mathrm{r}=\frac{3\mathrm{R}}{5}$ then $\mathrm{\Phi }=0$
4. If $\mathrm{h}>2\mathrm{R}$ and $\mathrm{r}=\frac{4\mathrm{R}}{5}$ then $\mathrm{\Phi }=\frac{\mathrm{Q}}{5{\in }_0}$

Solution

$\left(\mathrm{A}\right)$ $\Rightarrow h>2Rr>R$

$\varphi =\frac{Q}{{\epsilon }_0}$
Clearly from Gauss' Law $\Rightarrow$ Option $\left(\mathrm{A}\right)$ is correct.
$\Rightarrow$ for $h=2rR=\frac{4R}{5}$

$\mathrm{s}\mathrm{h}\mathrm{a}\mathrm{d}\mathrm{e}\mathrm{d}\text{\hspace{0.17em}charge}=2\pi (1-\mathrm{c}\mathrm{o}\mathrm{s}{53}^{\mathrm{o}})\times \frac{Q}{4\pi }=\frac{Q}{5}$
Where, $2\pi (1-\mathrm{c}\mathrm{o}\mathrm{s}53°)=\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{i}\mathrm{d}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{l}\mathrm{e}$
$\therefore q_{\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{l}\mathrm{o}\mathrm{s}\mathrm{e}\mathrm{d}}=\frac{2Q}{5}\left(\frac{Q}{5}\mathrm{a}\mathrm{b}\mathrm{o}\mathrm{v}\mathrm{e}\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{d}\frac{Q}{5}\mathrm{b}\mathrm{e}\mathrm{l}\mathrm{o}\mathrm{w}\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{e}\mathrm{r}\right)$
$\therefore \varphi =\frac{2Q}{5{\epsilon }_0}$
$\therefore forh>2R,r=\frac{4R}{5}$
$\therefore \varphi =\frac{2Q}{5{\epsilon }_0}\Rightarrow$ Option $\left(\mathrm{D}\right)$ is incorrect.
$\left(\mathrm{C}\right)\Rightarrow$ suppose $h=\frac{8R}{5}r=\frac{3R}{5}$

$\varphi =0$ (as no charge is enclosed inside the cylinder, as shown in diagram)
So for $h<\frac{8R}{5},\varphi =0\Rightarrow$ Option $\left(\mathrm{C}\right)$ is correct.
$\left(B\right)\Rightarrow$ like option $\left(\mathrm{D}\right)$ for $h=2R,r=\frac{3R}{5}$
$q_{\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{l}\mathrm{o}\mathrm{s}\mathrm{e}\mathrm{d}}=2\times 2\pi \left(1-\mathrm{c}\mathrm{o}\mathrm{s}{37}^{\mathrm{o}}\right)\frac{Q}{4\pi }=\frac{Q}{5}$
$\therefore \varphi =\frac{Q}{5{\epsilon }_0}\Rightarrow$ Option $\left(\mathrm{B}\right)$ is correct.

Asked in: JEE Advanced 2019 (Paper 1)

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