Question

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 centre same as that of the shell. Here center of 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? $$[\varepsilon_0$$ is the permittivity of free space$$]$$

Answer 1

Correct option is D. If $$h > 2R$$ and $$r > R$$ then $$\phi =\dfrac{Q}{\varepsilon_0}$$
(a) $$h > 2R$$
$$r > R$$
$$\phi =\dfrac{Q}{\varepsilon_0}$$ clearly the Gauss' Law.
(b) Suppose $$h=\dfrac{8R}{5}$$
$$r=\dfrac{3R}{5}$$
so for $$h < \dfrac{8R}{5} \phi =0$$
(c) for $$h=2R$$
$$r=\dfrac{4R}{5}$$
Shaded charge $$=2\pi (1-\cos 53^0)\times \dfrac{Q}{4\pi}=\dfrac{Q}{5}$$
$$\therefore q_{enclosed}=\dfrac{2Q}{5}$$
$$\therefore \phi =\dfrac{2Q}{5_{\varepsilon_0}}$$
$$\therefore$$ For $$h > 2R$$
$$r=\dfrac{4R}{5}$$
$$\therefore \phi =\dfrac{2Q}{5\varepsilon_0}$$
(d) like option C for $$h=2R$$ $$r=\dfrac{3R}{5}$$
$$q_{enclosed}=2\times 2\pi(1-\cos 37^o)\dfrac{Q}{4\pi}=\dfrac{Q}{5}$$
$$\therefore \phi =\dfrac{Q}{5\varepsilon_0}$$.