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A charge $q$ is surrounded by a closed surface consisting of an inverted cone of height $h$ and base radius $R$, and a hemisphere of radius $R$ as shown in the figure. The electric flux through the conical surface is $\frac{n q}{6 \epsilon_0}$ (in SI units). The value of $n$ is. . . .

$2$
$0$
$1$
$3$
Solution
From Gauss law,
$\phi_{\text {hemisplere }}+\phi_{\text {Cone }}=\frac{ q }{\varepsilon_0}$
Total flux produced from $q$ in $\alpha$ angle
$\phi=\frac{ q }{2 \varepsilon_0}[1-\cos \alpha]$
For hemisphere, $\alpha=\frac{\pi}{2}$
$\phi_{\text {hemisphere }}=\frac{ q }{2 \varepsilon_0}$
From equation $(i)$
$=\frac{ q }{2 \varepsilon_0}+\phi_{\text {cone }}=\frac{ q }{\varepsilon_0}$
$\phi_{\text {cone }}=\frac{ q }{2 \varepsilon_0}$
$\frac{4 q }{6 \varepsilon_0}=\frac{ q }{2 \varepsilon_0}$
$n =3$
Alternatively, $\phi \propto$ no of electric field lines passing through surface $q$ is point charge which has uniformly distributed electric field lines thus half of electric field lines will pass through hemisphere $\&$ other half will pass through conical surface.