Question

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 ______

Answer 1

From Gauss law,
$\phi_{\text {hemisphere }}+\phi_{\text {Cone }}=\frac{ q }{\varepsilon_0} \ldots . .$ (i)
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.