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 $\frac{nq}{6{ϵ}_0}$ (in SI units). The value of $n$ is ______

Answer 1

Answer: 3

From Gauss law,
${\varphi }_{\text{hemisphere }}+{\varphi }_{\text{Cone }}=\frac{q}{{\epsilon }_0}\dots ..$ (i)
Total flux produced from $q$ in $\alpha$ angle
$\varphi =\frac{q}{2{\epsilon }_0}[1-\cos \alpha ]$
For hemisphere, $\alpha =\frac{\pi }{2}$
${\varphi }_{\text{hemisphere }}=\frac{q}{2{\epsilon }_0}$
From equation (i)
$=\frac{q}{2{\epsilon }_0}+{\varphi }_{\text{cone }}=\frac{q}{{\epsilon }_0}$
${\varphi }_{\text{cone }}=\frac{q}{2{\epsilon }_0}$
$\frac{4q}{6{\epsilon }_0}=\frac{q}{2{\epsilon }_0}$
$n=3$
Alternatively, $\varphi \propto$ 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.