JEE Advance - Physics (2022 - Paper 2 Online - No. 6)
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
3
Explanation
Sol. From Gauss law,
$$ \phi_{\text {hemisphere }}+\phi_{\text {Cone }}=\frac{\mathrm{q}}{\varepsilon_0} $$ ............(1)
Total flux produced from $\mathrm{q}$ in $\alpha$ angle
$$ \phi=\frac{\mathrm{q}}{2 \varepsilon_0}[1-\cos \alpha] $$
For hemisphere, $\alpha=\frac{\pi}{2}$
$$ \phi_{\text {hemisphere }}=\frac{\mathrm{q}}{2 \varepsilon_0} $$
From equation (1)
$$ \begin{aligned} & =\frac{\mathrm{q}}{2 \varepsilon_0}+\phi_{\text {cone }}=\frac{\mathrm{q}}{\varepsilon_0} \\\\ & \phi_{\text {cone }}=\frac{\mathrm{q}}{2 \varepsilon_0} \\\\ & \frac{4 \mathrm{q}}{6 \varepsilon_0}=\frac{\mathrm{q}}{2 \varepsilon_0} \\\\ & \mathrm{n}=3 \end{aligned} $$
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