A charge q is surrounded by a closed surface | vedclass

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Question

From Gauss law,

$\phi_{	ext {hemisplere }}+\phi_{	ext {Cone }}=\frac{ q }{\varepsilon_0}$

Total flux produced from $q$ in $\alpha$ angle

$\

Vedclass · Accepted Answer

$3$

Vedclass · Answer

From Gauss law,

$\phi_{	ext {hemisplere }}+\phi_{	ext {Cone }}=\frac{ q }{\varepsilon_0}$

Total flux produced from $q$ in $\alpha$ angle

$\phi=\frac{ q }{2 \varepsilon_0}[1-\cos \alpha]$

For hemisphere, $\alpha=\frac{\pi}{2}$

$\phi_{	ext {hemisphere }}=\frac{ q }{2 \varepsilon_0}$

From equation $(i)$

$=\frac{ q }{2 \varepsilon_0}+\phi_{	ext {cone }}=\frac{ q }{\varepsilon_0}$

$\phi_{	ext {cone }}=\frac{ q }{2 \varepsilon_0}$

$\frac{4 q }{6 \varepsilon_0}=\frac{ q }{2 \varepsilon_0}$

$n =3$

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.

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. . . .

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