Question

Find the flux through the curved surface of a given cylindrical Gaussian surface around a point charge $+9\text{nC}$ kept at centre at the axis.
(Take ${ϵ}_o=9\times {10}^{-12}{\text{C}}^2/{\text{N-m}}^2$)

• A. $300\sqrt{3}{\text{N-m}}^2\text{/C}$
• B. $400\sqrt{3}{\text{N-m}}^2\text{/C}$
• C. $500\sqrt{3}{\text{N-m}}^2\text{/C}$
• D. $600\sqrt{3}{\text{N-m}}^2\text{/C}$

Answer 1

The correct option is C $500\sqrt{3}{\text{N-m}}^2\text{/C}$

Given that,
Charge, $q=9\text{nC}=9\times {10}^{-9}\text{C}$
Height of the cylinder, $h=2\sqrt{3}\text{m}$
Radius of the cylinder, $r=1\text{m}$


The flat top and bottom surface of the given cylinder can be assumed to be the base of a cone having semi-vertex angle $\theta$.
The electric flux through cone is given by
${\varphi }_{\text{cone}}=\frac{q}{2{ϵ}_0}(1-\cos \theta )$

The electric flux through curved surface only is given by
$\text{Electric flux through curved surface = Total flux through cylinder - Flux through top and bottom surface of cylinder}$
$\Rightarrow {\varphi }_{\text{curved surface}}={\varphi }_{\text{Total cylinder}}-2{\varphi }_{\text{cone}}$
$\Rightarrow {\varphi }_{\text{curved surface}}=\frac{q}{{ϵ}_0}-2\times \frac{q}{2{ϵ}_0}(1-\cos \theta )=\frac{q\cos \theta }{{ϵ}_0}$ ... (i)
Here from diagram
$\tan \theta =\frac{1}{\sqrt{3}}$
$\Rightarrow \theta ={30}^{\circ }$
Putting the value of $\theta$ and $q$ in (i), we get,
${\varphi }_{\text{curved surface}}=\frac{9\times {10}^{-9}\times \cos {30}^{\circ }}{9\times {10}^{-12}}$
$\Rightarrow {\varphi }_{\text{curved surface}}=500\sqrt{3}{\text{N-m}}^2\text{/C}$