Question

An infinitely long thin non-conducting wire is parallel to the $z$ - axis and carries a uniform charge of line charge density $\lambda$. It pierces a thin non-conducting spherical shell of radius $R$ in such a way that the arc $PQ$ sustends an angle of ${120}^{\circ }$ at the centre $\text{O}$ of the spherical shell as shown in the figure. The electric flux through the shell is

• A. $\sqrt{6}R\frac{\lambda }{{ϵ}_0}$
• B. $\sqrt{3}R\frac{\lambda }{{ϵ}_0}$
• C. $\sqrt{2}R\frac{\lambda }{{ϵ}_0}$
• D. $\sqrt{5}R\frac{\lambda }{{ϵ}_0}$

Answer 1

The correct option is B $\sqrt{3}R\frac{\lambda }{{ϵ}_0}$

Given that $\angle POQ={120}^{\circ }$


Let us draw a perpendicular bisector $OS$,
$\angle SOP=\angle SOQ={60}^{\circ }$

So looking at $\Delta OSQ$

$\angle SOP={60}^{\circ },\angle OSP={90}^{\circ }$

$\Rightarrow \angle SPO={30}^{\circ }$

From geometry,

$\cos {30}^{\circ }=\frac{PS}{PO}$

$\Rightarrow PS=R\times \frac{\sqrt{3}}{2}$

So, $PQ=2\times PS=\sqrt{3}R$

Now, charge enclosed by spherical shell,

$q=\lambda \times PQ=\lambda \times \sqrt{3}R=\sqrt{3}\lambda R$

Further, using Gauss law,
Electric flux, $\varphi =\frac{q}{{ϵ}_0}=\frac{\sqrt{3}\lambda R}{{ϵ}_0}$