Answer:
Option D
Explanation:
ANBP is cross-section of a cylinder of length L. The line charges pass through the centre O and perpendicular to paper
$AM=\frac{a}{2},MO=\frac{\sqrt{3}a}{2}$
$\therefore$ $\angle AOM=\tan ^{-1}\left(\frac{AM}{OM}\right)$
$=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=30^{0}$
Electric Flux passing from the whole cylinder
$\phi_{1}=\frac{q_{in}}{\epsilon_{0}}=\frac{\lambda L}{\epsilon_{0}}$
$\therefore$ Electric flux passing through ABCD plane surface (shown only AB)= Electric flux passing through cylindrical surface ANB
= $\left( \frac{60^{0}}{360^{0}}\right)(\phi_{1})=\frac{\lambda L}{6\epsilon_{0}}$
$\therefore$ n=6