Question

An infinitely long uniform line charge distribution of charge per unit length $\lambda$ lies parallel to the y-axis in the y-z plane at $z=\frac{\sqrt{3}}{2}a$. If the magnitude of the flux of the electric field through the rectangular surface ABCD lying in the x-y plane with its centre at the origin is $\frac{\lambda L}{n{\epsilon }_0}$ (${\epsilon }_0=$ permittivity of free space), then the value of $n$ is

Answer 1

flux through cyclender $\Phi =\frac{q_{inclosed}}{{\epsilon }_0}$
where, $q=\lambda L$
so, $\Phi =\frac{\lambda L}{{\epsilon }_0}$
Now, from the dig. the lenght of wire is $\frac{\sqrt{3}}{2}a$
from $\Delta APB:tan\theta =\frac{a/2}{\sqrt{3}a/2}\Rightarrow \frac{1}{\sqrt{3}}$
so, $\theta =ta{n}^{-1}\left(\frac{1}{\sqrt{3}}\right)\Rightarrow {30}^o$
so for $\Delta ABC,\theta ={60}^o$
where, $n=\frac{360}{60}=6$
So, $\Phi =\frac{\lambda L}{n{\epsilon }_0}$ (as given)
So, total flux passing through the given area $\Phi =\frac{\lambda L}{6{\epsilon }_0}$