An infinitely long uniform line charge distribution of charge per unit length λ lies parallel to the y-axis in the y-z plane at z=32a (see figure). 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 z=λLnε0 ( ε0 =permittivity of free space), then the value of n is

Solution:

Imagine , a cylindrical gaussian surface of length L and radius a.

$\begin{array}{l} \tan (∠ A P O) = \frac{\frac{a}{2}}{\frac{\sqrt{3} a}{2}} = \frac{1}{\sqrt{3}} \\ \Rightarrow ∠ A P O = 30^{0} \\ N o w, ∠ A P D = 2 ∠ A P O = 60^{0} \end{array}$

Given rectangular surface ABCD subscribes an angle $60^{0}$ at the infinity long uniform line of charge.

The total angle subscribe by cylinder is $360^{0}$ and charge enclosed is $λ L$ .

Flux through the rectangle is same as the flux through ${\frac{1}{6}}^{t h}$ of the hexagon since it subtends $60^{0}$ at P and field lines are symmetrically radially outwards.

$\begin{array}{l} ϕ_{c y l} = \frac{q}{\in_{0}} \\ ∴ ϕ_{A B C D} = \frac{q}{6 \in_{0}} = \frac{λ L}{6 \in_{0}} \\ \Rightarrow n = 6 \end{array}$

Solution:

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