Question

A long straight wire having uniform linear charge density lambda passes through an imaginary cube of edge length a the maximum electric flux through the cube will be

Answer 1

According to Gauss' law,
${\phi}_{cube}=\dfrac{Q_{in}}{{\epsilon}_{0}}$
So, in order to have maximum electric flux through the cube, we need to have maximum charge inside the cube, which is achieved by coinciding the wire with the space diagonal of the cube. So,
${\phi}_{cube}=\dfrac{Q_{in}}{{\epsilon}_0}=\dfrac{\sqrt3\lambda a}{{\epsilon}_0}$

Answer 2

Answer: let linear charge density be L

From Gauss Theorem,

Electric flux (say ∆) = Qin/€

or, ∆ = √3La/€

Explanation:

For maximum electric flux, the wire should be coincided with the space diagonal of the cube which has the coordinates (a,a,a) and the length of the diagonal becomes √(a^2 +a^2+ a^2) = √3a = k (say)

Again, Qin = L× k = √3All

Now use Gauss theory.

(The variables used are not the actual symbols)