Question

please help me in this question ​

Answer 1

Since the charge is kept on the top centre of the square the square can be considered as a face of a cube,

So electric flux through the square

1/6 (net electric flux)= q/6€o (by gauss law)

Answer 2

First draw a gaussian surface(in this case a cube)

Refer diagram.

Now, by gaussian theorem,

integration(E.dA)=q/(epsilon nought).

=>E. integration(dA)=q/e0.

E.(6*area square)=q/e0.

E.A=q/(6*e0).

Flux=q/(6*e0).