Question

a charge q is placed just above the centre of horizontal circle of radius r and a hemisphere about this is erected about the charge. compute the flux through the closed surface that consists of the hemisphere and the planar circle. do not use gauss law​

Answer 1

Alright, now just think about flux, it is defined as the surface integral

∫E.dS = Ф

Now since the charge is ket just above the circle and you want me to calculate the closed integral, gauss law is applicable! But since you mentioned not to use it i would go further.

The E is given by kq/r² all over the spheres surface and it is constant. so now if we compute total area it is 2πr² (remeber E.ds is dot product of E and area vector so that the circular surface's area gets cancelled out)

now it is pretty easier to see that flux is q/2∈