Cube! Not related to geometry this time

first we consider the flux when it is kept at vertex,the flux due to the given 3 sides will remain same as they are far apart from the charge and we have just given charge small displacement.(we get it q/8e(o)) now when it is inside flux is q/e(o) so change because of new flux generated on the three sides is 7q/8e(o)

Here goes my solution,

We know $\phi_{smaller cube}= \dfrac{Q}{\epsilon_{0}}$ .

Now for symmetry of the charge, let's add up 7 more cubes such that the charge is at the center for the larger cube.

We also get $\phi_{Larger cube} = \dfrac{Q}{\epsilon_{0}}$ .

Now we can see, that there are 24 small faces making up the larger cube. So we get,

$\phi_{each small face} = \dfrac{Q}{24 \epsilon_{0}}$

Now we see that the flux of EFDC,GHDC,ECGB are all equal to $\dfrac{Q}{24 \epsilon_{0}}$ .

Also we know, total flux for the small cube is = $\dfrac{Q}{\epsilon_{0}}$

So The sum of the flux through wanted sides are = $\dfrac{Q}{\epsilon_{0}} - 3 \times \dfrac{Q}{24 \epsilon_{0}} \implies \dfrac{7Q}{8 \epsilon_{0}}$ .

Anyone if thinks my solution isnt correct can correct it!

Thanks!

And i also dont know why is it said int the title No Geometry this time . I used pure geometry to get $\phi_{each small side}$ .

We consider the cube as our Gaussian surface. The total flux through the 6 sides of the cube is Q/€. Now consider a bigger cube formed by 8 given cubes such that the charge Q is at the center of the bigger cube. The total flux through the sides of the cube is Q/€. We say by symmetry the total flux through the sides HGCD, BGCE and DFEC is Q/8€. So the required flux is 7Q/8€