## Ans is Q/12ϵ

Feb 2, 2018

OK, flux, $\phi$ is defined as being flux density, $B \times A \cos \theta$ but as we aren’t told the angle we’ll assume it’s perpendicular so $\cos \theta = 1$

#### Explanation:

That means $\phi = B \times A$ so $\phi = B \times {Q}^{2}$ as the shape appears to be a cube.

You may have met B defined as mag field strength.

Feb 2, 2018

The flux through the shaded side is $\frac{Q}{12 {\epsilon}_{o}}$. I regret that the explanation is long -- a consequence of discussion 2 different situations.

#### Explanation:

I credit @Jane for this approach. She answered your similar question with an approach that I admire. Unfortunately, I think she misunderstood that question, thinking it was about a charge at a corner of a cube. It wasn't. But her approach to solving that question as she understood it inspires my answer to this question.
I have pasted her answer here and have changed its color to blue.

To be clear, the situation Jane analysed could be illustrated by the figure above if one of the charges were deleted.

color(blue)("From Gauss's law we know electric flux for a closed surface is" Q/(epsilon o)

$\textcolor{b l u e}{\text{But,here the charge is placed at one corner.}}$

color(blue)("So we need to create symmetrical build up of cubes"
$\textcolor{b l u e}{\text{so as to place this charge at the centre of one large cube.}}$

$\textcolor{b l u e}{\text{So,you can imagine 8 such cubes,making one large cube,}}$
$\textcolor{b l u e}{\text{and at the common corner of these the charge Q is present.}}$

$\textcolor{b l u e}{\text{So,for this large cube electric flux is} \frac{Q}{\epsilon o}}$

$\textcolor{b l u e}{\text{So,for 1 such cube its value will be} \frac{Q}{8 \epsilon o}}$

$\textcolor{b l u e}{\text{But, out of the 6 planes of the given cube,for 3 planes the only}}$
$\textcolor{b l u e}{\text{the direction of flux is only coming out radially outwards}}$
$\textcolor{b l u e}{\text{(two opposite sides and the one at the bottom).}}$

$\textcolor{b l u e}{\text{So, through each of those 3 faces electric flux associated is} \frac{Q}{24 \epsilon o}}$

$\textcolor{b l u e}{\text{It's very difficult to draw a diagram to understand this,}}$
$\textcolor{b l u e}{\text{so please use your books to create such thing and imagine the}}$
$\textcolor{b l u e}{\text{whole situation,specially about the creation of symmetry and}}$
$\textcolor{b l u e}{\text{why 3 planes are under consideration.}}$

I will apply this analysis to one of the 2 charges in this question and then show that it can also be applied to the other charge and the affect is putting a $2$ in the numerator of the answer.

I will take also the liberty of attempting to add to the explanation in Jane's paragraph beginning "But, out of the 6 planes ...".

Since Jane assumed the charge is at a corner of the cube, the charge is part of 3 of the planes of the cube. Therefore there would not be any flux passing through those 3 planes. Only the other 3 sides would have flux from the one charge being considered by Jane's analysis.

Now, to apply this to this question. The 2nd charge also has charge $Q$. Neither of the 2 charges is at a corner of the shaded side. That means that the fields from both charges have the shaded side in range for flux.

Therefore, the flux through the shaded side is $\frac{2 Q}{24 {\epsilon}_{o}} = \frac{Q}{12 {\epsilon}_{o}}$