Question

What is the electric flux through each face if a charge Q is present in the edge of a Cube?

Answer 1

`Q/(24 epsilon o)`

Explanation:

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

But,here the charge is placed at one corner.

So we need to create symmetrical build up of cubes so as to place this charge at the centre of one large cube.

So,you can imagine `8` such cubes,making one large cube,and at the common corner of these the charge `Q` is present.

So,for this large cube electric flux is `Q/(epsilon o)`

So,for `1` such cube its value will be `Q/(8epsilon o)`

But,out of the `6` planes of the given cube,for `3` planes the only the direction of flux is only coming out radially outwards(two opposite sides and the one at the bottom).

So, through each of those `3` faces electric flux associated is `Q/(24 epsilon o)`

It's very difficult to draw a diagram to understand this,so please use your books to create such thing and imagine the whole situation,specially about the creation of symmetry and why 3 planes are under consideration.