Consider a point charge `q=1 m C` placed at a corner of a cube of sides 10 cm . Determine the electric flux through each face of the cube.
Strategy `:` We will learn about the utility of symmetry in solving problems with the help of Gauss's law . Here we'll use the symmetry of the situation,which involves the faces joining at the corner at which the charge resides.

(a) A charge q is placed atthe corner of a cube. (b) By surrounding the charge with a series of cubes such that the charge is at the centre of a larger cube, we have created an arrangement sufficiently symmetric to be able to solve for desired flux values.
You can see from figure that for these faces `vec(E).hat(n)=0`. Since the normal is perpendicular to the surfaces while the electric field goes off in a spherically symmetric pattern and lies in the sides . In other words , the electric field that originates at the charge is tangential to the surface of these three sides. This means there is no flux through these sides. The electric flux through each of the remaining three faces of the be must be equal by symmetry . We'll referto these faces with the label F.
To fidn the flux through each of the sides F, we acan use a technique that puts the single charge in the middle of a larger cube. It takes seven other similarly placed cubes to surrounds the points charge q completely in figre. The charge is at the dead centre of the enw larger cube. So, the flux through each of the six sides of the large cube will now have an electric flux of one sixth of the total flux F. So given that the total structure is completely symmetric, the flux through a side F is one fourth of the flux through the larger side.

Explanation

The total flux is `(q)/(epsilon_(0))` . So that the flux through each of the sidesof the larger cube is `(q)/( 6 epsilon_(0))` and one quarter of that `(q)/( 24epsilon_(0))` goes through each of the far sides of the small cube. Numerical evaluation gives.
`phi= (q)/( 24epsilon_(0))=( 1 xx 10^(-3)C)/(24(8.85 xx 10^(-12)C^(2) //Nm^(2)))= 5 xx 10^(6) N .m^(2) //C`