Consider a cube. On 5 sides of the cube, there is an electric field that comes out perpendicular to that side, of value E. (a) If there is no charge inside the cube, what is the field (direction and magnitude) perpendicular to the remaining side? (b) If there is a charge Q inside the cube, how does your answer to (a) change?

Explanation:

There are six sides or faces  in a cube. We shall apply Gauss's law here to solve the problem.

a ) If there is no charge inside the cube , total net flux coming out of all the faces will be nil. Suppose area of each face is one squire metre.

If Ex be the required flux coming out of sixth face .

Total flux coming out of six faces of cube = Ex + 5 E

Since total flux coming out is nil

Ex + 5 E = 0

Ex =  - 5E

At the sixth face , electric field will be - 5E . The negative sign indicates that field will be in opposite direction ie it will be towards the inside of the cube.

b ) According to Gauss's theorem , total flux coming out this time will be

Q / ε₀

So this time total flux coming out will be equal to Q / ε₀.

Ex + 5 E  = Q / ε₀.

Ex = Q / ε₀ - 5 E.

If Q / ε₀ > 5 E

Ex will be positive , so its direction will be outward .

If Q / ε₀ < 5 E

Ex will be negative  , so its direction will be inward.