The flux of an electric field E¯ through an arbitrary closed surface S is equal to the algebraic sum of the charge enclosed by this surface divided by ε0.
Example 1. Figure shows a section of an infinite rod of charge having linear charge density λ which is constant for all points on the line. Find electric field E at a distance r from the line.
Solution : From symmetry, E¯ due to a uniform linear charge can only be radially directed. As a Gaussian surface, we can choose a circular cylinder of radius r and length l, closed at each end by plane caps normal to the axis.
The direction of E¯ is radially outward for a line of positive charge.
Example 2. Figure shows a sphericaly symmetric distribution of charge of radius R. Find electric field E for points A and B which are lying outside and inside the charge distribution respectively.
Solution: The spherically symmetric distribution of charge means that the charge density at any point depends only on the distance of the point from the centre and not on the direction. Secondly, the object can not be a conductor, or else the excess charge will reside on its surface.
Now, apply Gauss’s law to a spherical Gaussian surface of radius r ( r > R for