The electric field strength depends only on the `x` and `y` coordinates according to the law `E = a (x i + yj) /(x^(2) + y^(2))`, where `a` is a constant , `i` and `j` are the unit vectors of the `x` and `y` axes. Find the flux of the vector `E` through a sphere of raidus `R` with its centre at the origin of coordinates.

Explanation

Since the field is axissymmetric (as the field fo a uniformly changed filament), we conclude that the flux through the sphere fo radius `R` is we equal to the flux through the laterail surface of a cylindrical having the same radius and the height `2R`, as arrangement in the figure.
Now, `Phi = oint vec(E). vec(dS) = E_(r) S`
But `E_(r ) = (a)/(R )`
Thus `Phi = (a)/(R ) S = (a)/(R ) . 2R = 4pi aR`