Electric Field Intensity due to an Infinite Long Line Charge from Gauss’s Law
Let the linear charge density is λ and the intensity of electric field at point P is to be determined. To apply Gauss’s law, we have to decide the direction of electric field at P. For this purpose let us consider two small elements of length dl at A and B at equal distance from O as shown in the figure. The intensities of electric field due to these elements at P, dE1 and dE2 are equal in magnitude. If the electric fields are resolved in two normal components, then the components along OP provide the electric field \(\vec{E}\) along OP and normal components dE1 sin θ and dE2 sinθ cancel out each other. Similar will be the result for other pairs of length elements considered.
Thus the direction of electric field at P will be in OP direction i.e., normal to linear charge.
Calculation of Electric Field: For this purpose, we draw a cylindrical Gaussian surface of radius r and axis coinciding the linear charge AB. Point P lies on the curved surface of the cylinder. The enclosed charge Σq = λl,
where l is the length of the cylinder. Therefore from Gauss’s law,
Now equating equation (1) & (2), we have,
In vector form
where r̂ = unit vector in direction OP i.e., in perpendicular to linear charge.
From equation (3),
E ∝ 1/r
