9.2 Electric field due to a linear charge distribution

  Electrostatics

    1.0 Introduction
    2.0 Electric charge
    3.0 Coulomb's law
    4.0 Principle of superposition
    5.0 Continuous charge distribution
    6.0 Electric field
    7.0 Electric field lines
    8.0 Insulators and conductors
    9.0 Gauss's law
    10.0 Work done
    11.0 Electric potential energy
    12.0 Electric Potential
    13.0 Electric dipole

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9.2 Electric field due to a linear charge distribution

Consider a long line charge with a linear charge density (charge per unit length) $\lambda $. For calculating the electric field at a point $P$ which is at a distance $r$ from the line charge, we will construct a gaussian surface in the form of the cylinder of any arbitrary length $l$, radius $r$ and its axis coinciding with the axis of the line charge.

Since the cylinder has three surfaces. One is curved surface and the other two are plane-parallel surface. Electric field lines at plane parallel surfaces are tangential. So, the flux passing through these surfaces is zero.

The magnitude of electric field is having the same magnitude $(E)$ at the curved surface and simultaneously the electric field is perpendicular at every point to this surface.

So, from Gauss's law we can write, $$ES = \frac{{{q_{in}}}}{{{\varepsilon _0}}}$$
where,

$S = 2\pi rl:$ Area of curved surface
${q_{in}} = \lambda l:$ Net charge enclosed within the cylindrical gaussian surface

So, $$E\left( {2\pi rl} \right) = \frac{{\lambda l}}{{{\varepsilon _0}}}$$$$E = \frac{\lambda }{{2\pi r{\varepsilon _0}}}$$ or $$E \propto \frac{1}{r}$$

The above equation shows that the relation between $E-r$ is a rectangular hyperbola.