Electric Field due to Line Charge – Finite and Infinitely Long Charged Wire

Electric Field due to Line Charge – Finite and Infinitely Long Charged Wire

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The objective of this article is to calculate the Electric Field due to Line Charge Distribution. We will also calculate the Electric Field due to an Infinite Line Charge Distribution. However, if you are interested in a detailed study of the Electric Field, we recommend starting here. In our previous articles, we have discussed the calculations of electric potential due to different charge configurations and Electric Field due to Point Charge and System of Charges in details. You can browse those articles from our library for a better understanding of the subject.

Electric Field due to Line Charge Distribution

In our previous article, we have calculated the Electric Field due to a Point Charge. In this article, the electric field of a line of charge distribution can be calculated by superposing the point charge fields of infinitesimal charge elements. Let us consider a straight charged wire of length L. The line charge density of this charged wire is \lambda. We have to calculate the electric field at point P which is Z distance apart from the line charge distribution. The situation is shown in the figure below. electric field due to line chargeLet us consider an charge element of length dx at a distance x as shown in the figure. Charge contained within this element is \lambda dx. The electric field at point P due to this charge element is

    \[d\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{r^2}\hat{r}\]

and the radial part of the electric field from this charge element is,

    \[dE_z=\frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{r^2}\frac{z}{r}\]

    \[dE_z=\frac{1}{4\pi\epsilon_0}\frac{\lambda z dx}{(z^2+x^2)^\frac{3}{2}}\]

We will integrate over the whole charged wire to get the total radial electric field at point P due to this line charge distribution.

    \[E_z=\frac{\lambda z}{4\pi\epsilon_0}\left \int_{-a}^{b}\frac{dx}{(z^2+x^2)^\frac{3}{2}} \right \]

    \[= \frac{\lambda}{4 \pi \epsilon_0 z} \left [ \frac{b}{(z^2+b^2)^\frac{1}{2}}+\frac{a}{(z^2+a^2)^\frac{1}{2}} \right]\]

Please be careful about the limit of integration. It runs from -a to b. Similarly, we can calculate the axial component of the electric field.

    \[E_x=\frac{\lambda }{4\pi\epsilon_0}\left \int_{-a}^{b}\frac{x dx}{(z^2+x^2)^\frac{3}{2}} \right \]

    \[=\frac{\lambda}{4 \pi \epsilon_0}\left [ \frac{1}{(z^2+b^2)^\frac{1}{2}}-\frac{1}{(z^2+a^2)^\frac{1}{2}}\right]\]

In a symmetric case of a=b, this axial component vanishes and we are left with the radial component only.

Electric field due to an infinitely long line charge distribution can be considered as a limiting case of the above solution. In this case a and b approach to the infinity. The axial component of the electric field vanishes again. Thus the electric field due to an infinitely long line charge distribution is

    \[E_z= \frac{\lambda}{2 \pi \epsilon_0 z} \]

and it does not have any axial component. This becomes obvious if we look at the axial symmetry of the problem. In the next section, we will exploit this symmetry to calculate the electric due to an infinitely long charged wire.

Electric Field due to Infinitely Long Line Charge (Gauss’s Law Application)

We have to calculate the electric field at point P due to an infinitely long charged wire of charge density \lambda. The situation is shown in the diagram below. As already mentioned, the system has a cylindrical symmetry. This has significantly simplified the problem and we can use Gauss’s law to calculate the electric field. Let us imagine a hypothetical cylindrical Gaussian surface as shown in the figure. Since the field is pointing radially outwards, the flux through the two ends of the cylinder is zero. Also, at every point on the cylindrical surface, the electric field is constant and is pointing normal to the surface. The surface area of the curved surface of length l is 2\pi r l. Thus total flux crossing through the cylindrical Gaussian surface is \vec{E} \cdot 2\pi r l \hat{r}.

electric field due to infinitely long charged wire gauss law

Total charge enclosed within this Gaussian surface is \lambda l. Now according to Gauss’s law

    \[\vec{E}\cdot 2\pi r l \hat{r} = \frac{\lambda l}{\epsilon_0}\]

    \[\vec{E}=\frac{\lambda}{2 \pi \epsilon_0 r} \hat{r}\]