In this page, we are going to calculate the **electric field due to an infinite charged wire**. We will assume that the charge is homogeneously distributed, and therefore that the linear charge density λ is constant. We will also assume that the total charge q of the wire is positive; if it were negative, the electric field would have the same magnitude but an opposite direction.

The charged wire and the point P of space where we will calculate the electric field are represented in the following figure:

First, we will calculate the electric field due to a charge element *dq* of length *dy* at a point P of space. This charge element is located at a distance *r* of point P and its vertical coordinate is *y*. *dq* can be considered as a point charge, thus the electric field due to it at point P is:

And the total electric field due to the wire is given by the following integral:

Before evaluating this type of integrals, it is convenient to first analyze the symmetry of the problem to see if it can be simplified.

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Both the electric field *d E* due to a charge element

*dq*and to another element with the same charge located at coordinate

*-y*are represented in the following figure. The wire is positively charged so dq is a source of field lines, therefore

*d*is directed outwards. Furthermore, the electric field satisfies the superposition principle, so the net electric field at point P is the sum of the individual electric fields due to both charge elements:

**E**As you can see in the previous figure, the vertical component of the vector sum of both fields *d E* is zero. This is true for any charge element and their symmetrical, therefore the total electric field is the integral of the horizontal projections of

*d*. The magnitude of the electric field due to the wire at point P is therefore:

**E**The integral has unbounded intervals of integration because the wire is considered to be infinite.

As you can see in the figure, the cosine of angle α and the distance *r* are respectively:

After substituting in the expression of the total field, we have:

Where we use the following product of the vacuum permittivity constant to express the Coulomb constant:

After solving the integral we get:

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We can also rewrite the previous integral as a function of the angle α using *r* and *y*:

And as you can see, we get the same result as before.

The electric field vector is obtained by multiplying the calculated magnitude with a unit vector in the radial direction:

And the field lines are represented in the following figure:

You can see how to calculate the electric field due to an infinite wire using Gauss’s law in this page.

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