# Electric field due to an infinite wire

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 dE 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 dE 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: As you can see in the previous figure, the vertical component of the vector sum of both fields dE 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 dE. The magnitude of the electric field due to the wire at point P is therefore: 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: Knowledge is free, but servers are not. Please consider supporting us by disabling your ad blocker on YouPhysics. Thanks!

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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