# Electric field at a point due to an infinite line charge

**1. The problem statement, all variables and given/known data**

Charge is distributed uniformly along an infinite straight line with density

*λ*. Develop the expression for

**E**at the general point

*P*.

**2. Relevant equations**

The electric field at a point *P*, caused by *N* point charges *Q _{i}*, each a distance

**r**

_{i}from

*P*, is given by [itex]\mathbf{E} = \sum \limits_{i=1}^N \frac{Q_i}{4\pi \varepsilon_0 {r_i}^2} \hat{r}_i[/itex].

For continuous charge, the formula becomes an integral: [itex]\mathbf{E} = \int \frac{1}{4\pi \varepsilon_0 {r}^2}\hat{r}\:dQ[/itex], with the unit vector varying along the integral. (This probably isn’t quite correct but it should be the right idea.)

**3. The attempt at a solution**

I think I’m really just having big issues setting up the integral. There are a lot of things going into it, and I’m pretty sure I’m doing something wrong when dealing with the **r** vectors. Here is what I did. I’m not necessarily interested in evaluating the integral, more just setting it up correctly. After that point I could attempt to evaluate it or try it in cylindrical coordinates.

I want to try doing this using Cartesian coordinates, even if using cylindrical coordinates may be better (the solution I have uses those, but even using those I run into similar problems).

We can choose any coordinate system we like, so let’s have the line of charge run along the *z*-axis, and have the point *P* lie in the *x*,*y*-plane. The coordinates of *P* are then (*x*′, *y*′, 0) and the line charge is at (*x*,*y*) = (0,0).

If we consider an infinitesimal part of the line charge *dz*, and say that that piece has charge *dQ*, then we can make the equation *dQ* = *λ dz*.

Consider an infinitesimal piece of the line at the point *z* = *z*′. Its distance from *P* is [itex]r = \sqrt{(x’)^2 + (y’)^2 + (z’)^2}[/itex] along the vector [itex]\mathbf{r} = (x’, y’, z’)[/itex]. So [itex]\hat{r} = \frac{1}{r}\mathbf{r} = \frac{1}{\sqrt{(x’)^2 + (y’)^2 + (z’)^2}}\hat{x} + \frac{1}{\sqrt{(x’)^2 + (y’)^2 + (z’)^2}}\hat{y} + \frac{1}{\sqrt{(x’)^2 + (y’)^2 + (z’)^2}}\hat{z}[/itex].

So, this means my integral is [itex]\mathbf{E} = \int_{-\infty}^{\infty} \int_0^{y’} \int_0^{x’} \frac{\lambda}{4\pi \varepsilon_0 ((x’)^2 + (y’)^2 + (z’)^2)}\hat{r} \: dx \, dy \, dz[/itex]. Is this true? I don’t see unprimed *x*, *y*, or *z* anywhere in the integral, which worries me. I think they should be there. Reading the solution makes me think that every primed variable I have should actually be unprimed.

If it is true, I wouldn’t even know where to begin in solving it, since there are vectors in there and I’m still not sure what to do with vectors in integrals if the differential is also not a vector that is given a vector operation (meaning, the integral ends with something like · *d***r** or × *d***r**, and yes, I do think that *dx dy dz* = *d***r**.) Would I split it up component-wise? Does that even work? Where did the *dx* and *dy* even come from since *dQ* = *λ dz* only? I know they have to be in there but I don’t know from where.

Also, from looking at the solution which uses cylindrical coordinates, I think I’m missing a factor of [itex]\sqrt{(x’)^2 + (y’)^2 + (z’)^2}[/itex] in the integrand’s denominator. Am I?

Thank you very much!

(Also, I know that the *z*-components all will cancel due to symmetry, but I’m trying to see if I can do it the "complete" way this time. Also.. I probably wouldn’t know exactly where it was permitted to remove *z*-components and where they had to be left in.)

http://ift.tt/1eHi8nk

## Leave a comment