- #1
Uku
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Homework Statement
I know that around an infinitely long wire with even charge distribution the electric field is expressed as:
[tex]E=\frac{1}{2\pi\epsilon_{0}}\frac{\lambda}{r}[/tex] (1)
Where [tex]\lambda[/tex] can be expressed as [tex]\lambda=\frac{dq}{dl}[/tex]
Right, but I want to know where I get this formula from, I mean the E field.
The Attempt at a Solution
So I know that in general:
[tex]E=\frac{1}{4\pi\epsilon_{0}}\int\frac{\rho}{r^{2}}\widehat{r}dV[/tex]
In my case I don't have volume, I have a thread. I can also forget about the unit vector, since the field is radially pointed outward. The charge is evenly distributed so I can write:
[tex]E=\frac{1}{4\pi\epsilon_{0}}\lambda\int\frac{1}{r^{2}}dL[/tex]
Okay, but now... I can integrate the expression from minus infinity to infinity, but how do I get to that formula (1)