# Electric Field from a Finite Length, Uniformly Charged Wire (EM, not Gen Phys II)

jumi
So there's this situation going on:
http://imageshack.us/a/img826/7398/physicsforums.png [Broken]

Going from the definition of an electric field:
(1) $\vec{E} ( \vec{x} ) = \frac{1}{4\pi\epsilon_{0}} ∫ \frac{\vec{x} - \vec{x'}}{| \vec{x} - \vec{x'} | ^3} ρ( \vec{x'}) d^3x'$

(2) The $ρ(\vec{x'})d^3x'$ reduces to $λdz'$. And $\vec{x} - \vec{x'} = x \hat{i} - z' \hat{k} = \sqrt{x^2 + z'^{2}}$.

(3) Now, plugging this information into the electric field equation yields:
$\vec{E} ( \vec{x} ) = \frac{1}{4\pi\epsilon_{0}} \int^{l}_{-l} \frac{\sqrt{x^2 + z'^{2}}}{(\sqrt{x^2 + z'^{2}})^3} λdz'$

(4) However, the book (Electromagnetism by Pollack and Stump) shows:
$E_{x}(x,0,0) = \frac{1}{4\pi\epsilon_{0}} \int^{l}_{-l} \frac{x}{(\sqrt{x^2 + z'^{2}})^3} λdz'$

How do we get from (3) to (4)? Why is z' only removed from the numerator?

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## Answers and Replies

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Your formula (3) is obviously wrong, because on the left-hand side is a vector and on the right-hand side a scalar quantity.

From (1) and (2) you immediately write down (4). So your book is correct. There is no z' in the numerator of the x component!

andrien
so the charged wire is along z axis so by symmetry only x component will survive so
z will be absent in numerator and beware it is a vector.

jumi
Your formula (3) is obviously wrong, because on the left-hand side is a vector and on the right-hand side a scalar quantity.

From (1) and (2) you immediately write down (4). So your book is correct. There is no z' in the numerator of the x component!

I don't understand how I could just go from (1) and (2) to (4)...

so the charged wire is along z axis so by symmetry only x component will survive so
z will be absent in numerator and beware it is a vector.

Or why the numerator doesn't have a vector magnitude, whereas the demoninator does...

I was, however, able to get the correct answer, just not using the full notation from equation (1).

So I start with: $\vec{E} ( \vec{x} ) = \frac{1}{4\pi\epsilon_{0}} ∫ \frac{\vec{x} - \vec{x'}}{| \vec{x} - \vec{x'} | ^3} ρ( \vec{x'}) d^3x'$

And since $d\vec{E}$ generated by $dq$ on the line will be symmetric (i.e. only the x-component will survive), we can say $dE_{x} = d\vec{E}cos(\theta)$.

Therefore: $dE = \frac{1}{4\pi\epsilon_{0}} \frac{1}{r^2} dq$, where $r = | \vec{x} - \vec{x'}| = \sqrt{x^2 + z'^2}$

Therefore: $dE_{x} = \frac{1}{4\pi\epsilon_{0}} \frac{1}{x^2 + z'^2} \frac{x}{\sqrt{x^2 + z'^2}} λdz'$ from $cos(\theta) = \frac{x}{\sqrt{x^2 + z'^2}}$ and $dq = λdz'$

Which gives the result in the book: $E_{x}(x,0,0) = \frac{1}{4\pi\epsilon_{0}} \int^{l}_{-l} \frac{x}{(\sqrt{x^2 + z'^{2}})^3} λdz'$.

So was I supposed to recognize that I needed to use the cos(θ) relationship from the beginning? I'm also still confused on how the vector difference only acts on the denominator, if I were to just use equation (1).