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Let's say we have a uniformly charged rod of length $L$ at a distance $x$ ($\ne 0$) from the $(x, y, z)$ origin.

My question is most probably trivial as it's not completely clear to me how electric field lines work in 3D space, so thank you for your patience – given the rod is uniformly charged, I can go ahead and consider many tiny point charges along the rod's surface that generate an electrical field $\vec{E}$.

I know the electrical field lines, however, point outwards radially with respect to the rod so there should be no $x$ component ($\vec{E}_{x}$) such that there would be an electric field in $P$, the origin.

Why is it not the case and it's effectively possible to sum all the tiny charges' contributions (integrating along $x$) to find out the electric field in $P$? Thank you again!

Edit: image reference Graphic description of the rod

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It is not exactly true that the electric field lines point along the radial direction around the rod. It is true for the region of space that is located underneath or above the rod (i.e. from $x$ to $x+L$ and to be specific, at $x+L/2$).

So, at the origin (your point P), you will have an electric field that points towards the negative horizontal direction if the rod is positively charged, whereas in the case in which the rod is negatively charged, it will point towards the positive horizontal direction. You can think of it as a point charge located at $x+L/2$ if you like and ask yourself what would the electric field lines be at the origin... This is a good approximation if $L<<x$.

The reason, actually you do not have horizontal components in the electric field above the centre of the rod is because of symmetry. The left half of the rod negates the electric field caused by the right hald and vice versa.

If something is not clear, please let me know in the comments.

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    $\begingroup$ Well, actually above and underneath is a radial direction. If you think a circle with its centre being the $(x=a+l/2,y=0)$, then you can define a circle in 3D with all the points whose distance is $y$ (for example) from that point $\endgroup$
    – schris38
    Jun 30, 2022 at 10:08
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    $\begingroup$ I am not sure I understand the remaining of your questions... Could you please rephrase? $\endgroup$
    – schris38
    Jun 30, 2022 at 10:09
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    $\begingroup$ Or maybe I can say this: each infinitesimally small charge on the rod (i.e. with coordinates $(x,0)$ with $x\in[a,l+a]$) cause an electric field that has both horizontal and vertical components. Above the centre of the rod, the vertical components of the electric field caused from all the infinitesimally small charges on the left part of the rod, cancel the components of the electric field caused by the charges on the right. This does not happen above any of the points for which $x\ne a+l/2$, such as the origin. So, we expect to have horizontal component there. $\endgroup$
    – schris38
    Jun 30, 2022 at 10:13
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    $\begingroup$ Also at the beginning we do not have vertical component of the electric field due to the fact that there is no vertical distance ($y=0$) from the rod to the point P $\endgroup$
    – schris38
    Jun 30, 2022 at 10:14
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    $\begingroup$ Got it! Thank you so much! What was super confusing is that every single image I've been looking at online that attempts to describe the electric field of a charge has these vectors that represent the field lines... but on a single plane! I didn't quite grasp that there would be infinite field lines directed outwards (assuming positive) on all three dimensions. I would best represent an electric field using a sphere and field lines for all directions in the 3D space. Hopefully I got that right – it would make sense then for the charge in $x + L/2$ to be only radial on the $y,z$ plane! $\endgroup$ Jun 30, 2022 at 13:45

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