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I found the electric field of a uniformly charged thin rod to be, $E=\frac{Q}{4\pi \epsilon_0 (a^2-x^2)}$. Where $L_{rod}=2a$ and $x$ is the distance away from the centre of the rod, along the its axis (the horizontal or x axis in this case). My question is, according to this model the electric field at one of the ends of the rod is infinite for even a small charge. Is this really the case or do you say that this model is only valid for $x>a$ and at $x=a$ you can take all the charge to be at the centre or at $x=0$. However, I dont believe we can make this assumption as it should still hold for $x>a$ and if it did the electric field for a thin rod would be the same as a sphere.

So my question is, is the electric field at one of the ends really infinite, or is there an upper bound and if so, how do you calculate it?

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Yes, it is correct that the field goes to infinity; like the field near a point charge, it goes infinity when you go near.

Note that the rod is "infinitely thin", which is the source of trouble. Normally rods are a bit thick. The formula should be correct anywhere along the axis, and you get approximately spherical field for $x>>a$.

If the field can be actually, really, physically infinite, we must ask to philosophers.

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  • $\begingroup$ I considered the case of a point charge but any real point like object can always be treated as some sphere of a finite radius, the electric field of any point like charge always has an upper bound. I cant make the same link with a rod, are you saying that even if we include the rods finite dimensions, along the axis of the rod the electric field has no upper limit as you approach the road? $\endgroup$ – Vishal Jain Apr 8 at 17:53
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    $\begingroup$ I think if you make the rod big, with say uniform surface charge, the field should be finite. A cylinder rod is a bit tricky; the analog of the sphere around the point charge, here, is an ellipsoid with as focuses the ends (a,0,0); (-a,0,0) of the rod, such as it follows the electric field, and with uniform surface charge should give rise to the same external field $\endgroup$ – patta Apr 8 at 18:54

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