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?


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 '19 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 '19 at 18:54

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