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Looking at the formula of electric field for a finite charged rod (consider rod to be thin, which makes it a line charge with charge in only one dimension) at some equitorial distance $r$, $E=(λ/2πr)\sinθ$, it suggests that as $r$ goes to zero, or we can say as we move closer to the rod, the field blows up.

But on the other hand, if I take a solid or hollow sphere, and move closer and closer to its surface, even on the surface the field does not become infinite. And remains $KQ/R^2$. Why is this so?

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    $\begingroup$ The field does not become infinite at the rod if it is has a finite radius. Inside the rod the feld goes to zero at the center $|E|\propto r$ $\endgroup$
    – mike stone
    Commented Jun 20, 2021 at 17:33
  • $\begingroup$ @mikestone I meant a line charge , with linear charge density lambda.. sorry if the question lacked information. $\endgroup$
    – SHIKHAR CHAMOLI
    Commented Jun 20, 2021 at 17:42
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    $\begingroup$ But then the electric field of a point charge is also infinite at $r=0$. Is it at all surprising that at unphysically concentrated charge distribution leads to an unphysical answer? $\endgroup$
    – mike stone
    Commented Jun 20, 2021 at 17:43
  • $\begingroup$ @mikestone what is unphysical in linear charge distribution?, And what is not in a surface charge distribution like in hollow sphere, in which even on surface,(at the location of charge) field doesn't become large?? $\endgroup$
    – SHIKHAR CHAMOLI
    Commented Jun 20, 2021 at 17:46

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Consider the case where the rod has a very small but non-zero radius $R$, the length is $L$, and the total charge on the rod is $Q$. Then the amount of charge per unit area on the rod (ignoring the insignificant endpoints) would be $Q/(2\pi R L)$. If we take the limit where the $R \rightarrow 0$ this charge per unit area goes to infinity. That is why the field on a one-dimensional line charge goes to infinity.

However, for a finite sphere, the charge density is $Q/(4\pi R^2)$ so the charge density is finite as long as $R$ is non-zero. Note if you take the limit $R \rightarrow 0$ then the sphere becomes a point and the field at that point would approach infinity.

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