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We know that electric field due to a plane is given by,(skipping the directions for now)

$$ E=\frac{\sigma}{2\epsilon_{o}}. $$

Griffiths says, as the electric field is independent of the distance, so you can't escape the infinite plane even if you try to escape.

Also, electric field due to an infinite wire is equal to,

$$ E=\frac{\lambda}{2 \pi \epsilon_{0} d}. $$

Why we can't say the same for an infinite wire too. As farther we move, we will still see infinite wire in our eyes.

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But it gets smaller. At $r$ the thickness is $\frac 1 {\infty}$, while at $2r$ it's $\frac 1 {2\infty}$

The infinite plane is truly scale invariant.

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  • $\begingroup$ Hey JEB, your answer is not helping me much, if you can elaborate a bit more on this would make me easy to understand more. The concern is for infinite plane too, why can't the same argument hold. $\endgroup$
    – Anshul Sharma
    Commented Nov 5, 2022 at 4:52
  • $\begingroup$ Any direction you look in a small solid angle, the total $q/r^2$ is constant with the infinite plane $\endgroup$
    – JEB
    Commented Nov 5, 2022 at 5:39

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