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Electric field of infinite plane of charge is given by: $$E=\frac{\sigma}{2\varepsilon_0}$$ This can be derived using Gauss's Law or integrating the contribution by circular loops of charge enclosing one another. But if I use square loops instead, it becomes: $$E=\frac{2\sigma}{\pi\varepsilon_0}$$ enter image description here

Electric field of z component is $$\mathrm{d} \hat{E}_{z}=\frac{k \mathrm{d} q}{R^2}\cos\theta$$ Substituting $$\mathrm{d} q=\sigma (8r \mathrm{d} r)$$ $$R^2=z^2+r^2$$ $$\cos\theta=\frac{z}{\sqrt{z^2+r^2}}$$ Integrating $$E_{z}=8k\sigma z\int_{0}^{r}\frac{r}{({z^2+r^2})^{1.5}}\:\mathrm{d} r$$ When r=$\infty$, $$E=\frac{2\sigma}{\pi\varepsilon_0}$$ So difference is it will be a factor of $\frac{2}{\pi}$ instead of $\frac{1}{2}$, greater than the electric field that would be obtained using the correct formula. Why is this? Since it is an infinite plane, I don't think it matters regarding what enclosing shapes we are using, but it seems like different model yields different answers.

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  • $\begingroup$ What is $r$? Is it the apothem or the radius of the square? $\endgroup$
    – Sandejo
    Commented Feb 26, 2022 at 5:56
  • $\begingroup$ It is the apothem. $\endgroup$
    – radastro
    Commented Feb 26, 2022 at 6:00

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Your mistake is that you are assuming $E_z$ to have the same angular dependence as if you were considering circular loops. Note that as you move along the side of the square $E_z$ must vary because you are changing your distance with respect to the charge. The reason one takes circular loops is precisely due to this fact that $E_z$ is constant along circular loops, which makes the integration easier.

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  • $\begingroup$ Good point. I assumed a square as if it is a circle with radius r which is wrong, and a 'square loop' has a greater amount of charge than a circular loop, hence the bigger factor, $$\frac{2}{\pi}$$. $\endgroup$
    – radastro
    Commented Feb 28, 2022 at 2:54

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