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A charge $q$ is placed at shown at a height $a$ above the center of a square surface of side length $a$. I have to find the flux due to $q$ through the surface.

I do not find any symmetric shape about the charge $q$. Had the heigh been $\frac a2$ I could have easily drawn a body centered cube and be done with it using Gauss Law and utilizing the symmetry.

Next I tried integrating but then I ended up with an integral like this $$\phi=\frac{q}{\pi \epsilon_0} \int_{0}^a \frac{1}{\sqrt{x^2+a^2}} \tan^{-1}\left({\frac{a}{\sqrt{x^2+a^2}}}\right)dx$$

On plugging it in here it says "Antiderivative or integral could not be found".

I'm sure I did the calculations right, and it was lengthy so I am not posting them here yet (though I will if I am asked to). I'm just curious that what does it mean when I cannot mathematically get the value of a flux even when I know that it physically exists, if the situation is replicated in real life? And if we need to find the value of the flux, what other tools can we use to do it?

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  • $\begingroup$ Why is your integral one-dimensional rather than two-dimensional? $\endgroup$
    – Ghoster
    Commented Sep 25, 2023 at 20:24
  • $\begingroup$ @Ghoster I have solved it along one of them and arrived at that state $\endgroup$ Commented Sep 25, 2023 at 20:28
  • $\begingroup$ I strongly suspect that your original 2D integral, which you didn’t show, was incorrect. I think that once you set up the correct definite integral, it can be done analytically. But, if not, it can be done numerically. (Many/most integrals can’t be done analytically, but they tend not to appear in homework.) If you edit your question to show the 2D integral, be sure to explain how you got to it. $\endgroup$
    – Ghoster
    Commented Sep 25, 2023 at 20:45
  • $\begingroup$ There is also the possibility that the person who made the question simply made a mistake and meant the charge to be $a/2$ above the square. Sometimes teachers are careless. $\endgroup$
    – Ghoster
    Commented Sep 25, 2023 at 20:46
  • $\begingroup$ I'm sure I did the calculations right. The fact that you have run into this problem seems to indicate that you did not. $\endgroup$
    – Ghoster
    Commented Sep 25, 2023 at 20:52

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