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In the given figure, a charge q is placed at $\left(0,\dfrac{a}{\sqrt2},0\right)$. Find flux through the shown surface and represent it as $\dfrac{nq}{6m\epsilon_0}$. Find $m+n$.

Given ans:7

My attempt:

Enclose the charge through four other such triangles in each of the upper octants and a square of side $\sqrt2a$ in the XY plane. Then, the charge is placed at a distance equal to half of the square’s side. It is well known that flux in such a case through the square is $\dfrac{q}{6\epsilon_0}$.

The rest of the flux is distributed equally into each of the four triangular parts of the pyramid. Thus the flux Through each triangular face should be $\dfrac{1}{4}\cdot\dfrac{5q}{6\epsilon_0}=\dfrac{5q}{24\epsilon_0}$. But this gives me $m+n=9$. What did I do wrong?

I don’t have any background in multivariable calculus, but any solutions involving surface integral of $\vec E\cdot \vec{dS}$ are also welcome. I’d just like to confirm the answer.

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  • $\begingroup$ I tried it myself and get the same answer as yours, 9. Though I realized I actually used exactly the same method as yours. $\endgroup$
    – aystack
    Aug 26, 2022 at 13:01

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