# Electric Flux through a part of plane $x+y+z=a$ via symmetry arguments [closed]

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.

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