I was asked to find the *net* electric flux through an open-faced cone, if the uniform electric field is pointing horizontally to the right. (The circular face is open.)

What I attempted is having a Gaussian surface (closed cone) perfectly enclose this open cone. The net flux from the circular face is 0 on the Gaussian surface. The left half of the cone has inward flux of $-Erh$. The right half of the cone has outward flux of $Erh$. I expected to simply add the flux from both halves of the cone, to get 0 net electric flux.

However, the "correct" answer I was told was just that $Erh$ was the net electric flux. Being in an integral calculus based physics class, we used a simplification that $\int\_{}^{}\vec{E}\cdot{}\,d\vec{A} = EA$, where A was the corresponding shape the area elements "mapped" to. In this case, A was the area of a triangle through the center of the cone, which has area $rh$. The apparent logic was that since right and left half of the cone mapped to that same triangle, the net flux was $Erh$ for (E \* area of that triangle).

I can't understand how that answer could be correct. After examining my textbook's definition of flux (Young and Freedman University Physics 15th Ed), it seems the original question itself may be poorly defined. I couldn't find a convention for area vector direction on an open surface.

If we assume all the flux is positive, wouldn't the answer be $2Erh$ for left side flux and right side flux totaled? (That is, define the area vectors in a way to produce positive flux.)

If we assume that the area vector for all area elements points outward, wouldn't any amount of flux entering be symmetrically canceled by any flux leaving?

Should the correct answer really be $Erh$? Is the problem poorly defined to begin with?

[![enter image description here](https://i.stack.imgur.com/BCI18.jpg)](https://i.stack.imgur.com/BCI18.jpg)