chargeelectric-fieldselectrostaticsgauss-law
What would be the total electric flux $\Phi_E$ through an infinite plane due to a point charge $q$ at a distance $d$ from the plane?
I think it should be ${q/2\epsilon_0}$ but I cannot justify that. My teacher says the flux is infinite due to the infinite area of the sheet which I cannot believe.
The answer by @BrianMoths is correct. It's worth learning the language used therein to help with your future studies. But as a primer, here's a simplified explanation.
Start with your charge distribution and a "guess" for the direction of the electric field.
As you can see, I made the guess have a component upward. We'll see shortly why this leads to a contradiction.
Now do a "symmetry operation," which is a fancy phrase for "do something that leaves something else unchanged. In this case, I'm going to reflect everything about a horizontal line. I mean everything.
The "top" of the sheet became the "bottom." This is just arbitrary labeling so you can tell I flipped the charge distribution. The electric field is flipped too. (Imagine looking at everything in a mirror, and you'll realize why things are flipped the way they are.)
Hopefully, everything is okay so far. But now compare the original situation with the new inverted one.
You have exactly the same charge distribution. You can't tell that I flipped it, except for my arbitrary labeling. But if you have the same charge distribution, you ought to also have the same electric field. As you can see, this is not the case, which means I made a mistake somewhere.
The only direction for the electric field that does not lead to this contradiction is perpendicular to the sheet of charge.
Gauss' law is always true but not always useful; your example falls in the latter category. To infer the value of $\vec E$ from $\oint \vec E\cdot d\vec S$ you need a surface on which $\vert \vec E\vert $ is constant so that $$ \oint \vec E\cdot d\vec S= \oint \vert \vec E\vert \, dS = \vert \vec E\vert \oint dS = \vert \vec E\vert S \, . \tag{1} $$ $\vec E$ is not constant on your sphere, meaning you cannot use (1) and pull $\vert \vec E\vert$ out of the integral and recover $\vert\vec E\vert$ through $$ \vert \vec E\vert = \frac{q_{encl}}{4\pi\epsilon_0 S}\, . $$ In your specific example, this is why $\oint \vec E\cdot d\vec S=0$ even though $\vert \vec E\vert$ is never $0$ at any point on your Gaussian surface. The $0$ results from the geometry of $\vec E\cdot d\vec S$ everywhere on the sphere rather than $\vert \vec E\vert=0$.
To proceed you need to use a Gaussian pillbox with sides perpendicular to your sheet because, by symmetry, the field must also be perpendicular to your sheet. Thus $\vec E\cdot d\vec S$ for all sides of the pillbox is easy to compute. If your pillbox passes through the sheet, it will enclose non-zero charge and, using simple geometry, one easily shows that the flux through the back cap will add to the flux through the front cap and you can recover the usual result.
Best Answer
In general the flux through an oriented open or closed surface $\:\mathrm{S}\:$ due to a point charge $\:Q\:$ is
\begin{equation} \Phi_{\mathrm{S}}=\dfrac{\Theta}{4\pi}\dfrac{Q}{\epsilon_{0}} \tag{01} \end{equation} where $\:\Theta\:$ the solid angle by which the charge $\:Q\:$ sees the surface. In our case this solid angle is half the complete $\:4\pi\:$ solid angle, that is $\:2\pi\:$, so
\begin{equation} \Phi_{\mathrm{S}}=\dfrac{2\pi}{4\pi}\dfrac{Q}{\epsilon_{0}}=\frac12\dfrac{Q}{\epsilon_{0}} \tag{02} \end{equation}
You can find special cases for the solid angles by which a point sees rectangular parallelograms in my answer therein :What is the electric field flux through the base of a cube from a point charge infinitesimally close to a vertex?.