Electrostatics question help

A point charge Q is placed at the centre of a cube. What is the electric flux through each face of the cube.

ok so the answer is to say that since the cube is a closed surface, Gauss' Law tells us that the total flux through the cube is [itex]\frac{Q}{\epsilon_0}[/itex] and then from symmetry 1/6 of that is through each face. so the final answer is [itex]\Phi=\frac{Q}{\epsilon_0}[/itex].

I was wondering why i dont get the same answer when i do it this way:

set it up with cartesian coordinates. let's find the flux through the face with outward normal [itex]\vec{e_x}[/itex].

[itex]\vec{E}=\frac{Q}{4 \pi \epsilon_0 (\frac{a}{4})^2} \vec{r}[/itex]

now [itex]\vec{r}=(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},)[/itex] and [itex]\vec{e_x}=(1,0,0)[/itex]
so [itex]\Phi=\int_S \vec{E} \cdot \vec{dA} = \frac{Q}{\pi \epsilon_0 a^2} \vec{r} \cdot \vec{e_x} \int_S dA = \frac{Q}{\pi \epsilon_0 a^2} \frac{1}{\sqrt{3}} a^2[/itex]

which ends up as [itex]\Phi=\frac{Q}{\sqrt{3} \pi \epsilon_0}[/itex]

obviously the first way is much easier but surely it should be possible both ways?

 

Re: Electrostatics

ok yeah so i see now that its not a constant a/2 from each surface

what is [itex]\hat{\mathbf{r}}[/itex] then (i really should know this!)

is it possible to do this problem this way or is gauss' law the only way?

 

Re: Electrostatics

yeah so [itex]({x \over |x|},{y \over |y|},{z \over |z|})[/itex]

why doesn't the one i wrote earlier work though???