- #1

- 1,444

- 0

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?