latentcorpse
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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 \frac{Q}{\epsilon_0} and then from symmetry 1/6 of that is through each face. so the final answer is \Phi=\frac{Q}{\epsilon_0}.
I was wondering why i don't 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 \vec{e_x}.
\vec{E}=\frac{Q}{4 \pi \epsilon_0 (\frac{a}{4})^2} \vec{r}
now \vec{r}=(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},) and \vec{e_x}=(1,0,0)
so \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
which ends up as \Phi=\frac{Q}{\sqrt{3} \pi \epsilon_0}
obviously the first way is much easier but surely it should be possible both ways?
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 \frac{Q}{\epsilon_0} and then from symmetry 1/6 of that is through each face. so the final answer is \Phi=\frac{Q}{\epsilon_0}.
I was wondering why i don't 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 \vec{e_x}.
\vec{E}=\frac{Q}{4 \pi \epsilon_0 (\frac{a}{4})^2} \vec{r}
now \vec{r}=(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},) and \vec{e_x}=(1,0,0)
so \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
which ends up as \Phi=\frac{Q}{\sqrt{3} \pi \epsilon_0}
obviously the first way is much easier but surely it should be possible both ways?