# Electrostatics question help

1. Apr 13, 2009

### latentcorpse

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 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 $\vec{e_x}$.

$\vec{E}=\frac{Q}{4 \pi \epsilon_0 (\frac{a}{4})^2} \vec{r}$

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?

2. Apr 13, 2009

### Redbelly98

Staff Emeritus
Re: Electrostatics

There are two problems that I can find:

• The cube's surface is not a constant distance a/4 from the charge, as you appear to assume
• The unit vector r^ is not (1/√3,1/√3,1/√3) everywhere on the surface.

3. Apr 14, 2009

### latentcorpse

Re: Electrostatics

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

what is $\hat{\mathbf{r}}$ then (i really should know this!)

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

4. Apr 14, 2009

### Redbelly98

Staff Emeritus
Re: Electrostatics

The vector (x,y,z) points in the same direction as r^, everywhere. Normalize that vector, and you'll have r^.

It should be possible, just not necessarily easy.

5. Apr 14, 2009

### latentcorpse

Re: Electrostatics

yeah so $({x \over |x|},{y \over |y|},{z \over |z|})$

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

6. Apr 14, 2009

### Redbelly98

Staff Emeritus
Re: Electrostatics

Not quite, that vector just has components (±1,±1,±1). Try dividing by |r| instead of |x|, |y|, and |z|.

Not sure if this will help, but: the integral you're trying to do contains an r-hat term. What you wrote earlier is not equal to r-hat, therefore using it will give a wrong answer when you do the integral. (Or did I misunderstand your question here?)