Symmetric Gaussian Surface

1. Jan 27, 2013

CAF123

1. The problem statement, all variables and given/known data
A point charge of negative polarity is located at the centre of a cubic Gaussian surface with edges of length $0.5m$.

Calculate the electric flux through one of the faces of the surface.
What would happen if the charge was moved 10cm to the right?

2. Relevant equations
Gauss' Law for closed surfaces

3. The attempt at a solution
From intuition, I can see that the answer is simply $\frac{-q}{6\epsilon_o}$. However, I want to show this.

Denote the length of a side by $a$. The E-field lines end within the closed surface, so the net flux should be negative (as I get above). Through one face, I have $\Phi = -EA$, where $A$ is the area of a face and the negative is there because $E$ and $A$ are antiparallel. The E field of the charge a distance a/2 away is given by $$\frac{-q}{4\pi \epsilon_o (a/2)^2}$$ and so I get after some cancellation:$$\Phi = \frac{q}{\pi \epsilon_o}$$ which is not quite right. (I have a π still and the flux is positive). Where did I go wrong?

For the second question, the situation would no longer be symmetric. I am not entirely sure what the question wants me to do here though. Since the cube is a closed surface, I feel the net flux would still be simply $-q/\epsilon_o$, although I am not sure about the individual faces.

Many thanks

2. Jan 27, 2013

cepheid

Staff Emeritus
The E field is not constant over the face of the cube, because not every point on that surface is equidistant from the charge. It would be if it were a sphere of radius a/2, but it isn't. So what you need to do is find out how the radial distance between the charge and the surface varies as a function of x and y: r = r(x,y), and then you can compute the surface integral.

Hint, set the centre of the face as the origin, (0,0). Then x varies between +a/2 and -a/2, and so does y. The distance between the charge and the centre of the face is a/2. The distance "R" between the centre of the face and the point on the face where the E-field is being evaluated can be determined by using Pythagoras *in the plane of the face.* This distance R is the projection of r onto that plane. This means that you can use Pythagoras again to find r in terms of R and a/2.

3. Jan 27, 2013

CAF123

Thanks cepheid,
I see why I what I did initially was wrong. However, would it not also be wrong anyway since I was applying Gauss' Law to an open surface?!

Following your hints, denoting the position of some small dE element by x and y coordinates, from the origin this is $R = \sqrt{x^2 + y^2}$. Then the distance from the charge I think is then $r = \sqrt{R^2 + a^2/4} = \sqrt{x^2 + y^2 + a^2/4 }$. Hence I now compute:
$$\frac{-q}{4\pi \epsilon_o} \int_{-\frac{a}{2}}^{\frac{a}{2}} \int_{-\frac{a}{2}}^{\frac{a}{2}}\frac{1}{(x^2 + y^2 + \frac{a^2}{4})}dx dy,$$

Does it look okay? (It seems rather difficult to compute. On first attempt, involved the integral of arctan)

If I wanted to, could I simply enclose the cube within a sphere and then use Gauss' Law immediately?

Thanks

Last edited: Jan 27, 2013
4. Jan 27, 2013

cepheid

Staff Emeritus
What you did wrong was to assume that you could say $\Phi =\iint \vec{E}\cdot d\vec{A} = EA$. The second equality is not true, again because the E-field is not constant over the planar surface.

It's true that you can't apply Gauss' law to a non-closed surface, but that's NOT what you were doing anyway. You were just trying to compute the flux directly by multiplying (which was wrong, since you had to integrate). So I don't understand that part of your question.

This all looks great, and it is exactly what I was hinting at.

You DON'T need to enclose the cube in a sphere. Gauss' law is valid for ANY closed surface. So you can use Gauss's law directly to conclude that the net flux through the cube MUST be q/ε0, and then by symmetry, since the charge is in the middle, the flux through each of the six faces must be one sixth of the total. At this point the first part of the problem is DONE. No computation necessary! That's why Gauss' law is so useful.

The reason why I got you to set up the integral was because you said in your original post, "However, I want to show this," which I interpreted to mean, "I want to show this by direct computation."

Last edited: Jan 27, 2013
5. Jan 27, 2013

CAF123

The reason I suggested the sphere was because then, as you said, the E-field would be constant over the sphere. So I can do E(4πr2) = -q/ε. I see why this does not help me here though.

Yes, this is exactly what I wanted, so thanks. But how would I compute that integral to show that I get -q/6ε? I have tried but I end up with the integral of arctan. Is there an easier way?

6. Jan 27, 2013

cepheid

Staff Emeritus
I don't know if there is an easier way. You can look up the integral of arctan in a table of integrals. I don't think it's too bad, is it?

7. Jan 27, 2013

CAF123

So integrating wrt x first (choice does not make things any simpler) gives
$$\frac{1}{b}tan^{-1}\left(\frac{x}{b}\right),$$ where $b=y^2 +a^2/4$. Now I have to integrate this wrt y.

After evaluating the above at the endpoints, I now use integration by parts,
with $$u = tan^{-1}\left(\frac{a}{2b}\right) -tan^{-1}\left(-\frac{a}{2b}\right) , v' = 1/b$$ gives $$uv - \int_{-a/2}^{a/2} -\frac{2}{b}tan^{-1}(y/b)(4b^2ay)/(b^2(4b^2 + a^2)\,dy,$$ which looks like a mess and I don't feel like attempting to integrate that.

I haven't double checked this but do you think there is an easier way?

Many thanks.

8. Jan 27, 2013

CAF123

Actually I think I managed to do it via looking at the tables of integrals. The only problem I have now is that everything cancels (because I have y2 where i defined b) so I get zero as the final answer. I'll check it again tomorrow but if you have any other suggestions in the meantime that'd be great.

9. Jan 27, 2013

haruspex

Are you sure about this? Isn't the integrand the strength of the field at (x,y)? The field is a vector, and you can't integrate vectors like that when they don't all point the same way. There'll be some cancellation. You have to treat dxdy as the vector dA, and take the dot product of dA with the field.

10. Jan 27, 2013

cepheid

Staff Emeritus
Yeah, this is a good point, I missed that. Since we decided that the face of the cube we're considering is parallel to the x-y plane, area elements on this face point in the z direction. So only that component of the field contributes to the flux.

11. Jan 28, 2013

CAF123

So for some element dE on one surface, $d\Phi = dE(dxdy)cosθ$. I think I can still use what I derived earlier for the distance between dE and the charge. Then I integrate over all these $d\Phi$. Is there a way to express cosθ in terms of x,y ?

12. Jan 28, 2013

CAF123

Am I along the right lines above?

For part 2) of the question, I am not sure exactly what they want, but what I said was the following:

Net flux is still the same through the surface, I don't think much can be said about the individual faces. If the charge moves 10cm towards the right, then the face closest to the charge will experience the greatest negative flux because for the closest face, 1/r2 is the greatest and hence the flux is the largest negative.

However, I also thought about an argument that might counter the above. For the face closest to the charge, the angle between some dE and dA is constantly changing as we move 'across' the face. As we approach the centre of the face, this angle is decreasing which implies cosθ is increasing. However, in comparison for the angle between dE and dA for the face exactly opposite this one, the angle is a lot greater and so the effect lowers the flux in comparison to other faces.

(does that make sense?). So it's closeness to one face increases the flux (more negative, that is) because 1/r2 is greatest, but the angle is also the greatest so this implies cosθ is decreasing which acts to decrease the flux. I wander if these would cancel out? (I would say not since $1/r^2$ is dominant)

Last edited: Jan 28, 2013
13. Jan 28, 2013

cepheid

Staff Emeritus
Sure, I think so. Consider the right triangle that you used to arrive at this relation:

If you think about it, this is essentially the triangle that resolves the E-vector into components parallel to and perpendicular to the surface of the cube. And it is clear from this triangle that cosθ = (a/2)/r.

14. Jan 28, 2013

CAF123

Yes, that makes sense. I'll retry the integral ..
So what I have is:$$\Phi = \oint_S (Ecos \theta)\,dx\,dy = \frac{-qa}{8\pi \epsilon_o}\int_{-a/2}^{a/2} \int_{-a/2}^{a/2}\frac{1}{x^2+y^2+a^2/4}\frac{1}{r}\,dx\,dy= \frac{-qa}{8\pi \epsilon_o}\int_{-a/2}^{a/2} \int_{-a/2}^{a/2}\frac{1}{(x^2+y^2+a^2/4)^{3/2}}\,dx\,dy$$

I don't know where to begin with that integral. I have something of the form $$\int_{-a/2}^{a/2}(x^2 + b)^{-3/2}\, dx,$$ where b is a constant. Any ideas?

Last edited: Jan 28, 2013
15. Jan 28, 2013

haruspex

Despite the square range, this might work better in polar.

16. Jan 28, 2013

CAF123

Yeah, when I saw the $x^2 + y^2$, polars came to my mind but then I thought otherwise since the region of integration was a square.

Ok, so $dx dy\mapsto rdrdθ$, with θ going from 0 to 2π, but I am not sure about r. If I said r from 0 to a, that would just be a circle.

EDIT: Or might it be 0 to $\sqrt{r}$?

Last edited: Jan 28, 2013
17. Jan 28, 2013

haruspex

Try doing r < a and a < r < a√2 separately. For the latter range, you'll have something like a sec(θ) < r < a sec(π/4)

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