# I Gauss' law problem

1. Oct 3, 2016

### David112234

Lets consider a hollow sphere with charge +q and than a larger hollow sphere that encloses it with -q, so two spheres one inside the other. Now we use draw an even larger Gaussian sphere that encloses both of these,

∫EdA = Qinclosed/ ∈_0

the total enclosed charge is +q + -q = 0
so
∫EdA = 0
E ∫dA = E (area of sphere) =0
so the Electric field must be zero, but how could that be? So any test charge (black square in diagram) placed outside the two sphere will experience absolutely no electric force? Wouldn't it experience some force from the -q sphere since it is on the outside and always closer to any test charge?

2. Oct 3, 2016

### BvU

No it wouldn't. As you calculated.
To grasp this: where is the charge sitting on the outer sphere ?

By the way: your reasoning doesn't take into account that the negative charge 'on the left' is further away from the test charge ...

3. Oct 3, 2016

### David112234

I still don't get it, how would that effect it? According to Coulombs law a closer charge feels the force stronger,
what if we consider a situation where that issue is not present, such as a dipole?

The test charge (black) should feel a force on it since it is closer to the blue(+) charge but Gauss law says 0, My dilemma is still the same.

Last edited: Oct 3, 2016
4. Oct 4, 2016

### BvU

In this case you misinterpret Gauss law: there is no spherical symmetry so you can't just take $\vec E$ out of the integral: it does indeed not have the same value on every point of the Gaussian surface.

5. Oct 4, 2016

### David112234

Oh so in the integral some EdA will be positive and some negative so it will be a total of 0,
Lets go back to the two sphere, you said that "the negative charge 'on the left' is further away from the test charge"
I used a phet simulation for that

and it showed that it still felt an electric field from the closest (positive) charge

6. Oct 4, 2016

### BvU

I don't see two spheres anywhere. There are plenty proofs of Gauss' law around. A phet quadrupole simulation is no proof of the contrary.

7. Oct 4, 2016

### David112234

I know, I only selected those 4 as a slice of the spheres at the diameter to test what you said
"your reasoning doesn't take into account that the negative charge 'on the left' is further away from the test charge"

If I add more and keep on making it look like spheres the results don't change

I have seen how gauss law is derived and I do believe it works but for certain situations such as these its hard to accept these proofs when it goes against all my intuition.

8. Oct 4, 2016

### robphy

Have you viewed it using the "Electric Field" checkbox (showing magnitude [via brightness])... not just the sensors?

What happens further away from the centers?
Note that if your charged "spheres" are not placed precisely, the discrepancies will be more evident if you are not far enough away.

is a description of a different visualization ( http://www.glowscript.org/#/user/ma...older/matterandinteractions/program/13-fields )
that doesn't need precise placements.

9. Oct 5, 2016

### David112234

As you move further away E decreases, I would expect that anyway even if E was not supposed to be 0 everywhere
That tool is helpful to visualize the flux but it does not help me understand it.
My intuition keeps telling me that E would not be zero, but the force from the charge on the outer shell that is closer to the test charge,
The only way I could see it being 0 is if the charges on the left would negate it, but they do not.

To check that here is what I did, I take a slice of the sphere,

--- = distance = d
[] = test charge

(+) --- (-) --- (-) --- (+) --- []
1 2 3 4

Force from 1 = kq/ (4d)2
F_2 = -kq/ (3d)(4d)2
F_3 = -kq/ (2d)(4d)2
F_4 = kq/ (d)(4d)2

kq/d2 [ (1/16) - (1/9) - (1/4) + (1/1) ] = kq/d ( .7) , not zero

So from this I concluded that the points on the sphere that are further from the text charge would not balance out the force felt from the charge on the outer sphere closest to the test charge.

10. Oct 5, 2016

### robphy

11. Oct 6, 2016

### David112234

Yes that I understand, even tho the field close to one side is stronger, there are more of the weaker ones on the other side to counter act it, Since you bring it up I assume something similar will happen here? I cant figure out how. Here is my best case for the way I see it, the best way for me to understand this is to find out what is wrong with my current understanding.

lets consider the spheres as a stack of slices, as in an integral. Each point on each slice has one corresponding point bellow, and the forces of the test charge (green) will cancel in the Y direction, so only the X components will act,

So we can represent the forces as this

And If the spheres are just a stack of slices consisting of 4 points like the ones I drew
The total force from each slice is not zero, the sum of all these slices would also not be zero.

12. Oct 6, 2016

### robphy

In such proofs, I think the key is to cleverly exploit certain symmetries so that the individual terms sum nicely
(akin to the supposed story of Gauss adding up the numbers from 1 to 100 https://nrich.maths.org/2478 ).

I think your discrete examples don't exploit that symmetry.
There may be a discrete version of the approach taken here:
http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/sphshell.html#wtls
and https://en.wikipedia.org/wiki/Shell_theorem (see animation)

Try pg.85 of http://physics.wooster.edu/Lindner/Texts/Mechanics.pdf

13. Oct 7, 2016

### BvU

You test at 0 and $-{\pi\over 4}$ and find a field pointing away from the origin. Does it point towards the origin halfway ?
And is it very small ? Much smaller than in the previous (quadrupole) picture ?
If not, then that may be a numerical limitation of your simulation program. Gauss' law holds, period.

By the way, I don't see a conducting sphere ?

14. Oct 7, 2016

### vanhees71

I think this thread has very much deviated from the original question and is more confusing than helping. So let's go back to posting #1. To apply Gauss's Law in integral form as suggested here only works for systems of very high symmetry, and spherical symmetry is "symmetric enough". The point is that you can guess how the electric field should look like. For a spherically symmetric charge distribution the vector field schould be as spherically symmetric as possible. Taking the center of the spherical shell in the origin of the coordinate system thus the electric field should always point radially out, i.e., it should be of the form
$$\vec{E}(\vec{r})=E_r(r) \frac{\vec{r}}{r}, \quad r=|\vec{r}|.$$
To get the component in radial direction $E_r(r)$ in Gauss's Law you can now choose a spherical shell around the origin of radius $R$ with $R$ varying to get $E_r(R)$ at each distance from the center. The surface normal vector $\mathrm{d}^2 \vec{A}=R^2 \vec{r}/r \sin \vartheta \mathrm{d} \vartheta \mathrm{d} \varphi$ in the usual spherical coordinates. Thus your surface integral gives
$$\int \mathrm{d}^2 \vec{A} \cdot \vec{E}=\int_0^{2 \pi} \mathrm{d} \varphi \int_0^{\pi} \mathrm{d} \vartheta R^2 \sin \vartheta E_r(R) = 4 \pi R^2 E_r(R).$$
Now lets call the radius of the inner sphere $a$ and the radius of the outer sphere $b$. You have to read off the charge enclosed by the sphere with radius $R$, and thus you have to distinguish just three cases:

(1) $0 \leq R<a$

Then there is no charge inside the sphere and thus according to Gauss's Law
$$4 \pi R^2 E_r(R)=0 \; \Rightarrow \; E_r(R)=0 \quad \text{for} \quad 0 \leq R < a.$$

(2) $a<R<b$

Then there's the full positive charge $Q$ inside the sphere and thus
$$4 \pi R^2 E_r(R)=\frac{Q}{\epsilon} \; \Rightarrow \; E_r(R)=\frac{Q}{4 \pi R^2} \quad \text{for} a<R<b,$$

(3) $R>b$

Then the total charge is again 0 and thus again
$$E_r(R)=0 \quad \text{for} \quad R>b.$$
This means you have a Coulomb field for $a<R<b$ and otherwise no field at all.

15. Oct 7, 2016

### BvU

Dear Hendrik,

I don't think David has a problem with this derivation, but instead he is searching for a way to align the outcome with an intuitive expectation that is somewhat contrary.
At the moment all I can think of is to actually carry out the integration for two concentric shells of opposite charge. From symmetry it will be acceptable that the resulting field can only be radial, so all that remains is to integrate that component over the shell.

His choice of contributions in #11 is unfortunate and of course doesn't add up to zero -- it does not take into account all contributions.

 The integration is carried out in full on various youtubies, but I have a feeling Dave is better off convincing himself

Last edited: Oct 7, 2016
16. Oct 7, 2016

### vanhees71

I see! So you mean to use the Green's function to evaluate the potential. Of course, you can also try to evaluate directly the field, but that's more difficult since you have a vector rather than a scalar quantity. This calculation is not as trivial as the use of the integral Gauss's law.

So let's take the inner shell only. Say we have a total charge $Q$ on a sphere with radius $a$ around the origin. The surface-charge density is $\sigma=Q/(4 \pi a^2)=\text{const}$ along the surface. So what you have to evaluate is
$$\phi(\vec{r})=\sigma \int_{S_a} \mathrm d^2 A' \frac{1}{4 \pi \epsilon |\vec{r}-\vec{r}'|}.$$
The surface can be parametrized with spherical coordinates again,
$$\vec{r}=a (\cos \varphi \sin \vartheta,\sin \varphi \sin \vartheta,\cos \vartheta).$$
Again due to radial symmetry $\phi(\vec{r})=\phi(r)$, and thus we can choose $\vec{r}=r \vec{e}_z$. Then you get
$$|\vec{r}-\vec{r}'|=\sqrt{r^2+r^{\prime 2} + 2 r a \cos \vartheta}.$$
So the integral over $\varphi$ just gives a factor $2 \pi$, and the challenge is the integration over $\vartheta$. Since furter $\mathrm{d}^2 A'=\mathrm{d} \vartheta \mathrm{d} \varphi a^2 \sin \vartheta$, you get
$$\phi(r)=\frac{\sigma a^2}{2} \int_0^{\pi} \mathrm{d} \vartheta \frac{\sin \vartheta}{\sqrt{r^2+a^2+ 2 r a \cos \vartheta}}.$$
The next step is to substitute $u=\cos \vartheta$, $\mathrm{d} u=-\mathrm{d} \vartheta \sin \vartheta$, which leads to
$$\phi(r)=\frac{Q}{8 \pi} \int_{-1}^1 \mathrm{d} u \frac{1}{\sqrt{r^2+a^2 + 2 r a u}}.$$
Now the integral is
$$I(u)=\int \mathrm{d} u \frac{1}{\sqrt{r^2+a^2+2 r a u}}=\frac{\sqrt{r^2+a^2+2rau}}{ra}.$$
Now we have to plug in the boundaries $u=\pm 1$:
$$I(1)=\frac{r+a}{ra}, \quad I(-1)=\frac{|r-a|}{r a}$$
So we get
$$I(1)-I(-1)=\frac{1}{ra} \times \begin{cases} 2a & \text{for} \quad r>a,\\ 2r & \text{for} \quad r<a. \end{cases}$$
So we have
$$\phi(r)=\frac{Q}{4 \pi \epsilon_0} \times \begin{cases} 1/r & \text{for} \quad r>a,\\ 1/a & \text{for} \quad r<a. \end{cases}.$$
The field is thus
$$\vec{E}=-\vec{\nabla} \phi=-\frac{\vec{r}}{r} \phi'(r)=\frac{Q}{4 \pi \epsilon_0} \times \begin{cases} \vec{r}/r^3 & \text{for} \quad r>a,\\ 0 & \text{for} \quad r<a. \end{cases}.$$

17. Oct 7, 2016

### BvU

David ?

18. Oct 8, 2016

### David112234

That is correct, robphy provided me with lots of information It is just taking me a bit to digest all of it.

Can you please elaborate on this, I used 2 dimensional circles instead of spheres because it is easier to draw and It should, at least I think, behave the same way, in 2D the Y components from 2 points cancel out in 3D Y and Z would also cancel out.

I thought I did consider the contributions for all points, I treated the sphere (circle in this case) as a sum of slices and deduced that since the force from each pair of slices (top and bottom) is + than the sum of all the slices would no be 0.
What contributions am I missing? Or is my assertion that 2d can model 3d wrong?

I am not familiar with Green's Function and you brought parametric functions and potential into this, Id prefer to keep away from that for now, that is beyond me

I understand that a a test charge outside a hollow sphere will be modeled the same as single charge at its center, kq/ r^2, only the field will be weaker because r is larger,
So two spheres 1 q+ and 1 q- could be treated like two points +1 and -1 at the center, which is equivalent to 0
But that is the problem, it goes against all intuition,
the inner sphere is at a further distance from the test charge than the outer, doesn't shrinking down the two spheres to point charges just ignore that Δdistance?

Another question this raises for me, since robphy brought up Shell theorem for gravity,
If you have one sphere, and than place a smaller one inside of same mass like in this situation, would the gravitational force always increase 2x stronger? regardless of how smaller the inner sphere is?

Last edited: Oct 8, 2016
19. Oct 8, 2016

### BvU

There is no problem working it out in 2D; the reason for that is rotational symmetry around an axis through the center of the charge and the point of observation. In the links I hinted at in post #15 they integrate over $\phi$ from $0$ to $2\pi$ because $\phi$ does not appear in the expression for the $\vec E$ field. (Hendrik does idem in #16)

In post #11 you pick 8 points, but forget a few things:
• the black charges are smaller than the red
• there are no red charges 'above' the radius of the inner shell
so looking at it this way gets you nowhere. The good news is you only have to look at one of the two shells: if that gives a radial $|\vec E| = {1\over 4\pi\epsilon_0} {Q\over r^2}$ you're done (because the other gives the same but with a minus sign).

The bad news is the integral is a lot of work (at least for lazy dimwits like me).
And the very bad news is that you probably won't bother to work it out yourself because Hendrik has already done it for you. A bit of a spoiler, but fortunately for Hendrik PF only knows likes and not dislikes (probably fortunate for me too )
Still it's worth to look at it in detail. Do you want the links or did you find them already ?

It's not: that is exactly what the integration proves (I must correct myself a little here: it shows the average ${1\over r^2}$ is equal to the 1/distance2 to the center).

Provided it is concentric, yes !

(Coulomb's law and the gravitational force both go with ${1\over r^2}$).

For me the counter-intuitive result that makes me remember all this vividly is: Inside the entire inner radius sphere of a hollow planet there is no gravity from that planet itself ! (so for you this factor 2 will play that role, perhaps).

20. Oct 8, 2016

### Meir Achuz

In physics, you have obey the law. Otherwise, you get all those replies.