# When to use and not use Gauss's Law?

I'm not sure when and when you can't use Gauss' Law.

Question 1:

If I have the situation given in the first figure, where the red circle and the blue circle both represent 2 different Gaussian Surfaces. If I wanted to find the Electric Field at point (2,0). If use the red Gaussian Surface I find that

$E = Q/16\pi \epsilon _0$

However, if I use the blue Gaussian surface to calculate the electric field at (2,0), I get E = 0 because there is no charge enclosed inside the blue Gaussian Surface.

I know that there is an electric field at that point, how do I reconcile this?

Question 2: If I have the situation given by the figure below:

(Charge +Q is at (0,0.5) and Charge -Q is at (0,-0.5))

If I use Gauss' Law to calculate the electric field at Point A, I get that E = 0 (because the total charge contained within the Gaussian Surface is 0).

However, if I calculate it by calculating the net electric field using the equation:

E = k *[-Q/(2.5)^2 + Q/(1.5)^2] = k * 64*Q/225

I know I am using Gauss's Law incorrectly somehow in both of these instances but I am not sure how either cause is incorrect.

Help is really appreciated! Thank you!

Gauss' law is a statement about the net flux of $\vec E$ and does not allow you to infer $\vec E$ except in special circumstances. One such situation is when the flux is uniform on the Gaussian surface. In this case $$\oint \vec E\cdot d\vec S=\int \vert \vec E\vert dS \cos\theta =\vert \vec E\vert \int dS \cos\theta =\frac{Q_{encl}}{\epsilon}\tag{1}$$ because the flux is constant in magnitude on the surface. In the case of your red sphere this works because the spherical symmetry guarantees that $\vert \vec E\vert$ is constant and always in the $\hat r$ direction, so the flux computes to $$\oint \vec E\cdot d\vec S = \vert \vec E\vert 4\pi R^2=\frac{Q_{encl}}{\epsilon} \tag{2}$$

In the case of your blue sphere, the flux is NOT constant on the surface and so you cannot conclude that $\vec E=0$ using information from the flux of $\vec E$ through this surface. It is (intuitively) clear that the part of the sphere that is near-axis and closer to the source charge will have greater (and negative) flux than the part that is near-axis but on the far side of your sphere. Of course the net flux is $0$, but because $\vec E$ is not constant on the surface of your blue sphere you cannot write $\oint \vec E\cdot d\vec S=\vert \vec E\vert 4\pi R^2$ in (2) and so cannot deduce from the flux the value of the field.

The same type of analysis holds for your second example. The flux of $\vec E$ is not constant on that sphere, so knowledge of the flux is not useful in obtaining $\vec E$ on that sphere.

An analogy would be this. Imagine I give you the net amount of money in your pocket, and tell you all the coins have the same value. You could easily figure out the value of any one count. The net amount of money is the analogue of the net flux, the condition that all coins have the same value is the analogue of $\vert \vec E\vert$ constant on the surface, and the value of one coin is the value of $\vec E$ anywhere. Of course, if the coins to not have the same value you cannot (in general) find the value of one coin by knowing only total amount in your pocket.

• This makes sense, I knew symmetry was used somewhere and this helped me realize I was making the mistake of making E = 0 if the flux was 0. – user169315 Sep 16 '17 at 17:38

The other answers are correct. Just in case you don't see it very clearly, here's my attempt. There are few ideas.

1.- The flux of a field through a surface is defined as

$\iint \vec{E}\cdot d\vec{S}$, with $d\vec{S}=\hat{n} dS$

2.- The Gauss Law says that

$\iint \vec{E}\cdot d\vec{S}= \frac{Q_{int}}{\varepsilon_0}$

So the idea is that Gauss law doesn't talk about the electric field, but the flux of it through a closed surface. That's the key idea. For some reason, it seems it's not highlighted enough to new students, but it's simple.

So, for example, in your first picture... if you take the blue one, there's no charge inside. Does that mean there's no electric field? NO! It just means that the flux is 0. Of course, because there are as many flux lines coming in and going out. Every line arising from the charges that gets inside the blue circle needs to pass again through the circle (in the other side), because there're no sinks inside the blue circle.

Then, why does the red circle work?

The last idea is that, if you want to get the electric field, you need to find a way to take $\vec {E}$ out of the integral. For that, you have to choose the appropiate surface. In other words, Gauss law is ALWAYS true, but it is only useful to get the electric field in very particular cases. You have to be smart enough to choose the correct one. Keys:

1. Find a surface in which the angle between $\vec{E}$ and the normal $\hat{n}$ is constant, and preferably 0. That allows you to change $\vec{E}\cdot d\vec{S}$ for $E\ dS \cos \theta$.
2. Now, the surface must be such that $E$ is constant along the surface. Only along the surface. IF so, you can take $E$ outside the integral, and so

$E \cos \theta \cdot \iint dS = Q/\varepsilon_0$

$\Rightarrow \quad E \cos \theta = \frac{Q}{\varepsilon_0 S}$

And this is only a particular case.

• Thank you! Your explanation has been very helpful, I kept thinking that Gauss's Law related the electric field rather than the flux and kept getting confused. – user169315 Sep 16 '17 at 17:37

Gauss' Law states that the flux through a surface of the electric field is equal to the charge contained divided by $\epsilon_0$ For your first question, when you calculate the electric field using the blue surface, you are considering only the electric field produced by what's inside the blue surface, so of course you get 0. Calculating the field using the red surface gives you the right answer. What is actually in play here is the principle of sovrapposition. The field in the point you're interested in you could take as the sum of the field produced by what's inside the blue surface with what's inside the red surface, thus reconciliating the results Regarding your second question, remember that using Gauss' law you can state that the net flux through the net surface is zero, but not that the field is zero in the origin. Now, taking two surfaces surrounding both charges, you can calculate the field for each and use the principle of sovrapposition I mentioned earlier to calculate the field in the origin, you will find it's the same as you calculated starting from Coulomb's law.

Gauss law relates surface integral of normal component of electric field to charge inside. The key to resolving your 'paradox' is symmetry.

In the first example, you are correct that integral of normal component of electric field is zero across second surface. But just because integral is zero doesn't meant that the field is zero. Think about it, integral is zero just means that sum on different parts of the surface cancel each other. To give a one dimensional example $\int_0^{2\pi} sin(x)=0$ even when the function is non zero almost everywhere.

Using the first surface gives correct result because after placing the surface, the resulting figure is spherically symmetric. Thus we can conclude that magnitude of normal component of electric field will be same at all points on first surface. Thus $E\times 4\pi r^2 = q$. Note that this allows you to only conclude about the normal component. You still have to give a reason why tangential component is zero.

• Thank you! The sin(x) integral you showed was perfect, it helped me see that I was solving the formula incorrectly by saying that E = 0 from the integral(E*dA) =0. – user169315 Sep 16 '17 at 17:40

You actually find the total magnetic flux through a Gaussian surface due to the charges which are inside the surface.
So in you first diagram there are no charges inside the blue Gaussian surface and so you find that the total flux through the blue Gaussian surface is zero and by symmetry you then say that the electric field due to that region enclosed by the blue Gaussian surface is zero.

If I use Gauss' Law to calculate the electric field at Point A

This cannot be done directly because Gauss's law tells you about the net electric flux passing through the Gaussian surface.

So you have to work out $\displaystyle \int _{\rm S} \vec E \cdot d \vec A$.
With a carefully chosen Gaussian surface you can say that the dot product is alwats $E\, dA$ with $E$ a constant which you can take out of the integral.
If you look at the diagram below you will note that using the blue Gaussian surface you cannot do this.

I have draw the surface to be approximately symmetrical so that the electric flux passing through the surface on the right hand side is equal to the electric flux passing though the Gaussian surface on the left hand side.
So the net electric flux through the surface is zero which is what you would expect in that the net charge within the surface is zero.

• Can you elaborate? This looks very wrong. – lalala Sep 16 '17 at 9:53
• @lalala Thanks for point out my error due to my misreading the question. I have corrected my answer. – Farcher Sep 16 '17 at 10:14

## protected by Qmechanic♦Sep 16 '17 at 10:29

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