# When should the integral be used for Gauss's law?

1. Jun 15, 2010

### |<ings

I have difficulty when trying to decide which form of Gauss's law I should use for a problem.

Please, tell me when I should use the integral form:

$$\int E \cdot dA = \frac{Q_{in}}{\epsilon_{0}}$$

and when it's appropriate to use the other form:

$$E \cdot A = \frac{Q_{in}}{\epsilon_{0}}$$

Last edited: Jun 15, 2010
2. Jun 15, 2010

### FedEx

The Gauss's law is the truth of nature. It is always true. The question is whether we should apply it or not. Well that depends on the symmetry of the problem.

We always use the integral form. Its after that we apply the integral form and then we look at the problem and if it turns out that E and dA have a certain relationship.. Than we remove the dot product and bring the required trigonmetry and calculate the integral...

We can use the second form only if E has a constant magnitude and a constant direction for that given patch of area "A"...

3. Jun 15, 2010

### K^2

Technically, E can vary in second formula, as long as component of E normal to A stays constant. But it still isn't the most useful form.

4. Jun 15, 2010

### Staff: Mentor

A better way to put it: ...if E is normal (perpendicular) to the surface everywhere, and has the same magnitude everywhere on the surface.

5. Jun 16, 2010

### K^2

But that's the thing, it doesn't have to be normal. The component not normal to A can be anything. So the overall magnitude of E can also be just about anything. It's more flexible than it looks.

6. Jun 16, 2010

### |<ings

What you are saying is that if the component of E onto A (iow $$comp_{A} E$$) is constant for all the segments (patches) of area on the closed Gaussian surface, then the integral form "reduces" to the second form. Is this what you are saying?

BTW, vector A is the vector normal to the area.

Last edited: Jun 16, 2010
7. Jun 16, 2010

That allows you to take E outside of the integral, leaving you with the integral of dA - which is hopefully trivial or given.

8. Jun 16, 2010

### K^2

Hence the E*A form sans integral.

9. Jun 16, 2010

### |<ings

TNX, I think I get it now. I would rep you if I could.
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TNX to the others too. Feel free to add more insight if it will help.

10. Jun 16, 2010

### Sithdarth

I'm just wondering when you were taught Gauss's Law what was the justification given for choosing one Gaussian surface over another? When I was took E&M one of the first things explained when I learned Gauss's Law was that you choose a Gaussian surface specifically to keep the component of E perpendicular to dA constant, or at least as simple as possible. So that it could be pulled out of the integral when constant or easily computed when not. Of course making sure the integral of dA is computable or trivial and thereby reducing it to the other form when constant or keeping the integral as simple as possible. I did have a professor who was crazy about explaining the physical and mathematical reasoning behind essentially everything he put on the chalkboard though.

11. Jun 16, 2010

### K^2

First consideration is symmetry. If there is some symmetry in the problem, it should be reflected in the surface. If you have a spherical charge distribution, there is no question about using a spherical distribution.

And sometimes choosing part of the surface to be parallel to the field is easier. Think about solving infinite capacitor. You want a cylinder with bases parallel to plates. The sides of the cylinder make no contribution, as the field going to be parallel to these.

12. Jun 16, 2010

### |<ings

Well, TBH, we did only examples involving simple shapes (spheres, infinitely long cylinders, lines, planes).

When using spheres, the Gaussian surface is a sphere (concentric).

For cylinders and lines the surface we used was a cylinder of course.

And for the plane we used, a cylinder of cross-sectional area, A, (for the Gaussian surface) that crossed the plane. The plane intersects with the cylinder so that the charge enclosed by the Gaussian surface is lambda * A (lambda is the area density). You get something like this:

E * (A of the Gaussian surface that is normal to the E) = Q_in/epsilon

E (2A) = lambda * A/epsilon

E = lambda / (2*epsilon)

I guess since the areas eventually cancel out, you can use pretty much any type of prism as the Gaussian surface (please verify). I mean you could even use a prism whose bases were themselves planes(these should be parallel to the plane with charge distribution) The A would be infinity, but because they eventually cancel out, it's not a problem.

Last edited: Jun 16, 2010
13. Jun 16, 2010

### Sithdarth

That's roughly what I meant. Obviously the simplest form is going to take advantage of symmetry and that was addressed as well. But along side of that I was also taught very specifically the motivation for choosing the surface such the E perpendicular to dA was a constant (which includes 0 or the parallel case of course) was specifically to avoid the integral in the first place. This necessitates that basically any symmetry in your charge distribution shows up in your surface (for simple charge distributions). I just always thought everyone was taught that the main motivation behind choosing specific Gaussian surfaces in the first place was skipping the integral if at all possible. In which case the answer to the OP follows logically from that. You do the integral when you can't find a suitable Gaussian Surface that is easily integrable or trivial where the component of E perpendicular to dA is constant (including the 0 case). It just always seemed very logical to me that way.

Yeah the vast majority of what we did was that as well. My professor just felt the need to add that little bit about why (beyond simply matching symmetry) we choose those particular surfaces (to avoid the integral).

Pretty much though you want to stick with things with sides that are perpendicular and parallel to the field to make things easier on you. The area you choose for the top and sides is arbitrary to it can be arbitrarily large. In the case of the sides E*A is zero anyway so no worries there and the area of the top and bottom cancels with the area in the Qenc so no worries there either.

14. Jun 16, 2010

### |<ings

Much of my confusion at times came from wrongfully thinking that E varied according to dA, thereby necessitating the integral. I thought: if E varies (with respect to r), then it would also vary with respect to A; it's clear now that E can be a function of r, but not vary over the surface at a specific value of r, which is the case for the simple geometries.

More confusing though was the surface integral; this was the first time I had used and the teacher did not make an effort to even explain it.

Finally, thanks to all the posters.