# Gauss’ law and Enclosed Charges

• FS98
In summary: It is required for gauss's law to work that the charges on the outside of the Gaussian surface have no effect on the calculation. In other words, the symmetry has to be "simple". That is, the charges have to be the same at all points on the surface. Symmetry is not always required, but it is a necessary condition.
FS98
What exactly does the electric field as solved for by Gauss’s law tell us?

If you use a Gaussian surface that encloses no charge you find that the electric field is equal to 0. But if there is a charge outside of that Gaussian surface, it is not true that the electric field is 0 on the Gaussian sphere.

I want to say that the electric field calculated will only give the electric field due to the enclosed charges, but I don’t think that’s true either. If we solve for the electric field of an infinite line of charge, the charges on the outside of the Gaussian surface do have an impact on the calculation. If the line of charge we finite, we would have the same enclosed charge, but gauss’s law wouldn’t work so easily because there would no longer be the same symmetry?

So what exactly does the electric field calculated by gauss’s tell us?

FS98 said:
If you use a Gaussian surface that encloses no charge you find that the electric field is equal to 0.

No! It simply tells you that the same number of electric field lines that enter the surface leave it. This does not mean the electric field is zero. Remember, Gauss' law is always true, but it is only in cases where there is a simple symmetry that it is useful to calculate the electric field.

FS98
phyzguy said:
No! It simply tells you that the same number of electric field lines that enter the surface leave it. This does not mean the electric field is zero. Remember, Gauss' law is always true, but it is only in cases where there is a simple symmetry that it is useful to calculate the electric field.
But if the charge enclosed is equal to 0, doesn’t that mean that the surface integral of the electric field is equal to 0 as well? And then can’t you solve to find that the electric field is also equal to 0?

FS98 said:
But if the charge enclosed is equal to 0, doesn’t that mean that the surface integral of the electric field is equal to 0 as well? And then can’t you solve to find that the electric field is also equal to 0?

Actually, no. Again, let's start from the very beginning and look at Gauss's law equation (I'll use the integral form here):

∫E⋅dA=qencl0

where the integral is over a closed surface.

If you have enclosed charge being zero for ANY arbitrary Gaussian surface, the BEST that you can say is that the LHS is zero, i.e. the total electric flux is zero.

It is ONLY when you have a highly symmetric situation, where you can construct a highly-symmetric Gaussian surface, and allows for E to be pulled out of the integrand, can you end up with E being zero.

You have to be careful in interpreting what the mathematics is saying.

Zz.

FS98
FS98 said:
But if the charge enclosed is equal to 0, doesn’t that mean that the surface integral of the electric field is equal to 0 as well? And then can’t you solve to find that the electric field is also equal to 0?

No. You're forgetting the sign. Electric field lines that enter the Gaussian surface contribute with an opposite sign compared to electric field lines which exit the surface. It is only in highly symmetric cases, like the field of a point charge, that you know the direction of the field lines. Since Gauss' law is true for any surface, the whole key to using Gauss' law to calculate the E-Field is in cleverly choosing your surface to take advantage of known symmetries.

FS98
FS98 said:
But if the charge enclosed is equal to 0, doesn’t that mean that the surface integral of the electric field is equal to 0 as well?
Yes.
FS98 said:
And then can’t you solve to find that the electric field is also equal to 0?
Only in special cases which have enough symmetry that you can say that the field has to have the same value at all points on the surface.

FS98
ZapperZ said:
Actually, no. Again, let's start from the very beginning and look at Gauss's law equation (I'll use the integral form here):

∫E⋅dA=qencl0

where the integral is over a closed surface.

If you have enclosed charge being zero for ANY arbitrary Gaussian surface, the BEST that you can say is that the LHS is zero, i.e. the total electric flux is zero.

It is ONLY when you have a highly symmetric situation, where you can construct a highly-symmetric Gaussian surface, and allows for E to be pulled out of the integrand, can you end up with E being zero.

You have to be careful in interpreting what the mathematics is saying.

Zz.
What exactly is the symmetry required for gauss’s law to work out nicely?

FS98 said:
What exactly is the symmetry required for gauss’s law to work out nicely?

The same symmetric cases that you should have encountered in your general physics class: spherically-symmetric charge distribution, cylindrically-symmetric charge distribution, and infinite-plane charge distribution.

Zz.

## 1. What is Gauss' law?

Gauss' law is a fundamental law of electromagnetism that relates the electric flux through a closed surface to the total enclosed electric charge within that surface. It is named after the German mathematician and physicist Carl Friedrich Gauss.

## 2. How is Gauss' law used to calculate electric fields?

Gauss' law can be used to calculate electric fields in situations where there is symmetry in the distribution of charges. By choosing a closed surface that follows the symmetry of the charge distribution, the electric flux can be calculated and then used to determine the electric field at any point inside or outside the surface.

## 3. What does it mean for a surface to be "closed" in Gauss' law?

A closed surface is one that completely encloses a certain volume of space. In Gauss' law, the surface is used to create a boundary around the charges that are being considered in the calculation. This surface can be any shape as long as it is closed and follows the symmetry of the charge distribution.

## 4. Is Gauss' law applicable to all types of charge distributions?

Yes, Gauss' law is applicable to all types of charge distributions, as long as there is symmetry in the distribution. This includes point charges, line charges, and surface charges. However, for more complex charge distributions, it may be more difficult to choose a closed surface that follows the symmetry, making the calculation more challenging.

## 5. How is Gauss' law related to Coulomb's law?

Gauss' law is a more general form of Coulomb's law. While Coulomb's law gives the electric field due to a single point charge, Gauss' law can be used to calculate the electric field for any charge distribution with symmetry. Additionally, Gauss' law can also be used to calculate the electric field outside of a charged conducting surface, which Coulomb's law cannot do.

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