Gauss’ law and Enclosed Charges

[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?

 

[phyzguy]

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]

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?

 

[ZapperZ]

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=q_encl/ε₀

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.

 

[phyzguy]

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.

 

[jtbell]

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.

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]

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=q_encl/ε₀

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?

 

[ZapperZ]

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.