# I Question about Gauss' law

1. Mar 20, 2017

### swirlydragon

According to Gauss Law, we only consider charges which are inside the surface to produce net flux.
We disregard the charges outside the surface because they produce zero net flux (as the electric field lines enter the surface and leave the surface producing zero net flux)
My question is, since we don't consider the charges outside the surface to produce flux then why do we consider the electric field of those charges when solving Gauss Law problems?
In the formula, we consider only the charges inside the Gaussian surface but when it comes to electric field, we consider electric field caused by all the charges (whether they be inside or outside the Gaussian surface). Why?

2. Mar 20, 2017

### Diegor

You can divide the integral in two parts since the electric field can be considered as the sum of two components. One produced by the charge inside and the other produced by charges outside lets say Ein and Eout. The integral of Eout will be cero so Gauss law will only help you to get the value of Ein but it doesn't mean tha Eout doesn't exists.

3. Mar 20, 2017

### swirlydragon

@Diegor
Ok I understand that, but if what you say is true then if I don't consider electric field by the outside charge in the above formula then it should also not change the result because the result is zero anyway. So does that mean I can solve Gauss Law numericals without considering the electric field by the outside charge?

4. Mar 20, 2017

### mic*

Consider two point charges of the same sign that are separated by some small distance d.

These two points will repel each other in a way described by the coulombic interaction of their electric fields.

Draw an enclosed surface that surrounds one point charge but not the other. Call the enclosed charge point A.

Imagine the field lines. The lines of point A will exit the surface. No matter what shape the surface is or anything like that, they ALL exit, so long as the surface is enclosed.

Now consider the second point charge. Some of its field lines may enter the surface, but they will all exit the same surface too, eventually. So the NET FLUX for charges outside the surface is zero. Its not that there is zero flux.

I hope this helps.

5. Mar 20, 2017

### Diegor

You can use Gauss law to calculate the field produced by internal charges. Then you will have to use another method (for ex. Coulomb law or gauss law but with another surface) to calculate the field produced by external charges and finally sum both fields to get the total electric field. Usually Gauss law is usefull when there is certain simetry in the electric field.

An simple example where you can see this is getting E field between two parallel plates of a capacitor. There gauss law is used two times with two separated closed surfaces that share one side or face that is where the field is to be calculated. In the web you can find it it's a very common example.

Last edited: Mar 20, 2017
6. Mar 20, 2017

### JoePhysics

The point is, the flux of the electric field through a closed surface depends only on the enclosed charge. If you have charges outside the closed surface as well, the $\mathbf{E}$ field accommodates the outside charges as well as the ones inside, but the flux still remains only a function of the enclosed charge.

Now, to answer your question, what you can do is use the principle of superposition. Pretend that the charges outside the Gaussian surface don't exist, and then use Gauss's law to calculate the E-field solely due to the enclosed charge. Then add to this field the vector sum of whatever E-field the outside charges generate.