Gauss law and surface vector

In summary, the conversation discusses the concept of charge distribution within a sphere and how it relates to Gauss' law. The question is raised regarding the total charge and surface integral when considering a smaller and larger sphere within the original sphere. The issue of improper surface vector directions is also addressed.
  • #1
elis02
4
0
Hi all,

Lets suppose we have a sphere.
This sphere has density of [itex]\rho[/itex](r) from 0 to the sphere's final radius R.

Now, we all know that from Gauss law, the charge inside the sphere equals to the integral of the surface of the sphere.
If we set our normal vector to be as usual out of the sphere and our surface is the sphere than we have of course Qtot which is integral on the total volume.

But, let me ask you guys something else. what if our surface is a smaller sphere with radius R1 (let's symbol it S') and bigger sphere with radius R2, we'll symbol it S.
the total charge should be [itex]\int\rho[/itex](r)dv from R1 to R2
but how about the surface integral? I mean integral on the Surface is integral on S plus integral on S'.
now, if the normal to the surface is out of the surface what we get is ∫Eds+∫Eds'=Qtot+[itex]\int\rho[/itex](r)dv from 0 to R1.
yes, something is wrong here. can you tell what is it and explain?

* We all can solve it in another way, but that's not what I'm looking for. just want you guys to tell me if you see something wrong with the mathematics here... (I think I know the problem but wanted to be sure about it - I believe that if the field is outside you should say the surface vector is on the other direction, meaning the surface vector of the inner sphere is the opposite of the outer sphere)
 
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  • #2
"Outwards" in your case means "towards r=0" in the inner surface (it is out of the enclosed volume of your surface). You will get Q minus the charge in the innermost region if you do the surface integral properly.
 
  • #3
my following question

thanks.
So now I can continue to my following question (the real purpose of the thread)
it's attached as pdf file.
Edit: I got it. the professor did some mistakes with the directions of the surface vectors. said one thing and did another...
 

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1. What is Gauss law?

Gauss law, also known as Gauss's law, is a fundamental law of electromagnetism that relates the distribution of electric charges to the resulting electric field. It states that the total electric flux through a closed surface is equal to the enclosed electric charge.

2. How is Gauss law related to surface vector?

Gauss law is related to surface vector through the concept of electric flux, which is calculated by taking the dot product of the electric field and the surface vector. The surface vector represents the direction and orientation of the surface, and is used to determine the direction of the electric flux through that surface.

3. What is the significance of Gauss law in electrostatics?

Gauss law is significant in electrostatics because it allows us to calculate the electric field at any point caused by a distribution of charges, as long as we know the charge enclosed by a surface surrounding that point. It simplifies the calculation of electric fields for symmetric charge distributions and helps us understand the behavior of electric fields in different situations.

4. Can Gauss law be applied to all types of charge distributions?

Yes, Gauss law can be applied to all types of charge distributions, as long as the charge is enclosed by a closed surface. This includes point charges, line charges, surface charges, and volume charges.

5. How do we use Gauss law in practical applications?

Gauss law is used in practical applications, such as designing electronic circuits and devices, predicting the behavior of electric fields in different materials, and understanding the properties of capacitors. It is also used in the study of electromagnetism, including the behavior of light and other electromagnetic waves.

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